a-math-360-sol-29-sept-2014.pdf
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Additional Maths 360 solutions (Unofficial) [29 Sept 2014] Visit sleightofmath.com for up-to-date pdf and video solutions. Authors: Daniel and Samuel from Sleight of Math Disclaimer: Sleight of Math has no affiliations with the publisher of Additional Maths 360 i.e. Marshall Cavendish
Table of Contents Ex 1.1 ......................................................................................................... 3
Ex 6.1 ................................................................................................... 143
Ex 1.2 ......................................................................................................... 9
Ex 6.2 ................................................................................................... 149
Ex 1.3 .......................................................................................................20
Ex 6.3 ................................................................................................... 153
Ex 1.4 .......................................................................................................29
Ex 6.4 ................................................................................................... 162
Rev Ex 1..................................................................................................35
Rev Ex 6 .............................................................................................. 167
Ex 2.1 .......................................................................................................42
Ex 7.1 ................................................................................................... 174
Ex 2.2 .......................................................................................................51
Ex 7.2 ................................................................................................... 180
Ex 2.3 .......................................................................................................55
Ex 7.3 ................................................................................................... 184
Ex 2.4 .......................................................................................................60
Ex 7.4 ................................................................................................... 189
Rev Ex 2..................................................................................................63
Ex 7.5 ................................................................................................... 193 Rev Ex 7 .............................................................................................. 196
Ex 3.1 .......................................................................................................67 Ex 3.2 .......................................................................................................72
Ex 8.1 ................................................................................................... 202
Ex 3.3 .......................................................................................................78
Ex 8.2 ................................................................................................... 209
Ex 3.4 .......................................................................................................81
Rev Ex 8 .............................................................................................. 224
Ex 3.5 .......................................................................................................86 Ex 3.6 .......................................................................................................91
Ex 9.1 ................................................................................................... 233
Rev Ex 3..................................................................................................98
Ex 9.2 ................................................................................................... 242 Rev Ex 9 .............................................................................................. 252
Ex 4.1 .................................................................................................... 103 Ex 4.2 .................................................................................................... 108
Ex 10.1 ................................................................................................. 259
Rev Ex 4............................................................................................... 114
Ex 10.2 ................................................................................................. 266 Ex 10.3 ................................................................................................. 272
Ex 5.1 .................................................................................................... 118
Rev Ex 10............................................................................................ 280
Ex 5.2 .................................................................................................... 127 Rev Ex 5............................................................................................... 138
Ex 11.1 ................................................................................................. 288 Ex 11.2 ................................................................................................. 291 Ex 11.3 ................................................................................................. 298 Rev Ex 11............................................................................................ 319
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1
Additional Maths 360 solutions (Unofficial) [29 Sept 2014] Visit sleightofmath.com for up-to-date pdf and video solutions. Authors: Daniel and Samuel from Sleight of Math Disclaimer: Sleight of Math has no affiliations with the publisher of Additional Maths 360 i.e. Marshall Cavendish Ex 12.1 ................................................................................................. 326
Ex 18.1 ................................................................................................. 495
Ex 12.2 ................................................................................................. 330
Ex 18.2 ................................................................................................. 503
Rev Ex 12 ............................................................................................ 339
Ex 18.3 ................................................................................................. 509 Ex 18.4 ................................................................................................. 512
Ex 13.1 ................................................................................................. 347
Rev Ex 18............................................................................................ 517
Ex 13.2 ................................................................................................. 356 Ex 13.3 ................................................................................................. 366
Ex 19.1 ................................................................................................. 523
Rev Ex 13 ............................................................................................ 376
Ex 19.2 ................................................................................................. 530 Rev Ex 19............................................................................................ 535
Ex 14.1 ................................................................................................. 383 Ex 14.2 ................................................................................................. 390
Ex 20.1 ................................................................................................. 539
Ex 14.3 ................................................................................................. 396
Rev Ex 20............................................................................................ 548
Ex 14.4 ................................................................................................. 400 Rev Ex 14 ............................................................................................ 407 Ex 15.1 ................................................................................................. 410 Ex 15.2 ................................................................................................. 418 Ex 15.3 ................................................................................................. 421 Ex 15.4 ................................................................................................. 423 Rev Ex 15 ............................................................................................ 430 Ex 16.1 ................................................................................................. 436 Ex 16.2 ................................................................................................. 445 Rev Ex 16 ............................................................................................ 454 Ex 17.1 ................................................................................................. 460 Ex 17.2 ................................................................................................. 470 Ex 17.3 ................................................................................................. 479 Rev Ex 17 ............................................................................................ 488
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2
A math 360 sol (unofficial)
Ex 1.1 2(a)
Ex 1.1 1(a)
y = 2x + 1 −(1) 2 y = x + 2x − 3 −(2) sub (1) into (2): 2x + 1 = x 2 + 2x − 3 2 x −4 =0 (x + 2)(x − 2) = 0 x = −2 ✓ or x = 2 ✓ y|x=−2 = 2(−2) + 1 y|x=2 = 2(2) + 1 = −3 ✓ =5✓
1(b)
y=2+x y = 2x 2 − 5x − 6
y=2−x 2x 2 + xy + 1 = 0 sub (1) into (2): 2x 2 + x(2 − x) + 1 2x 2 + 2x − x 2 + 1 x 2 + 2x + 1 (x + 1)2 x = −1 y|x=−1 = 2 − (−1) ⇒ (−1,3) ✓
2(b)
−(1) −(2)
y = 1 − 3x x2 + y2 = 5
−(1) −(2)
=5 =5 =0 =0 =0 or
5
x=1
2
y|x=−2 = 1 − 3 (− )
y|x=1 = 1 − 3(1)
5
5
11
= −2
5 2 11
2x + y = 4 y = 4 − 2x −(1)
⇒ (− , 5
2(c)
5
)✓
⇒ (1, −2) ✓
3x + 2y = 1 2y = 1 − 3x y
=
1−3x 2
−(1)
3x 2 + 2y 2 = 11 −(2) sub (1) into (2): 3x 2 + 2 ( 3x 2 + 2 (
1−3x 2
)
2 1−6x+9x2 4
= 11 )
= 11
6x 2 + (1 − 6x + 9x 2 ) = 22 15x 2 − 6x − 21 =0 5x 2 − 2x − 7 =0 (5x − 7)(x + 1) =0 x=
7
or x = −1
5
y|x=7 =
7 5
1−3( )
5
=− 7
8
5
5
2 8 5
⇒ ( ,− ) ✓
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=3
2
x=−
=
y 2 − 4x = 0 −(2) sub (1) into (2): (4 − 2x)2 − 4x =0 2 (16 − 16x + 4x ) − 4x = 0 4x 2 − 20x + 16 =0 2 x − 5x + 4 =0 (x − 1)(x − 4) =0 x=1 or x = 4 ✓ y|x=1 = 4 − 2(1) y|x=4 = 4 − 2(4) =2✓ = −4 ✓
=0 =0 =0 =0
sub (1) into (2): x 2 + (1 − 3x)2 x 2 + (1 − 6x + 9x 2 ) 10x 2 − 6x − 4 5x 2 − 3x − 2 (5x + 2)(x − 1)
sub (1) into (2): 2+x = 2x 2 − 5x − 6 2 2x − 6x − 8 = 0 x 2 − 3x − 4 =0 (x + 1)(x − 4) = 0 x = −1 ✓ or x = 4 ✓ y|x=−1 = 2 + (−1) y|x=4 = 2 + (4) =1✓ =6✓ 1(c)
−(1) −(2)
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y|x=−1 =
1−3(−1) 2
=2 ⇒ (−1,2) ✓
3
A math 360 sol (unofficial) 3
x2
y
Ex 1.1
+ =3
−(1)
x + y= 8 y =8−x
−(2)
4
3
5(ii)
Perimeter 2x + 2y x+y y
sub (2) into (1): x2 4
+
(8−x)
(3x 2 ) + (32 − 4x) = 36 3x 2 − 4x − 4 =0 (3x + 2)(x − 2) =0 2 3
4(i) 𝑦
4x + 4y = 32 x+y =8 y =8−x ✓
−(1)
x 2 + y 2 = 34 ✓
−(2)
4(iii) sub (1) into (2): x 2 + (8 − x)2 = 34 2 2 x + (64 − 16x + x ) = 34 2x 2 − 16x + 30 =0 2 x − 8x + 15 =0 (x − 3)(x − 5) =0 x=3 or x=5 y|x=3 = 5✓ y|x=5 = 3✓ ∴ the length of the sides are 3cm & 5cm ✓ 5(i)
= 60 = 60 = 30 = 30 − x −(2)
sub (2) into (1): x(30 − x) = 216 2 (30x − x ) = 216 2 x − 30x + 216 = 0 (x − 12)(x − 18) = 0 x = 12 or y|x=12 = 30 − (12) = 18 ∴ 12 m by 18 m ✓
x = − or x = 2 ✓
4(ii)
−(1)
=3
3
𝑥
Area = 216 xy = 216
𝑦 𝑥 Area = xy ✓ Perimeter = 2x + 2y ✓
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6(a)
x = 18 y|x=18 = 30 − (18) = 12
3x − 2y = 1 2y = 3x − 1 y
=
3x−1 2
−(1)
(x − 2)2 + (2y + 3)2 = 26 sub (1) into (2): (x − 2)2
+ [2 (
3x−1 2
) + 3]
−(2) 2
= 26
x 2 − 4x + 4 + (3x − 1 + 3)2 = 26 x 2 − 4x + 4 + (3x + 2)2 = 26 2 2 (9x x − 4x + 4 + + 12x + 4) = 26 10x 2 + 8x + 8 = 26 10x 2 + 8x − 18 = 0 5x 2 + 4x − 9 =0 (5x + 9)(x − 1) = 0 x=−
9
or
5
y|x=−9 =
9 3(− )−1 5
5
=−
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2 16 5
✓
x=1✓ y|x=1 =
3(1)−1 2
=1✓
4
A math 360 sol (unofficial) 6(b)
Ex 1.1
x 2 − 2xy + y 2 = 1 x − 2y −2y
=2 = −x + 2
y
= x−1
7(a)
−(1)
xy + 20 = 5x
−(1)
x − 2y − 3 = 0 −2y = −x + 3
1
y
−(2)
2
1
3
2
2
= x−
−(2)
sub (1) into (2): 1
1
x 2 − 2x ( x − 1) + ( x − 1) 2
2
2
x − x + 2x 2x 1 2 x 4 1 2 x 4 2
sub (2) into (1):
=1
+x
=0
1
y|x=0 = (0) − 1
2 1 2 x 2 2
2
= −1 ✓
−
2 13 2
x + 20
1
3
2
2
y|x=5 = (5) −
=0
2
= −3 ✓ 7(b)
1
= x + 1 −(1)
=
⇒ (5,1) ✓
1
3
2 5
2
y|x=8 = (8) −
=1
1
= (−4) − 1
3y − x = 3 3y =x+3 y
= 5x
x − 13x + 40 =0 (x − 5)(x − 8) =0 x=5 or x = 8
x = −4 ✓ y|x=−4
2 3
1
=0 =0 or
3
2
( x 2 − x) +20 = 5x
=1 =1
1
x ( x − ) + 20
=1
+x+1
x + 4x x(x + 4) x=0✓
6(c)
2
2 1 + ( x 2 − x + 1) 4 1 2 +( x − x + 1) 4
2 5
⇒ (8, ) ✓ 2
2x − y = 4 −y = −2x + 4 y = 2x − 4
−(1)
2x 2 + 4xy − 3y = 0
−(2)
3
2
1
3y
− =2
−(2)
x
sub (1) into (2): 2
−
1 3
3( x+1) 2
−
x+3
1 1
=2
x
2x −(x + 3) 2x −x − 3 x−3 0 2x 2 + 5x + 3 (2x + 3)(x + 1) 3
x=− ✓
= 2x(x + 3) = 2x 2 + 6x = 2x 2 + 6x = 2x 2 + 5x + 3 =0 =0 or
2
1
3
3 1
2
y|x=−3 = (− ) + 1 2
sub (1) into (2): 2x 2 + 4x(2x − 4) − 3(2x − 4) 2x 2 + (8x 2 − 16x) −6x + 12 (10x 2 − 16x) −6x + 12 10x 2 − 22x + 12 5x 2 − 11x + 6 (5x − 6)(x − 1)
=2
x
= ✓ 2
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x=
6
or
5 6
y|x=6 = 2 ( ) − 4
x = −1 ✓
5
=−
1
y|x=−1 = (−1) + 1 3 2
6
8
5
5
5 8 5
⇒ ( ,− ) ✓
= ✓
=0 =0 =0 =0 =0 =0
x=1 y|x=1 = 2(1) − 4 = −2 ⇒ (1, −2) ✓
3
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5
A math 360 sol (unofficial) 7(c)
Ex 1.1
3x + y = 1 y = 1 − 3x
9 4𝑥
By Pythagoras Theorem, (4x)2 + (3x)2 = 152 16x 2 + 9x 2 = 225 25x 2 = 225 2 x =9 x = 3 or x = −3 (rej)
(x + y)(x + 2y) = 3 −(2) (1) (2): sub into [x + (1 − 3x)][x + 2(1 − 3x)] = 3 (1 − 2x)(x + 2 − 6x) =3 (1 − 2x)(2 − 5x) =3 (5x − 2)(2x − 1) =3 2 10x − 9x + 2 =3 2 10x − 9x − 1 =0 (10x + 1)(x − 1) =0 x=−
1
y|x=− 1 = 1 − 3 (− 10
= ⇒ (− 8
2x y
,
1
)
10
y|x=1 = 1 − 3(1)
13 10 13
)✓
⇒ (1, −2) ✓
10 10
y
−(1)
x
3x − y = 2 3x − 2 = y y = 3x − 2 −(2) sub (2) into (1): 2x
+
10(i) 𝑦
= −2
+ =3
3x−2 2
Width = 4(3) = 12 inch ✓ Height = 3(3) = 9 inch ✓
or x = 1
10
1
3𝑥
−(1)
3x−2
=3
x 2
2x + (3x − 2) = 3(3x 2 − 2x) 2x 2 + (9x 2 − 12x + 4) = 9x 2 − 6x 11x 2 − 12x + 4 = 9x 2 − 6x 2 2x − 6x + 4 =0 2 x − 3x + 2 =0 (x − 1)(x − 2) =0 x=1 or x = 2 y|x=1 = 3(1) − 2 y|x=2 = 3(2) − 2 =1 =4 ⇒ A(1,1) ✓ ⇒ B(2,4) ✓
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5 𝑥
By Pythagoras’ Theorem, x 2 + y 2 = 52 x 2 + y 2 = 25 ✓ −(1) 10(ii) y − x = 1 y =x+1 −(2) sub (1) into (2): x 2 + (x + 1)2 = 25 2 2 x + (x + 2x + 1) = 25 2x 2 + 2x + 1 = 25 2x 2 + 2x − 24 =0 2 x + x − 12 =0 (x + 4)(x − 3) =0 x = −4 ✓ or x=3✓ y|x=−4 = (−4) + 1 y|x=3 = (3) + 1 = −3 ✓ =4✓ 10(iii) ∵ length cannot be negative, x = 3, y = 4
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6
A math 360 sol (unofficial)
Ex 1.1
11(i) x 2 + xy + ay = b at {x = 2, y = 1} (2)2 + (2)(1) + a(1) = b 6+a =b b =a+6 −(1) 2ax + 3y = b at {x = 2, y = 1} 2a(2) + 3(1) = b 4a + 3 =b sub (1) into (2): 4a + 3 = 6 + a 3a =3 a =1✓ b|a=1 = 7 ✓
12(i) 1st eqn 12x 2 − 5y 2 = 7 At (1, p), 12(1)2 − 5(p)2 = 7 12 − 5p2 =7 2 −5p = −5 p2 =1 p = ±1
−(2)
2nd eqn 2p2 x − 5y = 7 At (1, p), 2p2 (1) − 5(p) = 7 2p2 − 5p − 7 = 0 (2p − 7)(p + 1) = 0
11(ii) Put a − 1, b = 7 into both equations, x 2 + xy + y = 7 −(1)
2
7
3
3
= − x + −(2)
2
7
3
3 7
x + x (− x + ) 2 x + (− x 2 + x) 3 3 1 2 7 x + x 3 3 1 2 5 7 2
x + x+
3 1 2 x 3 2
3 5
3 14
3
3
+ x−
or p = −1
2
∴ p = −1 (common sol) ✓
−(1)
2x − 5y = 7 −5y = −2x + 7
sub (2) into (1): 2
7
12(ii) At p = −1, 12x 2 − 5y 2 = 7
2x + 3y = 7 3y = −2x + 7 y
p=
y 2
7
3 2
3 7
3 2
3 7
3
3
2
7
5
5
= x−
+ (− x + ) = 7 + (− x + ) = 7
sub (2) into (1):
+ (− x + ) = 7
12x 2 − 5 ( x − )
=7
12x 2 − 5 (
=0
x + 5x − 14 = 0 (x + 7)(x − 2) = 0 x = −7 or x = 2 (taken) 2
7
3
3
y|x=−7 = − (−7) + =7 {x = −7, y = 7} ✓
2
7 2
5
5
4
x2 −
25
12x 2 − x 2 +
4
28
5
5
56 2 x 5
+
49
56 2 x 5
+
28 5 28 5
−(2)
x− x−
=7
28 25
x−
x+ 49 5
5
=0
(2x + 3)(x − 1)
=0
3
or
2
x = 1 (taken)
2
3
7
5
2
5
y|x=−3 = (− ) − 2
=7
=0
2x 2 + x − 3
x=−
) =7
=7
5 84
49
25
= −2 3
⇒ (− , −2) ✓ 2
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7
A math 360 sol (unofficial)
Ex 1.1
13(i)
14(ii) If k = 1, 𝑦 h
(7𝑥 − 20)2 + (6𝑦 + 10)2 = 200
r Total surface area = 32π 2πr 2 + 2πrh = 32π 2 r + hr = 16 [shown] ✓ −(1) 13(ii) h = 4 + r sub (2) into (1): r 2 + (4 + r)r = 16 2r 2 + 4r = 16 2 2r + 4r − 16 = 0 r 2 + 2r − 8 = 0 (r + 4)(r − 2) = 0 r = −4 or (rej ∵ r > 0)
=
y r=2✓ h|r=2 = 4 + (2) =6✓
6
2
) + 10]
(7x − 20) + (20 − 21x)
2
2
(x−2.86)2 (2.02)2
=0
49x 2 − 112x + 60
=0
−(−112)±√(−112)2 −4(49)(60) 2(49)
=
112±√784 98
6
or x = ✓ 7
✓
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6
+
+
5 2 62 (y+ ) 3 5 2 (y+ ) 3 200 62
=1 2
5 3
(y−(− ))
=1
2 200 (√ 2 ) 6 2
5 3
(y−(− )) (
+
=1
200
10√2 ) 6
(y−(−1.67))
=1
2
2
=1
(2.36)2 20 7
5
,− ) 3
⇒ horizontal radius = √
200 72
200
490x 2 − 1120x + 600
3
+
⇒ centre (
= 200 = 200
=−
1
2
5 2
+
20 2 7 2 10√2 ( ) 7
= 200
490x − 1120x + 800
6 10
7
3
(x− )
= 200
2
10 10−21( ) 7
7
20 2 7 2 200 (√ 2 ) 7
−(2)
(49x 2 − 280x + 400) +(400 − 840x + 441x 2 )
7
6
2
(x− )
(7x − 20)2 + [(10 − 21x) + 10]2 = 200
y|x=10 =
1
2
) + 62 (y + ) = 200
20 2 (x− ) 7 200 72
−(1)
10−21x
7
7
20 2
72 (x −
sub (1) into (2):
x=
3
Ellipse (7x − 20)2 +(6y + 10)2 = 200
200
6
10
,− )
=− x+
20 72 (x− ) 7
Ellipse (7x − 20)2 + (6y + 10)2 = 200
x=
7
5
Line 21x + 6y = 1 6y = −21x + 1
10−21x
(7x − 20)2 + [6 (
(
20
𝑦=− 𝑥+
−(2)
14(i) If k = 10, Line 21x + 6y = 10 6y = 10 − 21x y
𝑥
𝑂
y|x=6 =
⇒ vertical radius = √
62
Using the graphing calculator to plot curves, the line and ellipse don’t intersect. Hence, there are no solutions ✓ =
112±28 98
14(iii) Geometrically, a line and ellipse can only intersect twice at most. ✓
6 7
10−21( ) 6
7
=−
4 3
✓
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8
A math 360 sol (unofficial)
Ex 1.2 4(i)
Ex 1.2 1(a)
3x 2 + 9x =1 2 3x + 9x − 1 = 0 i.e. a = 3, b = 9, c = −1 Roots: α & β Sum of roots
1(b)
Sum of roots
b a
c
=
=
a
(−1) (3)
=− ✓ 3
1 2
=( )
α
2(ii)
3
b a
=−
(−4)
=4✓
(1)
(2)
c
2
2β+2α
a
1
= (4 )
β
αβ
+ =
=
αβ
=
5(i) b a
= (1)
a
2(−3) (1)
(3)
= − (1) = −3 (1)
c
= =
=1
= −6 ✓
√33 2
✓
x 2 − 4x + c = 0 i.e. a = 1, b = −4, c = c Roots: α & (α + 2)
=−
b a (−4) (1)
2α + 2 =4 2α =2 α =1✓ α+2 =3✓ 5(ii)
Sum of roots = α + β = −
=
4
=±
⇒ 2α + 2
40x − 138x + 119 = 0 i.e. a = 40, b = −138, c = 119 Roots: α & β are heights of two men
2
= −2
Sum of roots
2
Average height =
2
33
= α + (α + 2) = −
(2α − 1)(2β − 1) = 4αβ −2α − 2β +1 = 4αβ −2(α + β) +1 = 4(1) −2(−3) +1 = 11 ✓
α+β
1
−2(−2)
4
α−β
=α+β =−
2(α+β)
(2)
=
(α − β)2 = α2 − 2αβ + β2 = (α2 + β2 ) −2αβ
=2✓
= (1)
x + 3x + 1 = 0 i.e. a = 1, b = 3, c = 1 Roots: α & β
2
(2) (−4)
4
=
Product of roots = αβ
=
(−1)
=4 ✓
2
Sum of roots
a
=−
−2(−2)
2 1
=α+β =−
Product of roots = αβ
=
a
(α + β)2 = α2 + 2αβ + β2 (α + β)2 − 2αβ = α2 + β2 α2 + β2 = (α + β)2 −2αβ
1
4x + 2x 2 = 3x 2 + 2 x 2 − 4x + 2 = 0 i.e. a = 1, b = −4, c = 2 Roots: α & β Sum of roots
c
b
(9)
= − (3) = −3 ✓
4(ii)
2(i)
=α+β =−
Product of roots = αβ
=α+β =−
Product of roots = αβ
2x 2 − x − 4 = 0 i.e. a = 2, b = −1, c = −4 Roots: α & β
69 20
2
b a
=
=− 69 40
Product of roots c = α(α + 2) = a
⇒ α(α + 2) = c 1(1 + 2) =c ∵α=1 c =3✓
(−138) (40)
=
69 20
6(a)
✓
Roots: α = 2 & β = 5 Sum of roots = α + β = (2) + (5) Product of roots = αβ = (2)(5)
=7 = (10)
x 2 − (SOR)x + (POR) = 0 x 2 − 7x + 10 =0✓
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9
A math 360 sol (unofficial) 6(b)
Ex 1.2
Roots: α = −1 & β = 3 Sum of roots = α + β = (−1) + (3) = 2 Product of roots = αβ = (−1)(3) = −3
8(i)
x 2 − (SOR)x + (POR) = 0 x 2 − 2x − 3 =0✓ 7(i)
Sum of roots
1st equation 2x 2 − 4x + 5 = 0 i.e. a = 2, b = −4, c = 5 Roots: α & β
=−
= 5α
= −p ✓ =
= 4α2
=α+β =−
Product of roots = αβ
=
c a
b a
=−
(−4)
(5)
= (2)
(2)
=2 =
From (1): 5α = −p p α =−
a
−(1)
c a
=q✓
−(2)
−(3)
5
5
b
= α + 4α
Product of roots = α(4α)
8(ii) Sum of roots
x 2 + px + q = 0 i.e. a = 1, b = p, c = q Roots: α & 4α
2
sub (3) into (2): 2nd equation Roots: (α − 1) & (β − 1) Sum of roots Product of roots = (α − 1) + (β − 1) = (α − 1)(β − 1) = (α + β) −2 = αβ −α − β +1 (2) = −2 = αβ −(α + β) +1 5 =0 = ( ) −(2) +1
5
4p2
4p 9(i)
3 2
x − (SOR)x + (POR) = 0 3
2x 2 + 3 7(ii)
2x 2 − x − 2 = 0 i.e. a = 2, b = −1, c = −2 Roots: α & β =α+β =− =
b a
c
=− =
a
(−1)
(2) (−2) (2)
=
1 2
= −1
=0✓ α2 + β2 = (α + β)2 −2(αβ)
3rd equation Recall α + β = 2,
= 25q [shown] ✓
Product of roots = αβ
=0
2
2
Sum of roots
2
x 2 − 0x +
=q
25
2
=
2
1
4 (− p) = q
αβ =
5
1 2
=( )
2
−2(−1)
2 1
Roots: 2α & 2β
=2 ✓ 4
Sum of roots = 2α + 2β = 2(α + β) = 2(2) = 4
9(ii)
β α
+
α β
=
β2 +α2 αβ
1
(2 )
4 = (−1) = −2
1 4
✓
Product of roots = (2α)(2β) = 4(αβ)
5
= 4 ( ) = 10
9(iii)
2
α4 + β4
= (α2 + β2 )2 −2α2 β2 = (α2 + β2 )2 −2(αβ)2
x 2 − (SOR)x + (POR) = 0 x 2 − 4x + 10 =0 ✓
1 2
= (2 ) 4
=
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49 16
−2(−1)2
✓
10
A math 360 sol (unofficial) 10(i)
Ex 1.2
x2 = (k − 1)x + k 2 (1 x + − k)x − k = 0 i.e. a = 1, b = (1 − k), c = −k Roots: α & (α + 3)
12(i)
Sum of roots = α + (α + 3) = −
Sum of roots
b
=−
2α + 3 k
=k−1 = 2α + 4
=α+β =−
Product of roots = αβ
a (1−k)
⇒ 2α + 3
=
b
=−
a
c
=
a
−3 2
6
=
3 2
=3
2
(1)
2nd equation −(1)
Roots:
2α β
&
2β α
Sum of roots =
Product of roots c = α(α + 3) = ⇒ α(α + 3) =
1st equation 2x 2 − 3x + 6 = 0 i.e. a = 2, b = −3, c = 6 Roots: α & β
a (−k)
=
2α β
+
2β α
=
2α2 +2β2 αβ
2[(α+β)2 −2αβ]
(1)
αβ 2α
2β
β
α
=
=
2(α2 +β2 ) αβ
3 2 2[( ) −2(3)] 2
3
=−
5 2
Product of roots = ( ) ( ) = 4
= −k −(2) sub (1) into (2): α(α + 3) = −(2α + 4) 2 α + 3α = −2α − 4 2 α + 5α + 4 =0 (α + 1)(α + 4) = 0 α = −1 or α = −4 ✓ α+3=2 α + 3 = −1 ✓ (rej ∵ negative roots)
x 2 − (SOR)x + (POR) = 0 5
x 2 − (− ) x +4
=0
2
5
x2 + x 2
2
2x +5x
+4
=0
+8
=0✓
12(ii) 3rd equation 3
Recall α + β = , αβ = 3 2
10(ii) Put α = −4 into (1): k|α=−4 = 2(−4) + 4 = −4 ✓ 11
Roots: (3α + β) & (α + 3β) Sum of roots = (3α + β) + (α + 3β) = 4α + 4β = 4(α + β)
3x 2 − 3kx + k − 6 = 0 i.e. a = 3, b = −3k, c = k − 6 Roots: α & β
3
= 4( ) 2
Sum of roots
=α+β =−
Product of roots = αβ α2 + β2
=
2
(α + β) − 2αβ k
2
−2 (
k−6 3
)
= =
=
c a
b a
=− =
(−3k)
(3) (k−6) (3)
=6 =k =
k−6 3
20 3 20 3 20
3 2
= 3 [( ) − 2(3)] 2
3
3k 2 − 2k + 12 = 20 2 3k − 2k − 8 =0 (3k + 4)(k − 2) = 0 k=−
4 3
Product of roots = (3α + β)(α + 3β) = 3α2 + 10αβ + 3β2 = 3(α2 + β2 ) +10αβ = 3[(α + β)2 − 2αβ] +10αβ
=
75 4
x 2 − (SOR)x + (POR) = 0
or k = 2 ✓
x 2 −(6)x +
75 4
4x 2 −24x +75
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=0 =0✓
11
A math 360 sol (unofficial) 13(i)
Ex 1.2
1st equation 2x 2 = 8x + 3 2x 2 − 8x − 3 = 0 i.e. a = 2, b = −8, c = −3 Roots: α & β
13(iii) 4th equation Recall α + β = 4, αβ = −
=α+β =−
Product of roots
= αβ
=
b a
c
=− =−
a
(−8) (2) 3
Sum of roots = (α − β) + (β − α) =0
=4
Product of roots = (α − β)(β − α) = −(a − β)2 = −[(α + β)2 − 4αβ]
2
2nd equation 1 α2
&
1
= − [(4)2
β2
Sum of roots = =
1
+
α2
1
=
β2
(α+β)2 −2(αβ) (αβ)2
1
=
α2 +β2 α2 β2 3 2
(4)2 −2(− )
1
3
− 4 (− )] 2
= −22
3 2 (− ) 2
1
Product of roots = ( 2 ) ( 2 ) = (αβ)2 = α
2
Roots: (α − β) & (β − α)
Sum of roots
Roots:
3
β
=
1 3 2 (− ) 2
x 2 − (SOR)x + (POR) = 0 x 2 − (0)x + (−22) = 0 x 2 − 22 =0✓
76 9
=
4
14(i)
9
2x 2 − 71x + 615 = 0 i.e. a = 2, b = −71, c = 615 Roots: α & β are base and height
x 2 − (SOR)x + (POR) = 0 76
4
9
9
x2 − ( ) x + ( )
=0
Sum of roots = α + β = −
b
=−
a
2
9x − 76x +4 = 0 ✓
Product of roots = αβ
=
c a
−71
=
2 615 2
71
=
2
= 307.5
13(ii) 3rd equation Recall
α + β = 4, αβ = −
Roots:
α2 β & αβ2
Area = αβ = 307.5 ✓
3 2
71
Perimeter = 2(α + β) = 2 ( ) = 71 ✓ 2
Sum of roots = α2 β + αβ2 = αβ(α + β)
14(ii)
3
= (− ) (4) = −6
Base
2
Product of roots 3 3
27
2
8
= (α2 β)(αβ2 ) = (αβ)3 = (− ) = −
) =0
8x 2 + 48x − 27
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Base
Perimeter is not valid because the height is not the side of the quadrilateral ✓
27 8
Height
Area is valid because area = base × height
x 2 − (SOR)x + (POR) = 0 x 2 − (−6)x + (−
Height
=0✓
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A math 360 sol (unofficial) 15
Ex 1.2
kx 2 + (k − 1)x + 2k + 3 = 0 i.e. a = k, b = (k − 1), c = (2k + 3) Roots: α & 2α Sum of roots = α + (2α) = − ⇒ 3α
=−
3α
=
α
=
1st equation x 2 − 3x − 2 = 0 i.e. a = 1, b = −3, c = −2 Roots: α & β
b a k−1
Sum of roots
k
Product of roots = αβ
1−k k 1−k 3k
−(1)
2nd equation x 2 − 6x + p = 0 i.e. a2 = 1, b2 = −6, c2 = p
−(2)
Roots:
Product of roots c = α(2α) = 2
⇒ 2α
=
a 2k+3 k
2( 2(
)
3k k2 −2k+1
9k2 2k2 −4k+2 9k2 2
)
= = =
k
2k+3
= +
k 2k+3
⇒
1 16
k
α β kβ+kα
αβ k(α+β)
k 2k+3
&
c a
a
=− =
−3
−2 1
1
=3 = −2
k β
=−
b2 a2 −6 1
=6 = −4 ✓
k
or k = −2 ✓
=−
=6
αβ k(3) (−2)
k
2k − 4k + 2 = 18k 2 + 27k 16k 2 + 31k − 2 = 0 (16k − 1)(k + 2) = 0 k=
k α
=
b
Sum of roots
sub (1) into (2): 1−k 2
=α+β =−
Product of roots
rej ∵ k is a ( ) non − zero integer
k
k
c2
α
β
a2 (p)
= ( )( ) = ⇒
k2 αβ
k2 αβ
= (1) =p
k2 (−2)
=p
p
=− =−
k2
∵ k = −4
2 (−4)2 2
= −8 ✓
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A math 360 sol (unofficial)
Ex 1.2
17(a) 3x 2 + kx + 96 = 0 i.e. a = 3, b = k, c = 96 Roots: α & 2α
18
5x 2 − 102x + 432 = 0 i.e. a = 5, b = −102, c = 432 Roots: α & β are lengths of the two shorter sides of a ⊿
Sum of roots = α + 2α = −
b
Sum of roots
a k
⇒ 3α
=−
α
=−
−9α k
=k = −9α
Product of roots = αβ
3 k
⇒ 2α
=
a
a
=− =
(−102) (5)
=
102 5
432 5
√α2 + β2
α β
By Pythagoras Theorem, Hypotenuse = √α2 + β2
a 96 3
2
2α = 32 2 α = 16 α = 4 or α = −4 (rej ∵ positive roots) k|α=4 = −9(4) = −36 ✓ 2
=
c
b
9
Product of roots c = α(2α) = 2
=α+β =−
1 432
2
2
5
)= 43.2cm2 ✓
Perimeter = α + β + √α2 + β2 = α + β + √(α + β)2 − 2αβ =(
2
17(b) p + q = 13 pq = 6 Roots: p2 & q2 Sum of roots = p2 + q2 = 13 Product of roots = p2 q2 = (pq)2 = 62
1
Area = (αβ) = (
102 5
) + √(
102 2 5
) − 2(
432 5
)
= 36cm ✓
= 36
x 2 − (SOR)x + (POR) = 0 x 2 − 13x + 36 =0 (x − 4)(x − 9) =0 x=4 or x = 9 p2 = 4 p2 = 9 p = ±2 p = ±3 ✓
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A math 360 sol (unofficial)
Ex 1.2
19(a) 2x 2 + px + 24 = 0 i.e. a = 2, b = p, c = 24 Roots: α & β
19(b) 1st equation 3x 2 − x + 2 = 0 i.e. a = 3, b = −1, c = 2 Roots: α & β
α−β =4 α = β + 4 −(1)
Sum of roots
=α+β =−
Product of roots = αβ
Sum of roots =α+β
=−
⇒α+β
=−
(β + 4) + β = −
=
=−
a
c
=
a
(−1)
=
(3)
1 3
2 3
b a p
2nd equation Roots: α2 & β2
2 p 2 p
2β + 4
=−
p
= −4β − 8
Sum of roots = α2 + β2 = (α + β)2 −2(αβ)
2
1 2
2
=( )
=
−2 ( )
3
Product of roots c = αβ = ⇒ αβ
b
=−
a 24
3
11 9
2
(β + 4)β = 12 β2 + 4β = 12 2 β + 4β − 12 = 0 (β + 6)(β − 2) = 0 β = −6 or p|β=−6 = −4(−6) − 8 = 16 (rej ∵ p < 0)
2 2
4
3
9
Product of roots = α2 β2 = (αβ)2 = ( ) = x 2 − (SOR)x + (POR) = 0 x 2 − (−
11 9
)x +
4
=0
9
9x 2 + 11x + 4 β=2 p|β=2 = −4(2) − 8
20(i)
= −16 ✓
=0✓
1st equation 4x 2 − x + 36 = 0 i.e. a = 4, b = −1, c = 36 Roots: α2 & β2 = α2 + β2 = −
Sum of roots
2 2
Product of roots = α β
=
c a
b a
=− =
(−1)
(4) (36) (4)
=
1 4
=9
2nd equation Roots:
1 α2
&
1 β2
Sum of roots
=
1
1
α
1
β2 1
α
β
2 +
=
Product of roots = ( 2 ) ( 2 ) =
α2 +β2 α2 β2 1 α2 β2
= =
1 4
( ) (9) 1
=
1 36
9
x 2 − (SOR)x + (POR) = 0 x2 − 2
1 36
x+
1 9
36x − x + 4
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=0 =0✓
15
A math 360 sol (unofficial)
Ex 1.2
20(ii) 3rd equation: 2
1
2
Recall α + β = , 4
21(ii) 2nd equation Roots: (α + 2) & (β + 2)
2 2
α β =9
Roots: α & β Sum of roots =α+β
Product of roots = αβ
= ±√(α +
=
±√α2
β2
= ±√9 = ±3
+
+ 2αβ
1
= ±√( ) + 2αβ
=−
±√α2 β2
β)2
=
Sum of roots = (α + 2) + (β + 2) = (α + β) + 4 b
1
= [a(α + 2)(β + 2)] a 1
+4
a
= (4a − 2b + c) a
x 2 − (SOR)x
4
Product of roots = (α + 2)(β + 2)
+ (POR)
b
1
a
a 1
=0
x 2 − (− + 4) x + [ (4a − 2b + c)] = 0
For αβ = 3:
2
1
25
4
4
SOR = ±√ + 2(3) = ±√
=±
b
x + ( − 4) x
5
a
+ (4a − 2b + c) a
ax 2 + (b − 4a)x + (4a − 2b + c)
2
=0 =0✓
For αβ = −3: 1
23
4 5
24
SOR = ±√ + 2(−3) = ±√−
(rej)
∴ SOR = ± , POR = 3 2
x 2 − (SOR)x + (POR) = 0 5
x 2 − (± ) x + 3
=0
2
2
=0✓
2x ± 5x + 6 21(i)
1st equation ax 2 + bx + c = 0 Roots: α & β Sum of roots
=α+β =−
Product of roots = αβ
=
c
b a
a
a(α + 2)(β + 2) = a(αβ + 2α + 2β + 4) = a[αβ + 2(α + β) + 4] c
b
a
a
= a [( ) + 2 (− ) + 4] = c − 2b + 4a = 4a − 2b + c [shown] ✓
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A math 360 sol (unofficial)
Ex 1.2
1st equation 4x 2 + kx =1 4x 2 + kx − 1 = 0 i.e. a = 4, b = k, c = −1
22(ii) 3rd equation k
Recall α + β = − , Roots:
=α+β =−
Product of roots = αβ
=
b a
c
=− =−
a
k 4 1
=
4
Sum of roots
= 5(α + β)
9 9 = 1 αβ + 2(α + β) + 4 (− ) + 2 (− k) + 4 4 4 36 = 15 − 2k
= 5 (− ) 4
5
=− k 4
Product of roots
x2
= (2α + 3β)(3α + 2β) 2
= 6α + 13αβ + 6β
23(i)
+13αβ
1
1
4
4
1
k2 + )
−
2
8
Sum of roots
4
5
x 2 + kx 4
2
8x + 10kx
=α+β =−
Product of roots = αβ
x − (SOR)x +(POR) x − (− k) x
) =0 = 0✓
1st equation 2x 2 + 4x + 5 = 0 i.e. a = 2, b = 4, c = 5 Roots: α & β
4
4
5
15−2k
13
1
2 2
)x +(
+13αβ
4
= k −
15−2k
=0 36
(15 − 2k)x +(3k − 48)x +36
2)
k 2
2
−(
48−3k
2
= 6 [(− ) − 2 (− )] +13 (− )
3
−(SOR)x +(POR)
x2
2
= 6[(α + β)2 − 2αβ]
1
3β + 6 + 3α + 6 3(α + β) + 12 = αβ + 2(α + β) + 4 αβ + 2(α + β) + 4
=
k
16
β+2
3 3 9 )( )= =( (α + 2)(β + 2) α+2 β+2
= 5α + 5β
= 6(
4
3
Product of roots
= (2α + 3β) + (3α + 2β)
= 6(α + β
α+2
&
k 3 (− ) + 12 48 − 3k 4 = = 1 k (− ) + 2 (− ) + 4 15 − 2k 4 4
2nd equation Roots: (2α + 3β) & (3α + 2β)
2
3
1
Sum of roots 3 3 3(β + 2) + 3(α + 2) = + = (α + 2)(β + 2) α+2 β+2
Roots: α & β Sum of roots
αβ = −
4
3 + ( k2 8
=0 1
1
8
4
2
Roots:
4
3
+3k − 2
a
c
=−
4 2
= −2 ✓
5
= ✓
a
2
23(ii) 2nd equation
− )=0
+ k2 −
=
b
1 α
&
1 β
=0 Sum of roots
=0✓
1
1
α
β
= +
=
α+β αβ
1
1
1
α
β
αβ
Product of roots = ( ) ( ) =
= =
−2 5 2
1 5 2
=− =
4 5
2 5
x 2 − (SOR)x + (POR) = 0 4
2
5
5
x 2 − (− ) + 2
5x + 4 + 2
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=0 =0✓
17
A math 360 sol (unofficial)
Ex 1.2
23(iii) 1st equation ax 2 + bx + c = 0 Roots: α & β Sum of roots
25
=α+β =−
Product of roots = αβ
=
c
b a
Sum of roots
a
=α+β =−
2nd equation Roots:
1
&
α
2x 2 − kx + k =2 2 2x − kx + k − 2 = 0 i.e. a = 2, b = −k, c = (k − 2) Roots: α & β
1
⇒α+β =−
β
Sum of roots
1
1
α
β
= +
=
α+β αβ
1
1
1
α
β
αβ
Product of roots = ( ) ( ) =
= =
b − a c a
1
=− =
c a
b
b a −k 2
k
α+β
=
k
= 2α + 2β
2
a c
Product of roots c = αβ = ⇒ αβ =
x 2 − (SOR)x + (POR) = 0
a k−2 2
x 2 − (− ) x + ( )
=0
sub (1) into (2):
cx 2 − bx
=0✓
αβ =
b
a
c
c
+a
−(1)
c
−(2)
(2α+2β)−2 2
αβ = α + β − 1 24(i)
x(2 − x) =3 2 x − 2x + 3 = 0 α is root ⇒ α2 − 2α + 3 = 0 α2 = 2α − 3
If α < 0, β < 0 ⇒ LHS = αβ > 0 ⇒ RHS = α + β − 1 < 0 ⇒ LHS ≠ RHS ∴ both roots cannot be negative ✓
−(1)
(1) × α: α3 = 2α2 − 3α −(2) (1) (2): sub into α3 = 2(2α − 3) − 3α α3 = (4α − 6) − 3α α3 = α − 6 [shown] ✓ 24(ii) Roots: α & β Sum of roots
=α+β =−
Product of roots = αβ
=
c a
b a
=− =
(−2)
(3) (1)
(1)
=2 =3
Following the same manipulation in (i) ⇒ β3 = β − 6 α3 + β3 = (α − 6) + (β − 6) = (α + β) −12 = (2) −12 = −10 ✓
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A math 360 sol (unofficial) 26
Ex 1.2
x 2 + 4(c + 2) = (c + 4)x 2 (c x − + 4)x + 4(c + 2) = 0 i.e. A = 1, B = −(c + 4), C = 4(c + 2) Roots: a & b Sum of roots =a+b
=−
⇒a+b
=−
B A −(c+4) 1
a+b =c+4 2 (a + b) =c+4 2 2 a + 2ab + b = c 2 + 8c + 16
−(1)
Product of roots = ab = ⇒ ab = ab
C A 4(c+2) 1
= 4c + 8
−(2)
sub (2) into (1): a2 + 2(4c + 8) + b2 = c 2 + 8c + 16 a2 + (8c + 16) + b2 = c 2 + 8c + 16 a2 + b2 = c2 ∵ sides are related by pythagoras theorem, it is a right angle triangle & 90° is the largest angle ✓
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A math 360 sol (unofficial)
Ex 1.3 2(c)
Ex 1.3 1(a)
y = 3(x − 1)2 − 1 ⇒ turning pt (1, −1) ⇒ ∪ shape
𝑦 𝑦 = 3(𝑥 − 1)2 − 1 2 𝑂
𝑥 (1, −1)
y|x=0 = 2 ⇒ y − intercept = 2 1(b)
y = −2(x + 1)2 − 3 ⇒ turning pt (−1, −3) ⇒ ∩ shape y|x=0 = −5 ⇒ y − intercept = −5
1(c)
𝑂
1
(−2,1) 𝑂
y = −3(x − 2)2 ⇒ turning pt (2,0) ⇒ ∩ shape
𝑥
3(a) (2,0)
𝑂
𝑥
−12 y = −3(x − 2)2
Quadratic equation px 2 − 6x + p = 0 i.e. a = p, b = −6, c = p Discriminant For equal real roots: b2 − 4ac =0 (−6)2 − 4(p)(p) = 0 36 − 4p2 =0 2 p −9 =0 (p + 3)(p − 3) = 0 p = −3 or p = 3 ✓
✓
Quadratic equation 5x 2 − x − 2 = 0 i.e. a = 5, b = −1, c = −2
3(b)
Quadratic equation 3x 2 + 2x − p = 0 i.e. a = 3, b = 2, c = −p Discriminant For two distinct real roots: b2 − 4ac >0 (2)2 − 4(3)(−p) > 0 4 + 12p >0 12p > −4
Quadratic equation 9x 2 + 6x + 1 = 0 i.e. a = 9, b = 6, c = 1 Discriminant b2 − 4ac = (6)2 − 4(9)(1) =0✓ ⇒ 2 Real Roots ✓
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Quadratic equation (x − 2)2 = −6 2 x − 4x + 4 = −6 x 2 − 4x + 10 = 0 i.e. a = 1, b = −4, c = 10 Discriminant b2 − 4ac = (−4)2 − 4(1)(10) = −24 ✓ 0 ⇒ 2 Real roots ✓ 2(b)
2(d)
𝑦 1 𝑦 = (𝑥 + 2)2 + 1 4 2
4
y|x=0 = 2 ⇒ y − intercept = 2
2(a)
𝑥
✓
y = (x + 2)2 + 1
y|x=0 = −12 ⇒ y − intercept = −12
✓
−5 𝑦 = −2(𝑥 + 1)2 − 3
⇒ turning pt (−2,1) ⇒ ∪ shape
1(d)
Discriminant b2 − 4ac = (1)2 − 4(3)(1) = −11 ✓ − ✓ 3
20
A math 360 sol (unofficial) 3(c)
Ex 1.3
Quadratic equation 2x 2 + 3x + 2p = 0 i.e. a = 2, b = 3, c = 2p
4(c)
Discriminant For real roots: b2 − 4ac ≥0 2 (3) − 4(2)(2p) ≥ 0 9 − 16p ≥0 16p ≤9 p 3(d)
≤
9 16
Discriminant For no x − intercepts: b2 − 4ac ✓ 2
Quadratic Inequality −3x 2 + 6x + k − 1 is always negative i.e. a = −3, b = 6, c = k − 1 2 conditions (i) a < 0 −3 < 0 (ii) b2 − 4ac (6)2 − 4(−3)(k − 1) 36 + 12(k − 1) 36 + 12k − 12 12k k
Quadratic curve y = 9x 2 − px + 1 i.e. a = 9, b = −p, c = 1 Discriminant For one x − intercept: b2 − 4ac =0 (−p)2 − 4(9)(1) = 0 p2 − 36 =0 (p + 6)(p − 6) = 0 p = −6 or p = 6 ✓
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Quadratic Inequality 2x 2 + 2x + k is always positive i.e. a = 2, b = 2, c = k
(ii) b2 − 4ac 4
Discriminant For two x − intercepts: b2 − 4ac >0 2 (−4) − 4(4)(−p) > 0 16 + 16p >0 p > −1 ✓ 4(b)
3
2 conditions (i) a > 0 2>0
Discriminant For no real roots: b2 − 4ac 0 2 (−2) − 4(3)(1 − p) > 0 4 − 12(1 − p) >0 4 − 12 + 12p >0 12p >8 p
2
8
Quadratic equation p(x + 1)(x − 3) = x − 4p − 2 2 p(x − 2x − 3) = x − 4p − 2 px 2 − 2px − 3p = x − 4p − 2 2 px − (2p + 1)x + p + 2 = 0 i.e. a = p, b = −(2p + 1), c = p + 2 Discriminant For no real roots: b2 − 4ac [−(2p + 1)]2 − 4(p)(p + 2) (4p2 + 4p + 1) − 4p(p + 2) (4p2 + 4p + 1) − 4p2 − 8p −4p + 1 −4p 4p
2kx − k for all real values of x kx 2 − 2kx + k + 1> 0 i.e. a = k, b = −2k, c = k + 1 2 conditions (i) a > 0 k>0
Discriminant For no real roots: b2 − 4ac (−2)2 − 4(2)(2 + p) 4 − 8(2 + p) 1 − 2(2 + p) 1 − 4 − 2p −2p
0 4k 0 i.e. a = 1.2, b = −14.4, c = 53.7 − x 2 conditions (i) A > 0 1.2 > 0
(i) A > 0 1>0 (ii) B 2 − 4AC (−2)2 − 4(1)(−5 − c) 4 + 4(5 + c) 1 + (5 + c) 6+c c 19(i)
(ii) b2 − 4ac < 0 (−14.4)2 −4(1.2)(53.7 − x) 207.36 −4.8(53.7 − x) 43.2 −53.7 + x −10.5 + x x
|−0.2| > |−0.08| ⇒ A will produce the narrowest path ✓
−b+√b2 −4ac
α+β =
−b−√b2 −4ac
23(ii) Comparing y-intercepts, A> C >B 3 > 2.4 > 1.8 ⇒ A will send water the highest ✓
Incorrect. ✓ He forgets to include m ≠ 0 m = 0 ⇒ Linear equation α=
)(
𝑂 23(i)
✓
2a
𝑂 √10
(7,30.5) 6 𝑂
−b+√b2 −4ac
=( =
∴6< t x 2x 2 − 5x − 3 > 0 (2x + 1)(x − 3) > 0
+ − + −3 4
+
−
−
−3 ≤ x ≤ 4
1
+ 3
2
1
x < − or x > 3 ✓ 2
4
−3 1(c)
✓
3(a)
(2x + 3)(x − 2) > 0 +
−
−
3
x(x − 2) 2 2
− 1(d)
3(b)
3
2
2
✓
+ − + −2 6
x(x − 5) ≥ 0 +
x2 > 4x + 12 2 x − 4x − 12 > 0 (x − 6)(x + 2) > 0
− 0
x < −2 or x > 6 ✓
+ 5
3(c)
x ≤ 0 or x ≥ 5
4x(x + 1) ≤3 2 4x + 4x − 3 ≤0 (2x + 3)(2x − 1) ≤ 0 +
0 1(e)
5
x 2 − 4x x(x − 4) +
−
✓
0
3(d)
+
4
0≤x≤4
0
4
+ 1
2
2
3
1
2
2
− ≤x≤ ✓
≤0 ≤0 −
− 3
(1 − x)2 ≥ 17 − 2x 2 x − 2x + 1 ≥ 17 − 2x x 2 − 16 ≥0 (x + 4)(x − 4) ≥ 0 + − + −4 4
✓
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x ≤ −4 or x ≥ 4 ✓
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A math 360 sol (unofficial) 3(e)
(x + 2)2 x 2 + 4x + 4 2x 2 − 36 x 2 − 18 (x + √18)(x − √18) +
−
Ex 1.4 < x(4 − x) + 40 < 4x − x 2 + 40 0 >0
+ − + −14 8
≥ 3(x − 5) ≥ 3x − 15 ≥ −8 ≥ −4 ✓
Quadratic equation (k − 6)x 2 − 8x + k = 0 i.e. a = (k − 6), b = −8, c = k Discriminant For two distinct points: b2 − 4ac >0 2 (−8) − 4(k − 6)(k) > 0 64 − 4(k 2 − 6k) >0 16 − k 2 + 6k >0 2 k − 6k − 16 0 (2x − 3)(x + 2) > 0
+ − −2
−(2)
= −12 ✓ 17
−4
−(1)
sub (2) into (1): p = −(4 + 2(4))
r < −14 (rej ∵ r > 0) or r > 8 ✓ 15 5x − 7 (i)(a) 5x − 7 2x x
p = −(4 + 2k)
+ − + −2 8
+
−2 < k < 8
3 2
For minimum point, (k − 6) > 0 k >6
3
x < −2 or x > ✓ 2
−2
3 2
✓
Combine inequalities, −2 < k < 8 and k > 6 ⇒ 62 2 x −x−2 >0 (x − 2)(x + 1) > 0 + − + −2 0 + − + −1 2 −2 < x < 0 x < −1 or x > 2
x> ✓ 2
1st inequality 2x 2 + px + 16 < 0 2nd inequality 2 < x < k is solution ⇒ A(x − 2)(x − k) Compare x 2 : A = 2 2(x − 2)(x − k) 2(x 2 − 2x − kx + 2k) 2[x 2 − (2 + k)x + 2k] 2x 2 − 4(4 + 2k)x + 4k
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−2 −1
0 𝐱 + 𝟏 > 𝟎 or x − 3 > 0 ⇒ x > −1 x > 3 3rd step is incorrect ✓
and x 2 3
2
x=
x < −1 or x > 6
2
3
1 + 2x + (x − 1)x = 100 000 2 + 4x + 3(x − 1)x = 200 000 2 + 4x + 3x 2 − 3x = 200 000 3x 2 + x − 199 998 = 0
+ − + −1 6
x ≤ −2 or x ≥ 2
−2 −1
19(i)
Correct Solution x 2 − 2x − 3 > 0 (x + 1)(x − 3) > 0
−3 < x < 3
0
1
+ − + −1 3
3
∴ −3 < x ≤ 0 or 1 ≤ x < 3 ✓
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x < −1 or x > 3 ✓
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34
A math 360 sol (unofficial)
Rev Ex 1 A3(i)
Rev Ex 1 A1
x + 2y = 5 2y = −x + 5 y
1
5
2
2
=− x+
−(1) Sum of roots
2x + y = 2xy sub (1) into (2):
−(2)
2
x+
1
5
1
5
2
2
2
2
2
5 5
2
2
x − x+
2x − 7x + 5 (2x − 5)(x − 1) x=
5 2
or 1 5
5
2 2 5
2
y|x=5 = − ( ) +
a
c a
=− =−
1 2 5 2
5
−2 (− )
2
2
1
=5 ✓ 4
=0 =0
✓
2
1 2
= (− )
=0
2
=
b
a2 + β2 = (α + β)2 −2αβ
= −x 2 + 5x
2 7
=α+β =−
Product of roots = αβ
2x + (− x + ) = 2x (− x + ) 3
1st equation 2x 2 + x − 5 = 0 i.e. a = 2, b = 1, c = −5 Roots: α & β, where α < β
A3(ii) (α − β)2 = α2 − 2αβ + β2 = (α2 + β2 ) −2αβ
x =1✓ 1
5
2
2
y|x=1 = − (1) +
=− ✓
=2✓
4
=5 =
1
5
−2 (− )
4 41
2
4 1
= 10 ✓ 4
A2
Perimeter 2x + 2y = 36 x+y = 18 y = 18 − x
𝑦 𝑥 −(1)
A3(iii) α − β = √41 or α − β = − √41 ✓ 2
2
(rej ∵ α < β) A3(iv) 2nd equation
Square of diagonal length 2
1
Recall α + β = − , 2
αβ = −
5 2
(√x 2 + y 2 ) = 164 x2 + y2
= 164
−(2)
sub (1) into (2): x 2 +(18 − x)2 −164 =0 2 2 x +(x − 36x + 324) −164 = 0 (2x 2 − 36x + 324) − 164 =0 2 2x −36x + 160 =0 2 x − 18x + 80 =0 (x − 8)(x − 10) =0 x=8 or x = 10 y|x=8 = 18 − (8) y|x=10 = 18 − (10) = 10 =8 Dimensions are 8 m by 10 m ✓
Roots: 2α & − 2β Sum of roots = 2α + (−2β) = 2(α − β) = 2 (−
√41 ) 2
= −√41
Product of roots = (2α)(−2β) = −4αβ 5
= −4 (− ) = 10 2
x 2 − (SOR)x + (POR) = 0 x 2 − (−√41)x +10 = 0 x2 + √41x +10
=0✓
35
A math 360 sol (unofficial)
Rev Ex 1
A4(a) Quadratic equation x2 + 3 = 2x + p x 2 − 2x + 3 − p = 0 i.e. a = 1, b = −2, c = 3 − p Discriminant For real roots: b2 − 4ac (−2)2 − 4(1)(3 − p) 4 − 4(3 − p) 1+p−3 p
A5(a) x 2 − 5x + 3 > 5 − 4x x2 − x − 2 >0 (x − 2)(x + 1) > 0 + − + −1 2 x < −1 or x > 2 ✓
≥0 ≥0 ≥0 ≥0 ≥2✓
A4(b) Quadratic equation 2x 2 + 2√3x + p = p(x 2 + 2) 2x 2 + 2√3x + p = px 2 + 2p (2 − p)x 2 + 2√3x − p = 0 i.e. a = 2 − p, b = 2√3, c = −p
−1 2 A5(b) Quadratic inequality 3x 2 − 6x + c > 4 for all real values of x 2 3x − 6x + c − 4 > 0 i.e. A = 3, B = −6, C = c − 4 2 conditions (i) A > 0 3>0
a ≠0 (2 − p) ≠ 0 p ≠2 Discriminant For distinct real roots: b2 − 4ac 2
(2√3) −4(2 − p)(−p) 12 −4(p2 − 2p) 12 −4p2 + 8p p2 − 2p − 3 (p − 3)(p + 1)
(ii) B 2 − 4AC (−6)2 − 4(3)(c − 4) 36 − 12(c − 4) 3 − (c − 4) 3−c+4 7−c −c c
>0 >0 >0 >0 1, graph slopes up 𝑦 𝑦 = 6𝑒 𝑥
6 𝑂 2(b)
4(ii)
T|x=0 = 90(0.98)0 = 90° ✓
𝑥 ✓
5(ii)
T|x=10 = 90(0.98)10 = 73.5° ✓
5(iii)
T|x=60 = 90(0.98)60 = 26.8° ✓
−x
y = 2e = 2(e−1 )x 1 x
= 2( ) e
1
∵ 0 < base < 1, graph slopes down e 𝑦 𝑦 = 2𝑒 −𝑥 2 𝑥 𝑂 ✓
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A math 360 sol (unofficial) 6
Ex 2.4
y = 3(5x ) ∵ base 5 > 1, graph slopes up
9(ii)
y = 3(0.2)x ∵ 0 < base 0.2 < 1, graph slopes down 𝑦 𝑦 = 3(5𝑥 ) 3
O
𝑦 = 3(0.2) 𝑥
y|x=10 = 45 000(1.02)10 = 54 855 ✓
y = 3(2x ) ∵ base 2 > 1, graph slopes up 𝑦 𝑦 = 3(2𝑥 ) 6 3 𝑥
O
𝑦 =6−𝑥 ✓
7(iii)
One intersection ⇒ One solution ✓
8(i)
P = 600(2 + e−0.2t )
y = 45 000(1.02)x ∵ base 1.02 > 1, graph slopes up 𝑦 𝑦 = 45 000(1.02)𝑥
𝑥
✓ f(x) = 3(5x ) f(−x) = 3(5−x ) = 3(5−1 )x = 3(0.2)x y = f(x) and y = f(−x) are reflection of each other in the y-axis ✓ 7(i) 7(ii)
45 000
P|t=12 = 600(2 + e−0.2(12) ) = 1254 ✓
8(iii)
t → ∞, e−0.2t → 0, P ⇒ 600(2 + 0) = 1200 ✓
9(i)
y = 45 000(1.02)
𝑥
𝑂 10(i)
✓
C = 4.86e−0.047t C|t=0 = 4.86e0 = 4.86μg/ml ✓
10(ii) C|t=10 = 4.86e−0.047(10) = 3.04μg/ml ✓ 10(iii) C|t=24 = 4.86e−0.047(24) = 1.57μg/ml ✓ 10(iv) t ⇒ ∞, e−(0.047t) ⇒ 0, C⇒ 0✓ 10(v) C = 4.86e−0.047t = 4.86(e−0.047 )t = 4.86 (
P|t=0 = 600(2 + e0 ) = 1800 ✓ 8(ii)
end of 2021: x = 2021 − 2011 = 10
1 e0.047
∵ 0 < base
)
t
1
e0.047
< 1, graph slopes down
𝐶 4.86 𝑂
𝐶 = 4.86𝑒 −0.047𝑡 𝑡 ✓
x
end of 2015: x = 2015 − 2011 =4 y|x=4 = 45 000(1.02)4 = 48 709 ✓
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61
A math 360 sol (unofficial) 11
Ex 2.4 1 x
12(iv) m
y = 2−x = (2−1 )x = ( ) 1
405 ( ) = 80
∵ 0 < base < 1, graph slopes down 2
2 t
3 2
𝑦 = 8(2𝑥 )
8
13(i)
𝑦 = 2−𝑥
1
𝑥
𝑂
3
✓
16
( )
=
( ) 3 ⇒t
=( ) 3 =4
3 2 t
y = 8(2x ) ∵ base 2 > 1, graph slopes up 𝑦 (− , 2√2)
= 80 2 t
2
81 2 4
V = Ae−kt Initial value of the car is $60 000: V|t=0 = 60 000 −k(0) Ae = 60 000 A = 60 000
−x
y=2 −(1) y = 8(2x ) −(2) sub (1) into (2): 2−x = 8(2x ) 2−2x = 8 2−2x = 23 ⇒ −2x = 3 3 x =− −(3)
Value of the car is $39 366 after 4 years: V|t=4 = 39366 Ae−k(4) = 39366 −4k 60 000e = 39366 ∵ A = 60 000 39 366 −4k e = (e
2
60 000 9 4
−k 4
=( )
)
−k
⇒e
=
sub (3) into (1): 3 −(− )
V = 60 000 ( ) 10
= 60 000(0.9)t [shown] ✓
2
13(ii) V = 60 000(0.9)t ∵ 0 < base 0.9 < 1, graph slopes down
2 t
m = 405 ( ) 3
2 0
𝑉
m|t=0 = 405 ( ) 3
= 405g ✓ 12(ii)
10
9 t
y = 2 2 = 21.5 = 2√2 3 ⇒ (− , 2√2) 12(i)
10 9
60 000 𝑂
2 3
m|t=3 = 405 ( ) 3
𝑉 = 60 000(0.9)𝑡 𝑡 ✓
= 120g ✓ 12(iii) Amount decayed in 3 years = m|t=0 −m|t=3 = 405 −120 = 285g ✓
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A math 360 sol (unofficial)
Rev Ex 2 A2(b)
Rev Ex 2 A1(i)
1
(√5 − 2)x = √5 + 2 x
=
√5+2
=
√5+2
=
×
√5−2
√5+2 √5+2
(√5) +2(√5)(2) +(2)2 (√5) 5
2
−(2)2
+4√5
+4
5−4
= 9 + 4√5 ✓ A1(ii) x
+
= 9 + 4√5
+
1 x 1 9+4√5 1
= 9 + 4√5
+
= 9 + 4√5
+
= 9 + 4√5
+
= 9 + 4√5
+9 − 4√5
9+4√5
×
9−4√5 9−4√5
9−4√5 (9)2 −(4√5)
2
= 4√2
9−4√5
−√50
−
−√25 × 2
−
3
+
√2 3 √2
×
3
−5√2
− √2 2
= 4√2
−5√2
− √2 2
= 4√2
−5√2
− √2 2
= 4√2
−5√2
− √2 2
23 12
√96 3
−
√2 2
×(
√16×6 3
−
√2 2
×(
√2 2
× ( √6 − √6 3 3
√2 2
× ( √6 − √6 3 3
√2 2
× ( √6 − √6 3 3
√2 2
× ( √6 − √6 3 3
√2 2
× (10√6)
√2 √2
4√6 3
2 √6
√6
×
√6
2√6
+
6 1
− √6 3
4 4
+
1
+
1
4
1
4
1
+
+
3√216 2
)
= k√3
3√23 ×33 2 3√23 √33 2
)
= k√3
)
= k√3
3(21.5 )(31.5 ) 2
)
3(2√2)(3√3) 2 18√6
)
= k√3 = k√3
)
= k√3
+9√6)
= k√3
+9√6)
= k√3
2
= k√3
5√12
= k√3
5(√4 × 3)
= k√3
5(2√3)
= k√3
10√3
= k√3
⇒ k = 10 ✓
= 4√2
=−
×(
×
+
√6
81−16(5)
= 18 ✓ A2(a) 4√2
−
1
2
=
√12√8 3
√2
√5−2
2
×(
√2
3 3 3
√2 √2
+ + + + +
A3(a) √x 2 − 7 =3 2 x −7 =9 2 x − 16 =0 (x + 4)(x − 4) = 0 x = −4 or x = 4 ✓
7 √72 7 √36×2 7 6√2 7 6√2 7 6(2) 7 12
×
√2 √2
√2
√2
A3(b) 2x + √3 − 4x √3 − 4x 3 − 4x 4x 2 + 4x − 3 (2x + 3)(2x − 1)
√2 ✓
3
1
2
2
x = − or x = A3(c)
x √1−8x
=0 = −2x = 4x 2 =0 =0 (rej) ✓
=
1 3
3x = √1 − 8x 2 9x = 1 − 8x 2 9x + 8x − 1 = 0 (9x − 1)(x + 1) = 0 x=
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1 9
or x = −1 (rej) ✓
63
A math 360 sol (unofficial) A3(d)
4x
A5(b) 4x−1 + 16x = 66 (4x )(4−1 ) + (42 )x = 66
= √2(2x−1 )
8 22x
1
= 22 (2x−1 )
23
1 4
1 x− 2
2x−3
2
= 66
4
2
4u2 + u − 264 =0 (4u + 33)(u − 8) = 0
5 2
u=−
1st eqn 8 × 4y 23 × 22y 23+2y ⇒ 3 + 2y 2x x
= 66
sub u = 4 : u + u2
1
= ✓
x
(4x ) + (4x )2 x
=2
⇒ 2x − 3 = x −
A4
Rev Ex 2
2x−1
=2 = 22x−1 = 22x−1 = 2x − 1 = 4 + 2y =y+2
33
or
4
sub u = 4x : 4x = −
u=8 sub u = 4x :
33
4x = 8
4 x
4x = 23 22x = 23 ⇒ 2x = 3
(rej ∵ 4 > 0) −(1)
3
= ✓
x 2nd eqn 3y √3x
A6(i)
= 81 x
x
= 34 =4
2
−(2)
sub (1) into (2): y+2
y+
y
=4
2
2
2
y
A6(ii) m|t=2 = 300e−0.85(2) = 54.8mg ✓ A6(iii) y = 300e−0.85t = 300(e−1 )0.85t
y + + 1= 4 3
1 0.85t
=3
y
m = 300e−0.85t m|t=0 = 300e0 = 300mg ✓
(3y ) (32 ) = 34 3y+2 x ⇒y+
2
= 300 ( ) e
=2
1
∵ 0 < base < 1, graph slopes down e
Put y = 2 into (1): x|y=2 = (2) + 2
𝑚
=4✓ x)
A5(a) 2(4
x)
2(4
x+2
+4 +
(4x )(42 )
2(4x ) + 16(4x ) x)
18(4
=
= 9(4 = 9(
2
sub u = 4x : 18u
=
u
=
1 4 0.5 1 √4
B1(a) √x − 8 × √x √(x − 8)x
9
4 1
4x ⇒x
= 4−1 = −1 ✓
4
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=3 =3
√x 2 − 8x =3 x 2 − 8x =9 2 x − 8x − 9 = 0 (x − 9)(x + 1) = 0 x = 9 or x = −1 (rej) ✓
sub u = 4x : =
✓
A6(iv) p = 300 − 300e−0.85t = 300(1 − e−0.85t ) ✓
2 1
4x
𝑡
)
= 9( )
9
300 𝑂
−0.5 )
𝑚 = 300𝑒 −0.85𝑡
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64
A math 360 sol (unofficial)
Rev Ex 2
B1(b) √18 − √x − 1 = 4 18 − √x − 1 = 16 =2 √x − 1 x−1 =4 x =5✓ B1(c)
√11+√4 x 2
=
B3(i)
1 2 1 2
x
33 (32 )
=
(BC)h
√11−√4
= =
3
x 2
3x 3
√9 3x
=3
⇒3+
2
x
x B2
=x− =
2
=
ℎ B C (8√3 − 2√2)
92 8√3−2√2
92 8√3 − 2√2
×
8√3 + 2√2 8√3 + 2√2
92(8√3 + 2√2) 2
2
=
92(8√3 + 2√2) 64(3) − 4(2)
=
92(8√3 + 2√2) 184
=
8√3 + 2√2 2
1
x−
x
= 46
(8√3) − (2√2)
(32 )3 3+
A
(8√3 − 2√2)h = 46
h=
x
x = −√7 or x = √7 ✓ =
= 46
(8√3 − 2√2)h = 92
x = 11 − 4 2 x −7 =0 (x + √7)(x − √7) = 0
B1(d) 27(√3)x
Area
2 3
2 3
11 3 22 3
= 4√3 + √2 ✓
✓
(a − 6√5)(2 + b√5)
B3(ii) By Pythagoras’ Theorem,
= −82
1
AC = √( BC)
a(2 + b√5) −6√5(2 + b√5) = −82
2
+ h2
2
2a + ab√5 −12√5 − 6b(5) = −82 2a − 30b +(ab − 12)√5 Equate rational terms: 2a − 30b = −82 a − 15b = −41 a = 15b − 41
1
= √[ (8√3 − 2√2)]
= −82
2
2
2
= √(4√3 + √2)
2
+ (4√3 − √2) + (4√3 − √2)
2
−(1)
2
2
[(4√3) + 2(4√3)(√2) + (√2) ]
=√
2
2
+ [(4√3) − 2(4√3)(√2) + (√2) ] Equate irrational terms: ab − 12 = 0 −(2) sub (1) into (2): (15b − 41)b − 1 = 0 15b2 − 41b − 12 = 0 (15b + 4)(b − 3) = 0 b=−
4
or
15
a|b=− 4 = 15 (− 15
4
) − 41
15
= −45 ✓
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= √16(3) + 8√6 + 2
+ (16(3) − 8√6 + 2)
= √100 = 10 b=3✓ a|b=3 = 15(3) − 41 =4✓
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Perimeter = AB +AC +BC = 10 +10 +8√3 − 2√2 = 20 +8√3 −2√2 ✓
65
A math 360 sol (unofficial) B4
At (3,4):
Rev Ex 2 = a(3)n − 23 = a(3)n = a(3)n =a = 33−n
4 27 33 33−n a
At (9,220): 220 243 243 a(32n )
−(1)
= a(9)n − 23 = a(9)n = a(32n ) = 243 −(2)
At (−1, k): k = a(−1)n − 23
B5(b) 9x + 10(3x ) (32 )x + 10(3x ) (3x )2 + 10(3x ) sub u = 3x : u2 + 10u u2 + u − 12 (u + 4)(u − 3) u = −4 or sub u = 3x : 3x = −4 (rej ∵ 3x > 0)
−(3)
= 3x+2 + 12 = (3x )(32 ) + 12 = 9(3x ) + 12 = 9u + 12 =0 =0 u=3 sub u = 3x : 3x = 3 3x = 31 ⇒x=1✓ t
sub (1) into (2): (33−n )32n = 243 33+n = 35 ⇒3+n =5 n =2✓
B6(i)
Put n = 2 into (1): a|n=2 = 33−2 = 3 ✓
B6(ii) θ| = 20 + 100(0.8)86 t=8 ≈ 94.3 ✓
θ = 20 + 100(0.8)6 θ|t=0 =X 0
20 + 100(0.8)6 20 + 100(1) X
Put a = 3, n = 2 into (3): k = 3(−1)2 − 23 = −20 ✓
B6(iii) θ
= 84
20 + 100(0.8) x
x−1 )
B5(a) 5 + 5 = 30(5 5x + 5 = 30(5x )(5−1 )
100(0.8)
1
5x + 5 = 30(5x ) ( ) 5
5x + 5 = 6(5x ) sub u = 5x : u + 5= 6u 5 = 5u u =1 sub u = 5x : 5x = 50 ⇒ x =0✓
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=X =X = 120°C ✓
t 6
= 84
t 6
= 64
(0.8)
t 6
= 0.64
(0.8)
t 6
= (0.8)2
⇒
t
=2
6
= 12s ✓
t B6(iv)
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t
t → ∞, 100(0.8)6 → 0 θ → 20°C ✓
66
A math 360 sol (unofficial)
Ex 3.1 3(b)
Ex 3.1 1(a)
Yes ✓ Are exponents Non-negative? Integers?
= 15x 4 +5x 3 −10x 2 +6x 3 +2x 2 −4x 2 −12x −4x +8 = 15x 4 + 11x 3 − 20x 2 − 8x + 8 ✓
True True 4(a)
∵ both conditions are met, 2 x − 1 is a polynomial 3
1(b)
(2x 2 − x + 1)(3x − 2) Coefficient of x 2 = (2)(−2) + (−1)(3) = −7 ✓
Yes ✓ Are exponents Non-negative? True Integers? True
4(b)
∵ both conditions are met, 4x 2 − 2x is a polynomial . 1(c)
(5x 2 + 2x − 4)(3x 2 + x − 2)
(x 2 + 3x + 2)(8x 2 − 5x − 4) Coefficient of x 2 = (1)(−4) + (3)(−5) +(2)(8) = −4 −15 +16 = −3 ✓
No ✓ Are exponents Non-negative? True Integers? False
4(c)
(2x 2 − 2x + 5)(−x 2 − 3x + 1)
1
√x has non-integer power of 2.
Coefficient of x 2 = (2)(1) + (−2)(−3) +(5)(−1) =2 +6 −5 =3✓
∵ both conditions are not met, 4x 3 + √x + 3 is not a polynomial. 1(d)
No ✓
4(d)
Are exponents Non-negative? False Integers? True 3 x2
Coefficient of x 2 = (−1)(6) + (−2)(3) = −6 − 6 = −12 ✓
has a negative power of −2.
∵ both conditions are not met, 3 1 + 2 is not a polynomial.
5(a)
x
2(i)
Q(x) − P(x) = (2x 2 − 3x + 2) −(x 2 + x + 1) = x 2 − 4x + 1 ✓
2(ii)
P(x) + 2Q(x) = (x 2 + x + 1) +2(2x 2 − 3x + 2) = (x 2 + x + 1) +(4x 2 − 6x + 4) = 5x 2 − 5x + 5 ✓
3(a)
(4x 3 − x 2 + 7x − 2)(2x 3 + 3x 2 + 6)
a(x − 2) + b = 5 − 3x sub x = 2: a(2 − 2) + b = 5 − 3(2) b = −1 ✓ sub x = 0: a(0 − 2) + b = 5 − 3(0) −2a + b =5 −2a + (−1) = 5 a = −3 ✓
∵ b = −1
(7x − 3)(2x 2 + 4x − 1) = 14x 3 +28x 2 −7x −6x 2 −12x +3 3 = 14x + 22x 2 − 19x + 3 ✓
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67
A math 360 sol (unofficial) 5(b)
Ex 3.1
a(x − 1) + b(x + 3) = 3x + 1
6(i) (3x 2 + 2)
sub x = 1: a(1 − 1) + b(1 + 3)= 3(1) + 1 4b =4 b =1✓
(5x + 10) 1
A(x) = (5x + 10)(3x 2 + 2) 2 1
= [5x(3x 2 + 2) +10(3x 2 + 2)] 2 1
sub x = −3: a(−3 − 1) + b(−3 + 3) = 3(−3) + 1 −4a = −8 a =2✓ 5(c)
= (15x 3 + 10x +30x 2 + 20) 2 1
= (15x 3 + 30x 2 + 10x + 20) =
2 15 3 x 2
+ 15x 2 + 5x + 10
a(x − 2) + b(x − 4) = x + 2 Yes. Exponents are non-negative integers ✓ sub x = 2: a(2 − 2) + b(2 − 4)= 2 + 2 −2b =4 b = −2 ✓
6(ii)
= √(3x 2 + 2)2
sub x = 4: a(4 − 2) + b(4 − 4)= 4 + 2 2a =6 a =3✓ 5(d)
5(e)
= √3x 4 + 37x 2 + 100x + 104 P(x) = (5x + 1) + (3x 2 + 2) +√3x 4 + 37x 2 + 100x + 104
a =4✓ b =0✓ c = −1 ✓ 3 = −c + d 3 = −(−1) + d d=2✓
= 3x 2 + 5x + 3 +√3x 4 + 37x 2 + 100x + 104 No because √3x 4 + 37x 2 + 100x + 104 will have non-integer exponent ✓
x 3 − 6x 2 + 14x − 8 = (x − 2)3 + ax
7(i) (a)
sub x = 2: (2)3 − 6(2)2 + 14(2) − 8 = (2 − 2)3 + 2a 4 = 2a a =2✓ 5(f)
7(i) (b)
= a(2 − 2)2 + b(2 + 1)3 +
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2P(x) − [Q(x)]2 = 2(x + 1) − (x 2 − x + 1)2 = 2(x + 1) − (x 2 − x + 1)(x 2 − x + 1) Coefficient of x 2 = −[(1)(1) +(−1)(−1) +(1)(1)] = −3 ✓
= 27b + 8 = 27b = −1 ✓
sub x = −1: −2(−1)2 − 7(−1) + 3 1)3 + (−1)3 8 9 a
M(x) = P(x)Q(x) = (x + 1)(x 2 − x + 1) Coefficient of x 2 = (1)(−1) +(1)(1) =0✓
−2x 2 − 7x + 3 = a(x − 2)2 + b(x + 1)3 + x 3 sub x = 2: −2(2)2 − 7(2) + 3 (2)3 −19 −27 b
+ (5x + 10)2
= √(3x 4 + 12x 2 + 4) + (25x 2 + 100x + 100)
ax 3 − x + 3 = 4x 3 + bx 2 + c(x − 1) + d Compare x 3 : Compare x 2 : Compare x: Compare x 0 :
By Pythagoras Theorem Hypotenuse
= a(−1 − 2)2 + b(−1 +
7(i) (c)
= 9a − 1 = 9a =1✓
Q(x)[3x 2 + P(x)] = (x 2 − x + 1)(3x 2 + x + 1) Coefficient of x 2 = (1)(1) +(−1)(1) +(1)(3) =1 −1 +3 =3✓
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68
A math 360 sol (unofficial) 7(ii)
Ex 3.1 9(c)
Deg (M(x)) = 3 ✓ Deg [2P(x) − [Q(x)]2 ] = 4 ✓ Deg [Q(x)[3x 2 + P(x)]] = 4 ✓
sub x = 2: 3(2)2 − 5(2) + 4 = a(2 − 2)2 + b(2 − 2) + c 6 =c c =6✓
8(i)
V(x) = (2x + 3)(3x − 1) (2x + 5) = (6x 2 + 7x − 3) (2x + 5)
sub x = 0: 3(0)2 − 5(0) + 4 4 4
= 12x 3 +14x 2 −6x +30x 2 +35x −15 3 = 12x + 44x 2 + 29x − 15 ✓ 8(ii)
9(a)
A(x) = 2[(2x + 3)(3x − 1) +(2x + 3)(2x + 5) +(3x − 1)(2x + 5)] = 2[(6x 2 + 7x − 3) +(4x 2 + 16x + 15) +(6x 2 + 13x − 5)] = 2(16x 2 + 36x + 7) = 32x 2 + 72x + 14 ✓
4 2b b 9(d)
x 3 − 6x 2 − x + c = (x − 3)(ax 2 − 3x + b) 3
Compare x : a = 1 ✓ sub x = 3: (3)3 − 6(3)2 − 3 + c = (3 − 3)(a(3)2 − 3(3) + b) −30 + c =0 c = 30 ✓ sub x = 0: (0)3 − 6(0)2 − (0) + c b) c 30 b 9(b)
= (0 − 3)(a(0)2 − 3(0) + = −3b = −3b ∵ c = 30 = −10 ✓
x 3 + cx 2 + x + 6 = (x + 1)(x − 2)(ax + b) Compare x 3 : a = 1 ✓ sub x = 0: (0)3 + c(0)2 + (0) + 6 = (0 + 1)(0 − 2)(a(0) + b) 6 = −2b b = −3 ✓ sub x = 2: (2)3 + c(2)2 + 2 + 6 4c + 16 c+4 c
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3x 2 − 5x + 4 = a(x − 2)2 + b(x − 2) + c Compare x 2 : a = 3 ✓
= a(0 − 2)2 + b(0 − 2) + c = 4a − 2b + c = 4(3) − 2b + 6 ∵ a = 3, c = 6 = 18 − 2b = 14 =7✓
x 3 + 3x 2 − 2x + 16 − c(x + 2) = ax 2 (x − 1) + b(x − 2)2 (x − 1) sub x = 1: (1)3 + 3(1)2 − 2(1) + 16 − c(1 + 2) = a(1)2 (1 − 1) + b(1 − 2)2 (1 − 1) 18 − 3c =0 −3c = −18 c =6✓ sub x = 0: (0)3 + 3(0)2 − 2(0) + 16 − c(0 + 2) = a(0)2 (0 − 1) + b(0 − 2)2 (0 − 1) 16 − 2c = −4b 16 − 2(6) = −4b ∵ c = 6 4 = −4b b = −1 ✓ sub x = 2: (2)3 + 3(2)2 − 2(2) + 16 − c(2 + 2) = a(2)2 (2 − 1) + b(2 − 2)2 (2 − 1) 32 − 4c = 4a 32 − 4(6) = 4a ∵ c = 6 8 = 2a a =2✓
= (2 + 1)(2 − 2)(a(2) + b) =0 =0 = −4 ✓
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69
A math 360 sol (unofficial) 10(i)
𝐱 𝟑 yz 2
Ex 3.1
+(𝐱 − z)2 y
2
+(2𝐱 − √y)
+y 2 z 4
= 𝐱 𝟑 yz 2 +(x 2 − 2𝐱z + z 2 )y +4𝐱 𝟐 − 4𝐱√y + y +y 2 z 4 = 𝐱 𝟑 yz 2 +𝐱 𝟐 y − 2𝐱zy + z 2 y +4𝐱 𝟐 − 4𝐱√y + y +y 2 z 4 = yz 2 𝐱 𝟑 +y𝐱 𝟐 − 2yz𝐱 + yz 2 +4𝐱 𝟐 − 4√y𝐱 + y +y 2 z 4 = yz 2 𝐱 𝟑 +y𝐱 𝟐 −2yz𝐱 +yz 2 +4𝐱 𝟐 −4√y𝐱 +y +y 2 z 4 = yz 2 𝐱 𝟑 +(y + 4)𝐱 𝟐 −(2yz + 4√y)𝐱 +yz 2 + y + y 2 z 4 Yes, all exponents of x are non-negative integers ✓ Degree is 3 ✓ 10(ii) x 3 𝐲z 2
+(x − z)2 𝐲 +(2x − √𝐲)
2
+𝐲 𝟐 z 4
= x 3 z 2 𝐲 +(x − z)2 𝐲 +4x 2 − 4x√𝐲 + 𝐲 +z 2 𝐲 𝟐 = x3z2 𝐲 +(x − z)2 𝐲 +𝐲
+4x 2 −4x√𝐲 +z 2 𝐲 𝟐
= [x 3 z 2 + (x − z)2 + 1]𝐲 +4x 2 −4x√𝐲 +z 2 𝐲 𝟐 = z 2 𝐲 𝟐 +[x 3 z 2 + (x − z)2 + 1]𝐲
−4x√𝐲 +4x 2
No, exponents of y include fractions. √y has non − integer exponent ✓ 10(iii) x 3 y𝐳 𝟐
+(x − 𝐳)2 y
2
+(2x − √y)
+y 2 𝐳 𝟒 2
+y 2 𝐳 𝟒
2
+y 2 𝐳 𝟒
2
+y 2 𝐳 𝟒
= x 3 y𝐳 𝟐 +(x 2 − 2x𝐳 + 𝐳 𝟐 )y +(2x − √y) = x 3 y𝐳 𝟐 +x 2 y − 2xy𝐳 + 𝐳 𝟐 y +(2x − √y) = x 3 y𝐳 𝟐 +x 2 y − 2xy𝐳 + y𝐳 𝟐 +(2x − √y) = x 2 y𝐳 𝟐 +x 2 y +y𝐳
𝟐
−2xy𝐳 2
+(2x − √y)
+y 2 𝐳 𝟒 2
= [x 2 y + y]𝐳 𝟐 +x 2 y + (2x − √y) −2xy𝐳 = y 2 𝐳 𝟒 +[x 2 y + y]𝐳 𝟐
−2xy𝐳
+y 2 𝐳 𝟒
+x 2 y + (2x − √y)
2
Yes, all exponents of z are non-negative integers ✓ Degree is 4 ✓
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70
A math 360 sol (unofficial) 11(i)
Ex 3.1
A(t) = −t 3 + at 2 + p B(t) = bt(t − 2)2 + 2t 2 + ct + q
12(ii) (x − a)(x − b)(x − c) (b) = (x 2 − (a + b)x + ab)(x − c)
Yes. Exponents of t are non-negative integers ✓
= x 3 −(a + b)x 2 +abx −cx 2 +c(a + b)x −abc 3 2 (a = x − + b + c)x + [ab + c(a + b)]x − abc = x 3 − (a + b + c)x 2 + (ab + ac + bc)x − abc ✓
11(ii) A(0) = 300 p = 300 ✓ B(0) = 0 q=0✓ 11(iii) A(t) would decrease until it becomes empty at zero. A(t) cannot be negative ✓ 11(iv) 300 − A(t) 300 − (−t 3 + at 2 + p) t 3 − at 2 − p + 300 Sub p = 300, q = 0: t 3 − at 2 − 300 + 300 t 3 − at 2
12(ii) (x − a)(x − b) … (x − z) = (x − a)(x − b) … (x − x) … (x − z) = (x − a)(x − b) … (0) … (x − z) =0✓ 13
= B(t) = bt(t − 2)2 + 2t 2 + ct + q = bt(t − 2)2 + 2t 2 + ct + q = bt(t − 2)2 + 2t 2 + ct + 0 = bt(t − 2)2 + 2t 2 + ct
A = 3, B = −2, C = 5 and D = −2 3(x − 1) − 2(x − 1)(x + 1) + 5x(x 2 − 1) − 2 𝑦 𝑦 = 3(𝑥 − 1) − 2(𝑥 − 1)(𝑥 + 1) +5𝑥(𝑥 2 − 1) − 2
𝑦 = 3𝑥 3 − 2𝑥 2 + 𝑥 − 4
Compare t 3 : b = 1 ✓ t 3 − at 2 = t(t − 2)2
The graphs are not aligned therefore James’ answers are wrong.
+2t 2 + ct
t 3 − at 2 = t(t 2 − 4t + 4) +2t 2 + ct
3x 3 − 2x 2 + x − 4 = A(x − 1) + B(x − 1)(x + 1) + Cx(x 2 − 1) + D
t 3 − at 2 = t 3 − 4t 2 + 4t +2t 2 + ct t 3 − at 2 = t 3 − 2t 2 + (4 + c)t −at 2
Compare x 3 : C = 3 ✓
= −2t 2 + (4 + c)t
Compare t 2 : −a = −2 a =2✓
sub x = 1:
D = −2 ✓
sub x = −1:
−10 −10 2A A
Compare t: 4 + c = 0 c = −4 ✓ 12(i) (a)
𝑥
𝑂
(x − a)(x − b) = x 2 − (a + b)x + ab ✓
sub x = 0:
= −2A + D = −2A + (−2) =8 =4✓
∵ D = −2
−4 = −A − B + D B =4−A+D = 4 − (4) + (−2) ∵ A = 4, D = −2 = −2 ✓
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71
A math 360 sol (unofficial)
Ex 3.2 2(b)
Ex 3.2 1(a)
1(b)
3x +4 2 3x −2 9x +6x +2 −(9x 2 −6x) 12x +2 −(12x −8) 10 Quotient = 3x + 4 ✓ Remainder = 10 ✓ +3x 2x 2 3 2x +1 4x +8x 2 −(4x 3 +2x 2 ) 6x 2 −(6x 2
1(c)
+2 +7x −5
+8x 5x 4 x −2x −3 5x −2x 3 −(5x 4 −10x 3 8x 3 −(8x 3 2
4x 2
+37 +6x 2 −15x 2 ) +21x 2 −16x 2 37x 2 −(37x 2
−4x
−4
−x 2 −3x 2 ) +8x 2x 2 2 +2x) −(2x 6x −4 −(6x +6) −10
+
2x 2 +2x −1 8x 4 +0x 3 −(8x 4 +8x 3 −8x 3 −(−8x 3
11 2
−x 2 −4x 2 ) +3x 2 −8x 2 11x 2 −(11x 2
+0x
+5
+0x +4x) −4x +5 +11x − 11) 2
−15x + 21 2
+4x
−3
8x 4 − x 2 + 5 = (2x 2 + 2x − 1) (4x 2 − 4x +
+4x −24x) +28x −3 −74x −111) 102x +108
+1 2x 2 3 2 x −2 2x −4x +x −2 −(2x 3 −4x 2 ) +x −2 −(x −2) 0 2x 3 − 4x 2 + x − 2 = (x − 2)(2x 2 + 1) ✓
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+6 +8x
2(c)
Quotient = 5x 2 + 8x + 37 ✓ Remainder = 102x + 108 ✓ 2(a)
+2x −x 2
3x 4 − x 2 + 8x − 4 = (x + 1)(3x 3 − 3x 2 + 2x + 6) − 10 ✓
+7x +3x) 4x −5 −(4x +2) −7 2 Quotient = 2x + 3x + 2 ✓ Remainder = −7 ✓ 2
3x 3 −3x 2 x +1 3x 4 +0x 3 4 −(3x +3x 3 ) −3x 3 −(−3x 3
11
21
2
2
) − 15x +
✓
3(i) Deg(Dividend) = Deg(Divisor) +Deg(Q(x)) 4 3 2 Deg (x + 2x − 2x ) = Deg(x + 2) +Deg(Q(x)) −2x + 4 (4) = (1) +Deg(Q(x)) Deg (Q(x))
=3✓
Deg(R(x)) < Deg(Divisor) < Deg(x + 2) 0) ✓
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124
A math 360 sol (unofficial) 15(i)
Ex 5.1
(1 + x)5 (1 − 4x)4 = (1 + x)5 [1 + (−4x)]4 4 4 4 5 5 5 = [( ) (x)0 + ( ) (x)1 + ( ) (x)2 + ⋯ ] [( ) (−4x)0 + ( ) (−4x)1 + ( ) (−4x)2 + ⋯ ] 0 1 2 0 1 2 2) 2) [ ] [ (5)(x) (10)(x (4)(−4x) (6)(16x = 1 + + +⋯ 1 + + +⋯] 2 2 (1 − 16x + 96x + ⋯ ) = (1 + 5x + 10x + ⋯ )
= 1 −16x +96x 2 +5x −80x 2 +100x 2 + ⋯ ≈ 1 − 11x + 26x 2 [shown] ✓ 15(ii) Factor & multiply polynomials (a) (1 + x)5 (1 − 4x)5 = [(1 + x)5 (1 − 4x)4 ] (1 − 4x) = [1 − 11x + 26x 2 + ⋯ ] (1 − 4x) = 1 −11x +26x 2 −4x +44x 2 + ⋯ = 1 − 15x + 70x 2 + ⋯ ✓ 15(ii) Substitute (1 − x)5 (1 + 4x)4 = [1 + (−x)]5 [1 − 4(−x)]4 (b) ≈ 1 −11(−x) +26(−x)2 ≈ 1 +11x
+26x 2 ✓
15(ii) Expand & substitute (1 + x 2 )5 (1 − 2x)4 (1 + 2x)4 = (1 + x 2 )5 (1 − 4x 2 )4 (c) = [1 + (x 2 )]5 [1 − 4(x 2 )]4 ≈ 1 −11(x 2 ) +26(x 2 )2 ≈ 1 −11x 2
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+26x 4 ✓
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125
A math 360 sol (unofficial)
Ex 5.1
16 28x 6 − 30x 4 − 1 = 0 sub x = 1 + h: 28(1 + h)6 − 30(1 + h)4 − 1 =0 6 (h)0 6 (h)1 6 (h)2 28 [( ) +( ) +( ) +⋯] 0 1 2 =0 4 4 4 −30 [( ) (h)0 + ( ) (h)1 + ( ) (h)2 + ⋯ ] − 1 0 1 2 28(1 + 6h + 15h2 + ⋯ ) −30(1 + 4h + 6h2 + ⋯ ) − 1
18(ii) Selective expansion (1 + y)7 sub y = x(1 + x): [1 + x(1 + x)]7 = 1 +7x(1 + x) +21[x(1 + x)]2 +35[x(1 + x)]3 +35[x(1 + x)]4 +21[x(1 + x)]5 +7[x(1 + x)]6 +[x(1 + x)]7 (1 + x)4 = 35x 4 +21x 5 (1 + x)5 +7x 6 (1 + x)6 (1 + x)7 + ⋯ +x 7
=0
28 + 168h + 420h2 =0 −30 − 120h − 180h2 − 1 + ⋯ 240h2 + 48h − 3 80h2 + 12h − 1 (20h − 1)(4h + 1) h=
1
1
20
or h = − (rej ∵ root greater than 1) 4
⇒ root x = 1 + 17(i)
4 = 35x 4 [… ( ) (x)3 … ] 3 5 [… (5) (x)2 +21x …] 2 6 +7x 6 [… ( ) (x)1 … ] 1 7 7 [… ( ) (x)0 ] + ⋯ +x 0
≈0 ≈0 ≈0
1 20
= 1.05 ✓
= [35(4) + 21(10) + 7(6) + 1]x 7 + ⋯ = 393x 7 + ⋯
(1 + 2x)2n 2n 2n 2n = ( ) (2x)0 + ( ) (2x)1 + ⋯ + ( ) (2x)r 0 1 r 2n (2x)2n +⋯+ ( ) ✓ 2n
17(ii) 22n = [1 + (1)]2n =(
=(
2n (1)0 2n 2n ) + ( ) (1)1 + ( ) (1)2 0 1 2 2n + ⋯ + ( ) (1)2n 2n 2n ) 0
⇒ coefficent of x 7 = 393 ✓ 18(iii) 1st Observation (1 + y)7 = 1 + 7y + 21y 2 + 35y 3 + 35y 4 + 21y 5 + 7y 6 + y7 ✓
2n 2n 2n + ( ) + ( ) + ⋯ + ( ) [proven] 1 2 2n
✓ 18(i) (1 + y)7 7 7 7 7 = ( ) (y)0 + ( ) (y)1 + ( ) (y)2 + ( ) (y)3 0 3 1 2 7 (y)4 7 (y)5 7 (y)6 7 +( ) +( ) +( ) + ( ) (y)7 5 6 4 7 2 3 4 5 = 1 + 7y + 21y + 35y + 35y + 21y + 7y 6 + y 7 ✓
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All the coefficients of the terms from the 2nd term onwards are divisible by 7 except the last term. 2nd Observation Last Term of y = [x(1 + x)]7 = x 7 (1 + x)7 7 7 7 7 ( ) (x)0 + ( ) (x)1 + ( ) (x)2 + ( ) (x)3 0 3 1 2 = x7 [ ] 7 7 7 7 (x)4 +( ) + ( ) (x)5 + ( ) (x)6 + ( ) x 7 5 6 7 4 = x 7 [(1)x + (7)x 2 + (21)x 2 + (35)x 3 + (35)x 4 + (21)x 5 + (7)x 6 + (1)x 7 ]
126
A math 360 sol (unofficial)
Ex 5.2
Ex 5.2 1(a)
(2 − x)3 = [2 + (−x)]3 = 23
3 3 3 + ( ) (2)3−1 (−x)1 + ( ) (2)3−2 (−x)2 + ( ) (2)3−3 (−x)3 3 1 2
=8
+(3)(4)(−x)
+(3)(2)(x 2 )
+(−x)3
= 8 − 12x + 6x 2 − x 3 ✓ 1(b)
(x + 2y)4 = x4
4 4 4 4 + ( ) (x)4−1 (2y)1 + ( ) (x)4−2 (2y)2 + ( ) (x)4−3 (2y)3 + ( ) (x)4−4 (2y)4 3 1 2 4
= x4
+(4)(x 3 )(2y)
+(6)(x 2 )(4y 2 )
+(4)(x)(8y 3 )
+(16y 4 )
= x 4 + 8x 3 y + 24x 2 y 2 + 32xy 3 + 16y 4 ✓ 1(c)
(2 + x 2 )5 = 25
5 5 5 5 5 + ( ) (2)5−1 (x 2 )1 + ( ) (2)5−2 (x 2 )2 + ( ) (2)5−3 (x 2 )3 + ( ) (2)5−4 (x 2 )4 + ( ) (2)5−5 (x 2 )5 3 5 1 2 4
= 25
+(5)(16)(x 2 )
+(10)(8)(x 4 )
+(10)(4)(x 6 )
+(5)(2)(x 8 )
+x10
= 32 + 80x 2 + 80x 4 + 40x 6 + 10x 8 + x10 ✓ 2(a)
x 5
1
2
2
(4 − ) = [4 + (− x)]
5
1 1 5 + ( ) (4)5−1 (− x) 2 1
= 45
1
= (1024) +(5)(256) (− x) 2
2 1 5 + ( ) (4)5−2 (− x) 2 2
3 1 5 + ( ) (4)5−3 (− x) + ⋯ 2 3
1
1
+(10)(64) ( x 2 )
+(10)(16) (− x 3 )
4
8
+⋯
= 1024 − 640x + 160x 2 − 20x 3 + ⋯ ✓ 2(b)
1
12
( + x2 ) 2
1 12 2
= = 2(c)
(
1 2x
12 1 12−1 (x 2 )1 12 1 12−2 (x 2 )2 +( )( ) +( )( ) 2 1 2 2
=( ) 1
+(12) (
4096 1 4096
8
− 2x 2 ) = [
+
1 2x
3 512
1 8 1 1
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x4 +
55 128
+(66) (
1 1024
) (x 4 )
−
1 8x5
+
12 1 12−3 (x 2 )3 )( ) +⋯ 3 2
+(220) (
1
) (x 6 )
512
+⋯
x6 + ⋯ ✓
8
+(8) (
256x8 256x8
) (x 2 )
8 1 8−1 (−2x 2 )1 8 1 8−2 (−2x 2 )2 +( )( ) +( )( ) 1 2x 2 2x
2x
=
33 512
+ (−2x 2 )]
=( ) =
x2 +
1 2048
+(
1 128x7
7 4x2
) (−2x 2 )
+(28) (
1 64x6
) (4x 4 )
8 1 8−3 (−2x 2 )3 +( )( ) +⋯ 3 2x +(56) (
1 32x5
) (−8x 6 ) + ⋯
− 14x + ⋯ ✓
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127
A math 360 sol (unofficial) 3(a)
Ex 5.2
For (x + 2y)5
5
5 Tr+1 = ( ) (x)5−r (2y)r ✓ r 3(b)
3(c)
Tr+1 4(a)
1 13 2y
)
= [3x 2 + (−
1 2y
13
4
For coefficient of x 3 ,
1 r
13 = ( ) (3x 2 )13−r (− ) ✓ 2y r
r=6
6(a)
10 (2)10−6 (x)6 ) 6 = (210)(16)(x 6 ) = 2260x 6 ✓
(rej)
18
1
For ( − x 2 )
For term independent of x, 3r − 18 = 0 r =6
r=3
Term independent of x 18 = ( ) (−1)6 (x)0 6 = (18 564)(1)(x)0 = 18 564 ✓
9 4th term = ( ) (3x)9−3 (−2)3 3 = (84)(729x 6 )(−8) = 489 888x 6 ✓ For (y − 2x)10 = [y + (−2x)]10
6(b)
10 (y)10−r (−2x)r ) r
For middle term,
r=
10 2
= [x −1 + (−x 2 )]18
18 (x −1 )18−r (−x 2 )r ) r 18 (−1)r x 2r = ( ) x r−18 r 18 = ( ) (−1)r x 3r−18 r
For (3x − 2)9 = [3x + (−2)]9
Tr+1 = (
2
Tr+1 = (
9 Tr+1 = ( ) (3x)9−r (−2)r r
4(c)
3
2r = 6 ⇒ r = 3 1 3 15 10 Coefficient of x 6 = [( ) (− ) ] = (− ) ✓ 4 8 3
7th term = (
For 4th term,
⇒r=
For coefficient of x 6 ,
x
4(b)
2r = 3
Coefficient of x 3 = 0 ✓
10 = ( ) (2)10−r (x)r r
For 7th term,
10
1
= [1 + (− x 2 )]
)]
For (2 + x)10 Tr+1
4
10
)
r 1 10 ) (− x 2 ) 4 r 1 r 10 = ( ) (− ) (x 2 )r 4 r 1 r 10 = ( ) (− ) x 2r 4 r
10 (2x)10−r (−3)r ) ✓ r
For (3x 2 −
x2
Tr+1 = (
For (2x − 3)10 = [2x + (−3)]10 Tr+1 = (
For (1 −
For (x +
1 2x2
12
)
1
= [x + (2 x−2 )]
12 12−r )x r 12 = ( ) x12−r r 12 1 r = ( )( ) r 2
Tr+1 = ( =5
10 Middle term = ( ) (y)10−5 (−2x)5 5 = (252)(y 5 )(−32x 5 ) = −8064x 5 y 5 ✓
For middle term,
1
( x −2 )
12
r
2
1 r
( ) x −2r 2
x12−3r r=
12 2
=6
12 1 6 (x)12−3(6) )( ) 6 2 231 = 6 ✓
Middle term = (
16x
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128
A math 360 sol (unofficial) 7
Ex 5.2
(px − 3)n = [px + (−3)]n n = ( ) (px)n−0 (−3)0 0 = (px)n
n + ( ) (px)n−1 (−3)1 + ⋯ 1 +(n)(px)n−1 (−3) + ⋯
= pn x n
−3npn−1 x n−1 + ⋯
= 512x 9 −qx 8 + ⋯ [given] Compare 1st term: Compare 2nd term: −3npn−1 x n−1 = −qx 8 pn x n = 512x 9 −3(9)29−1 x 9−1 = −qx 8 ⇒ n =9✓ ∵ n = 9, p = 2 8 )x 8 −27(2 = −qx 8 ⇒ pn = 512 ⇒q = 6912 ✓ ∵ n = 9, p9 = 512 p =2✓ 8(i)
x 7 (2 − ) 2
7 1 = [2 + (− x)] 2 1 1 7 = (2)7 + ( ) (2)7−1 (− x) 2 1 1 = 128 +(7)(64) (− x) 2 2
2 1 7 + ( ) (2)7−2 (− x) 2 2 1 2 +(21)(32) ( x ) 4
3 1 7 + ( ) (2)7−3 (− x) + ⋯ 2 3 1 3 +(35)(16) (− x ) + ⋯ 8
3
= 128 − 224x + 168x − 70x + ⋯ ✓ 8(ii)
(1.995)7 = (2 − 0.005)7 = (2 −
0.01 7 2
)
= 128 − 224(0.01) + 168(0.01)2 − 70(0.01)3 + ⋯ ≈ 125.7767 ✓ 9(i)
(1 − 2x)9 = [(1) + (−2x)]9 9 9 =1 + ( ) (−2x)1 + ( ) (−2x)2 + ⋯ 1 2 =1 +(9)(−2x) +(36)(4x 2 ) + ⋯ = 1 − 18x + 144x 2 + ⋯ ✓ 5 5 + ( ) (2)5−1 (x)1 + ( ) (2)5−2 (x)2 1 2 = 32 +(5)(16)(x) +(10)(8)(x 2 ) = 32 + 80x + 80x 2 + ⋯ ✓
(2 + x)5 = 25
9(ii)
(1 − 2x)9 (2 + x)5 = (1 − 18x + 144x 2 + ⋯ )(32 + 80x + 80x 2 + ⋯ )
= 32 +80x +80x 2 −576x −1440x 2 +4608x 2 + ⋯ = 32 −496x +3248x 2 + ⋯ ✓
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129
A math 360 sol (unofficial) 10(i)
Ex 5.2
(2x − 1)6 = [(2x) + (−1)]6 6 6 6 = (2x)6 + ( ) (2x)6−1 (−1)1 + ( ) (2x)6−2 (−1)2 + ( ) (2x)6−3 (−1)3 3 1 2 = (64x 6 ) +(6)(32x 5 )(−1) +(15)(16x 4 )(1) +(20)(8x 3 )(−1) = 64x 6 −192x 5 +240x 4 −160x 3 + ⋯ ✓
10(ii) (2x − 1)6 (x 2 − 2x + 3) = (64x 6 − 192x 5 + 240x 4 − 160x 3 + ⋯ )(x 2 − 2x + 3) = (−192x 5 )(3) +(240x 4 )(−2x) +(−160x 3 )(x 2 ) + ⋯ = −576x 5 −480x 5 = −1216x 5 + ⋯
−160x 5 + ⋯
∴ Coefficient of x 5 = −1216 ✓ 11(i)
1
( − 2x)
5
1
= [ + (−2x)]
2
5
2
1 5
5 1 5−1 (−2x)1 +( )( ) 1 2 1 +(5) ( ) (−2x)
=( ) = =
2 1
32 1 32
5 1 5−2 (−2x)2 +( )( ) 2 2 1 +(10) ( ) (4x 2 )
16
5
2
8
5 1 5−3 (−2x)3 +( )( ) +⋯ 3 2 1 +(10) ( ) (−8x 3 ) + ⋯ 4
3
− x + 5x − 20x + ⋯ ✓ 8
11(ii) Expansion 1
(1 + ax + 3x 2 ) ( − 2x) 2
= (1 + ax + 3x 2 ) ( = 5x 2 5a 2 x 8 3 + x2 32 5a
= (5 − 163 32
5
− x + 5x 2 − 20x 3 + ⋯ ) 8
−20x 3 +5ax 3
−
=(
1 32
5
+
8
−
15 3 x +⋯ 8 3 ) x 2 + (5a 32
−
5a 8
) x2
− 20 −
+ (5a −
175 8
15 8
) x3 + ⋯
) x3 + ⋯
Coefficients Coefficient of x 2 = (
163
32 5a
−
5a 8
)
2
=
[given] 13 2
=−
8
a
=−
Coefficent of x = 5a −
32 9 4
3
8
= 5 (− ) − 4 265 8
45
175 9
=−
13
175 8
∵a=−
9 4
✓
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130
A math 360 sol (unofficial) 12(i)
Ex 5.2
For (x + my)8 8 Tr+1 = ( ) x 8−r (my)r ✓ r
13(iii) For constant term,
Constant Term = (
12(ii) if m = 2: 8 (a) Tr+1 = ( ) x 8−r (2y)r r 8 8−r r r = ( )x 2 y r 8 = ( ) 2r x 8−r y r r
14(i)
For (x 3 −
2 10 x2
)
x
k r 9 Tr+1 = ( ) x 9−r ( ) x r 9 9−r (kx −1 )r = ( )x r 9 9−r r −r = ( )x k x r 9 r 9−2r = ( )k x r For coefficient of x 3 , For coefficient of x 3 , 9 − 2r = 3 9 − 2r = 3 −2r = −6 −2r = −6 r =3 r =3
r=5 8 5 = ( )2 5 = 1792 ✓
= −1512x 5 y 3
Coefficient of x 3 Coefficient of x 3 9 9 = ( ) k3 = ( ) k3 3 3 3 = 84k = 84k 3 Equate coefficients Coefficient of x 3 = Coefficient of x [given] 84k 3 = 126k 4 2k 3 = 3k 4 3 4 2k − 3k =0 3 k (2 − 3k) =0
= −1512x 5 y 3 = −1512x 5 y 3 = −1512x 5 y 3 = −1512 = −3 ✓ = [x 3 + (−2x −2 )]10
10 (x 3 )10−r (−2x −2 )r ) r 10 = ( ) x 30−3r (−2)r x −2r r 10 (−2)r =( ) x 30−5r r
2
k = 0 (rej ∵ k is positive)or k = ✓
Tr+1 = (
13(i)
Specific coefficients k 9
For term (−1512x 5 y 3 ), r=3
13
= 13 440
For (x + ) ,
8 8−r (my)r 12(ii) Recall: T r+1 = ( ) x r (b)
T4 8 ( ) x 8−3 (my)3 3 8 ( ) x 5 m3 y 3 3 56m3 x 5 y 3 56m m
10 (−2)6 30−5(6) ) x 6
✓
For coefficient of x 3 y 5 , Coefficient of x 3 y 5
30 − 5r = 0 −5r = −30 r =6
3
For term in x10 ,
Term in x10
30 − 5r = 10 −5r = −20 r =4 10 (−2)4 10 =( ) x = 3360x10 ✓ 4
13(ii) For coefficient of 1 , 5
30 − 5r = −5
x
−5r r Coefficient of
1 x5
=(
= −35 =7
10 (−2)7 ) = −15 360 ✓ 7
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131
A math 360 sol (unofficial)
Ex 5.2
14(ii) Put k = 2, 3
(1 − 6x
2)
2 3
9
(x + ) x
= (1 − 6x 2 )[(term in x 3 ) + (term in x 5 ) + ⋯ ] [to be continued] Specific terms 2
9
9
2
For (x + 3 ) = [x + ( x −1 )] , x
3
r 2 9 Tr+1 = ( ) x 9−r ( x −1 ) 3 r 9 9−r 2 r −r ( ) x = ( )x 3 r r 9 2 = ( ) ( ) x 9−2r r 3
For term in x 3 , 9 − 2r = 3 2r =6 r =3 2 3
Term in x 3 = 84 ( ) x 3 = 3
224 3 x 9
For term in x 5 , 9 − 2r = 5 2r =4 r =2 2 2 9 Term in x 5 = ( ) ( ) x 5 = 16x 5 2 3 Expansion (1 − 6x
2)
2 3
9
(x + ) x
= (1 − 6x 2 ) (
224 3 x 9
+ 16x 5 + ⋯ )
= (1)(16x 5 ) + (−6x 2 ) ( =
400 5 − x 3
224 3 x ) 9
+⋯
+⋯
∴ Coefficient of x 5 = −
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400 3
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132
A math 360 sol (unofficial) 15(i)
Ex 5.2
x n
(2 − ) 2
1
= [2 + (− x)]
n
2
0 1 n n 1 1 = ( ) (2)n−0 (− x) + ( ) (2)n−1 (− x) 2 2 0 1 1
2 n 1 + ( ) (2)n−2 (− x) + ⋯ 2 2 n(n−1)
1
= 2n
+n(2n−1 ) (− x)
= 2n
−n(2n−1 ) ( ) x
+
= 2n
−n(2n−1 )(2−1 )x
+n(n − 1)(2−3 )2n−2 x 2 + ⋯
= 2n
−n(2n−2 )x
+n(n − 1)2n−5 x 2 + ⋯ ✓
2
1 2
+
2
(2n−2 ) x 2 + ⋯
n(n−1) n−2 2 2 x 8
4
+⋯
15(ii) (1 + 2x) (2 − x)n 2
= (1 + 2x)[2n −n(2n−2 )x +n(n − 1)2n−5 x 2 + ⋯ ] = 2n
−n(2n−2 )x +(2n+1 )x
+n(n − 1)2n−5 x 2 −(n)2n−1 x 2 + ⋯
= 2n +[2n+1 − n(2n−2 )]x = a + bx 2 + ⋯ [given]
+[n(n − 1)2n−5 − (n)2n−1 ]x 2 + ⋯
Compare x: 2n+1 −n(2n−2 ) =0 n−2 3 n−2 (2 )2 −n(2 ) = 0 2n−2 (23 − n) =0 n =8✓ 15(iii) Compare x 0 : a = 2n = 28 ∵ n = 8 = 256 ✓ Compare x 2 : b = n(n − 1)2n−5 − (n)2n−1 = 8(8 − 1)28−5 − (8)28−1 ∵ n = 8 = −576 ✓
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133
A math 360 sol (unofficial) 16
Ex 5.2
Expansion (1 + x)(a − bx)12 = (1 + x)[(term in x 7 ) + (term in x 8 ) + ⋯ ] [to be continued]
17(i)
Specific terms For (a − bx)12 = [a + (−bx)]12 , 12 12−r (−bx)r )a r 12 = ( ) a12−r (−b)r x r r
For coefficient of x 2 , r = 2 n n Coefficient of x 2 = ( ) 32 = 9 ( ) 2 2 Equate coefficients (Coefficient of x 3 ) = 6(Coefficient of x 2 ) n n 27 ( ) = 6 ⋅ 9( ) 3 2 n n ( ) = 2 ( ) [shown] ✓ 3 2
Tr+1 = (
For term in x 7 ,
Find specific coefficients For (1 + 3x)n n Tr+1 = ( ) (3x)r r n r r = ( )3 x r For coefficient of x 3 , r = 3 n n Coefficient of x 3 = ( ) 33 = 27 ( ) 3 3
r=7
12 Term in x 7 = ( ) a5 (−b)7 x 7 7 = −792a5 b7 x 7
17(ii)
For term in x 8 , r = 8 12 4 (−b)8 8 )a x 8 = 495a4 b8 x 8
Term in x 8 = (
n! (n−3)!3! n(n−1)(n−2)(n−3)! (n−3)!3! n(n−1)(n−2) 3! (n)(n−1)(n−2) 6
Expansion (1 + x)(a − bx)12 = (1 + x)[(−792a5 b7 x 7 ) + (495a4 b8 x 8 ) + ⋯ ] = 1(495a4 b8 x 8 ) + x(−792a5 b7 x 7 ) + ⋯ = 495a4 b8 x 8 − 792a5 b7 x 8 + ⋯
n!
= 2 ⋅ (n−2)!2! =2⋅ =2⋅ =2⋅
n(n−1)(n−2)! (n−2)!2! n(n−1) 2! (n)(n−1) 2
(n)(n − 1)(n − 2) = 6(n)(n − 1) (n)(n − 1)[(n − 2) − 6] = 0 (n)(n − 1)(n − 8) =0 n=0 or n = 1 or n = 8 ✓ (rej ∵ n ≥ 2) (rej ∵ n ≥ 2)
= (495a4 b8 − 792a5 b7 )x 8 + ⋯ Coefficient of x 8 = 0 [given] 4 8 5 7 495a b − 792a b = 0 5a4 b8 − 8a5 b7 =0 5a4 b8 = 8a5 b7 a b
5
= ✓
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134
A math 360 sol (unofficial) 18(i)
Ex 5.2
(2 + p)5 5 5 5 + ( ) (2)5−1 (p)1 + ( ) (2)5−2 (p)2 + ( ) (2)5−3 (p)3 + ⋯ 3 1 2 2 = (32) +(5)(16)(p) +(10)(8)(p ) +(10)(4)(p3 ) +⋯ 2 3 = 32 +80p +80p +40p +⋯✓ = 25
18(ii) (2 + x − 2x 2 )5 = [2 + (x − 2x 2 )]5 = 32 +80(x − 2x 2 ) +80(x − 2x 2 )2 = 32 +80(x − 2x 2 ) +80(x 2 − 4x 3 + ⋯ ) = 32 +80x −160x 2 2 +80x −320x 3 +40x 3 + ⋯ = 32 +80x 19
−80x 2
+40(x − 2x 2 )3 + ⋯ +40(x 3 + ⋯ )
−280x 3 + ⋯ ✓
(a + bx + cx 2 )4 = 81 + 216x + 108x 2 + dx 3 + ⋯ [given] (a + bx + cx 2 )4 = [a + (bx + cx 2 )]4 4 4 = a4 + ( ) (a)4−1 (bx + cx 2 )1 + ( ) (a)4−2 (bx + cx 2 )2 1 2
4 + ( ) (a)4−3 (bx + cx 2 )3 + ⋯ 3
= a4 + 4a3 (bx + cx 2 )
+6a2 (b2 x 2 + 2bcx 3 + ⋯ )
+4a(b3 x 3 + ⋯ ) + ⋯
= a4 + 4a3 bx +4a3 cx 2
+6a2 b2 x 2 +12a2 bcx 3
+4ab3 x 3 + ⋯
= a4 + 4a3 bx + (4a3 c + 6a2 b2 )x 2 + (12a2 bc + 4ab3 )x 3 + ⋯ = 81 + 216x + 108x 2 + dx 3 + ⋯ [given] Compare x 0 : a4 = 81 a = 3 ✓ or a = −3 Compare x:
(rej ∵ a > 0)
4a3 b = 216 a3 b = 54 ∵ a = 3, (3)3 b = 54 b =2✓
Compare x 2 : 4a3 c + 6a2 b2 = 108 3 2 2 2a c + 3a b = 54 ∵ a = 3, b = 2, 2(3)3 c + 3(3)2 (2)2 = 54 54c + 108 = 54 c = −1 ✓ Compare x 3 : 12a2 bc + 4ab3 =d ∵ a = 3, b = 2, c = −1, 12(3)2 (2)(−1) + 4(3)(2)3 = d d = −120 ✓
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135
A math 360 sol (unofficial) 20(i)
For (2x 2 −
1 n √x
Ex 5.2 1
) = [2x 2 + (−x −2 )]
n
21(i)
1 r n Tr+1 = ( ) (2x 2 )n−r (−x −2 ) r 1 n n−r 2n−2r (−1)r x −2r = ( )2 x r 5 n n−r = ( ) 2 (−1)r x 2n−2r r
For (a + b)n , n TR+1 = ( ) an−R bR R For 2nd term, R = 1 n 2nd term: p = ( ) an−1 b1 [given] 1 = nan−1 b For 3rd term: R = 2 n 3rd term: q = ( ) an−2 b2 [given] 2
For term independent of x: 5
n!
2n − r = 0
= (n−2)!2! an−2 b2
2
5 2
r
= 2n 5
n
4
n(n−1) n−2 2 a b 2
n!
= (n−3)!3! an−3 b3
For coefficient of x 7 √x: 5
10 − r = 7.5 2
r
=
For 4th term: R = 3 n 4th term: r = ( ) an−3 b3 [given] 3
5 5−r (−1)r 2(5)−52r 20(ii) T x r+1 = ( ) 2 r 5 5 = ( ) 25−r (−1)r x10−2r r
2
n(n−1)(n−2)! n−2 2 a b (n−2)!2!
= r
∵ r is non − negative integer, smallest n = 5 ✓
5
=
r
=
5 2
=
n(n−1)(n−2)(n−3)! n−3 3 a b (n−3)!3!
=
n(n−1)(n−2) n−3 3 a b 6
pr q2
=1
nan−1 b
Coefficient of x 7 √x 5 = ( ) (2)5−1 (−1)1 1 = (5)(16)(−1) = −80 ✓
=
n(n − 1)(n − 2) n−3 3 [ a b ] 6 n(n − 1) n−2 2 [ a b ] 2
2
1 2 n (n − 1)(n − 2)a2n−4 b4 6 = 1 2 n (n − 1)2 a2n−4 b 4 4 =
2(n − 2) [shown] 3(n − 1)
✓ 21(ii) when p = 8, q = 24, r = 36: (8)(36) (24)2 1 2
= =
2(n−2) 3(n−1) 2(n−2) 3(n−1)
3(n − 1) = 4(n − 2) 3n − 3 = 4n − 8 n =5✓
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A math 360 sol (unofficial)
Ex 5.2
22 3
3
Given: a = √2 + √5 + √2 − √5 Prove: a3 = 4 − 3a LHS = a3 3
3
= [( √2 + √5) + ( √2 − √5)] 3
= ( √2 + √5)
3
3
3−1 3 1 3−2 3 2 3 3 3 3 ( √2 − √5) + ( ) ( √2 + √5) ( √2 − √5) + ( ) ( √2 + √5) 1 2 3
2
3
3
+3 ( √2 + √5) ( √2 − √5)
= 2 + √5 3
3
3
3
+3 ( √2 + √5) ( √2 − √5)
3−3 3 3 3 3 ( √2 − √5) + ( ) ( √2 + √5) 3
2
+2 − √5
3
= 4 +3 ( √2 + √5) ( √2 − √5) [( √2 − √5) + ( √2 + √5)] 3
= 4 +3( √4 − 5)
(a)
= 4 +3(−1)
(a)
= 4 −3
(a)
3
3
∵ a = √2 + √5 + √2 − √5
= 4 −3a = RHS [proven] ✓ a3 = 4 − 3a 3 a + 3a − 4 =0 (a − 1)( + + ⬚) (a − 1)(a2 + + ⬚) 2 (a − 1)(a + + 4) 2 (a − 1)(a + a + 4) = 0 a = 1 or a =
−(1)±√(1)2 −4(1)(4) 2(1)
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=
−1±√−15 2
(rej ∵ discriminant < 0) ✓
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A math 360 sol (unofficial)
Rev Ex 5 A2(b)
Rev Ex 5 A1(i) (a)
(1 + 3x)6 6 6 6 = 1 + ( ) (3x)1 + ( ) (3x)2 + ( ) (3x)3 + ⋯ 3 1 2 = 1 + (6)(3x) +(15)(9x 2 ) +(20)(27x 3 ) + ⋯ = 1 + 18x + 135x 2 + 540x 3 + ⋯ ✓
A1(i) (b) (1 − 4x)5
For (x −
2 12 x2
)
= [x + (−2x −2 )]12
12 12−r (−2x −2 )r )x r 12 = ( ) x12−r (−2)r x −2r r 12 = ( ) (−2)r x12−3r r
Tr+1 = (
For coefficient of x 3 , 12 − 3r = 3 −3r = −9 r =3
= [1 + (−4x)]5 5 5 5 = 1 + ( ) (−4x)1 + ( ) (−4x)2 + ( ) (−4x)3 + ⋯ 3 1 2
Coefficient of x 3 12 = ( ) (−2)3 3 = −1760 ✓
= 1 + (5)(−4x) +(10)(16x 2 ) +(10)(−64x 3 ) + ⋯ = 1 − 20x + 160x 2 − 640x 3 + ⋯ ✓ A1(ii) (1 + 3x)6 (1 − 4x)5 = (1 + 18x + 135x 2 + ⋯ )(1 − 20x + 160x 2 + ⋯ ) = (1)(160x 2 ) +(18x)(−20x)
+(135x 2 )(1) + ⋯
= 160x 2
+135x 2 + ⋯
−360x 2
= −65x 2 + ⋯ Coefficient of x 2 = −65 ✓ A2(a)
x 8
For (1 + ) , 2
8 x r Tr+1 = ( ) ( ) r 2 8 1 r = ( ) ( x) r 2 8 1 r = ( ) ( ) xr r 2 For coefficient of x 3 , r=3 Coefficient of x 3 8 1 3 = ( )( ) 3 2 =7✓
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138
A math 360 sol (unofficial) A3(i)
Rev Ex 5
2 8
(x 2 − ) x
2 8 = [x 2 + (− )] x = (x 2 )8
2 1 2 2 8 8 + ( ) (x 2 )8−1 (− ) + ( ) (x 2 )8−2 (− ) x x 1 2
= x16
+(8)(x14 ) (− )
2 x
2 3 8 + ( ) (x 2 )8−3 (− ) x 3
4
8
+(28)(x12 ) ( 2 )
+(56)(x10 ) (− 3 )
x
x
= x16 − 16x13 + 112x10 − 448x 7 + ⋯ ✓ A3(ii)
2 8
(x 3 + 1)2 (x 2 − ) x = (x 6 + 2x 3 + 1)(x16 − 16x13 + 112x10 − 448x 7 + ⋯ ) = (x 6 )(−448x 7 ) +(2x 3 )(112x10 ) = (−448x13 ) +(224x13 ) 13 = −240x + ⋯
+(1)(−16x13 ) + ⋯ +(−16x13 ) + ⋯
Coefficient of x13 = −240 ✓ A4(i)
x n
1
2
2
(1 − ) = [1 + (− x)]
n
0 1 n n 1 1 = ( ) (− x) + ( ) (− x) 2 2 0 1 1
= (1)
+(n) (− x)
=1
− nx
2
1
2
+( +
2
A4(ii) (1 − x)n = 1 + ax + 7x 2 + ⋯
2 n 1 + ( ) (− x) + ⋯ 2 2 n(n−1)
1
2
4
) ( x2 ) + ⋯
n(n−1) 2 x 8
+⋯✓
[given]
Compare x 2 : n(n−1)
=7
8 2
n −n = 56 2 n − n − 56 =0 (n − 8)(n + 7) = 0 n = 8 or n = −7 (rej ∵ n > 2) ✓ A4(iii) Compare x: 1
− n 2 1
=a
− (8) = a 2
a
= −4 ✓
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139
A math 360 sol (unofficial) A5(i)
Rev Ex 5
(1 + p)4 4 4 4 4 = ( ) (p)0 + ( ) (p)1 + ( ) (p)2 + ( ) (p)3 0 3 1 2 =1
+6p2
+4p
+4p3
B2
4 + ( ) (p)4 4 4 +p ✓
B2(i)
8
1
1 1
= [a2 x −2 + (− x 2 )]
8
a
r
8
For middle term,
r= =4 2
B2(ii) For coefficient of 1, x
= 1 + 4x + 10x 2 + 16x 3 + ⋯ ✓
Coefficient of
A5(iii) (1.11)4 = [1 + (0.1) + (0.1)2 ]4 ≈ 1 + 4(0.1) +10(0.1)2 + 16(0.1)3 ≈ 1 + 0.4 +0.1 +0.016 ≈ 1.516 ✓ B1(i)
√x ) a
8 Middle term = ( ) a16−12 (−1)4 x 4−4 4 = 70a4 ✓
= 1 +4x +4x 2 +6(x 2 + 2x 3 + ⋯ ) +4(x 3 + ⋯ ) + ⋯ = 1 +4x +4x 2 +6x 2 +12x 3 +4x 3 + ⋯
√x
−
8−r
= [(1) + (x + x 2 )]4 +4(x + x 2 )3 + ⋯
a2
1 1 1 8 (− x 2 ) Tr+1 = ( ) (a2 x −2 ) a r 1 1 8 = ( ) a16−2r x 2r−4 (−1)r a−r x 2r r 8 = ( ) a16−3r (−1)r x r−4 r
A5(ii) (1 + x + x 2 )4 = 1 +4(x + x 2 ) +6(x + x 2 )2
For (
1 x
r=3
8 = ( ) a16−3(3) (−1)(3)−4 3 = (56)a7 (−1) = −56a7 ✓
(2 − x)7 = [2 + (−x)]7 7 7 = ( ) (2)7−0 (−x)0 + ( ) (2)7−1 (−x)1 0 1 7 + ( ) (2)7−2 (−x)2 + ⋯ 2 = 128
+(7)(64)(−x) +(21)(32)(x 2 ) + ⋯ = 128 − 448x + 672x 2 + ⋯ ✓ B1(ii) 1.997 = (2 − 0.01)7 = 128 −448(0.01) +672(0.01)2 + ⋯ = 128 −4.48 +0.0672 + ⋯ ≈ 123.587 ✓ B1(iii) (k − x)(2 − x)7 = (k − x)(128 − 448x + 672x 2 + ⋯ ) = k(672x 2 ) +(−x)(−448x) + ⋯ = 672kx 2 +448x 2 + ⋯ = (672k + 448)x 2 + ⋯ Coefficient of x 2 = 616 [given] 672k + 448 = 616 672k = 168 k
1
= ✓ 4
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A math 360 sol (unofficial)
Rev Ex 5
B3(i) 5
(2 + √3) 0 5 = ( ) (2)5−0 (√3) 0
1 2 3 4 5 5 5 5 5 5 + ( ) (2)5−1 (√3) + ( ) (2)5−2 (√3) + ( ) (2)5−3 (√3) + ( ) (2)5−4 (√3) + ( ) (2)5−5 (√3) 3 5 1 2 4
= (32)
+(5)(16)(√3)
+(10)(8)(3)
+(10)(4)(3√3) +(5)(2)(9)
+(1)(1)(9√3)
= (32 + 240 + 90) + (80 + 120 + 9)√3 = 362 + 209√3 ✓ B3(ii) (2 − √3)5 = 362 − 209√3 ✓ 5
Show: (2 − √3) =
1 (2+√3)
5 (2+√3)
LHS = (2 − √3) B4(i)
(2+√3)
5
5 5
=
(4−3)5 (2+√3)
5
=
1 (2+√3)
5
= RHS [shown] ✓
(a − x)(1 + 2x)n = 3 + 47x + bx 2 + ⋯ sub x = 0: n
(a − 0)(1 + 2(0)) = 3 + 47(0) + b(0)2 a
=3✓
B4(ii) (3 − x)(1 + 2x)n = 3 + 47x + bx 2 + ⋯ [given] (3 − x)(1 + 2x)n
= (3 − x) [1
n n + ( ) (2x)1 + ( ) (2x)2 + ⋯ ] 1 2 n(n−1) (4x 2 ) + ⋯ ] + (n)(2x) +
= (3 − x)(1
+ 2nx
= (3 − x) [1
= 3 +6nx −x
2
+ 2n(n − 1)x 2 + ⋯ )
+6n(n − 1)x 2 −2nx 2
= 3 +(6n − 1)x +(6n2 − 6n − 2n)x 2 + ⋯ = 3 +(6n − 1)x +(6n2 − 8n)x 2 + ⋯ = 3 +47x + bx 2 + ⋯ [given] Compare x: (6n − 1) = 47 6n = 48 n =8✓ B4(iii) Compare x 2 : (6n2 − 8n) = b 6(8)2 − 8(8) = b ∵ n = 8 b = 320 ✓
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A math 360 sol (unofficial) B5
Rev Ex 5
Expansion (1 + ax)6 (2 + bx)5 6 6 = [1 + ( ) (ax)1 + ( ) (ax)2 + ⋯ ] 1 2 5 (2)5−1 (bx)1 5 5 [2 + ( ) + ( ) (2)5−2 (bx)2 + ⋯ ] 1 2 = (1 + 6ax + 15a2 x 2 + ⋯ )(32 + 80bx + 80b2 x 2 + ⋯ )
+80b2 x 2 +480abx 2 +480a2 x 2 + ⋯ +(80b2 + 480ab + 480a2 )x 2 + ⋯
= 80bx +192ax = (80b + 192a)x
Equate coefficients Coeff. of x = −112 [given] 80b + 192a = −112 b
= =
−112−192a 80 −7−12a
−(1)
5
Coeff. of x 2 = 80 [given] 2 2 80b + 480ab + 480a = 80 b2 + 6ab + 6a2 =1 −(2) sub (1) into (2): (
−7−12a 2
)
5 144a2 +168a+49 25
+6a ( +
−7−12a
5 −42a−72a2
) + 6a2 = 1
5
+6a2 = 1
(144a2 + 168a + 49) −210a − 360a2 + 150a2
= 25
(144a2 + 168a + 49) −210a − 210a2
= 25
−66a2 −42a +24 11a2 +7a +4 (11a + 4)(a + 1)
=0 =0 =0
a=−
4 11
(rej ∵ a ∈ ℤ)
or
a = −1 ✓ b|a=1 =
−7−12(−1) 5
=1✓
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A math 360 sol (unofficial)
Ex 6.1 4(ii)
Ex 6.1 1(a)
Point:
D(1,4) or E(−3,1) (4)−(1)
Gradient: mDE = (1)−(−3) =
A(4,5) B(6,9) MAB = (
4+6 5+9
,
2
2
y − (4)=
= (5,7) ✓ 1(b)
−5+11 3+(−7)
,
2
2
y
2(b)
4
4
✓
C = MAB 9+b (−6)+b2 (3, −4) = ( 1 , )
2
3 = 2 a1 = 0 ⇒ A(0,3) ✓
4 13
A(9, −6) C(3, −4) B(b1 , b2 )
M = MAB a +6 a +7 (3,5) = ( 1 , 2 ) 2 (a1 )+(6)
3
4 3
5(i) A(a1 , a2 )
M(3,5) B(6,7)
[x − (1)]
4 3
= x+
)
= (3, −2) ✓ 2(a)
3
y−4 = x−
A(−5,3) B(11, −7) MAB = (
4
y − y1 = mDE (x − x1 )
DE:
)
3
9+b1
(a2 )+(7)
and
5 = a2 = 3
2
(−6)+b2
3 = and −4 = 2 2 b1 = −3 b2 = −2 ∴ B(−3, −2) ✓
2
5(ii)
M(−2,6) B(−4, −8) A(a1 , a2 )
2
Radius = |BC| = √[(−3) − 3]2 + [(−2) − (−4)]2
M = MAB a +(−4) a2 +(−8) (−2,6) = ( 1 ) , 2 (a1 )+(−4)
2
−2 = 2 a1 = 0 ∴ A(0,20) ✓ 3(a)
and
= √40
(a2 )+(−8)
6 = a2 = 20
= √4 × 10 = 2√10 ✓
2
6(i)
A(2a, −a) B(4a, 5a) MAB = (
= √36 + 4
|AC| = √(5 − 6)2 + [3 − (−4)]2
2a+4a −a+5a
,
2
)
2
= √1 + 49 = √50
= (3a, 2a) ✓ 3(b)
|BC| = √[5 − (−2)]2 + (3 − 4)2
A(2t, 5) B(4,1 − 2t) MAB = (
= √49 + 1 = √50
(2t)+4 5+(1−2t)
,
2
2
)
∵ |AC| = |BC|, △ ABC is isosceles ✓
= (t + 1,3 − t) ✓ 4(i)
A(−1,6) B(3,2)
C(−5, −4)
6(ii)
MAB = (
6+(−2) (−4)+4 2
,
2
)
= (2,0) ✓
D = MAB =(
A(6, −4) B(−2,4) C (5,3)
(−1)+3 6+2
,
2
2
)
= (1,4) ✓ E = MAC =(
(−1)+(−5) 6+(−4) 2
,
2
)
= (−3,1) ✓
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143
A math 360 sol (unofficial) 6(iii)
Ex 6.1
area of △ ABC
8
=
1 |AB||MAB C| 2
=
1 [6 − (−2)]2 √(5 − 2)2 + (3 − 0)2 √ 2 +[(−4) − 4]2 1
= √64 + 64 2
y = x 2 + 2x − 3
√9 + 9
1 √128√18 2 1 = √2304 2 1 = (48) 2
x= =
= 24 unit 2 ✓
(
2 p2 +q2 2 2
2
=
−4±4√2
2
= −2 ± √2
2
y|x=−2+√2 = 1 − 2(−2 + √2)
) = (5,1)
=5
−4±√16×2
−4±√32
⇒ A(−2 − √2, −3 + 2√2)
= (5,1) 2
=
2(1)
= 5 + 2√2
p2 +q2 p+q
,
−(4)±√(4)2 −4(1)(−4)
y|x=−2−√2 = 1 − 2(−2 − √2)
A(p2 , p) B(q2 , q) MAB
−(2)
sub (1) into (2): 1 − 2x = x 2 + 2x − 3 2 x + 4x − 4 = 0
=
7
Points A & B At A & B, 2x + y = 1 intersects y = x 2 + 2x − 3. 2x + y = 1 y = 1 − 2x −(1)
and
2
p + q = 10 −(1) sub (2) into (1): p2 + (2 − p)2 = 10 2 2) (4 p + − 4p + p = 10 2p2 − 4p − 6 =0 2 p − 2p − 3 =0 (p + 1)(p − 3) =0
p+q 2
= 5 − 2√2 =1
⇒ B(−2 + √2, −3 − 2√2)
q = 2 − p −(2)
p = −1 or q|p=−1 = 3 A = (p2 , p) = ((−1)2 , −1) = (1, −1)
p=3 q|p=3 = −1 A = (p2 , p) = ((3)2 , 3) = (9,3) ✓
B = (q2 , q) = ((3)2 , 3) = (9,3)
B = (q2 , q) = ((−1)2 , −1) = (1, −1) ✓
Midpoint of AB MAB = (
(−2−√2)+(−2+√2) (5+2√2)+(5−2√2)
,
2
2
)
= (−2,5) ✓ 9
B(−2,5) C(c1 , c2 )
A(3,7) MAC = (
(3)+(c1 ) (7)+(c2 ) 2
,
2
)
MAC lies on x-axis (y = 0): 7+c2 2
c2
=0 = −7
MBC = (
(−2)+(c1 ) (5)+(c2 ) ) , 2 2
MBC lies on y − axis (x = 0): −2+c1 2
c1
=0 =2
∴ C(2, −7) ✓
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A math 360 sol (unofficial) 10(i)
Ex 6.1
A(6,8) B(8, −4) C(−2, −2) Point M M = MAB = (
6+8 8+(−4) 2
,
2
12(i)
O(0,0) P(4, r) Q(q1 , q2 ) R(3,4) For rhombus OPQR: |OP| = |OR|
) = (7,2) ✓
√(4 − 0)2 + (r − 0)2 = √(3 − 0)2 + (4 − 0)2
Point N N = MBC = ( Line MN Points:
8+(−2) (−4)+(−2)
,
2
2
M(7,2) or N(3, −3) (2)−(−3)
Gradient: mMN =
=
(7)−(3)
= √9 + 16 √16 + r 2 r2 =9 r = 3 ✓ or r = −3 (NA)
) = (3, −3) ✓
12(ii) O(0,0) P(4,3) Q(q1 , q2 ) R(3,4)
5 4
MPR = (
y − y1 = mMN (x − x1 )
MN:
5
35
4
4
+2
12(iii) O(0,0) P(4,3) Q(q1 , q2 ) R(3,4) MOQ
Point P 5
27
4
4
At P, MN (y = x − y 5
x−
4 5
4
=
x
=
⇒ P(
27 5
) cuts x-axis (y = 0).
27 4 27
2
q2
7
= 2 =7
For parallelogram PQRS, MPR = MQS
27 5
(
M = MAC = (
1+6 3+2
2 1+6
, 0) N(3, −3)
2
A(3, −2) B(b, 3) C(6,2) 3+6 (−2)+2 2
,
2
,
=
) =(
2 3+s1 2
3+s1 5+s2
and
s1 = 4 ∴ S(4,0) ✓
2
,
2 3+2 2
s2
) 5+s
2 = 2 =0
D(7, d) 9
) = ( , 0) ✓ 2
=M
b+7 3+d
2
and
2 2 0+q2
13(a) P(1,3) Q(3,5) R(6,2) S(s1 , s2 )
By similar triangles (using x-coordinates) MP: PN = (2 − 0) : [0 − (−3)] = 2: 3 ✓
2 b+7
2
7 7
)=( , )
5
M(7,2) P (
(
=
2 7
q1 = 7 ∴ Q(7,7) ✓
, 0) ✓
11(ii) MBD
,
2
10(ii) Ratio 𝐌𝐏: 𝐏𝐍
11(i)
= MPR
0+q1 0+q2
2 0+q1
=0
x
4
(
=0 27
)
2
2 2
5 27 x− 4 4
=
,
7 7
4
= x−
2
=( , )✓
5
y − (2)= ( ) [x − (7)] y
4+3 4+3
,
=
2 9 2
9
) = ( , 0) 2
and
b =2✓
3+d 2
=0
d = −3 ✓
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A math 360 sol (unofficial)
Ex 6.1
13(b) P(1,3) Q(3,5) R(6,2) S(4,0) T(t1 , t 2 )
15(i)
T(t1 , t 2 ) lies on y = 2x: t 2 = 2t1 ⇒ T(t1 , 2t1 ) PT
M = MPR = ( N = MQS = ( 15(ii) M
= √(3 − t1
)2
+ [5 − (2t1
)]2
(
5(t1 )2 − 14t1 + 10 12t1 t1
2
= 5(t1 )2 − 26t1 + 34 = 24 =2
r
r−4 7
6+0 (−2)+s
,
2
2
s−2
) = (3,
2
, )✓
2
2
)✓
2
) =(
,
=
2 0+d1
(−1)+d1 6+d2
=
,
2
2 6+1 2
) =(
2 0+3 2
(−1)+0 6+0
,
=
2
) 7
and
2
= 10 ✓
s−2
=
2
s =9✓
𝑦
=
2
0+b a+0
,
2
b a
) =( , )
2
2 2
b 2
a 2
2
2
=√
b2 4
+
a2 4
0+d2 2 b 2
a 2
2
2
b 2
a 2
2
2
|AM| = √(0 − ) + (a − )
=√
|BM| = √(b − ) + (0 − )
=√
b2 4
b2 4
+
+
a2 4
a2 4
0+3 0+1
and
) =(
2 3+d1
𝑥
|OM| = √(0 − ) + (0 − )
)
2
,
2 6+d2
) =
2
d2 d1 =4 ∴ D(4, −5) ✓ Case 3 For parallelogram ACBD, MAB = MCD 2 (−1)+0
s−2
𝑂 B(b, 0)
d1 =2 ∴ D(2,7) ✓ Case 2 For parallelogram ABDC, MAD = MBC
2
=3
M = MAB = (
0+d1 0+d2
and
2
,
2
16
d2 = 7
2 (−1)+d1
, ) = (3,
D(d1 , d2 )
C(3,1)
(−1)+3 6+1
2 (−1)+3
=N
PQRS is a parallelogram ✓
Case 1 For parallelogram ABCD, MAC = MBD
2
)=(
2
A(0, a)
A(−1,6) B(0,0)
(
,
2
S(0, s)
15(iii) ∵ MPR = MQS ,
⇒ T(2,4) ✓
(
R(r, 5)
(−4)+r 2+5
r−4 7
2 r−4
(t1 )2 − 2t1 + 1 (t1 )2 − 6t1 + 9 = + 4(t1 )2 − 12t1 + 9 +4(t1 )2 − 20t1 + 25
(
Q(6, −2)
= QT
(1 − t1 )2 √ +[3 − (2t1 )]2
14
P(−4,2)
and
d1 = −4 ∴ D(−4,5) ✓
,
17(i)
O(0,0)
P(2a, 0) Q(2b, 2c)
A = MOP = (
(0)+(2a) (0)+(0)
,
2
2
R(2d, 2e)
)
= (a, 0) [shown] ✓
2 6+0 2
2
= −5
3+d1 1+d2 2
∵ |OM| = |AM| = |BM|, M is equidistant from the three vertices ✓
0+1
)
=
B = MPQ
1+d2 2
=(
(2a)+(2b) (0)+(2c)
,
2
2
)
= (a + b, c) [shown] ✓
d2 = 5
C = MQR = (
(2b)+(2d) (2c)+(2e)
,
2
2
)
= (b + d, c + e) [shown] ✓ D = MRO = (
(2d)+(0) (2e)+(0) 2
,
2
)
= (d, e) [shown] ✓ © Daniel & Samuel A-math tuition 📞9133 9982
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146
A math 360 sol (unofficial)
Ex 6.1
17(ii) A(a, 0) B(a + b, c) C(b + d, c + e) D(d, e) MAC = ( MBD = (
a+(b+d) 0+(c+e)
,
2 2 (a+b)+d c+e
,
2
2
)= (
)= (
a+b+d c+e
,
2 2 a+b+d c+e 2
,
2
19 O(0,0) B(2,4) (a)(ii) Let A be (x, y)
) OA
)
√ ∵ MAC = MBD , ABCD is a parallelogram ✓ 18
𝑦 C(0, n)
B(m, n)
O
A(m, 0)
𝑥
MAC = (
0+m 0+n
,
2 2 m+0 0+n 2
,
2
=√
x2 + y2 4x x
= x 2 + y 2 − 4x − 8y + 20 = −8y + 20 = −2y + 5 −(1)
m n 2 2 m n
√
) =( , ) 2 2
∵ MOB = MAC , diagonals of rectangle OABC bisect each other ✓
= OM 1)2
+ (y −
(x 2 − 2x + 1) +(y 2 − 4y + 4)
2)2
= √(0 − 1)2 + (0 − 2)2 = √12 + 22
(x 2 − 2x) + (y 2 − 4y) + 5 (x 2 − 2x) + (y 2 − 4y)
=5 =0
−(2)
sub (1) into (2): (−2y + 5)2 − 2(−2y + 5) +y 2 − 4y = 0 (4y 2 − 20y + 25) + (4y − 10) +y 2 − 4y = 0 (4y 2 − 16y + 15) +y 2 − 4y = 0 5y 2 − 20y + 15 = 0 y 2 − 4y + 3 =0 (y − 3)(y − 1) =0 y=3 or y=1 x = −2(3) + 5 x = −2(1) + 5 = −1 =3 A(−1,3) or A(3,1) C(3,1) C(−1,3)
19 As OABC is a square, 0+2 0+4 (a)(i) M ) = (1,2) ✓ , AC = MOB = (
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(x 2 − 4x + 4) +y 2 − 8y + 16
√x 2 + y 2
AM
) =( , )
2
= AB
(x − 0)2 = √(x − 2)2 + (y − 4)2 +(y − 0)2
√(x − MOB = (
M(1,2)
2
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147
A math 360 sol (unofficial) 19(a) (ii)
Ex 6.1 19 (a) (iii)
𝑦 𝐵(2,4) 𝐴
𝐶
𝑥
𝑂
19(b) M(1,2) ⇒ (m + 1, m + 2) ✓ A(−1,3) ⇒ (m − 1, m + 3) ✓ C(3,1) ⇒ (m + 3, m + 1) ✓
⊥ 𝐛𝐢𝐬𝐞𝐜𝐭𝐨𝐫 𝐨𝐟 𝐎𝐁 OB⊥ ≡ ⊥ bisector of OB Point: MOB (1,2) Gradient: mOB⊥ =
−1
=
mOB
−1
=−
4−0 2−0
20
1 2
y − y1 = mOB⊥ (x − x1 )
OB⊥ :
y − (2)= (− ) (x − 1)
(
2
1
2 a+b
2 1
2 5
2
2
2
=− x+
2
5 2
1
5
2
2
⇒ A (a1 , − a1 + ) |AM|
= |OM|
√(a1 − 1)2 + [(− 1 a1 + 5) − 2] 2
2
1
5
2
2
(a1 − 1)2 + [(− a1 + ) − 2] 1
1 2
2
2
2
(a1 − 1)2 + (− a1 + )
1
1
1
4
2 5
4
5
5 a − a1 + 4 1 2 4 5 5 15 (a1 )2 − a1 − 4 2 4 5(a1 )2 − 10a1 − 15 (a1 )2 − 2a1 − 3
(a1 + 1)(a1 − 3) a1 = −1 or 1
5
2
2
a2 = − (−1) + =3 ⇒ A(−1,3) ⇒ C(3,1)
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2
) = (2,6)
=2
2
=√
(0 − 1)2 +(0 − 2)2
=5
and
2a+b+3 2
=6
2a + b + 3 = 12 2a + b =9 b = 9 − 2a −(2) sub (1) into (2): b = 9 − 2(4 − b) b = 9 − 8 + 2b b = 1 + 2b −b = 1 b = −1 a|b=−1 = 4 − (−1) =5 ⇒ A(5,10) ⇒ B(−1,2)
=5
(a1 )2 − 2a1 + 1 + (a1 )2 − a1 +
,
a + b= 4 a =4−b −(1)
Point A A(a1 , a2 ) lies on OB⊥ : 1
= (2,6)
a+b 2a+b+3
1
y−2 =− x+
a2 = − a1 +
A lies on y = 2x ⇒ A(a, 2a) B lies on y = x + 3 ⇒ B(b, b + 3) MAB
1
y
M(1,2) ⇒ (4,5) ✓ A(−1,3) ⇒ (2,6) ✓ C(3,1) ⇒ (6,4) ✓
=5 =5 =0 =0 =0 =0 a1 = 3 1
5
2
2
a2 = − (3) + =1 ⇒ A(3,1) ✓ ⇒ C(−1,3) ✓
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148
A math 360 sol (unofficial)
Ex 6.2 3
Ex 6.2
P(−1,2) Q(5,0)
R(7,4)
S(1,6)
𝑆 1(a)
A(−3,5) B(2,7)
𝑄
(5)−(7)
mAB = (−3)−(2) tan θ =
1(b)
(2)−(0)
mPQ = (−1)−(5) = −
2
(6)−(4)
5
mSR = (1)−(7) = −
A(4, −3) B(4,6)
mPS = (−1)−(1) = 2 (0)−(4)
(4)−(4)
∵ mPQ = mRS and mPS = mQR , P, Q, R and S are vertices of a parallelogram ✓
A(5, −4) B(7, −4)
A(1,0)
4
(1)−(−5)
B(4,2)
(0)−(2) 6
C(0, −3) D(3, −1)
−2
−3 c
2 3
(−3)−(−1) (0)−(3)
=
B(2,7)
(1)−(c)
= (0)−(−1) =
1−c 1
=1−c =4✓
2 3
5(a)
Point: A(−1,3) Gradient: ∵ line ∥ y = 4x − 1, m=4 Line: y − y1 = m (x − x1 ) y − (3)= (4)[x − (−1)] y − 3 = 4x + 4 y = 4x + 7 ✓
5(b)
Point:
∵ mAB = mCD , AB ∥ CD ✓ A(1,5)
B(2, −5) C(−1, c)
A,B,C lie on same straight line, mAB = mAC
5−7
(0)−(2)
2(b)
A(0,1)
(−4)−(−4)
mAB = (1)−(4) = mCD =
3
mQR = (5)−(7) = 2
(−3)−(6)
tan θ = 0 θ = 0° ✓ 2(a)
3
(2)−(6)
tan θ → ∞ θ = 90° ✓
mAB =
1
1
θ ≈ 21.8° ✓
mAB =
1(c)
𝑅
𝑃
C(0,4)
D(1,3)
(5)−(7)
mAB = (1)−(2) = 2 (4)−(3)
A(0,1)
mCD = (0)−(1) = −1 Gradient: ∵ line ∥ 2x + y = 3 y = −2x + 3 m = −2
∵ mAB ≠ mCD , AB is not ∥ CD ✓
Line:
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y − y1 = m (x − x1 ) y − (1)= (−2)[x − (0)] y − 1 = −2x y = −2x + 1 ✓
149
A math 360 sol (unofficial) 5(c)
Ex 6.2
Point:
A(−2,0)
7
Gradient:
∵ line ∥ x + 3y = 12 3y = 12 − x
Line:
b
1
y m=−
=− x+4
2nd line dx + ey + f = 0 ey = −dx − f
3
= m(x − x1 )
y
1
y − (0)= − [x − (−2)]
5(d)
1
2
3
3
=− x− ✓
a
− =− b
A(−2,0)
Gradient:
∵ line ∥ BC, m = mBC =
Line:
6(a)
Point:
a
8 3−6
=
−6 −3
e
d e
[shown] ✓
P(k 2 , 3k) Q(k, k − 2) R(k, k + 2)S(1,1) For PQ ∥ RS: mPQ
y − y1 = m (x − x1 ) y − (1)= (2)[x − (3)] y − 1 = 2x − 6 y = 2x − 5 ✓
= mRS
3k−(k−2)
=
k2 −k 2k+2
=
k2 −k 2(k+1)
(1,2)
=
k(k−1) k+1 2
( − 1)
k+1 2−k
y − y1 = m(x − x1 ) y − (2)= (4)[x − (1)] y − 2 = 4x − 4 y = 4x − 2 ✓
(
k
(k+2)−1 k−1 k+1 k−1 k+1 k−1
=0
k−1 k k−1
)
=0
(k + 1)(2 − k) = 0 k = −1 or k = 2 ✓ 9
6(b)
e
=2
Gradient: ∵ line ∥ y = 4x − 3, m=4 Line:
d
=
b
(−2)−4
f
e
Lines are parallel: m1 = m2
A(3,1) B(3, −2) C(6,4) Point:
d
=− x−
3
y
b
3
1
y − y1
1st line ax + by + c = 0 by = −ax − cy a c y =− x−
P(a + b, a) Q(a − b, 2a)
R(b, c)
y − intercept: c = 3 Gradient: ∵ line ∥ y = 3x + 4, m=3
P, Q, R are collinear points, mPQ = mQR
Line:
(a+b)−(a−b)
(a)−(2a)
y = mx + c = 3x + 3 ✓
−a 2b 2ab−a2 2b
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(2a)−(c)
= (a−b)−(b) =
2a−c a−2b
= 2a − c 2ab−a2
c
= 2a −
c
= 2a − a +
c
=a+
a2 2b
2b a2 2b
✓
150
A math 360 sol (unofficial) 10(i)
Ex 6.2
Gradient of AP: 3−(−5) (−2)−p
mAP =
=
11(ii) Line Point:
8
Gradient: ∵ Line ∥ 3x + 2y − 6 = 0 2y = −3x − 6
Line & its gradient: 4x + 3y − 5 = 0 3y = −4x + 5 y
A(−1, −1)
−2−p
4
5
3
3
=− x+
3
y ⇒m=−
=− x−3 2
3 2
4
⇒ mline = −
3
Line:
y − y1
= m (x − x1 ) 3
y − (−1) = (− ) [x − (−1)]
∵ AP ∥ line mAP = mline 8 −2−p
24 p
=−
2
= 8 + 4p =4✓
10(ii) Line AP Point:
12(i) A(−2,3) or P(4, −5) 8
mAP =
AP:
y − y1 = mAP (x − x1 )
−2−(4)
=−
4
Gradient:
4 3
y−3 =−
4 3
(x + 2)
4
8
3
3
4
1
3
3
y−3 =− x− y
=− x+ ✓
Point A At A, x + y + 2 = 0 intersects 3x − 2y + 1 = 0: x+y+2 =0 y = −x − 2 −(1) 3x − 2y + 1 = 0
= − (x + 1)
y+1
=− x−
y
=− x− ✓
2 3
3
2 3
2 5
2
2
A(3, −1) or P
Gradient:
∵ AP ∥ y = 2x + 3, mAP = 2
AP:
y − y1 = mAP (x − x1 ) y − (−1) = 2 (x − 3) y+1 = 2x − 6 y = 2x − 7 ✓
12(ii) Point P At P, AP (y = 2x − 7) intersects y = 3x − 11. 2x − 7 = 3x − 11 −x = −4 x =4 y|x=4 = 2(4) − 7 =1 ⇒ P(4,1) ✓
−(2) 13(i)
sub (1) into (2): 3x − 2(−x − 2) + 1 = 0 3x + 2x + 4 + 1 =0 5x = −1 x = −1 −(3) y|x=−1 = −(−1) − 2 = −1 ⇒ A(−1, −1) ✓
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Line AP Point:
3
y − (3)= (− ) [x − (−2)]
11(i)
y+1
4 3
3
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Point B B = (6 − 3,2) = (3,2) ✓
151
A math 360 sol (unofficial) 13(ii) Line BC Point: Gradient:
Ex 6.2
B(3,2) or C ∵ BC ∥ 2y + x = 0 2y = −x
BC:
C or D(0,5)
Gradient:
∵ CD ∥
=− x 2
1
y
2
1
mCD =
y − y1 = mBC
1
3
2
2
1
7
2
2
y−2 =− x+
y − 5 = (x − 0) 2 1
= x+5✓
y
2
=− x+ ✓ Point C 1
1
7
2
2
2
At C, CD (y = x + 5) meets BC (y = − x + )
A(6,2)
1 2
∵ AD ∥ BC,
1
7
2 3
2
x+5=− x+
x
mAD = mBC = − AD:
2
1
2
Gradient:
2
y − y1 = mCD (x − x1 )
CD:
1
13(iii) Line AD Point:
1
= x
(x − x1 )
y − (2)= − (x − 3)
y
2y − x = 0 2y =x
1
y mBC = −
13(iv) Line CD Point:
=−
1
2
2 1
3
2 17
2
y|x=−3 = (− ) + 5
y − y1 = mAD (x − x1 )
2
=
1
4 3 17
y − (2)= (− ) (x − 6) 2
⇒ C (− , 2
1
y−2 =− x+3
4
)✓
2
y
1
=− x+5✓ 2
Point D At D, AD cuts y − axis (x = 0): y|x=0 = 5 ⇒ D(0,5) ✓
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152
A math 360 sol (unofficial)
Ex 6.3 4(ii)
Ex 6.3 1
A(3,7) B(6,1) C(20,8) (7)−(1)
6
mAB = (3)−(6) =
−3
(1)−(8)
mBC = (6)−(20) =
=
−14
1 2
−5
3
) (− )
= −1 = −1
2t − t 2 = −15 t 2 − 2t − 15 = 0 (t − 5)(t + 3) = 0 t = 5 or t = −3 ✓
2
∴ AB ⊥ BC ✓ A(2, −1) B(5,4) C(15, −2) 5 mAB = mBC =
(−1)−(4)
−5
=
(2)−(5) (4)−(−2)
−3 6
=
(5)−(15)
5
=
3
=−
−10
5
3
3
5
3
mBC =
a−2 (−3)−1 2−10
= =
1
6
(− ) (k)= −1 2
a−2 −4
1
−8
2
=
k
AB ⊥ BC: (mAB )(mBC ) = −1 (
6
1
)( )
a−2 6
2
6
= −1 = −1
2a−4
6 2a a
mBC =
2−0 1−9 0−t 9−6
= =
2 −8 −t 3
=− =−
t
4
3
1st line & its gradient y = ax + b m1 = a
(1, −2) lies on y = ax + b −2 = a + b −(1)
1 4 t
lines are ⊥: m1 ⋅ m2 = −1 a ⋅ (3) = −1
3
∡ABC = 90°: (mAB )(mBC ) = −1 1
=2✓
2nd line & its gradient y − 3x = 4 y = 3x + 4 m2 = 3
= 4 − 2a = −2 = −1 ✓
A(1,2) B(9,0) C(6, t) mAB =
2
Lines are ⊥: m1 m2 = −1
A(a, 3) B(2, −3) C(10,1) 3−(−3)
1
=− x+1
2nd line y − kx + 4 = 0 y = kx − 4
∴ AB ⊥ BC ∡ABC = 90° ✓
mAB =
1st Line x + 2y − 2 = 0 2y = −x + 2 y
5
(mAB ) (mBC ) = ( ) (− ) = −1
4(i)
t
−5 2t−t2 15
(mAB )( mBC ) = (−2) ( ) = −1
3
2−t
(
1
2
2−t
AC ⊥ BC: (mAC )(mBC ) = −1
= −2 −7
(2)−(t)
mAC = (1)−(6) =
a
1
= − ✓ −(2) 3
(− ) (− ) = −1 t 12
t
sub (2) into (1):
= −1
1
−2 = (− ) + b
= −12 ✓
3
5
b =− ✓ 3
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153
A math 360 sol (unofficial) 7
Ex 6.3
(4,5)
Point:
10
Gradient: ∵ Line ⊥ x + 2y − 4 = 0 2y = −x + 4 1
y ⇒m=
1 (− ) 2
=− x+2 2
y 2 = 6x − 32
=2
Point A At A, y = x − 1 intersects y = x 2 − x x−1 = x2 − x x 2 − 2x + 1 = 0 (x − 1)2 =0 x =1
⊥ bisector of AB AB⊥ ≡ ⊥ bisector of AB
y|x=1 = (1) − 1 =0 ⇒ A(1,0) Line Point: A(1,0) Gradient: ∵ line ⊥ y = x − 1
9
11(i)
A(3,3) B(7,3) AB⊥ ≡ ⊥ bisector of AB AB⊥ :
x=
3+7
Gradient:
mAB⊥ =
AB⊥ :
y − y1 = mAB⊥ (x − x1 ) y − (−3) = 1(x − 11) y+3 = x − 11 y = x − 14 ✓
2
,
2 −1 mAB
2
=
=1
A(3,6) or F
Gradient:
∵ F is foot of ⊥ from A to BC,
AF:
−1 mBC
=
−1 (−1)−(7) ( (2)−(6) )
=−
1 2
y − y1 = mAF (x − x1 ) 1
y − 6 = − (𝑥 − 3) 2 1
3
2 1
2 15
2
2
y−6 =− x+ y
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−1 (2)−(−8)
( (6)−(16) )
Line AF Point:
mAF =
=5✓
) = (11, −3)
MAB = (
1
y − y1 = m (x − x1 ) y − 0 = −1(x − 1) y = −x + 1 ✓
6+16 2+(−8)
Point:
m = − (1) = −1 Line:
−(2)
sub (1) into (2): (8 − x)2 = 6x − 32 x 2 − 16x + 64 = 6x − 32 x 2 − 22x + 96 = 0 (x − 6)(x − 16) = 0 x=6 or x = 16 y = 8 − 6 y = 8 − 16 =2 = −8 ⇒ A(6,2) ⇒ B(16, −8)
y − y1 = m(x − x1 ) y − 5 = 2 (x − 4) y = 2x − 3 ✓
Line:
8
−1
Points A & B At A & B, x + y = 8 meets y 2 = 6x − 32: x + y= 8 y = 8 − x −(1)
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=− x+
✓
154
A math 360 sol (unofficial) 11(ii) Line BC Point:
Ex 6.3
B(2, −1) or C(6,7) (−1)−7
Gradient:
mBC =
BC:
y − y1 = mBC (x − x1 ) y − (−1) = 2 [x − (2)] y+1 = 2x − 4 y = 2x − 5
2−6
=2
12(ii) Point P At P, AP (y = −4x + 13) cuts x − axis (y = 0). y =0 −4x + 13 = 0 x ⇒ P(
1
15
2
2
).
1
15
2
2
2x − 5 = − x + 5 2
x
x
=
13 4
P(
2
=5
, 0) ✓
13 4
, 0) A(3,1)
PA: AQ = (
11(iii) Perpendicular distance AF A(3,6) F(5,5) 13(i)
= √4 + 1 = √5 units ✓
Line AB Point:
AB: A(3,1) or P
y
y − y1 y−1 y−1 y
4
− 3) : (3 − 0)
1
:3
4
A(4,13) or B(9,3) (13)−(3)
Gradient: mAB =
Gradient: ∵ AP ⊥ x − 4y = 8 −4y = −x + 8
AP:
13
= 1: 12 ✓
|AF| = √(3 − 5)2 + (6 − 5)2
mAP =
Q(0,13)
By similar triangles (using x-coordinates) [diagram?]
=
Line AP Point:
4
12(iii) Ratio 𝐏𝐀: 𝐀𝐐
25
y|x=5 = 2(5) − 5 =5 ⇒ F(5,5) ✓
12(i)
13
Point Q At Q, AP(y = −4x + 13) cuts y-axis (x = 0). y|x=0 = 13 ⇒ Q(0,13) ✓
Point F At F, BC (y = 2x − 5) intersects AF (y = − x +
=
−1 1 4
( )
1
= x−2 4
= −4
13(ii) Line Point:
(4)−(9)
=
= −2
= mAB (x − x1 ) = −2(x − 4) = −2x + 8 = −2x + 21 ✓
y − y1 y − 13 y − 13 y
C(10,8)
Gradient: ∵ Line ⊥ y − 4x = 5 y = 4x + 5
= mAP (x − x1 ) = −4(x − 3) = −4x + 12 = −4x + 13 ✓
m=− Line:
1 4
y − y1 = m (x − x1 ) 1
y − 8 = − (x − 10) 4 1
5
4 1
2 21
4
2
y−8 =− x+ y
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10 −5
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=− x+
✓
155
A math 360 sol (unofficial)
Ex 6.3
13(iii) Point P 1
21
4
2
At P, y = − x + 1
21
4
2
− x+ 7 4
meets AB (y = −2x + 21):
= −2x + 21
x
=
x
14(iii) Ratio 𝐀𝐌: 𝐌𝐃 A(3,1) M(4,3) D(−2,6) AM: MD = √(3 − 4)2 + (1 − 3)2
21
: √[4 − (−2)]2 + (3 − 6)2
2
=6
= √5 : √45
y|x=6 = −2(6) + 21 =9 ⇒ P(6,9)
= √5 : √9 × 5 = √5 : 3√5 =1
14(i)
Line DM Point: Gradient:
D(−2,6) ∵ DM ⊥ AB (y = 2x − 5) −1
mDM =
=−
mAB
15(i)
2
AB⊥ ≡ ⊥ bisector of AB
1
y − 6 = − [x − (−2)] 2 1
y−6 =− x−1 2 1
=− x+5✓
y
⊥ bisector of AB A(5,4) B(3, −2)
1
y − y1 = mDM (x − x1 )
DM:
:3✓
2
5+3 4+(−2)
) = (4,1)
Point:
MAB = (
Gradient:
mAB⊥ =
AB⊥ :
y − y1 = mAB⊥ (x − x1 )
2
,
−1 mAB
2
=
−1 (4)−(−2)
( (5)−(3) )
=
−1 6 2
( )
=−
1 3
1
y − 1 = − (x − 4)
14(ii) Point M 1
3 1
4
3 1
3 7
3
3
At M, DM (y = − x + 5) intersects
y−1 =− x+
AB (y = 2x − 5).
y
2
=− x+ ✓
1
− x + 5 = 2x − 5 2 5
− x
15(ii) Point P
= −10
2
x
1
7
3
3
At P, AB⊥ (y = − x + ) intersects y = x + 5.
=4
1
7
3 4
3
− x+ =x+5 1
8
y|x=4 = − (4) − 5
− x
=
=3 ⇒ M(4,3) ✓
x
= −2
2
3
y|x=−2 = (−2) + 5 =3 ⇒ P(−2,3) ✓
Point B Let B be (b1 , b2 ) M
= MAB
(4,3) = ( 4 =
3
3+b1 1+b2 2
3+b1 2
b1 = 5 ∴ B(5,5) ✓
,
2
and
) 3 =
1+b2 2
b2 = 5
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156
A math 360 sol (unofficial) 16(i)
Ex 6.3
⊥ bisector of AB A(5,2) B(3,6) AB⊥ ≡ ⊥ bisector of AB MAB = (
Point:
−1
18(i)
2
=
,
) = (4,4)
2 −1
m=
AB⊥ :
y − y1 = m (x − x1 )
(2)−(6)
((5)−(3))
=
−1 −4 ) 2
(
=
2
1
Point C At C, BC(y = 5x + 6) cuts y − axis (x = 0). y|x=0 = 6 ⇒ C(0,6)
2 1
y−4 = x−2 2 1
= x+2✓ 2
16(ii) Point P At P, AB⊥ cuts x-axis (y = 0) 1 2
Point Q At P, AB⊥ cuts y-axis (x = 0)
18(ii) Line AC Point:
∵ BD is ⊥ bisector of AC, mAC =
2
x = −4 ⇒ P(−4,0) ✓
A or C(0,6)
Gradient:
1
y|x=0 = (0) + 2
x+2=0
y − y1 = mBC (x − x1 ) y − (1)= (5) [x − (−1)] y − 1 = 5x + 5 y = 5x + 6
BC:
1
y − 4 = (x − 4)
y
B(−1,1) or C ∵ BC ∥ y = 5x mBC = 5
5+3 2+6
Gradient:
mAB
Line BC Point: Gradient:
=2 ⇒ Q(0,2) ✓
AC:
−1 mBD
=
−1 (1)−(7)
((−1)−(8))
=−
3 2
y − y1 = mAC (x − x1 ) 3
y − 6 = − (x − 0)
17(a) ⊥ bisector of PQ P(3,5) Q(5,9) PQ ⊥ ≡ ⊥ bisector of PQ
2 3
y−6 =− x
3+5 5+9
MPQ = (
Gradient:
m=
PQ ⊥ :
y − y1 = m (x − x1 )
−1 mPQ
=
,
2 −1
(5)−(9) ((3)−(5))
=−
1
18(iii) Point A 1
2
At A, AC (y = − x + 6) cuts x − axis (y = 0): 3
y
=0 3
1
y − 7 = − (x − 4)
− x + 6= 0
y−7 =− x+2
− x
2 3
2 1
y
2
) = (4,7)
Point:
2
2 3
=− x+6✓
y
2
2 1
= −6
x =4 ⇒ A(4,0) ✓
=− x+9✓ 2
17(b) Point At point where y = 6x intersects
Midpoint of AC A(4,0) C(0,6)
1
PQ ⊥ (y = − x + 9). 2
M = MAC = (
1
6x = − x + 9
4+0 0+6 2
,
2
) = (2,3) ✓
2
13 2
x
x= 9 =
18 13 18
y|x=18 = 6 ( ) 13
=
13 108
⇒ Point (
13 18 108 13
,
13
)✓
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157
A math 360 sol (unofficial)
Ex 6.3
18(iv) Area of quadrilateral 𝐀𝐁𝐂𝐃 A(4,0) B(−1,1) C(0,6) D(8,7)
20(i)
|AC| = √(4 − 0)2 + (0 − 6)2 = √52 = √4 × 13 = 2√13
Line AB Point:
A(1, −1) or B(5,3)
Gradient:
mAB =
AB:
y − y1 = mAB (x − x1 ) y − (−1) = (1) [x − (1)] y+1 =x−1 y =x−2✓
|BD| = √[(−1) − 8]2 + (1 − 7)2 = √117 = √9 × 13 = 3√13
M = MAB = ( 20(ii) Line PQ Point:
Line CD A(2,3) B(6,7) C(7, t) MAB = (
Point:
2+6 3+7 2
,
2
) = (4,5)
−1
CD:
y − y1 y−5 y−5 y
mAB
1
1
(2)−(6)
−4
= − (3)−(7) = − −4
= mCD (x − x1 ) = (−1)[x − (4)] = −x + 4 = −x + 9 ✓
(1)+(5) (−1)+(3) 2
−4 −4
= −1
,
2
) = (3,1) ✓
Gradient:
mPQ =
PQ:
y − y1 y−1 y−1 y
−1 mAB
=
−1 1
= −1
= mPQ (x − x1 ) = (−1)(x − 3) = −x + 3 = −x + 4 ✓ Value of q Q(7, q) lies on PQ(y = −x + 4), (q) = −(7) + 4 q = −3 ✓
20(iii) Area of quadrilateral APBQ A(1, −1) P(0.5,3.5) B(5,3) Q(7, −3)
19(ii) Value of t C(7, t) lies on y = −x + 9: (t) = −(7) + 9 t =2✓
|AB| = √(1 − 5)2 + [(−1) − 3]2
19(iii) Point D C(7,2) D(d1 , d2 )
area of APBQ = |AB||PQ|
= √32 = √16 × 2 = 4√2 169
|PQ| = √(0.5 − 7)2 + [3.5 − (−3)]2 = √
4 =
7+d1 2+d2 2
7+d1 2
,
2
2
=
13 √2
1 2 1
13
2
√2
= (4√2) ( )
AB is ⊥ bisector of CD, MAB = MCD (4,5) = (
=1
M(3,1) or P(p, 3.5) or Q(7, q)
Value of p P(p, 3.5) lies on PQ(y = −x + 4), (3.5) = −(p) + 4 p = 0.5 ✓
Gradient: ∵ CD is ⊥ bisector of AB, mCD =
=
(1)−(5)
Midpoint of AB
Area of ABCD 1 = |AC||BD| 2 1 = (2√13)(3√13) 2 = 3(13) = 39 unit 2 ✓ 19(i)
(−1)−(3)
= 26 ✓
)
and
d1 = 1
5 =
2+d2 2
d2 = 8
∴ D(1,8) ✓
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158
A math 360 sol (unofficial) 21(i)
Ex 6.3
⊥ 𝐛𝐢𝐬𝐞𝐜𝐭𝐨𝐫 𝐨𝐟 𝐀𝐁 A(−4,3) B(8, −3) AB⊥ ≡ perpendicular bisector of AB
21(iii) C(p, q) lies on AB⊥ : q = 2p − 4 ⇒ C(p, 2p − 4)
C A(−4,3)
MAB B(8, −3) C
Point:
MAB = (
(−4)+(8) (3)+(−3)
Gradient: mAB⊥ = − =− AB⊥ :
,
2 1 mAB 1 6 −12
=−
2
) = (2,0)
MAB = (
−4+8 3+(−3)
,
2
1 (3)−(−3)
|CMAB | = √(p − 2)2 + [(2p − 4) − 0]2
((−4)−(8))
= √(p2 − 4p + 4) + (4p2 − 16p + 16)
=2
= √5p2 − 20p + 20
y − y1 = mAB⊥ (x − x1 ) y − 0 = 2 (x − 2) y = 2x − 4 ✓
|AB| = √[(−4) − 8]2 + [3 − (−3)]2 = √180 = √36 × 5 = 6√5
21(ii) Show If (10,16) lies on AB⊥ (y = 2x − 4), (16) = 2(10) − 4 16 = 16 [consistent] ∴ (10,16) lies on AB⊥
△ ABC area 1 2 1 2
=6
|CMAB ||AB|
3√5p2 − 20p + 20√5
=6
√5p2 − 20p + 20√5
=2
√25p2 − 100p + 100
=2
25p2 − 100p + 100
=4
25p2 − 100p + 96
=0
(5p − 8)(5p − 12)
=0
8
or
5
2p − 4 = − 8
4
5
5
4 5
⇒ C( ,− )
sleightofmath.com
=6
√5p2 − 20p + 20(6√5) = 6
p=
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2
) = (2,0)
p=
12 5
2p − 4 = ⇒ C(
4 5
12 4 5
, )✓ 5
159
A math 360 sol (unofficial)
Ex 6.3
21(iii) Shoelace formula A C B C(p, q) lies on AB⊥ (y = 2x − 4) q = 2p − 4 ⇒ C(p, 2p − 4)
22
P(0,4)
−4 | | 2 3
2
−4 8 || 3 −3
p 2p − 4 p 2p − 4
Q
−4 || 3
Line PQ Pt: P(0,4) or Q
=6
Grad: mPQ =
−4 || 3
= 12
|30p − 60|
p
=
PQ:
= 12
p
5 12
⇒ C(
5
4
=−
5
12 4
8
4
5
5
5
5 4
1 2
(− )
=2
y − y1 = mPQ (x − x1 ) y − (4)= (2) [x − (0)] y = 2x + 4
Line PR Pt: P(0,4) or R
5
, ) or C ( , − ) ✓ 5
−1
y|x=−4 = −4 ⇒ Q(−4, −4) ✓
2p − 4 = 2 ( ) − 4
5
=
8
8
2p − 4 = 2 ( ) − 4 =
=
−1 ml3
Point Q At Q, PQ meets y = x 2x + 4 = x x = −4
=2 or 5p − 10 = −2 5p =8
12
R
l1 : x = 0
|12 + (16p − 32) + 3p = 12 −24 − (−3p) − (−8p + 16)| |5p − 10| 5p − 10 = 2 5p = 12
𝑥
𝑂 =6
8 −3
l2 : y = x
1
l3 : y = − x
△ ABC area 1
𝑦
Grad: mPR = −
1 ml2
1
= − (1) = −1
PR: y − y1 = mPR (x − x1 ) y − (4)= (−1)[x − (0)] y = −x + 4 Point R 1
At R, PR meets y = − x 2
1
−x + 4 =− x 2
1
− x
= −4
x
=8
2
1
y|x=8 = − (8) 2
= −4 ⇒ R(8, −4) ✓
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160
A math 360 sol (unofficial) 23(i)
Ex 6.3
∵ AO = OC, (radius) △ AOC is isosceles △ ∵ BO = OC, (radius) △ BOC is isosceles △ ✓
A
23(ii) α = ∡AOC = ∡OCA β = ∡OBC = ∡OCB sum of ∡s in △
= 180°
23(iii) A(1,0) B(5,2) C(a, b)
C
A
O = MAB = ( B
O
C 𝛼 𝛽 𝛼 O
= 180°
2α + 2β
= 180°
2(α + β)
= 180°
α+β
= 90°
∡ACB
= 90°
∴AC ⊥ BC ✓
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,
2
) = (3,1)
= radius
|OC|
= (diameter AB)
|OC|
= |AB|
1 2 1
3)2
+ (b −
1)2
2 1
= √(1 − 5)2 + (0 − 2)2 2 1
√(a − 3)2 + (b − 1)2 = √20 2
B
∡AOC + ∡OCA = 180° +∡OBC + ∡OCB α +α +β + β
2
|OC|
√(a − 𝛽
1+5 0+2
1
(a − 3)2 + (b − 1)2
= (20)
(a − 3)2 + (b − 1)2
= 5 [shown] ✓
4
Note: Alternate approach is to use right angle triangle in semicircle. By Pythagoras’ Theorem, |AC|2 + |BC|2 = |AB|2 . 23(iv) At (2,3), [(2) − 3]2 +[(3) − 1]2 = 5 (−1)2 +22 =5 1 +4 =5 5 = 5 [consistent] ∴ (2,3) lies on circle ✓
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161
A math 360 sol (unofficial)
Ex 6.4 3(i)
Ex 6.4 1(a)
A(2,3) B(5,6) C(−1,4)
△ ABC area
△ ABC area
=| |
1 2 = | 2 3
5 6
1 2
−1 4
2 | 3
2 1
= (12) 2
A(5,2) B(1,6) C(−2, −3) △ ABC area 1 −2 6 −3
A, B and C are collinear ✓
4(a)
O(0,0) A(4,1) B(6,4) C(−2,4)
5 | 2
Area of OABC =
1 2
2
2 4
−2 4
0 | 0
2
=
3 6 5 2
4(b) 2 || 4
1 = |10 + 6 + 24 2 1 = |−6| 2
1 (42) 2
= 21 unit 2 ✓
△ ABC area 1
4 6 1 4
= [0 + 16 + 24 + 0 −0 − 6 − (−8) − 0]
A(2,4) B(3,5) C(6,2)
=| |
1 0 | 2 0 1
1 (48) 2
= 24 unit 2 ✓ 2(a)
− 4 − 0 − (−10)|
3(ii)
= [30 + (−3) + (−4) −2 − (−12) − (−15)] =
−5 || −2
= 0 unit 2 ✓
= 6 unit 2 ✓
5 2
−2 1 0 2
1 |0 + (−4) + (−2) 2 1 = |0| 2
= [12 + 20 + (−3) −15 − (−6) − 8]
=|
−5 −2
=
1
1(b)
A(−5, −2) B(−2,0) C(1,2)
P(1,4) Q(−4,2) R(1, −2) S(4,0) Area of PQRS 1 1 −4 1 4 1 | | 2 4 2 −2 0 4 1 = [2 + 8 + 0 + 16 − (−16) − 2 − (−8) − 0] 2 1 = (48) 2 =
− 12 − 30 − 4|
= 3 unit 2 ✓
= 24 unit 2 ✓ 2(b)
A(−4, −2) B(−2,4) C(6,0) △ ABC area 1
=| | 2
−4 −2
−2 6 4 0
−4 || −2
1 |−16 + 0 + (−12) 2 1 = |−56| 2 =
− 4 − 24 − 0|
= 28 unit 2 ✓
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162
A math 360 sol (unofficial) 5(i)
Ex 6.4
A(2, −3) B(3, −1) C(2,0) D(−1,1) E(−2, −1)
6(iii)
Area of △ ABC 1 2 3 2 2 | = | 2 −3 −1 0 −3 1 = [−2 + 0 + (−6) −(−9) − (−2) − 0] =
2 3
1 2 1 2
2 9
7(i)
unit 2 ✓
2
√5(5) sin ∡BAC
A(3,5)
1
| | 2
||
= 5 unit 2 ✓
9
2
2
= 11 unit 2 ✓ 𝑦
3 5
3 5
7(ii) 𝑥
𝑂
A
C(9, −1)
1 2
4 −1
B(−5,9) C(k, k + 2)
−5 9
= 18
k k+2
−5 9
k k+2
3 || 5
= 18
3 || 5
= 36
5 9 1 −1
= 36
F
B
CF ≡ perpendicular distance of C from AB Area of △ ABC = 18
Area of △ ABC =| |
✓
C
B(5,1)
A(4, −1)
2 √5
|−2k + 6| =6 −2k + 6 = 6 or −2k + 6 = −6 −2k =0 −2k = −12 k =0✓ k =6✓
Area of pentagon 3
=
|−12k + 36|
= + +5
1
4 || −1
2
|AB||CF|
|CF|
1
=
= [4 + (−5) + (−9) −(−5) − 9 − (−4)] 2 1
=
= |−10| 2
= 5 unit 2 ✓ 6(ii)
=5
|27 + (−5k − 10) + 5k = 36 −(−25) − 9k − (3k + 6)|
2
6(i)
(AB)(AC) sin ∡BAC = 5
Area of △ ABC
Area of △ ADE 1 2 −1 −2 2 | = | 2 −3 1 −1 −3 1 = [2 + 1 + 6 −3 − (−2) − (−2)]
5(ii)
=5
sin ∡BAC
Area of △ ACD 1 2 2 −1 2 | = | 2 −3 0 1 −3 1 = [0 + 2 + 3 −(−6) − 0 − 2] =
Area of △ ABC
unit 2 ✓
2
AC = 9 − 4 =5
8(i)
AB = √(4 − 5)2 + [(−1) − 1]2
= 18
36 |AB| 36 √80
= =
36 √[3−(−5)]2 +(5−9)2 36 √5×16
=
36 4√5
=
9 √5
9
= √5 ✓ 5
A(2, t) B(3 + t, 2) C(3,4) Area of △ ABC 1 2 3+t 3 2 | = | 2 t 2 4 t 1 = (4 + 12 + 4t + 3t − 3t − t 2 − 6 − 8)
= √1 + 4 = √5 units [shown] ✓
2 1
= (−t 2 + 4t + 2) unit 2 ✓ 2
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163
A math 360 sol (unofficial) 8(ii)
Ex 6.4
Area of △ ABC = 2 1
(−t 2 + 4t + 2) =
2
1
11(i)
2
5
Area of △ ABC 1 3 −4 6 3 | = | 2 4 2 −1 4 1 = [6 + 4 + 24 − (−16) − 12 − (−3)] 2 1 = (41) 2 = 20.5 unit 2 ✓
2
−t 2 + 4t + 2 =5 2 t − 4t + 3 =0 (t − 1)(t − 3) =0 t = 1 or t = 3 ✓ 9(i)
A(1,3) B(5,1) C(3, r) Area of △ ABC =4 1 1 5 3 1 | | =4 2 3 1 r 3 1 (1 + 5r + 9 − 15 − 3 − r) = 4 2 1
(4r − 8)
2
A(3,4) B(−4,2) C(6, −1) D(p, 3)
=4
2r − 4 2r r
=4 =8 =4✓
11(ii) Area of ABCD 1 3 −4 6 p 3 | = | 2 4 2 −1 3 4 1 = [6 + 4 + 18 + 4p − (−16) − 12 − (−p) − 9] 2 1 = (23 + 5p) unit 2 ✓ 2
11(iii) (Area of ABCD) = 3(Area of △ ABC) 1 2
9(ii)
Area of △ ACB =4 1 1 3 5 1 | | =4 2 3 r 1 3 1 (r + 3 + 15 − 9 − 5r − 1) = 4 2 1
(−4r + 8)
2
=4
−2r + 4 −2r r 10(i)
12(i)
2
=9 = −3x + 9
y
= x−
3
9
2
2
3 2
AD ⊥ line ⇒ mAD =
−1 mline
=
−1
= MBD
2+1 1+4
2 2+1
= 123 = 100 = 20 ✓
3x − 2y −2y mline =
A(2,1) B(b1 , b2 ) C(1,4) D(0,2)
(
= 3(20.5)
23 + 5p 5p p
=4 =0 =0✓
MAC
(23 + 5p)
,
=
) =(
2 b1 +0 2
b1 +0 b2 +2 2
,
and
b1 = 3 B(3,3) ✓
Line BC Point: Gradient:
)
2 1+4 2
=
b2 +2 2
b2 = 3
BC:
=−
2 3
B(4,7) or C ∵ BC ∥ AD mBC = mAD =
10(ii) Area of rhombus 1 2 3 1 0 2 | = | 2 1 3 4 2 1 1 = (6 + 12 + 2 + 0 − 3 − 3 − 0 − 4)
3 2
−1 3 ( ) 2
=−
2 3
y − y1 = mBC (x − x1 ) 2
y − (7)= (− ) (x − 4) 3
2
8
3 2
3 29
3
3
y−7 =− x+
2 1
= (10)
y
2
=− x+
✓
2
= 5 unit ✓
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164
A math 360 sol (unofficial) 12(ii) Line DE Point:
Ex 6.4 12(iv) Area of quadrilateral ABFD 1
D(−3,3) or E
A (1, ) 2
∵ DE ⊥ BC (y = − x +
Gradient:
3
−1
mDE =
=
mBC
−1 2 3
(− )
=
29 3
Area of ABFD 0 1 1 4 = | 1 7 29 2
3 2
3
3 2
3
2
2
y
3
15
2
2
= x+
= 13(i)
Point E 1
15
3 29
2
At E, DE (y = x + 2
(y = − x + 3
3
x+
2 13 6
15 2
x
6
=1 29
3
3
=9 ⇒ E(1,9) ✓
3
104
unit 2 ✓
3
𝐕𝐚𝐥𝐮𝐞 𝐨𝐟 𝐚 A(3a, 4a + 1), a > 0 B(0,1) =5 1]2
⇒ A(3,5) 13(ii) Line BC Point:
12(iii) Point F 2
At F, BC (y = − x + 3
2
29
3
3
y|x=0 = − (0) + =
)
2
√(3a − + [(4a + 1) − =5 2 2 9a + 16a = 25 2 25a = 25 a2 − 1 =0 (a + 1)(a − 1) =0 a = −1 (rej ∵ a > 0) or a = 1 ✓
3
2
3
0)2
13
y|x=1 = − (1) +
4
+ 0 + (−1) − − 0 − (−29) − 3]
|AB|
29
3
3
3
) intersects BC
)
=− x+ =
x
3 2
116
2
= (
2
9
1
−3 1 1| 3
1 208
y − 3 = (x + 3) 3
3
= [7 +
y − (3)= ( ) [x − (−3)]
y−3 = x+
3
)
y − y1 = mDE (x − x1 )
DE:
29
F (0, ) D(−3,3)
B(4,7)
3
29 3
B(0,1) or C
) cuts y-axis (x = 0) Gradient: ∵ BC ⊥ AB, mBC =
29 3 29
−1 mAB
1
1
(0)−(3)
4 3
= − (1)−(5) = −
=−
3 4
⇒ F (0, ) 3
y − y1 = mBC (x − x1 )
BC:
3
y − 1 = − (x − 0)
Point A A(a1 , a2 ) B(4,7)
29 3
(
2
a1 +0 2
a1
,
=
2
)=(
4+(−3) 2
3
At C, BC (y = − x + 1) cuts x-axis (y = 0) 4
3
4+(−3) 7+3
and
=1
,
2
2
29 a2 +( ) 3
2
a2
4
Point C
For Parallelogram ABFD, MAF = MBD 29
=− x+1✓
y
F (0, ) D(−3,3)
a1 +0 a2 +( 3 )
4 3
− x + 1= 0
) = =
4 3
− x
= −1
x
=
4
7+3 2 1
4 3
4
3
⇒ C ( , 0) 3
1
A (1, ) ✓ 3
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165
A math 360 sol (unofficial)
Ex 6.4
13(iii) ⊥ bisector of AB AB⊥ ≡ ⊥ bisector of AB MAB = (
Point:
13(iv) Area of quadrilateral ABCD
(3)+(0) (5)+(1) 2
,
2
A(3,5)
3
3
C ( , 0)
B(0,1)
4
) = ( , 3)
D(
11 2
, 0)
2
Area of ABCD Gradient: mAB⊥ = mBC = −
3
y − y1 = mAB⊥ (x − x1 )
AB⊥ :
3 4 3
9
4 3
8 33
y
4
8
=− x+
3
0
2
5
1
4
11
3
2
0
0
2
=
3
|
5
1
55
2
2
= (3 + 0 + 0 +
3
y − 3 = − (x − ) y−3 =− x+
1
= |
4
4
− 0 − − 0 − 0) 3
1 175 ( ) 2 6
= 14
7 12
unit 2 ✓
Point D At D, AB⊥ cuts x-axis (y = 0). y =0 3
33
4 3
8
− x+ − x
=−
4
x ⇒ D(
=0
= 11 2
33 8
11 2
, 0) ✓
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166
A math 360 sol (unofficial)
Rev Ex 6 A2(ii) Point D at k = 2: A(3,1) B(5,3)
Rev Ex 6 A1(i) A(−2,1) B(10,6) C(a, −6) AB
D on x − axis: D(d, 0)
= BC 10]2
√[(−2) − + (1 − 169 169 a2 − 20a + 75 (a − 5)(a − 15) a = 5 or a = 15 ✓
6)2
= √(10 − a)2 + [6 − (−6)]2 = (100 − 20a + a2 ) + 144 = 244 − 20a + a2 =0 =0
BD ⊥ AC: mBD ⋅ mAC = −1 3−0 5−d 3 5−d
Let D be (d1 , d2 ) For ABCD is rhombus, MAC = MBD (−2)+a 1+(−6)
2 (−2)+a 2
d1
,
=
) =(
2 10+d1
and
2
= a − 12
,
)
2 1+(−6) 2
d2
A(k + 1,1) B(2k + 1,3)
3−6 −3 −3
mBD =
= −1 = −1
=
3−0 5−8
tan θ = −1 α = 45°
6+d2 2
= −11
a = 5: D(−7, −11) ✓ a = 15: D(3, −11) ✓
A2(i)
⋅
1−4
Obtuse angle that BD makes with x-axis
10+d1 6+d2 2
⋅
3 =d−5 d =8 ⇒ D(8,0) ✓
A1(ii) A(−2,1) B(10,6) C(a, −6)
(
C(6,4)
∴ obtuse ∡= 180 − 45 = 135° ✓ A3(i)
A(6,7)
B(0,1)
C(9,4)
|BC| = √(0 − 9)2 + (1 − 4)2
C(2k + 2,2k)
= √81 + 9 ∵ A, B, C are collinear, mAB = mBC (1)−(3) (k+1)−(2k+1) −2 −k 2
= √9 × 10 = 3√10 units ✓
(3)−(2k)
= (2k+1)−(2k+2) =
3−2k −1
= 2k − 3
k
= √90
2 = 2k 2 − 3k 2k 2 − 3k − 2 = 0 (2k + 1)(k − 2) = 0 1
k = − or k = 2 ✓ 2
A3(ii) Area of △ ABC 1 6 0 9 6 | = | 2 7 1 4 7 1 = (6 + 0 + 63 − 0 − 9 − 24) 2 1 = (36) 2 = 18 ✓ A3(iii) Area of △ ABC = 18 1 2 1 2
|AF||BC|
|AF|(3√10) = 18
|AF|
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= 18
=
36 3√10
=
12 √10
=
12√10 10
=
6√10 5
✓
167
A math 360 sol (unofficial)
Rev Ex 6
A4(a) y = ax + b ⇒ m1 = a
A5(i)
2x + y = 8 y = −2x + 8 ⇒ m2 = −2
Line BC Point:
B(−1, −3) or C
Gradient:
mBC =
BC:
y − y1
1 2
= mBC (x − x1 ) 1
y − (−3) = ( ) [x − (−1)] 2
y = ax + b ⊥ 2x + y = 8 ⇒ m1 m2 = −1 a(−2) = −1 a
1
= ✓
A5(ii) Line AC Point: Gradient:
2
y = ax + b on y − axis (x = 0): y = a(0) + b =b ⇒ (0, b)
A4(b) AB: y = 3x + 1 BC: y = x − 1 F(3,2)
2
2
y
= x− ✓
−1
=−
mAB
=−
1 −8 ( ) −4
1 2
B
y
C F(3,2)
1
3
2 1
2 13
2
2
1 2
x−
5 2
1
13
2
2
=− x+
1
5
2
2
−1 −3 −1 −3
9 2
=9 1
13
2
2
y|x=9 = − (9) +
Gradient: ∵ AF ⊥ BC, −1
=2 ⇒ C(9,2) ✓
= (1) = −1
A5(iv) Height of triangle ABC A(3,5) B(−1, −3) C(9,2) Equating area of △ ABC, 1
Point A At A, AF intersects AB (y = 3x + 1) −x + 5 = 3x + 1 −4x = −4 x =1
2
|BC||AD|
|BC||AD| √
=4
1
= | 2
=|
sleightofmath.com
3 5
3 5
9 2
3 | 5 3 | 5
[(−1) − 9]2 −9 − 2 + 45 |AD| = [ ] 2 −(−50) − (−27) − 6 [( ] + −3) − 2
√125|AD| 5√5|AD|
= 60
|AD|
=
= 60
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✓
At C, AC intersects BC (y = x − )
𝑥
y=x−1
y − y1 = mAF (x − x1 ) y − (2)= (−1)[x − (3)] y − 2 = −x + 3 y = −x + 5
=− x+
A5(iii) Point C
x
mBC
2
y − (5)= (− ) [x − (3)]
y = 3x + 1
−1
1
y − y1 = mAC (x − x1 )
𝑦 A
mAF =
1 (−3)−(5)
((−1)−(3))
=−
y−5 =− x+
Line AF Point: A or F(3,2)
y|x=1 = −(1) + 5 ⇒ A(1,4) ✓
2 5
= x+
A(3,5) or C ∵ ∡BAC = 90°,
AC:
𝑂
AF:
1
2 1
y+3
mAC =
(0, b) lies on x + y + 3 = 0, 0+b+3 =0 b = −3 ✓
1
12 √5 12√5 5
✓
168
A math 360 sol (unofficial) A6(i)
Rev Ex 6 𝑦
AD: 3x + 2y = 6
E
C
B AB: 5y + 6 = 3x 𝑥
D
𝑂
A
A6(ii) Line BC ∵ BC ∥ x − axis, BC is y = 6 Point B At B, BC (y = 6)cuts AD (5y + 6 = 3x), 5(6) + 6 = 3x 36 = 3x x = 12 ⇒ B(12,6) ✓ Line CD Pt: C or D(0,3) Grad: ∵ CD ∥ AB,
Point A At A, AD cuts x − axis (y = 0): 3x + 2(0) = 6 3x =6 x =2 ⇒ A(2,0) ✓ Point D At D, AD cuts y − axis (x = 0): 3(0) + 2y = 6 y =3 ⇒ D(0,3) ✓
mCD = mAB = CD: y − y1
5 3
0 =
2
3
2+e1 0+e2
2+e1
2
= x+3 5
Point C At C, CD cuts BC (y = 6),
= MAE ,
= mCD (x − x1 ) 3
y
(0,3) = (
5
y − (3)= [x − (0)]
A6(ii) Point E Let E be (e1 , e2 ) D
3
2
5 3
)
and
e1 = −2 ∴ E(−2,6) ✓
3 =
0+e2
5
2
x+3=6 x
=3
x =5 ⇒ C(5,6) ✓
e2 = 6
Area of trapezium ABCD Area of ABCD 1 2 12 5 0 2 | = | 2 0 6 6 3 0 1 = (12 + 72 + 15 + 0 − 0 − 30 − 0 − 6) 2 1 = (63) 2 = 31.5 unit 2 ✓ B1(a) A(−1,2) B(3,10) C(p, 8) A, B and C lie on the same line: mAB = mBC 2−10 (−1)−3 −8 −4
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= =
10−8 3−p 2 3−p 2
2
=
3−p p
=1 =2✓
3−p
169
A math 360 sol (unofficial) B1(b) AC ⊥ BC mAC ⋅ mBC 2−8 10−8 ⋅ (−1)−p 3−p −6 2
⋅
−1−p 3−p 6 2
⋅
Rev Ex 6 B2(iii) 𝐑𝐚𝐭𝐢𝐨 𝐀𝐏: 𝐏𝐐 3
= −1
A ( , 0) P(1, −1) B(0, −3) 2
= −1
By similar triangles (using x-coordinates) [diagram?]
= −1 = −1
p+1 3−p
3
AP: PB = ( − 1) : (1 − 0) 2
12 = (p + 1)(p − 3) 12 = p2 + 2p − 3 2 p − 2p − 15 = 0 (p + 3)(p − 5) = 0 p = −3 or p = 5 ✓ B2(i)
=
:1
2
= 1: 2 ✓ B3(i)
Point P At P, x − y − 2 = 0 intersects 2x − 5y − 7 = 0. x−y−2 =0 y = x − 2 −(1)
Show 𝐚 = 𝟑 A(2,6) B(6, −2) C(a, 8 − 3a) 8 − 3a < 0 −3a < −8 a
2x − 5y − 7 = 0 −(2) sub (1) into (2): 2x − 5(x − 2) − 7 = 0 2x − 5x + 10 − 7 = 0 −3x = −3 x =1
>
8 3
Area of △ ABC
= 10
6 a 2 || = 10 −2 8 − 3a 6 2 6 a 2 | | = 20 6 −2 8 − 3a 6 | − 4 + (48 − 18a) + 6a = 20 −36 − (−2a) − (16 − 6a)| |−8 − 4a| = 20 −8 − 4a = 20 or −8 − 4a = −20 −4a = 28 −4a = −12 a = −7 a =3 8 (rej ∵ a > ) 3 Point C C(3, −1) 1
| | 2
y|x=1 = (1) − 2 = −1 ⇒ P(1, −1) ✓ B2(ii) Line AB Point: P(1, −1) or A or B Gradient: mAB = 2 y − y1 = mAB (x − x1 ) y − (−1) = (2) [x − (1)] y+1 = 2x − 2 y = 2x − 3 ✓ Point A Point B At A, AB cuts x-axis At B where cuts y-axis (x = 0). (y = 0). y =0 y|x=0 = 2(0) − 3 2x − 3 = 0 = −3 3 ⇒ B(0, −3) ✓ x = AB:
1
2 6
B3(ii) Line AB Point:
2
A(2,6) or B(6, −2) (6)−(−2)
mAB =
AB:
y − y1 = mAB (x − x1 ) y − (6)= (−2)[x − (2)] y − 6 = −2x + 4 y = −2x + 10 ✓
(2)−(6)
=
8
Gradient:
−4
= −2
3
⇒ A ( , 0) ✓ 2
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170
A math 360 sol (unofficial)
Rev Ex 6
B3(iii) Line (through C & ⊥ to AB) Point: C(3, −1)
B4(iv) Line CD Point: C(8,2) or D Gradient: ∵ CD ∥ AB, (4)−(9)
Gradient: ∵ line ⊥ AB (y = −2x + 10) m=−
1 mAB
=−
1 −2
=
mCD = mAB = (2)−(7) =
1
= m (x − x1 )
y − y1
1
y − (−1) = ( ) [x − (3)] 2
1
3
2 1
2 5
2
2
y+1
= x−
y
= x− ✓
B3(iv) Point F (Foot of ⊥) C A
2 5 2
x− x
x
2
1
5
2
2
y|x=3 = (3) − 6 = −3 ⇒ D(3, −3) ✓
= −2x + 10 =
B5(i)
25 2
=5 1
5
2
2
y|x=5 = (5) −
Show △ 𝐏𝐐𝐑 𝐢𝐬 𝐢𝐬𝐨𝐬𝐜𝐞𝐥𝐞𝐬 P(1,0) Q(0,2) R(2,3) |PQ| = √[(1) − (0)]2 + [(0) − (2)]2 = √1 + 4 = √5
=0 ⇒ F(5,0) ✓ B4(i)
= mCD (x − x1 ) = (1) [x − (8)] =x−8 =x−6✓
y − y1 y − (2) y−2 y
B
F
5
=1
B4(v) Point D At D, BM (y = 3x − 12) intersects CD (y = x − 6). 3x − 12 = x − 6 2x =6 x =3
At F, y = x − meets AB (y = −2x + 10). 1
−5
2
CD: Line:
−5
|QR| = √[(0) − (2)]2 + [(2) − (3)]2 = √4 + 1 = √5
Show 𝐀𝐁 = 𝐁𝐂 A(2,4) B(7,9)
C(8,2)
∵ |PQ| = |QR|, PQR is isosceles ✓
AB = √[(2) − (7)]2 − [(4) − (9)]2 = √50 BC = √[(7) − (8)]2 + [(9) − (2)]2 = √50
B5(ii) Area of △ 𝐏𝐐𝐑 1
area of △ PQR = | |
∴ AB = BC [shown] ✓
2
1 0
0 2
2 3
1 || 0
1
= |2 + 0 + 0 − 0 − 4 − 3|
B4(ii) Point M M = MAC = (
(2)+(8) (4)+(2)
,
2
2
2 1
= |−5|
) = (5,3) ✓
= B4(iii) Line BM Point:
2 5 2
unit 2 ✓
B(7,9) or M(5,3) (9)−(3)
6
Gradient: mBM = (7)−(5) = = 3 2
BM:
y − y1 = mBM (x − x1 ) y − (9)= (3) [x − (7)] y − 9 = 3x − 21 y = 3x − 12 ✓
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171
A math 360 sol (unofficial)
Rev Ex 6
B5(iii) Line QS
B6(ii) Angle that BC makes with the positive x-axis 2−30
mBC = (−12)−12
R Q
tan β = S
6
β ≈ 49.4° ✓
P Point: Q(0,2) or S Gradient: ∵ QS ⊥ PR (using kite property), 1
mQS = −
mPR
=−
1 (0)−(3) ((1)−(2))
=−
1 −3 ( ) −1
=−
y − y1 = mQS (x − x1 )
QS:
7
1 3
B6(iii) Angle ACB ∡ACB = α − β = 69.4 − 49.4 = 20.0° ✓ B6(iv)
𝑦
C(12,30)
1
y − (2)= (− ) [x − (0)] 3
1
y−2 =− x 3 1
y
=− x+2 3
=− x+2
𝛾 Point D D(d1 , d2 ) BC = BD: B = MDC
−(1)
3
(−12,2) = (
x − y= 5 sub (1) into (2):
𝑥
A(−3, −10)
D
B5(iv) Point S At S, QS (x + 3y = 6) intersects x − y = 5. x + 3y = 6 3y = −x + 6 1
α 𝑂
3y = −x + 6 x + 3y = 6 ✓
y
β
B(−12,2)
−(2)
−12 = d1
1
d1 +12 d2 +30 2
d1 +12 2
,
2
)
and
2 =
= −36
d2 +30 2
d2 = −26
x − (− x + 2) = 5 3
4 3 4 3
x−2
=5
x
=7
x
=
∴ D(−36, −26) ✓
B6(v) Line AD Point:
21 4
A(−3, −10) or D(−36, −26)
1 21
y|x=21 = − ( ) + 2 3
4
= ⇒ S( B6(i)
4
tan α =
AD:
y − y1
4
, )✓ 4
(−10)−30 (−3)−12 8
(−10)−(−26) (−3)−(−36)
=
16 33
= mAD (x − x1 ) 16
y − (−10) = ( ) [x − (−3)]
Angle that AC makes with the positive x-axis A(−3, −10) 𝑦 𝐶(12,30) B(−12,2) C(12,30) mAC =
mAD =
1
21 1 4
Gradient:
𝐵(−12,2)
3
α ≈ 69.4° ✓
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𝛽
𝛼 𝑂
y + 10
=
y
=
33 16
48
33 16
33 282
33
x+ x−
33
✓
𝑥
𝐴(−3, −10)
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172
A math 360 sol (unofficial)
Rev Ex 6
B6(vi) Angle ADB mAD = tan γ =
16 33 16 33
γ ≈ 25.9° ∡ADB = β − γ = 29.4 − 25.9 ≈ 23.5° ✓
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173
A math 360 sol (unofficial)
Ex 7.1 2(e)
Ex 7.1 1(a)
ln 4 + lg 6 ≈ 2.16 ✓
1(b)
3 lg 2 − ln 2 ≈ 0.210 ✓
1(c)
ln 7.1 2 lg 5
1(d)
= log √2 (5 − 2√2) Is base √2 > 0 ? True Is base √2 ≠ 1 ? True Is input 5 − √2 > 0 ? True
≈ 1.40 ✓
lg 9−ln 3 ln(e2 −1)
∴ Yes ✓
≈ −0.0778 ✓ 3(a)
2(a)
log x (5 − 2x)|x=√2
log x (5 − 2x)|x=0.5 = log 0.5 (4)
3−2 log 3
Is base 0.5 > 0 ? True Is base 0.5 ≠ 1 ? True Is input 4 > 0 ? True
3(b)
10n
= 1 9
1 9
= −2 ✓ =5
log10 5 = n lg 5 =n✓
∴ Yes ✓ 3(c) 2(b)
log x (5 − 2x)|x=3 = log 3 (−1)
ex
=4
log e 4 = x ln 4 =x✓
Is base 3 > 0 ? True Is base 3 ≠ 1 ? True Is input −1 > 0 ? False
3(d)
2x
=p
log 2 p = x ✓ ∴ No ✓ 2(c)
3(e)
log x (5 − 2x)|x=2.5 = log 2.5 (0) Is base 2.5 > 0 ? Is base 2.5 ≠ 1 ? Is input 0 > 0 ?
=y
log a y = 3 ✓ True True False
∴ No ✓ 2(d)
a3
4(a)
53 4(b)
log x (5 − 2x)|x=1 = log1 (3) Is base 1 > 0 ? True Is base 1 ≠ 1 ? False Is input 3 > 0 ? True ∴ No ✓
log 5 125 = 3
lg 100 =2 log10 100 = 2 102
4(c)
sleightofmath.com
=x✓
log x 3 = 4 x4
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= 100 ✓
ln x = 2 log e x = 2 e2
4(d)
= 125 ✓
=3✓
174
A math 360 sol (unofficial) 4(e)
log 4 x = 2 4 x x y
=
log 2 𝑦 = 3 3
=x = 16 16 8
2 y
=y =8
=b
6(a)
log 4 4 −3 log 2 2 = 1 −3(1) = −2 ✓
6(b)
log 2 1 +4 log 5 5 = 0 +4(1) =4✓
6(c)
(3 − log 3 3)3 = [3 − (1)]3 = 23 =8✓
6(d)
(
8(a)
3 logx x+2 2
3(1)+2 2
4−2 log5 1
4−2(0) 5 2
) =[
8(b)
= 6(e)
6(f)
7(a)
16
8(c)
8(d)
9(a)
=x =8✓
ln x = lg 2 log e x = lg 2 =x ≈ 1.35 ✓
lg(3x) = 9 log10 3x = 9 109
= 3x
x
= (109 )
x
≈ 3.33 × 108 ✓
e2x
1 3
=k
log e k = 2x ln k = 2x ✓ 9(b)
10x−4 = 9 log10 9 = x − 4 lg 9 =x−4✓
log 2 x = 3 23 x
= e±√3 ≈ 5.65 or 0.177 ✓
elg 2 x
✓
log 2 (4 − 2 lg 10) = log 2 [4 − 2 log10 10] = log 2 [4 − 2(1)] = log 2 2 =1✓
= 100.61 ≈ 4.07 ✓
(ln x)2 = 3 ln x = ±√3
x x
]
log 2 (6 − 5 log 7 7) = log 2 [6 − 5(1)] = log 2 (1) =0✓
lg x = 0.61 log10 x = 0.61
log e x = ±√3
=( ) 4 25
2
x x
log y b = 2
x1 =a y2 x =a 2 xy = (a)(b) = ab ✓
3
= ✓
x
=2✓
log x a = 1
log 4 8 = x 4x =8 22x = 23 ⇒ 2x = 3
=y✓
2
5(b)
7(c)
log 3 y = n 3n
5(a)
Ex 7.1
9(c)
x4
=2−k
log x (2 − k) = 4 ✓ 7(b)
log x 9 = 2 x2 =9 x = −3 or x = 3 ✓ (rej)
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175
A math 360 sol (unofficial) 9(d)
em+5
=x−2
Ex 7.1 12
log 3 p = a p
log e (x − 2) = m + 5 ln(x − 2) = m+5✓
= 3a
log 27 q = b 10(a) log 3 y = n + 1 3n+1
e
33b a−3b
= 3c
13(a) log 2 (2x + 1) = −3
=x−y✓
2x + 1
= 2−3
2x
=−
x
=−
= 4y ✓ 13(b) log 3 (x 2 − 1)
log 4 y = a = 4a
= 21+2a = 21+2a = 1 + 2a = 3b − 1 [shown] ✓
7 8 7 16
✓
=1
x2 − 1 = 31 2 x −4 =0 (x + 2)(x − 2) = 0 x = −2 or x = 2 ✓
−(1)
log 8 (2y) = b −(2) sub (1) into (2): log 8 [2(4a )] = b log8 [2(22a )] = b log 8 (21+2a ) = b 8b 23b ⇒ 3b 2a
= 3c
3 = 3c ⇒ c = a − 3b ✓
10(d) log 2 (4y) = p + 1
y
= 3c
27b 3a
=k✓
10√2
11
p q 3a
10(c) lg(x − y) = √2 log10 (x − y) = √2
2p+1
= 27b
=y✓
10(b) ln k = x − 3 log e k = x − 3 x−3
q
13(c)
log x (6x − 8) = 2 x2 = 6x − 8 x 2 − 6x + 8 =0 (x − 2)(x − 4) = 0 x = 2 or x = 4 ✓
13(d) log x 64 = 3 2
3
x2
= 64
3 2
= 26
x
2
x x x
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= (26 )3 = 24 = 16 ✓
176
A math 360 sol (unofficial)
Ex 7.1
13(e) log 9 √27 = x + 1 9x+1
15(b) log y x = 2
= √27
2x+2
3
= y2
x
3 2
−(1)
=3
⇒ 2x + 2 =
xy = 8 −(2) sub (1) into (2): (y 2 )y = 8 y3 = 23 ⇒ y =2✓
3 2 1
2x
=−
x
=− ✓
2 1 4
14(a) ln 2 × ln(4x)= 3 3
ln(4x)
=
log e (4x)
=
4x
= eln 2
x|y=2 = (2)2
ln 2 3
16
ln 2
=4✓
log 2 (log 3 x) = ln e
3
1
log 2 (log 3 x) = 1 log 3 x = 21
3 ln 2
x
= e
x
≈ 18.9 ✓
4
= 32 =9✓
x x
14(b) lg(x − 2) = (lg 3)2 log10 (x − 2) = (lg 3)2
17(a) For lg(x + 2) to be defined, 2
= 10(lg 3)
x−2
= 2 + 10(lg 3) ≈ 3.69 ✓
x x
x + 2> 0 x > −2
2
17(b) For ln(x 2 − 2x) to be defined, 14(c) ln(4x) = lg 3 × lg 5 log e 4x = lg 3 × lg 5 4x
= elg 3×lg 5
x
= e(lg 3×lg 5)
x
≈ 0.349 ✓
x 2 − 2x x(x − 2) +
1
17(c) For log x (3 − x) to be defined, x > 0, x ≠ 1 and 3 − x > 0 3 >x x 2 ✓
14(d) lg(x − 1) = lg(e2 − 1) ⇒ x − 1 = e2 − 1 x = e2 x ≈ 7.39 ✓
log 2 y = x sub (1) into (2): log 2 y = 2
− 0
4
>0 >0
−(2)
18(a) ln(y + 1) − x = 0 ln(y + 1) =x log e (y + 1) = x y+1 y
= ex = ex − 1 ✓
= 22 =4✓
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177
A math 360 sol (unofficial) 18(b) 2 lg y lg y
Ex 7.1
=x−2
20(ii) log 9 (31 + log x 16) = log 2 √8
x−2
=
log 9 (31 + log x 16) =
2 x−2
log10 y =
log 9 (31 + log x 16) =
2 x−2 2
y
= 10
y
= 102x−1 ✓
= (9)2
31 + log x 16 31 + log x 16 31 + log x 16 log x 16
= (32 )2 = 33 = 27 = −4
x −4 x −4
= 16 = 24
x
= (24 )−4 = 2−1
3
=x =x−4
1
= ln(x − 4) ✓
y
2 3
= log e (x − 4)
2y
2 3
31 + log x 16
1
18(c) e2y + 4 e2y
3
2
1
18(d) ln(x + y) − 4x = 0 ln(x + y) = 4x log e (x + y) = 4x
2
= e4x = e4x − x ✓
x+y y 19
1
= ✓
ln(x 2 + 1 − e3 lg 10 )
21
For equal real roots: b2 − 4ac =0 2 (−4) − 4(1)(log 2 p) = 0 16 = 4 log 2 p log 2 p =4 p = 24 p = 16 ✓
=3
log e (x 2 + 1 − e3 lg 10 ) = 3 x 2 + 1 − e3 lg 10 x 2 + 1 − e3(1) x2
= e3 = e3 = 2e3 − 1
x = ±√2e3 − 1 x = ±6.26 ✓ [textbook answer is wrong] 20(i)
p = log 2 √8 3
p = log 2 22 3
p= ✓ 2
x 2 − 4x + log 2 p = 0 i.e. a = 1, b = −4, c = log 2 p
22(i) (a) (b) (c)
lg 546 ≈ 2.74 ✓ lg 12 458 ≈ 4.10 ✓ lg 464 777 399 ≈ 8.67 ✓
22(ii) Number of digits in integer k from lg k lg k + 1 if lg k ∈ ℤ ={ ✓ ⌈lg k⌉ if lg k ∉ ℤ 22(iii) Number of digits in 342 = ⌈lg 342 ⌉ = ⌈20.03⌉ = 21 ✓ Number of digits in 732 = ⌈lg 732 ⌉ = ⌈27.02⌉ = 27 ✓
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178
A math 360 sol (unofficial)
Ex 7.1
22(iv) As we use the base-10 numeral system, The ‘index of base 10’ is very indicative of the number of digits. 1 digit: k = 1 to 9 ⇒ lowest k = 1 = 100 2 digits: k = 10 to 99 ⇒ lowest k = 10 = 101 3 digits: k = 100 to 999 ⇒ lowest k = 100 = 102 Observe that the lowest k for each digit range is always 10integer Every integer increase in index results in increase in digits but increase in index smaller than 1 will not result in increase in digits and still be registered as the same number of digits. ∴ for k with lg k is an integer, the no. of digits is lg k + 1 1 digit: k = 100 ⇒ no. of digits = lg 100 + 1 = 0 + 1 = 1 2 digit: k = 10 = 101 ⇒ no. of digits = lg 101 + 1 = 1 + 1 = 2 ∴ for k with lg k is a non-integer, the no. of digits is ⌈lg k⌉ (round to the next higher integer) 1 digit: k = 5 ≈ 100.699 ⇒ no. of digits = ⌈lg 100.699 ⌉ = ⌈0.699⌉ = 1 2 digit: k = 55 ≈ 101.74 ⇒ no. of digits = ⌈lg 101.74 ⌉ = ⌈1.74⌉ = 2
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179
A math 360 sol (unofficial)
Ex 7.2 4(a)
Ex 7.2 1(a)
1(b)
1
9
9
log 3 27 + log 3 = log 3 (27 ⋅ )
4(b)
log5 4×log2 10 log25 √10
=
= log 6 (6) =1✓
=
3
=
log 5 4 +2 log 5 3 −3 log 5 2 = log 5 4 + log 5 32 − log 5 23 23
=
]
=
9
= log 5 ✓ 2
2(b)
lg
8
=
−2 lg
75
3 5
+4 lg
3
8
3 2
3 4
75
5
2
8 3 4 ( ) 75 2 3 2 ( ) 5
= lg [
5(a)
]
3
= lg ✓ 2
1
+2 log 3 5 − log 3
3(52 ) 3−3
]
5(b)
= log 3 2025 ✓ 3(a)
log 5 7 =
3 ln 3 ln 3
lg 4 lg 10 × lg 5 lg 2 lg √10 lg 25
lg 4 lg 10 lg 5
⋅
lg 22 lg 5 2 lg 2 lg 5 2 lg 2 lg 5 2 lg 5
lg 2
⋅ ⋅ ⋅ ⋅
⋅
1 lg 2 1 lg 2 1 lg 2
lg 25 lg √10
⋅
lg 52 1
lg 102 2 lg 5
⋅1 2
lg 10
2 lg 5
⋅1 2
lg 10
2 lg 5 1 2
4 1 2
y = 100x1.5 lg y = lg(100x1.5 ) = lg 100 + lg x1.5 = 1.5 lg x + lg 102 = 1.5 lg x +2 ✓ m = 1.5 ✓ X = lg x ✓ c =2✓
1 27
= log 3 3 + log 3 52 − log 3 3−3 = log 3 [
ln 3
=8✓
2
= lg ( ) − lg ( ) + lg ( )
2(c)
=
= log 6 ( 2 )
= log 5 [
ln 33
=3✓
log 6 54 −2 log 6 3 = log 6 54 − log 6 32
4(32 )
ln 5
ln 3
=
54
2(a)
ln 27
ln 27
=
1
×
ln 3
=
log 2 8 = log 2 23 =3✓
= log 3 (3) =1✓ 1(c)
ln 5
log 3 5 × log 5 27 =
ln 7 ln 5
y = 0.1(1000)x y = (10−1 )(103 )x y = 103x−1 lg y = 3x − 1 ✓
≈ 1.21 ✓ 3(b)
log 1 5.3 = 2
m = 3✓ X= x✓ c = −1 ✓
ln 5.3 ln
1 2
≈ −2.41 ✓ 6(a)
log a 8 −2 log a 4 = log a 8 − log a 42 8
= log a ( 2 ) 4
1
= log a ✓ 2
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180
A math 360 sol (unofficial) 6(b)
2 log x 5 −3 log x 2 + log x 4 = log x 52 − log x 23 + log x 4 = log x [ = log x
6(c)
Ex 7.2
52 (4) 23
25 2
8(iv)
log a 3 = 0.477 log a 5 = 0.699 loga 25 loga 3a
= =
loga 52 loga 3+loga a 2 loga 5 (0.477)+1
2
9 5a
log 4 3 = a 3 = 4a
3
= (4a )(4b ) = 4a+3b ✓ 9 2
= log a 9
− log a 5a
= log a 32 −(log a 5 + log a a) = 2 log a 3 − log a 5 − log a a = 2(0.477) −0.699 −1 = −0.745 ✓ log 4 3 = a log 4 5 = b
= 2(lg 3 + lg √x)
− (lg − lg x 2 )
1
4
+
3 4
− (lg − 2 lg x)
= 2 lg 3 + lg x
− lg + 2 lg x
= 3 lg x
+2 lg 3 − lg
2
+ lg 32 − lg + lg
9
= lg x
+ lg
2 9
= lg x
+ lg
2
log 4 20= log 4 × 5) 2 = log 4 2 + log 4 5 =1 +b ✓
4
+
3 4
2 lg 22 +3 lg x 2 2 lg 2+3 lg x 2 2
32
+ lg 2
4 3
32 ×2 4 3
27
✓
2
x = 10p
y = 10q
y2
lg 4+lg x3
3
lg y =q log10 y = 𝑞
100√x
lg 4x3 lg 100
+ lg 2 + lg x
3
lg x =p log10 y = 𝑝
lg (
sleightofmath.com
+ +
3
9
(22
3
4
2
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3x2
= 2 (lg 3 + lg x)
10(i) 8(ii)
+ log100 4x 3
− lg
= lg x
log 4 45= log 4 (32 × 5) = log 4 32 + log 4 5 = 2(log 4 3) + log 4 5 = 2a +b ✓
4
2 lg 3√x
= 3 lg x 8(i)
✓
375 = 3 × 53
log a (5a ) = log a 5 + log a a = 0.699 +2 = 2.699 ✓ log a
a+2b
log 4 5 = b 5 = 4b
2(0.699) = 1.477 ≈ 0.947 ✓
7(iii)
2
=
]
= log a 8a ✓
7(ii)
log 4 42 log 4 (3 × 52 ) 2 = log 4 3 + log 4 52 2 = log 4 3 + 2 log 4 5
✓
a6 2
log4 75
=
]
23 (a6 )
log4 16
log 75 16 =
3 log a 2 −4 +2 log a a3 = log a 23 − log a a4 + log a (a3 )2 = log a [
7(i)
8(iii)
) = lg 100 + lg √x − lg y 2 1
=2
+ lg x
−2 lg y
=2
+ p
−2q ✓
2 1 2
181
A math 360 sol (unofficial)
Ex 7.2
10(ii) lg(y x ) = x lg y = 10p (q) = q(10p ) ✓ 11
14
log 8 5 =
ln K = ln a − ln b + ln c − t − bc
1
ln K = ln (a ⋅ ⋅ c ⋅ e b
K
=
lg(y + 1) = lg ⇒y+1
=
y
=
⇒
√x 100 √x
34
15(ii) lg(√3 − √2) = lg [
(1)−(m) 3(m)
=
1−m 3m
✓
√3+√2 √3+√2
1 √3+√2
]
= lg 1 − lg(√3 + √2)
15(iii) k = log 2 (√9 + √5) log √2 (√9 − √5) = log √2 [(√9 − √5) = log √2 (
9−5 √9+√5
√9+√5 √9+√5
]
)
= (x + 2)3
= log √2 4 − log √2 (√9 + √5)
= 81(x + 2)3
=
3 2
or y = −9(x + 2) ✓ (rej ∵ y > 0)
12(c) 3 + log 2 (x + y) 3 log 2 2 + log 2 (x + y) log 2 8(x + y) ⇒ 8(x + y) 8x + 8y = x − 2y 10y = −7x 7
=
√3+√2
= 3 log 3 (x + 2) = log 3 (x + 2)3
y2
10
3 lg 2
√3 + √2 1 = [shown] ✓
= log 3 (x + 2)3
=−
lg 10−lg 2
=
3−2
=
)
−1✓
34
=
= log 2 (x − 2y) = log 2 (x − 2y) = log 2 (x − 2y) = x − 2y 16(i)
x✓
y = 3, x = 2: log 2 (3 + 1) = 2 log 2 (2) + c log 2 22 =2+c 2 =2+c c =0
log2 22 1 log2 22
2 1 2
−
log2 (√9+√5) 1
log2 22
−
log2 (√9+√5) 1 2
=4
−2 log 2 (√9 + √5)
=4
−2k
= 2(2 − k) ✓
log 2 (y + 1) = 2 log 2 x + c
∴ log 2 (y + 1) log 2 (y + 1) ⇒y+1 y
10 2 lg 23
lg( )
= − lg(√3 + √2) ✓
y2
3 2
=
√3 − √2 = (√3 − √2)
√x 100
y = 9(x + 2)
13
lg 8
100
y2
y
15(i)
t bc
− lg √x − lg √x
12(b) 2 log 3 y −4 log 3 y 2 − log 3 34 log 3
−
[shown] ✓
12(a) lg(y + 1) = 2 lg(y + 1) = lg 102
lg 5
t bc
ln K = ln a − ln b + ln c + ln e ac − t e bc b
m = lg 2
16(ii)
p log a p log a p ⇒p 1 q
= aloga x = log a aloga x = log a x =x✓
3 =3
1 log4 3
=3
1 1 log3 4
= 3log3 4 = 4 ✓
= 2 log 2 x = log 2 x 2 = x2 = x2 − 1 ✓
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182
A math 360 sol (unofficial) 17
Ex 7.2
Method 1 (prime factorization) (lg 5)2 + lg 2 lg 50 = (lg 5)2 + lg 2 lg(2 × 52 ) = (lg 5)2 + lg 2 (lg 2 + 2 lg 5) = (lg 5)2 + 2 lg 5 lg 2 + (lg 2)2 = (lg 5 + lg 2)2 = (lg 10)2 =1✓
1 2 18 un = (1 − ) n (ii)(a) 1 2
lg u4 = lg (1 − ) 4
3
= 2 lg [shown] ✓ 4
lg u Method 2 (selective factorization) (lg 5)2 + lg 2 lg 50 = (lg 5)2 + lg 2 lg(5 × 10) = (lg 5)2 + lg 2 (lg 5 + lg 10) = (lg 5)2 + lg 2 (lg 5 + 1) = (lg 5)2 + lg 2 lg 5 + lg 2 = lg 5 (lg 5 + lg 2) + lg 2 = lg 5 lg 10 + lg 2 = lg 5 + lg 2 = lg 10 =1✓ 18(i)
1
2
3
998
2
3
998
999 999
999
lg + lg + lg + ⋯ + lg 1
2
4 3
2 1
3
4
= lg [( ) ( ) ( ) … ( = lg (
)(
+ lg
1000
= lg (1 − = 2 lg (
1 10k
)
2
10k −1 10k
)✓
lg u2 + lg u3 + lg u4 + ⋯ + lg u10k 18 (ii)(b) 1 2 1 2 3 = lg (1 − ) + lg (1 − ) +2 lg 2
3
4
+ ⋯ + 2 lg = 2 lg
1 2 1
= 2(lg
2
+2 lg + lg
2
2 3
+2 lg
+ lg
3
3
3
+ ⋯ + 2 lg
4
+ ⋯ + lg
1
2
3
4 10k −1
2
3
4
10k
= 2 lg [( ) ( ) ( ) … (
10k −1 10k 10k −1 10k
10k −1 10k
)
)]
1 = 2 lg ( k ) 10 = 2 lg(10−k ) = 2(−k) = −2k ✓
999 1000
)]
)
1000 −3 )
= lg(10 = −3 ✓
(10k )
19
1 lg ≈ −0.301 < 0 2 When you multiply both sides of an inequality with a negative number, switch the reverse the inequality sign from 2nd line onwards
20
lg[(−2)(−5)] ≠ lg(−2) + lg(−5) ✓
20(i)
x = −2, y = −5 ✓
20(ii) x = −2, r = 2 ✓
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183
A math 360 sol (unofficial)
Ex 7.3
Ex 7.3 1(a)
log 5 (x + 1) = log 5 3 ⇒x+1 =3 x = 2✓
1(b)
log 2 (x − 1) ⇒x−1 3x x
1(c)
2(a)
2(b)
3(a)
3(b)
= log 2 (4x − 7) = 4x − 7 =6 = 2✓
3(c)
log 2 (x − 1)2 =2 + log 2 (x + 2) 2 2 (x log 2 − 1) = log 2 2 + log 2 (x + 2) log 2 (x − 1)2 = log 2 4(x + 2) 2 ⇒ (x − 1) = 4(x + 2) 2 x − 2x + 1 = 4x + 8 2 x − 6x − 7 =0 (x − 7)(x + 1) = 0 x = 7 or x = −1 ✓
4(a)
log 3 x = 9 log x 3 log 3 x =
log 3 x + log 3 (x + 2)= 1 log 3 [x(x + 2)] = log 3 3 ⇒ x(x + 2) =3 x2 + 2 =3 2 x + 2x − 3 =0 (x + 3)(x − 1) =0 x = −3 (rej) or x = 1 ✓ log x 25 + log x 5 log x (125) ⇒ 125 x
=3 = log x x 3 = x3 =5✓
9 log3 x
sub u = log 3 x: u=
9 u
u2 = 9 u =3 log 3 x = 3 x x 4(b)
or
u = −3 log 3 x = −3
= 33 = 27 ✓
x
= 3−3
x
=
1 27
✓
log 3 x + 2 = 3 log x 3 3
log 3 x + 2 =
log3 x
3 log x 2 + log x 18 = 2 log x 23 + log x 18 = log x x 2 log x (23 × 18) = log x x 2 log x 144 = log x x 2 ⇒ 144 = x2 x = 12 or x = −12 (rej) ✓
sub u = log 3 x:
lg[(x + 2)(x − 2)] = lg(2x − 1) ⇒ (x + 2)(x − 2) = (2x − 1) x2 − 4 = 2x − 1 2 x − 2x − 3 =0 (x − 3)(x + 1) =0 x = 3 or x = −1 (rej) ✓
x
u+2
=
3 u
u2 + 2u − 3 =0 (u + 3)(u − 1) = 0 u = −3 or u =1 log 3 x = −3 log 3 x = 1
x
= 3−3 =
1 27
x
=3✓
✓
log 2 [(x − 2)(8 − x)] − log 2 (x − 5)= 3 log 2 [ ⇒
(x−2)(8−x)
(x−5) (x−2)(8−x)
]
(x−5)
= log 2 23 = 23
−(x − 2)(x − 8) = 8(x − 5) 2 −(x − 10x + 16) = 8x − 40 x 2 − 2x − 24 =0 (x − 6)(x + 4) =0 x = 6 or x = −4 (rej) ✓
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184
A math 360 sol (unofficial) 5
Ex 7.3
ln(3x − y) = ln 36 − ln 9
7(a)
36
ln(3x − y) = ln ( )
2 log 5 x + (
⇒ 3x − y = 4 y = 3x − 4 −(1)
2 log 5 x + (
(ex )2
2 log 5 x + (
9
=e
ey
5
e2x−y = e1 ⇒ 2x − y = 1
2
= 55 2
log 2 x − log 4 (x + 6) = 0 log 2 x = log 4 (x + 6) =[ =[ =[
=2
6
)
x
2
= 5q − p = 5q − 1
log2 (x+6) 2
] ]
= √x + 6
2
=x+6 =0
x = 3 or x = −2 (rej) ✓ 7(c) −(2)
sub (1) into (2): 2 + 3q = 5q − 1 2q =3 3
log 5 (5 − 4x) = log √5 (2 − x) log 5 (5 − 4x) =
log5 (2−x)
log 5 (5 − 4x) =
log5 (2−x)
log 5 (5 − 4x) =
log5 (2−x)
log5 √5 1
log5 52 1 2
log 5 (5 − 4x) = 2 log 5 (2 − x)
= ✓
q
log2 22
]
(x − 3)(x + 2) = 0
=2
1 p
log2 (x+6)
x2 − x − 6
=2
5q−p
log2 4
⇒x
= log 2 2
15q−3p
log2 (x+6)
= log 2 √x + 6
2p+2 = 24 (23q ) 2p+2 = 24+3q ⇒ p + 2 = 4 + 3q p = 2 + 3q −(1) log 2 6 − log 2 (15q − 3p) = 1
⇒
= log 5 55 = (55 )5 = 52 = 25
3
15q−3p 6
= log 5 55
)
2
x
4 [(22 )2q ]
log 2 (
) = log 5 55
= log 5 55
= 16 (4 )
2
log5 52 log5 x
5 2
3 q 2
p+2
) = log 5 55
log5 25 log5 x
log 5 x
⇒x
Put x = 3 into (1): y|x=5 = 3(3) − 4 =5✓ 2
=5
log 5 x 2
7(b)
p+2
log5 x
5
−(2)
sub (1) into (2): 2x − (3x − 4) = 1 −x + 4 =1 −x = −3 x =3✓
6
2 log 5 x + log 25 x
2
log 5 (5 − 4x) = log 5 (2 − x)2 Put q =
3 2
into (1): 3
p|q=3 = 2 + 3 ( ) 2
2
=
13 2
✓
⇒ 5 − 4x
= (2 − x)2
5 − 4x
= 4 − 4x + x 2
−x 2 + 1
=0
x2 − 1
=0
(x + 1)(x − 1) = 0 x = −1 or x = 1 ✓
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185
A math 360 sol (unofficial) 8(b)
log 9 y + log 3 y log3 y
+ log 3 y
log3 9 log3 y
+ log 3 y log3 32 1 log 3 y + log 3 y 2 3 log 3 y 2 3
Ex 7.3
= 2 log 3 x +3 log 3 2 = log 3 x
2
9(c)
+ log 3 2
3
= log 3 [(x 2 )(23 )] = log 3 (8x 2 )
1
= log 3 (8x 2 )
log 3 √3x + 1 = log 3 (2x) ⇒ √3x + 1 = 2x 3x + 1 = 4x 2 4x 2 − 3x − 1 = 0 (4x + 1)(x − 1) = 0
2
⇒ y2
= 8x 2
y
= (8x 2 )3 2
=
2 4 (23 )3 x 3
=
4 22 x 3
log 3 (3x + 1) = log 3 (2x)
1
2
x = − (rej) or x = 1 ✓
= (8)3 (x 2 )3
4
9(d)
3 log 4 x − log16 x = 3.75 log 4 x 3 −
4
= 4x 3 ✓ log 4 (6 − x) − log 2 8
log 4 x 3 −
= log 9 3
log 4 x 3 −
1
log 4 (6 − x) − log 2 23 = log 9 92
log4 x
= log 4 43.75
log4 16 log4 x log4
log4 x
= log 4 4
2
1
log 4 x 2
2
5
15
x2
= 44
log 4 (6 − x)
=
6−x
= 42
x
= (22 )2 = 27 = −122 ✓
5
5
15
= (22 ) 4
5
15
x2
= 22 2 15 5 2
= (2 )
x
= 23 =8✓
log 7 (9x + 38) − log 7 (x + 2) = log 9 81 log 7 ( log 7 ( ⇒
9x+38
)
= log 9 92
)
=2
x+2 9x+38 x+2 9x+38
x+2 9x+38
x+2 9x+38 x+2
)
9(e)
log 5 x − log 25 (x + 10) = 0.5 log 5 x −
2
= log 7 7 = 72
log 5 x −
= 49
log 5 x −
9x + 38 −40x
= 49x + 98 = 60
x
=− ✓
15 4
( )
= log 4 4
x2 7
15 4
( )
2 7
=
log 7 (
15 4
( )
1
log 4 (6 − x) −3
15 4
( )
= log 4 4
42
log 4 x 3 − log 4 x 2 = log 4 4
7
9(b)
= log 3 (2x)
log3 32
2
9(a)
= log 3 (2x)
log3 9 log3 (3x+1)
= log 3 (8x 2 )
log 3 y 2
log 9 (3x + 1) = log 3 x + log 3 2 log3 (3x+1)
3
log5 (x+10) log5 25 log5 (x+10) log5 52 log5 (x+10) 2
log 5 x − log 5 √x + 10
3
log 5 (
2
⇒
x
)
√x+10
x √x+10
x2 x+10
= log 5 50.5 = log 5 √5 = log 5 √5 = log 5 √5 = log 5 √5 = √5 =5
x2
= 5x + 50
x 2 − 5x − 50
=0
(x − 10)(x + 5)
=0
x = 10 or x = −5 (rej ∵ x > 0) ✓
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186
A math 360 sol (unofficial) 10
Ex 7.3 + log a (7x − 10a)
2 log a x
=1
log a x 2
= log a a + log a (7x − 10a)
1st eqn log 2 (2y − 3x) − log 2 3 = 4 log 4 2
= log a [a(7x − 10a)]
log 2 (
= a(7x − 10a)
log 2 (
log a x ⇒x x
2
2
2
12
2
log 2 (
= 7ax − 10a
x 2 − 7ax + 10a2 = 0
log 2 (
(x − 2a)(x − 5a) = 0
⇒
x = 2a or x = 5a ✓ 11
27 × 3lg x
= 91+lg(x−20)
(33 )(3lg x )
= (32 )1+lg(x−20)
33+lg x
= 32+2 lg(x−20)
⇒ 3 + lg x
= 2 + 2 lg(x − 20)
1 + lg x
= 2 lg(x − 20)
lg 10 + lg x
= lg(x − 20)2
lg(10x)
= lg(x − 20)2
⇒ 10x
= (x − 20)2
2y−3x 3 2y−3x 3 2y−3x 3 2y−3x 3
2y−3x
1
)
= 4 (log 4 42 )
)
= 4( )
)
=2
)
= log 2 22
1 2
= 22
3
2y − 3x 2y
= 12 = 3x + 12
y
= x + 6 −(1)
3 2
2nd eqn log 3 6 + log 3 (x + y) log 3 6 + log 3 (x + y) log 3 [6(x + y)] ⇒ 6(x + y) 6x + 6y
= log 3 (−x) + log 3 (1 − x) = log 3 [(−x)(1 − x)] = log 3 (x 2 − x) = x2 − x = x2 − x −(2)
sub (1) into (2): 3
2
10x
= x − 40x + 400
6x + 6 ( x + 6)
0
= x 2 − 50x + 400
x 2 − 50x + 400
=0
6x + 9x + 36 = x2 − x 15x + 36 = x2 − x x 2 − 16x − 36 =0 (x − 18)(x + 2) =0 x = 18 or x = −2 ✓
= x2 − x
2
(x − 40)(x − 10) = 0 x = 40 or x = 10 (rej) ✓
3
y|x=−2 = (−2) + 6
(rej ∵ −x > 0)
2
=3✓ 12(ii) log 3 6 + log 3 (x + y) = log 3 (x 2 − x) ⇒ x + y > 0 x2 − x > 0 ∴ On top of x = −2, y = 3, the solution includes 3
x = 18, y|x=18 = (18) + 6 = 33 ✓ 2
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187
A math 360 sol (unofficial)
Ex 7.3
13(i) log 2 (x 3 + 1) − 2 log 2 x = log 2 (x 2 − x + 1) − 2 log 2 (x 3 + 1) − log 2 x 2 = log 2 (x 2 − x + 1) − log 2 22 log 2 ( ⇒
x3 +1 x2
)
= log 2 (
x3 +1
=
x2 (x+1)(x2 −x+1)
=
x2 x+1
=
x2
22
)
x = 2 − 2√2 or (rej ∵ x > 0)
+ log a2 x
x
1 x
loga ( ) 1 x
x2 −x+1
loga ( )
22 x2 −x+1
1 2
+ log a x
+
+ log a x
+
1
7
x 1 2
4
loga x loga a2 loga x 2
+ log a4 x = c + +
loga x loga a4 loga x 4
1 4
log a ( ) x
7
1 2
7
= log a ac
x
4±√16+16 2
=
4±√32 2
=
4±4√2 2
x = 2 + 2√2 ✓
(x 3
7
log a [(x −1 )2 x 4 ]
= log a ac
7
log a [(x −2 )x 4 ]
= log a ac
1
2
(x 2
13(ii) log 2 + 1) − log 2 x = log 2 − x + 1) − 2 3 2 log 2 (x + 1) − 2 log 2 x = log 2 (x − x + 1) − 2 ⇒ x = 2 ± 2√2 ✓
log a (x −4 )
= log a ac
1
⇒ x −4 x 15
= log a ac
+ log a x 4 = log a ac
log a [( ) x 4 ] =
= log a ac
= log a ac
2 log a ( ) + log a x
4
−(−4)±√(−4)2 −4(1)(−4) 2(1)
1
log √a ( ) + log a x loga √a
= x2 =0
4x + 4 x 2 − 4x − 4 x=
x2 −x+1
14
= ac = a−4c ✓ 2 log a b + 4 log b a = 9 2 log a b + 4 (
1 loga b
)=9
sub u = log a b: 2u +
4
=9
u
2u2 − 9u + 4 = 0 (2u − 1)(u − 4) = 0 u=
1 2
sub u = log a b: log a b = b b
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1
u=4 sub u = log a b: log a b = 4
2 1 2
=a = √a ✓
b = a4 (rej ∵ a > b > 1)
188
A math 360 sol (unofficial)
Ex 7.4 3(ii)
Ex 7.4 1(a)
1(b)
4x
= 9(5x )
4 x
( ) 5
5x = 9 lg 5x = lg 9 x lg 5 = lg 9
lg ( ) = lg 9 5
4
x lg ( ) = lg 9 5
lg 9
x
=
x
≈ 1.37 ✓
4e2x
lg 5
= 21
2x
=9
4 x
4(a)
21
e
=
2x
= ln
x
= ln
x
≈ 0.829 ✓
4 21
1
4 21
2
4
lg 9
x
=
x
≈ −0.985 ✓
2x . 3x 6x lg 6x x lg 6
= 10 = 10 = lg 10 =1
x
=
lg
4 5
1 lg 6
≈ 1.29 ✓ 1(c)
4 − 72x −72x 72x lg 72x 2x lg 7
=1 = −3 =3 = lg 3 = lg 3
x
=
x
≈ 0.282 ✓
4(b)
2x+1 = 3x (2x )(21 ) = 3x 2x
=
3x 2 x
( )
lg 3
3
2 lg 7
=
2 x
lg ( )
2
3(i)
3x+1 − 12 3x+1 lg 3x+1 (x + 1) lg 3
=0 = 12 = lg 12 = lg 12
x+1
=
x
=
x
≈ 1.26 ✓
lg 3
2
x lg ( ) = lg 3
x
=
1 2 1 2
1 2 2 lg( ) 3
lg
≈ 1.71 ✓
lg 12 lg 3 lg 12
2 1
= lg
3
2
1(d)
1
4(c) −1
3x+1 . 2x−2 = 21 x x −2 (3 )(3) (2 )(2 ) = 21 32
(3x )(2x )
= (3)(2−2 )
y = 5e
6x lg 6x x lg 6
= 28 = lg 28 = lg 28
y = 12: 5e0.2x = 12
x
=
0.2x
=
0.2x
= ln
x
= 5 ln
x
≈ 4.38 ✓
4
4x 5x 4 x
lg 6
≈ 1.86 ✓
12
e0.2x
x
lg 28
5 12 5
5(i)
9x − 4 = 3x+1 (32 )x − 4 = (3x )(3) (3x )2 − 3(3x ) − 4 = 0 ✓
5(ii)
(3x − 4)(3x + 1) = 0 3x = 4 or 3x = −1 (rej) lg 3x = lg 4 x lg 3 = lg 4
12 5
x)
= 9(5 =9
( ) =9✓ 5
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lg 4
x
=
x
≈ 1.26 ✓
lg 3
189
A math 360 sol (unofficial) 6(i)
6(ii)
7(a)
Ex 7.4 3
8x + 11(2x )
= 4x+2 − 20
(23 )x + 11(2x ) (2x )3 + 11(2x ) (2x )3 + 11(2x ) (2x )3 − 8(2x )2 + 11(2x ) + 20 sub u = 2x u3 − 8u2 + 11u + 20 (u + 1)(u2 + + 20)
= (22 )x+2 − 20 = (22x )(23 ) − 20 = 8(2x )2 − 20 =0
(u + 1)(u2 − 9u + 20) = 0 (u + 1)(u − 5)(u − 4) = 0 u = −1 or u = 5 or sub u = 2x sub u = 2x x 2 = −1 2x = 5 (rej ∵ 2x > 0) x lg 2 = lg 5
3
=0✓
u =4 sub u = 2x 2x = 22 ⇒ x =2✓
7(d)
=
x
≈ 2.32 ✓
1 2
lg 2
sub y = ex : ex = −
1 2
(NA ∵ ex > 0) ex x
e
2y − 7√y + 3 1
y=
1 4
sub y = ex ex =
8(a)
1 4 1
x
= ln
x
≈ −1.39 ✓
4
x 3 = e6x−1 ln x 3 = ln e6x−1 3 ln x = 6x − 1 ln x = 2x −
12 ex
sub y = ex y
=7−
12
a=2✓
y
b=− ✓
y2 = 7y − 12 2 y − 7y + 12 = 0 (y − 3)(y − 4)= 0 y = 3 or y=4 x sub y = e sub y = ex x e =3 ex = 4 x = ln 3 x = ln 4 x ≈ 1.10 ✓ x ≈ 1.39 ✓
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=0 √y = 3 y=9 sub y = ex ex = 9 x = ln 9 x ≈ 2.20 ✓
√y = 2 or
= 7 − 12e−x =7−
2ex = 7√ex − 3
(2√y − 1)(√y − 3) = 0
y=2 sub y = ex : ex = 2 x = ln 2 x ≈ 0.693 ✓
or
y=2 sub y = ex ex = 2 x = ln 2 x ≈ 0.693 ✓
1
lg 5
x
e3x + 2ex = 3e2x (ex )3 + 2ex = 3(ex )2 x sub y = e y 3 + 2y = 3y 2 y 3 − 3y 2 + 2y = 0 y(y 2 − 3y + 2) = 0 y(y − 1)(y − 2) = 0 y=0 or y = 1 or x sub y = e sub y = ex x e = 0 (NA) ex = 1 x =0✓
2ex = 7(ex )2 − 3 sub y = ex 2y = 7√y − 3
2e2x − 3ex =2 x )2 x 2(e − 3e =2 x sub y = e 2y 2 − 3y =2 2y 2 − 3y − 2 = 0 (2y + 1)(y − 2) = 0 y=−
7(b)
7(c)
1 3
1 3
8(b)
xe−x ln xe−x ln x + ln e−x ln x + (−x) ln x
= 2.46 = ln 2.46 = ln 2.46 = ln 2.46 = x + ln 2.46
a=1✓ b = ln 2.46 ≈ 0.9 ✓
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190
A math 360 sol (unofficial) 8(c)
Ex 7.4
(xex )2 x 2 e2x x2 ln x 2 2 ln x 2 ln x
= 30e−x = 30e−x = 30e−3x = ln(30e−3x ) = ln 30 + ln e−3x = ln 30 + (−3x)
ln x
= − x + ln 30
3
1
2
2
9(d)
3x × 102x
= 4 × 20x−2
(3)x (102 )x = 4 (20)x (20)−2 (3)x (102 )
x
= 4(20)−2
(20)x
[
3
a=− ✓ 2
1
b = ln 30 2
(3)(102 ) (20)
x
]
= =
lg(15)x
= lg
x lg 15
= lg
x
5x−1 × 3x+2 = 10 (5x )(5−1 ) × (3x )(32 ) = 10 =
15x
=
x lg 15
= lg =
=
100 1 100 1 100
1 100
lg
lg 15
≈ −1.70 ✓
10 (5−1 )(32 ) 50
(5x )(3x )
x
1
(15)x
≈ 1.70 ✓ 9(a)
4 202
9 50 9
50 9
lg
lg 15
≈ 0.633 ✓ 9(b)
22x × 5x+1 =7 2 x x (2 ) × (5) (5) = 7 (4)x × (5)x
=
x
7 5 7
20
=
x lg 20
= lg
x
=
5 7 5
lg
7 5
lg 20
≈ 0.112 ✓ 9(c)
4(32x ) 4(32 )x (32 )
= ex = ex
x
ex 9 x
( ) e
9 x
= =
1 4 1 4
ln ( )
= ln
x ln
= ln
x
e 9
e
=
1 4 1 4
1 4 9 ln e
ln
≈ −1.16 ✓
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191
A math 360 sol (unofficial) 10
Ex 7.4
1st eqn 4x+3
= 32(2x+y )
(22 )x+3
= (25 )(2x+y )
22x+6
= 25+x+y
⇒ 2x + 6
=5+x+y
x
=y−1
−(1)
2nd eqn 9x + 3y = 10
−(2)
sub (1) into (2): 9y−1 + 3y = 10 (32 )y−1 32y−2
+3y
= 10
+3y
= 10
(32y )(3−2 ) +3y 1
(3y )2 ( ) +3y
= 10
9
1 9
(3y )2
= 10
+3y −10 = 0
sub u = 3y 1 2 u 9
+ u − 10
=0
u2 + 9u − 90
=0
(u + 15)(u − 6)
=0
u = −15 or
u
=6
3y = −15
3y
=6
(rej)
lg 3y = lg 6 y lg 3 = lg 6 lg 6
y
=
y
≈ 1.631 ✓
lg 3
x|y=1.631 ≈ (1.631) − 1 ≈ 0.631 ✓
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192
A math 360 sol (unofficial)
Ex 7.5
Ex 7.5 1(a)
y = log 2 x ∵ base = 2 > 1, graph slopes up
y O
2(ii)
pH = − lg(7.8 × 10−6 ) ≈ 5.11 ✓ ∴ Blue
3(i)
T = 85(0.96)x T|x=0 = 85(0.96)0 = 85°C ✓
3(ii)
T|x=15 = 85(0.96)15 ≈ 46.1 ✓
3(iii)
T = 30 85(0.96)x = 30
y = log 6 x ∵ base = 6 > 1, graph slopes up y O
x
1
✓ 1(c)
pH = −lg(5.6 × 10−7 ) ≈ 6.25 ✓ ∴ Pink
x
1
✓ 1(b)
2(i)
y = log 0.2 x ∵ 0 < base 0.2 < 1, graph slopes down y O
x
1
✓ 1(d)
6
(0.96)x
=
x lg 0.96
= lg
x
=
x
≈ 25.5 ✓
17 6 17
lg
6 17
lg 0.96
4(i)
N = 100(1.65)t N|t=0 = 100(1.65)0 = 100 ✓
4(ii)
N|t=4 = 100(1.65)4 ≈ 741 ✓
4(iii)
N 100(1.65)t (1.65)t t lg(1.65)
= 400 = 400 =4 = lg 4
t
=
t
≈ 2.77 ✓
y = log 1 x 4 1
∵ 0 < base < 1, graph slopes down 4
y O
1
x ✓
1(e)
y = lg x ∵ base = 10 > 1, graph slopes up y O
1
x
5(i)
R = 100e−0.000427 9t R|t=10 000 = 100e−0.0004279(10 000) = 100(1 − e−4.279 ) ≈ 1.39g ✓
5(ii)
R
✓ 1(f)
y = ln x ∵ base = e > 1, graph slopes up
y
lg 4 lg(1.65)
1
= (100) 2
100e(−0.0004279)t = 50 O
1
x ✓
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1
e−(0.0004279)t
=
−0.0004279t
= ln
t
≈ 1620 yrs ✓
2 1 2
193
A math 360 sol (unofficial) 6(i)
6(ii)
Ex 7.5
I = 0.87T I|T=1.5 = (0.87)1.5 ≈ 0.811 ✓
9(ii)
I = 0.5 T 0.87 = 0.5 T lg 0.87 = lg 0.5
≈
ln(t + 1)
≈ ln e 6
⇒t+1
≈ e6
t t
≈ e6 − 1 ≈ 7.72
6
13
lg 0.5
13
=
T
≈ 4.98 mm ✓
lg 0.87
∴t= 8✓
y
O
13
ln(t + 1)
13
T
7(i)
S ≈ 62 75 − 6 ln(t + 1) ≈ 62 6 ln(t + 1) ≈ 13
𝑦 = ln 𝑥 𝑦 = lg 𝑥 x
1
9(iii)
S = 75 − 6 ln(t + 1) 6 ln(t + 1) = 75 − S 1
ln(t + 1) = (75 − S)
✓
6
1
(75−S)
7(ii) (a)
Range of values of x: ⇒x>1✓
lg x > 0
Range of values of x: ⇒01✓
ln x > lg x
8(ii)
1
10(i)
A = 5000(1.04)t A|t=5 = 5000(1.04)5 ≈ $6083 ✓
>
t
𝐴 5000
t 9(i)
1500 1+1499e−0.85(6) 1500 1+1499e−5.1
≥ 0.4(1500) 1500
1+1499e−0.85t 1500 5 3
≥ 1 + 1499e−0.85t ≤ 1499e−0.85t
2 3
≥ e−0.85t
2998
ln ( t
3 2998
)
3 ) 2998
ln(
(1.04)𝑡
−0.85
5000
)✓
S = 75 − 6 ln(t + 1) 0 ≤ t ≤ 12 S|t=4 = 75 − 6 ln(4 + 1) = 75 − 6 ln(5) = 75 − 6 ln 5 ≈ 65.3 ✓
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≥ 600 ≥ 1 + 1499e−0.85t
600
lg 1.04
A
−1✓
1+1499e−0.85t
10(ii) y
2
= log1.04 (
(75−S)
1500
y=
lg 1.6
= 5000(1.04) =
= e6
(75−S)
≈ 148 ✓
> 11.98
A
t
=
∴ t = 12 ✓ 8(iii)
= e6
y|t=6 =
A > 8000 t 5000(1.04) > 8000 (1.04)t > 1.6 t lg(1.04) > lg(1.6) t
⇒t+1
1
7(ii) (b)
8(i)
ln(t + 1) = ln e6
≥ −0.85t ≤t
8.13
≤t
t ∴t= 9✓
≥ 8.13
194
A math 360 sol (unofficial) 11
t = −2.35 ln (
Ex 7.5
T−24.5 12.4
13(ii)
)
3.5 = Ce
t|T=28.8 = −2.35 ln (
28.8−24.5 12.4
)
3.5
− 12.4
)
2h 58min before 7:29am ⇒ 4: 31am [verified]
log 4 x
=
29 60
4.3
[shown] ✓ 35
k = ln ( ) 43
1
13(iii) T = T0 + Ce−kt T0 = 24.5, C = 12.4, k ≈ 0.426: T = 24.5 + 12.4e−0.426t
2 1 2
x x
3.5
k ≈ 0.426 ✓ t = 0, T = 36.9°C (assumption of normal body temperature of 36.9°C at death in Q11) 36.9 = 24.5 + Ce−0.426k(0) 36.9 = 24.5 + C C = 12.4 ✓
≈ 2.97h ≈ 2h 58min
=1
29 60
=e
e−60k =
7: 29am: T = 28.0°C t|T=28.0 = −2.35 ln (
Ce−kt1
29
2h 29min before 7:00am ⇒ 4: 31am ✓
=4 =2✓
T−24.5
ln (
y
𝑦 = 2 log 4 𝑥 𝑦=1 x
1 2
t 14(i)
✓
12(iii) 2 log x 4 ≤ 1 ⇒0 0] C2 : y = ln x 2 [x 2 > 0] 2 C3 : y = (ln x) [x > 0] All different. C1 ≠ C2 because of the different domain C1 ≠ C3 Evident by algebriac observation ✓ C2 ≠ C3 Evident by algebraic observation
T = T0 + Ce−kt 7: 00am ⇒ T = 28.8°C, T0 , 24.5°C, t = t1 28.8 = 24.5 + Ce−kt1 4.3 = Ce−kt1 −(1) 7: 29am ⇒ T = 28.0°C, T0 , 24.5°C, t = t1 +
12.4
y
29 60
✓
C3 : y = (ln x)2 C2 : y = ln x 2
C1 : y = 2 ln x 𝑥 ✓
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195
A math 360 sol (unofficial)
Rev Ex 7 A2(b) 2 log 5 x
Rev Ex 7 A1(a) log x 27 = 1.5 x1.5 = 27 3
= 33
x2
= (33 ) = 32 =9✓
x x x
A1(b) log 2 x log 8 x log 2 x ( log 2 x (
log2 x
3
) = 12
1 64
A1(c) log 3 (x − 2) log 3 (x − 2) log 3 (x − 2) ⇒x−2
2 log 5 x
=5−(
log5 x log5 52 log5 x 2 log5 x
)
)
)
2
=5−( ) u
2
x
✓ =3 − log 3 (x + 4) 3 = log 3 3 − log 3 (x + 4) = log 3 ( =
= 52 = 25 ✓
x
1 2
=5
= √5 ✓
33
)
A3(a) log 3 (x − 19) x − 19 x x x lg x
x+4
=4 = 34 = 34 + 19 = 81 + 19 = 100 = lg 102 =2✓
33 x+4
A3(b) lg(x + 2) + 7 lg 2
(x − 2)(x + 4) = 33 x 2 + 2x − 8 = 27 2 x + 2x − 35 = 0 (x + 7)(x − 5) = 0 x = −7 or x = 5 ✓ (rej ∵ x − 2 > 0) A2(a) log 3 xy − log 3 (x − 1) log 3 xy − log 3 (x − 1) xy log 3 ⇒
=5−(
= 12
= 2±6
x = 64 or
2 log 5 x
log5 25
2u2 = 5u − 2 2 2u − 5u + 2 = 0 (2u − 1)(u − 2) = 0 1 u =2 u = or 2 log 5 x = 2 1 log 5 x =
(log 2 x)2 = 36 log 2 x = ±6 x
=5−(
2u
) = 12
log2 23
2 log 5 x
sub u = log 5 x:
= 12
log2 8 log2 x
(log2 x)2
2 3
= 5 − log x 25
x−1 xy
x−1
xy y
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= log 3 9
+ lg(2x + 1)
lg(x + 2) + lg 27
= log 3 32 + lg(2x + 1)
lg[(x + 2)27 ]
=2
+ lg(2x + 1)
lg[(x + 2)27 ]
= lg 102
+ lg(2x + 1)
lg[128(x + 2)]
= lg[100(2x + 1)]
⇒ 128(x + 2)
= 100(2x + 1)
= log 3 6x − 1 = log 3 6x 2 − log 3 3 = log 3 2x 2
32(x + 2)
= 25(2x + 1)
32x + 64
= 50x + 25
= 2x 2
−18x
= −39
= 2x 2 (x − 1) = 2x(x − 1) ✓
x
=
2
A3(c) e2x + ex − 6 = 0 (ex )2 + ex − 6 = 0 sub u = ex : u2 + u − 6 =0 (u + 3)(u − 2) = 0 u = −3 sub u = ex ex = −3 (NA ∵ ex > 0)
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13 6
✓
u=2 ex = 2 sub u = ex : x = ln 2 x ≈ 0.693 ✓ 196
A math 360 sol (unofficial) A4(i)
Rev Ex 7 A5(ii) P > 90 000 0.07n 12 000e > 90 000
log 2 x = a = 2a
x
log 8 y = b = 8b
y
x 2 y = (2a )2 (8b ) = 22a+3b ✓ x
=
2a 8b
=
2a 23b
A6
= 2a−3b ✓
=
x
5−3b 2
2a−3b = 2−1 ⇒ a − 3b = −1
5−3b 2
>
n
> 28.78
0.07
log b (xy 2 ) =m log b x + log b y 2 = m log b x +2 log b y = m −(1)
=
2n−m 5
[shown]✓ −(3)
sub (3) into (1): (
2n−m
) + 2 log b y = m
5
2 log b y
7
= ✓ 9
=m+
2 log b y
=
a|b=7 = ✓
log b y
=
Let P be the population of a town after n years from the beginning of 1990 P = 12000e0.07n
log b
4
A5(i)
n
log b x
5 − 3b − 6b = −2 9b =7
9
2
15 2
−(2)
) − 3b = −1
b
ln
15
sub (2) into (1): log b x +2(n − 3 log b x) = m log b x +2n − 6 log b x = m −5 log b x = m − 2n
sub (1) into (2): (
> ln
−(1)
= 0.5
y
0.07n
2
log b (x 3 y) =n log b x 3 + log b y = n 3 log b x + log b y = n log b y = n − 3 log b x −(2)
A4(ii) x 2 y = 32 2a+3b 2 = 25 ⇒ 2a + 3b = 5 a
>
∴ 1990 + ⌊28.78⌋ = 2018 ✓
= 22a (33b )
y
15
e0.07n
3
1990 ⇒ n = 0 2005 ⇒ n = 2005 − 1990 = 15
y x
m−2n 5
6m−2n 5 3m−n 5
✓
= log b y − log b x =
3m−n
=[ =
−
2n−m
5 5 (3m−n)−(2n−m) 5
4m−3n 5
]
✓
1
log b √xy = log b xy P|n=15
2 1
= (log b x + log b y)
0.07(15)
= 12000e ≈ 34 292 ✓
2 1 2n−m
= (
2 5 1 2m+n
= [ =
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2 5 2m+n 10
+
3m−n 5
)
]
✓
197
A math 360 sol (unofficial)
Rev Ex 7 B1(c) log 2 x 2 − log 2 (2x + 5) = 2
A7 y = 2 lg x (i)(a) y|x=4 = 2 lg 4 ≈ 1.20 ✓
log 2 ( ⇒
A7 y = 1.5 (i)(b) 2 lg x = 1.5 lg x = 0.75 log10 x = 0.75
A7 (ii)
x2
= 22
2x+5
=4
2x+5
x2
= 8x + 20
x 2 − 8x − 20
=0
(x − 10)(x + 2) = 0 x = 10 or x = −2 ✓ B1(d) log 2 x
y = 2 lg x y x
1
✓ A7 (iii)
4
x√10 = √10x 1 2
x (10 ) = 10
= log 2 22
)
2x+5
x2
= 100.75 ≈ 5.62 ✓
x x
x2
B2(i)
x 4
= 4 log x 2 1
log 2 x
= 4(
log 2 𝑥
= ±2
x
2±2 = 4 or
log2 x
)
1 4
log 2 x = p log 4 y = q
x 1
= 104−2
x
x
1
4 1
2
lg x
= −
2 lg x
= x−1
log 2 xy = log 2 x + log 2 y = log 2 x + = log 2 x +
2
= log 2 x +
1
∴y= x−1✓ 2
B1(a) log 9 (3𝑥+1 ) 9x
= 3x+1 2
32x = 3x+1 ⇒ 2x 2 =x+1 2 2x − x − 1 =0 (2x + 1)(x − 1) = 0
= = =
2
=
43x + log 2 2 43x −3 43x (22 )3x 26x ⇒ 6x x
=5 =5 =5 =8 = 23 = 23 =3
log2 x
− log 4 y
log2 4 p log2 22 p 2 p
− log 4 y − log 4 y −q ✓
2
B2(iii) log 4y = log2 4y = log2 [4(22q )] x log2 x
=
log2 [22 (22q )]
B2(iv) log 2 x = p
p
p
=
log2 (22+2q ) p
=
2+2q p
✓
log 4 y = q
y = 4q y = 22q x 2 y = (2p )2 (22q ) = 22p+2q ✓ x
1
= ✓
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1 2
y
1
8 −3
1
log4 4 2 log4 y
B2(ii) log 4 x = log 4 x − log 4 y
x = − or x = 1 ✓ B1(b) 43x + log (1) 2
log4 2 log4 y
= log 2 x +2 log 4 y =p +2q ✓
= x2
2
log4 y
2
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= 2p
198
A math 360 sol (unofficial)
Rev Ex 7
B3(a) 2x = 128(4y ) 2x = 27 (22y ) 2x = 27+2y ⇒ x = 7 + 2y −(1) ln(4x + y) = ln 40 − 2 ln 2 ln(4x + y) = ln 40 − ln 22 ln(4x + y) = ln 10 ⇒ 4x + y = 10 −(2)
B4(i)
ln 2 = a ln 5 = b 1
3
ln √10e = ln 10e 3 1
= (ln 10 + ln e) 3 1
= [ln(2 × 5) + ln e] 3 1
= [ln 2 + ln 5 +1) 3 1
= (a + b +1)
sub (1) into (2): 4(7 + 2y) + y = 10 (28 + 8y) + y = 10 28 + 9y = 10 9y = −18 y = −2
3 1
= (a + b + 1) ✓ 3
B4(ii) ln x = b−2a = =
x|y=−2 = 7 + 2(−2)
=
=3✓
2 (ln 5)−2(ln 2) 2 ln 5−ln 22 2 ln 5−ln 4 1
2 5
2
4
= ln B3 lg(1 + 2x) − lg x 2 = 1 − lg(2 + 5x) 2 (b)(i) lg(1 + 2x) − lg x = lg 10 − lg(2 + 5x) lg ⇒
1+2x
= lg
x2 1+2x
= lg
x2
5
= ln √
4
10
⇒x =
2+5x 10
√5 ✓ 2
2+5x
(2x + 1)(5x + 2) = 10x 2 10x 2 + 9x + 2 = 10x 2 9x + 2 =0 2
=− ✓
x
9
B3 3y+2 = 5y y 2 (b)(ii) (3) (3 ) = (5)y (3)y
=
(5)y 3 y
( ) 5
=
3 y
lg ( ) 5
1 32 1 9
= lg
3
y lg ( ) = lg 5
1 9 1 9
1 9 3 lg 5
lg
y
=
y
≈ 4.30 ✓
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199
A math 360 sol (unofficial) B5
Rev Ex 7
y = ln(2x + e2 ) At (0, h), h = ln[2(0) + e2 ] h = ln e2 h =2✓
B6(b) log 3 5 + log 3 (4 + x) = log 3 (10 − x) + 2 log 4 2 1
log 3 5 + log 3 (4 + x) = log 3 (10 − x) + 2 log 4 42 1
log 3 5 + log 3 (4 + x) = log 3 (10 − x) + 2 ( ) 2
y = ln(2x + e2 ) At (k, 0), 0 = ln[2(k) + e2 ] 0 = log 𝑒 (2𝑘 + 𝑒 2 ) 0
k
=
1−e2 2
✓
1
y = − eax 2
At (h, k) i.e. (2, 1−e2 2 2
= log 3 (10 − x) + 1 = log 3 (10 − x) + log 3 3 = log 3 [3(10 − x)] = 3(10 − x) = 30 − 3x = 10
x
= ✓
5 4
2
e = 2k + e 1 = 2k + e2 1 − e2 = 2k
1
log 3 5 + log 3 (4 + x) log 3 5 + log 3 (4 + x) log 3 [5(4 + x)] ⇒ 5(4 + x) 20 + 5x 8x
1−e2 2
).
1
= − ea(2)
B6(c) 102x+1 +7(10x ) = 26 x 2 x (10 ) (10) +7(10 ) = 26 x )2 x) 10(10 +7(10 −26 = 0 sub u = 10x : 10u2 +7u −26 =0 (10u − 13)(u + 2) =0 13 u = −2 u= or 10 10x = −2 (rej) 13 x 10 = 10
2
1
x
2
2 1 2 e 2 1 ln ( e2 ) 2 1 1 2
sub x = lg a:
e2a
=
2a
=
a
= ln ( e ) 2 1
2 1
2 1
2
= lg
13
1
− e2 = − e2a
lg a = lg a
=
13 10
10
13 10
✓
= (ln + ln e2 ) = (ln 2−1 + 2) 2 1
= (− ln 2 + 2) 2
1
= 1 − ln 2 ✓ 2
B6 m = 32e−0.02t (a)(i) m|t=20 = 32e−0.02(20) ≈ 21.5 ✓ 1 B6 m = m|t=0 2 (a)(ii) 1 −0.02t 32e = [32e−0.02(0) ] 2
32e−0.02t = 16 1
e−0.02t
=
−0.02t
= ln
t
= −50 ln
t
≈ 34.7 ✓
2 1 2
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1 2
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200
A math 360 sol (unofficial) B7(i)
x 0.2 0.4 0.6 0.8 1.0 1.2 y -1.61 -0.92 -0.51 -0.22 0.00 0.18
Rev Ex 7 B7(ii) xe2x
= √e3 3
xe2x
= e2
ln(xe2x )
=
ln x + 2x = ln x B7(iii)
3 2
3 2 3
= −2x + ✓
y = −2x +
2
3 2
B7(iv) Intersection pt (0.84,0.16) ⇒ x = 0.84 ✓ There is only 1 intersection (and thus only 1 solution) as both curves would not turn around. ✓
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201
A math 360 sol (unofficial) 1(b)
Ex 8.1 1(a)
Ex 8.1
Method 1 (No action) xy = 3x −5 Y = 3X −5 ⇒m=3✓ ⇒ c = −5 ✓
x
Y = X −2 ⇒m=1✓ ⇒ c = −2 ✓ Method 2 (Divide by 𝐱 𝟐 ) y = x 2 −2x
Method 2 (Divide by x) xy = 3x −5 −
y x2 y
5
y
=3
y
= −5 ( ) + 3
x
xy 1
x
Y = −2X +1 ⇒ m = −2 ✓ ⇒c=1✓
xy
x
Method 3 (Divide by y) y = x 2 −2x 1 =
y 3
xy
x2 y
= −1 =
= −2 ( ) + 1
5
y 3
−
x2
1
y
5y 5 3 1 1 5 y 5 3 1
Y = X−
x2
−
y
=1 +
2x y 2x y
x
= 2 ( ) +1 y
Y = 2X +1 ⇒m=2✓ ⇒c=1✓
= ( )− 5
x
1
3
1 = −
2
−
x2
Method 3 (Divide by 𝐱𝐲) xy = 3x −5
xy 1
=1
1
Y = −5X +3 ⇒ m = −5 ✓ ⇒c=3✓
5
Method 1 (Divide by x) y = x 2 −2x y = x −2
5
3
⇒m= ✓ 5
1
1(c)
⇒c=− ✓
Method 1 (Multiply denominator) y
5
=
3 x−2
xy − 2y = 3 xy = 2y +3 Y = 2X +3 ⇒m=2✓ ⇒c=3✓ Method 2 (Swap y and 𝐱 − 𝟐) y
=
x − 2=
3 x−2 3 y 3
x
=
+2
x
= 3 ( ) +2
y 1 y
Y = 3X ⇒m=3✓ ⇒c=2✓
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+2
202
A math 360 sol (unofficial) 1(d)
Ex 8.1
Method 1 (Multiply x) 5
3y = 2x
−
3xy= 2x 2
−5
2
xy = x 2
−
3 2
xy = (x 2 ) − 3 2
Y = X− 3
2(a)
x
y 2 = 2x + 3 Y = 2X + 3 𝑦2
5 3 5
3
3
𝑥
𝑂
5
✓
3
2
⇒m= ✓ 3
2(b)
5
⇒c=− ✓ 3
xy = 2y 2 − 5 Y = 2X − 5 𝑥𝑦
Method 2 (Divide by x) 3y
= 2x
−
y
3( )= 2
−
x
y x y
=
2
−
3 5
1
x
= − ( 2)
+
Y
=− X
+
3 x 5
3 5
5
x2 5 3x2 2
✓ 3(a)
3 2 3
Method 1 (Divide by x) xy 2 = 2x +5y y 2 y =2 +5 x
⇒m=− ✓
y
3
2
y
= 5 ( ) +2 x
Y = 5X +2 ✓
Log both sides y = 10 × 2x lg y = lg(10 × 2x ) = lg 10 + lg 2x =1 +x lg 2 = lg 2 (x) +1 ✓ Y = (lg 2)X +1 ⇒ m = lg 2 ✓ ⇒c=1✓
Method 2 (Divide by y) xy 2 = 2x +5y x
xy = 2 ( ) +5 y
+5 ✓
Y = 2X 3(b)
Method 1 (Divide by x) 3xy= 5y −2x y
3y = 5 ( ) −2 y
1(f)
2
⇒c= ✓ 3
1(e)
𝑦2
𝑂 -5
x 5
Log both sides y = 5x 7 lg y = lg(5x 7 ) = lg 5 + lg x 7 = lg 5 +7 lg x = 7 lg x + lg 5 Y = 7X + lg 5 ⇒m=7✓ ⇒ c = lg 5 ✓
x 5 y 3 x 5
3 2
− ✓
Y = X 3
3
Method 2 (Divide by y) 3xy= 5y −2x x
−2 ( )
3x = 5 5
y
2 x
− ( )
x
=
x
=− ( ) +
3
3 y 2 x
5
3 y 2
3 5
Y =− X 3
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2
= ( ) −
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+ ✓ 3
203
A math 360 sol (unofficial) 4(i)
Ex 8.1
3
y
y
5 Non-linear eqn: (c)(ii)
= 5x 2 3 2
lg y = lg (5x ) lg y = lg 5 + lg x
1
= − (x + y) +2
x
2
y
3 2
y x
3
lg y = lg 5 + lg x 2
1
lg 5 𝑂 5(a) (i)
3 lg x + lg 5 2 lg 𝑥
(0,1) (4,9)
Gradient:
m=
Y-intercept:
c=1
1−9 0−4
−8
=
2
1
1
x
2
1
−4
2+x 2x
2
) =
y 5 Points: (d)(i) Gradient:
✓
Points:
= − x +2
2
y( lg y =
2
1
+ y
2
lg 𝑦
1
2
y ( + ) = − x +2
3
lg y = lg x + lg 5 [shown] ✓ 4(ii)
1
= − x − y +2
x
2 x(4−x) x+2
✓
(0, −2) (4,1) m=
Y − intercept: Linear eqn:
=2
=
4−x
(−2)−(1) (0)−(4)
=
−3 −4
=
3 4
c = −2 Y = mX +c 3
= ( ) X + (−2) 4
3
= X −2 ✓ 4
Linear eqn:
5(a) (ii)
Non-linear eqn:
Y = mX + c = 2X + 1 ✓
Y − intercept: Linear eqn:
5 Non-linear eqn: (b)(ii) 5 Points: (c)(i) Gradient: Y-intercept: Linear eqn:
x
A(−4,5) (0,1) (5)−(1)
m = (−4)−(0) =
4 −4
1
y = −x 3 + ✓
Point:
(1,3) or (2,2)
Gradient:
m=
Linear eqn:
Y − Y1 = m (X − X1 ) Y − (3) = (−1)[X − (1)] Y − 3 = −X +1 Y = −X +4
x
0−4
c=2 Y = mX
2 −4
3−2 1−2
=
lg y = lg (
(0,2) (4,0) =
−2x ✓
1
= −1
−1
Non-linear eqn: lg y = − lg x +4 lg y = − lg x + lg 104
xy = −x 3 + 1
2−0
4 3 2 x 4
= −1
c=1 Y = mX +c = (−1)X + 1 = −X +1 ✓
m=
3
= x −2
y=
y = 2x 2 + 1 ✓ 6(a)
5 Points: (b)(i) Gradient:
y
5 Non-linear eqn: (d)(ii)
=−
⇒y =
1 2
6(b)
+c
104
x 10 000 x
Point:
(1,3) or (3,4)
Gradient:
m=
Linear eqn:
Y − Y1
1
3−4 1−3
=
−1 −2
)
✓
=
1 2
= (− ) X +(2) 2
1
=− X 2
+2 ✓
= m (X − X1 ) 1
Y − (3) = ( ) [X − (1)] 2
Non-linear eqn:
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1
1
2 1
2 5
2 1
2 5
Y−3
= X−
Y
= X+
x+y x
= x+
x+y
=
y
=
2 2 1 2 5 x + x 2 2 1 2 3 x + x 2 2
✓
204
A math 360 sol (unofficial) 7(i)
Ex 8.1
Method 1 (non-linear to linear) Non-linear eqn: y = pqx ln y = ln(pqx ) = ln p + ln qx = ln p +(x ln q) = (ln q)x + ln p
8(i)
Points:
(1,8) or (4,2)
Gradient:
m = (1)−(4) =
Linear eqn:
Y − Y1 Y − (8) Y−8 Y (y√x)
= m (X − X1 ) = (−2)[X − (1)] = −2X +2 = −2X +10 = −2(x) + 10
y
= −2√x +
(8)−(2)
Non-linear eqn:
Linear eqn: Y = (ln q)X + ln p 8(ii)
y|x=16 = −2√16 +
Y-intercept = 2 ln p =2 p = e2 ≈ 7.4 ✓
= −2(4) + =− 8(iii)
Gradient = − ln q
=−
7(i)
2 3 2
m=−
10
√16 10 4
✓
C(9, −8) lies on graph of y√x against x
y√x = −8 y(√9) = −8 3y = −8 y
3
8
=− ✓ 3
2
9(i)
3
Y = − X +2 2
Non-linear eqn:
Points:
(−1, −3) (0,0) (1,3)
Gradient:
m = (−1)−(0) =
y-intercept:
c=0
Linear eqn:
Y = mX + c = (3)X + (0) = 3X
Non-linear eqn:
y − x = 3x 3 y = 3x 3 + x ✓
3
ln y = − x + 2 2
3 2
− x+2 3
= e2 (e−2x ) 2
✓
Equate Y(y√x):
Linear eqn: Y = mX +c
=e
√x
3
Method 2 (linear to non-linear) Point: (0,2) Y-intercept: c = 2
y
10
Equate X(x): x=9
=e 2 ≈ 0.2 ✓
Gradient:
2
= −2
3
−
q
11
6 −3
3
= (ee 2 ) (e−2 )
x
9(ii)
(−3)−(0)
−3
=3
−1
y|x=2 = 3(2)3 + (2) = 24 + 2 = 26 ✓
= pqx [given] ⇒ p = e2 ≈ 7.4 ✓ 3
⇒ q = e−2 ≈ 0.2 ✓ 7(ii)
y = pqx = 7.4(0.2)x ✓
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205
A math 360 sol (unofficial) 9(iii)
Ex 8.1
m + 2n = 18 m = 18 − 2n
11(b) y = ab1−x lg y = lg(ab1−x ) = lg a = lg a = lg a = − lg b (x) lg y = − lg b (x)
−(1)
n2 − m = 17 sub (1) into (2): n2 − (18 − 2n) = 17 n2 + 2n − 18 = 17 2 n + 2n − 35 =0 (n − 5)(n + 7) = 0 n=5 or m|n=5 = 18 − 2(5) =8 ⇒ C(8,5)
−(2)
n = −7 m|n=−7 = 18 − 2(−7) = 32 ⇒ C(32, −7) (rej ∵ graph slopes up)
11(c) aey = b2x ln(aey ) = ln(b2x ) ln a + ln ey = (2x) ln b ln a + y = 2x ln b y = 2x ln b − ln a y = 2 ln b (x) − ln a ✓ 12
Equate X (x 3 ) − coordinate: x3 = 8 x =2✓ Equate Y (y − x) − coordinate: y−x =5 y − (2)= 5 y =7✓ 10(i)
Gradient:
m=
Linear eqn:
Y − Y1 = m(X − X1 ) Y − (10) = 3 [X − (3)] Y − 10 = 3X − 9 Y = 3X + 1
Non-linear eqn:
y
1
y|x= 1 = 3 ( ) √2
√2
= = = = 11(a) x
=
y−b =
3 √2 3
×
√2 3√2
√2
2 3√2+1 2
−18 −6
=3
1 x
= 3x + x 2 ✓ 1
+( )
2
√2
+ √2
=
= 3( ) + 1
x2
y 10(ii)
⇒m=q✓ ⇒ c = lg p ✓
(3,10) or (9,28) (3)−(9)
y = p(x + 1)q lg y = lg[p(x + 1)q ] = lg p + lg(x + 1)q = q lg(x + 1) + lg p = q lg(x + 1) + lg p Plot lg y against lg(x + 1) ✓
Point:
(10)−(28)
+ lg b1−x +[(1 − x) lg b] + lg b − lg b (x) + lg a + lg b + lg ab ✓
+ +
1 2 1 2 1 2
✓
a y−b a x a
y
= +b
y
= a( ) + b ✓
x
1 x
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206
A math 360 sol (unofficial) 13(i)
Method 1 (non-linear to linear) Non-linear eqn: ax y =
Ex 8.1 14(i)
x−b
yx − yb = ax yx = yb +ax y y = (b) +a x
1
Point:
A(3,1) or B (8,3 ) 2
Gradient:
mAB =
Linear eqn:
Y − Y1
Linear eqn: Y = bX +a
Non-linear eqn:
=
1 2
= m (X − X1 ) 3
2 1
2 1
2
2
Y−1
= X−
Y
= X− 1 1
1
2 x 1
2
xy
= ( )−
2xy
= − 1 (shown) ✓ x
2
1
4y
=−
y
=− ✓
2 1 8
−(1)
x
y=x
1
2(2)y = − 1
14(iii) 2xy = 1 − 1 2
−(2)
sub (2) into (1):
Linear eqn: Y = mX +c = 3X +2
1
2x(x 2 )
= −1
2x 3
= −1
x 1 x
2x 4 =1−x 4 2x + x − 1 = 0
Non − linear eqn: y
= 3 ( ) +2 x
y−
3y
let f(x) ≡ 2x 4 + x − 1 f(−1) = 2(−1)4 + (1) − 1 =0 ∴ x = −1 ✓
=2
x 3
y (1 − ) = 2 x
y(
−5
1
14(ii) when x = 2,
Y-intercept = 2 a = 2✓
y
5 2
2
x
Method 2 (linear to non-linear) Gradient: m = 3 Y-intercept: c = 2
−
1
y
Gradient = 3 b =3
(3)−(8)
=
Y − (1) = ( ) [X − (3)]
= (b) +a
y
1 2
(1)−(3 )
x−3
)
x
y
=2 = =
2x x−3 ax x−b
⇒ a=2✓ ⇒ b=3✓ 13(ii) y =
2x x−3
x=y
−(1) −(2)
sub (2) into (1): (x)
=
2x x−3
x 2 − 3x = 2x x 2 − 5x = 0 x(x − 5) = 0 ∴ x = 0 or x = 5 ✓
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207
A math 360 sol (unofficial) 15
Ex 8.1
He connected the dots with a straight which is not necessarily the correct shape as verified by the graphing calculator
𝑦 𝑦 = 𝑥 7 − 14𝑥 5 + 49𝑥 3 − 35𝑥
𝑂
𝑥
y|x=−4 = −5044 ≠ −4 (−4, −4) does not lie on graph ✓ y|x=4 = 5044 ≠ 4 (4,4) does not lie on graph ✓
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208
A math 360 sol (unofficial)
Ex 8.2
Ex 8.2 1(i)
Linearization y = ax 2 + b x2 y
1(ii)
1 4 9 16 25 6.2 5.6 4.6 3.2 1.4
Gradient & Y-intercept (X1 , Y1 ) = (0,6.4) (X 2 , Y2 ) = (20,2.4) Gradient ≈
(6.4)−(2.4) (0)−(20) 1
≈− ✓
a
5
Y − intercept ≈ 6.4 b ≈ 6.4 ✓
2(i)
Linearization 1 y
= ax + b x 1 y
2(ii)
1
2
3
4
5
0.40 0.90 1.41 1.89 2.38
Gradient & Y-intercept (X1 , Y1 ) = (0, −0.1) (X 2 , Y2 ) = (4.5,2.15) Gradient ≈ a
(−0.1)−(2.15) (0)−(4.5)
≈ 0.50 ✓
Y − intercept ≈ −0.1 b ≈ −0.1 ✓
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209
A math 360 sol (unofficial) 3(i)
Ex 8.2
Linearization y = ax 2 + x − b y − x = ax 2 − b 1 4 9 16 25 x2 y − x 40.5 36 28.5 18 4.5
3(ii)
Gradient & Y-intercept (X1 , Y1 ) = (0,42) (X 2 , Y2 ) = (20,12) Gradient ≈
(42)−(12) (0)−(20)
≈ −1.5 ✓
a
Y − intercept ≈ 42 −b ≈ 42 b ≈ −42 ✓ 4(i)
Linearization a
+
N a
t
=4 =−
N 1
=−
N 1
b
+4
t b at b 1
+
=− ( ) +
N
a
1 t 1 N
4(ii)
b
t
4 a 4 a
1.00 0.50 0.30 0.25 0.20 0.75 0.99 1.09 1.12 1.15
Gradient & Y-intercept (X1 , Y1 ) = (0,1.25) (X 2 , Y2 ) = (0.88,0,80) Gradient ≈ −
b a
(1.25)−(0.80) (0)−(0.88)
≈ −0.51
−(1)
Y − intercept ≈ 1.25 4
≈ 1.25
a
≈ 3.2 ✓
a
−(2)
sub (2) into (1): −b 3.2
b
≈ −0.51 ≈ 1.6 ✓
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210
A math 360 sol (unofficial) 5(i)
Ex 8.2
Linearization xy = h(x + k) xy = hx + hk hk
y
=h+
y
= hk ( ) + h
x 1 x
1 x
5.00 2.50 1.67 1.25 1.00
y 5.25 3.38 2.75 2.44 2.23 5(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,1.5) (X 2 , Y2 ) = (4,4.5) Gradient ≈ hk
(1.5)−(4.5) (0)−(4)
≈ 0.75
Y − intercept ≈ 1.5 h ≈ 1.5 ✓
−(1)
−(2)
sub (2) into (1): (1.5)k ≈ 0.75 k ≈ 0.5 ✓
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211
A math 360 sol (unofficial) 6(i)
Ex 8.2
Linearization y
= ax +
b x
xy = ax 2 + b x 2 0.25 1.00 2.25 4.00 xy 7.3 6.8 6 4.8 Scale (optional) Y1 = mX1 + c c = Y1 − mX1 (7.3)−(4.8)
≈ (7.3) − [(0.25)−(4.00)] (0.25) ≈ 7.47 Domain = [0,4] X-interval =
(4)−(0) 10
= 0.4 ⇒ 0.25
X − scale: 1 cm to 0.25 units Range
= [4.8,7.47]
Y-interval = 6(ii)
(7.47)−(4.8) 12
≈ 0.23 ⇒ 0.2
Y-scale: 1 cm to 0.2 units Gradient & Y-intercept (X1 , Y1 ) = (0,7.5) (X 2 , Y2 ) = (3,5.48) Gradient = a
(7.5)−(5.48) (0)−(3)
≈ −0.67 ✓
Y − intercept ≈ 7.5 b ≈ 7.5 ✓ 6(iii) Substitution y ≈ −0.67x +
7.5 x
y|x=1.7 ≈ −0.67(1.7) +
7.5 1.7
≈ 3.3 ✓
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212
A math 360 sol (unofficial) 7(i)
Ex 8.2
Linearization y = ax 2 + bx y = ax + b x
x y x
7(ii)
1
2
3
4
1.6 3.6 5.6 7.6
Gradient & Y-intercept (X1 , Y1 ) = (0, −0.4) (X 2 , Y2 ) = (3.7,7) Gradient ≈
(−0.4)−(7) (0)−(3.7)
≈2✓
a
Y − intercept ≈ −0.4 b ≈ −0.4 ✓ 7(iii)
Intersection 𝑦 Put 𝑎 = 2, 𝑏 = −0.4 into = 𝑎𝑥 + 𝑏, 𝑦 𝑥
𝑥
= 2𝑥 − 0.4
2𝑥 2 − 0.4𝑥 = 10 2𝑥 − 0.4 𝑦
= =
𝑥
x y x
2.0
2.5
10 𝑥 10 𝑥
3.0
3.5
4 .0
5.00 4.00 3.33 2.86 2.50
Intersection Pt (2.35,4.3) ⇒ X ≈ 2.35 𝑥 ≈ 2.35 ✓
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213
A math 360 sol (unofficial) 8(i)
Ex 8.2
Linearization P
=
En R
ln P = ln
En R
ln P = ln E n − ln R ln P = n ln E − ln R ✓ ln E 1.61 2.30 2.89 3.00 3.22 3.40 ln P 0.92 2.30 3.48 3.69 4.14 4.50 Scale (optional) Y1 = mX1 + c c = Y1 − mX1 (0.92)−(4.50)
≈ (0.92) − [(1.61)−(3.40)] (1.61) ≈ −2.3 Domain = [0,3.4] X-interval =
(3.4)−(0) 10
= 0.34 ⇒ 0.25
X − scale: 1 cm to 0.25 units Range
= [−2.3,4.5]
Y-interval = 8(ii)
(4.5)−(−2.3) 12
≈ 0.57 ⇒ 0.5
Y-scale: 1 cm to 0.5 units The graph produces a straight line ⇒ true ✓
8(iii) Gradient & Y-intercept (X1 , Y1 ) = (0, −2.3) (X 2 , Y2 ) = (3.25,4.2) Gradient ≈ n
(−2.3)−(4.2) (0)−(3.25)
≈2✓
Y − intercept ≈ −2.3 − ln R ≈ −2.3 R ≈ 10.0 ✓
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214
A math 360 sol (unofficial) 9(i)
Linearization f = kλn lg f = lg[kλn ] = lg k + lg λn = lg k +n lg λ = n(lg λ) + lg k lg λ 2.40 2.67 2.81 3.08 3.18 lg f 3.08 2.81 2.67 2.40 2.30
9(ii)
Gradient & Y-intercept (X1 , Y1 ) = (0, 5.48) (X 2 , Y2 ) = (3,2.48) Gradient = n
Ex 8.2
(5.48)−(2.48) (0)−(3)
≈ −1 ✓
Y − intercept = 5.48 lg k = 5.48 k ≈ 301 995 ✓ 9(iii) Intersection kλn = 10−2.5n 𝑓 = 10−2.5𝑛 lg 𝑓 = −2.5𝑛 𝑌 = −2.5𝑛 𝑌 = 2.5 ∵ 𝑛 = −1 Intersection Pt (2.97,2.5) ⇒ X = 2.97 lg λ = 2.97 λ ≈ 933 ✓
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215
A math 360 sol (unofficial) 10(i)
Ex 8.2
Linearization y
= Cx +
D x
xy = Cx 2 + D x 2 0.04 0.16 0.36 0.64 xy 1.55 1.69 1.93 2.27 10(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,1.5) (X 2 , Y2 ) = (0.6,2.22) Gradient ≈ C
(1.5)−(2.22) (0)−(0.6)
≈ 1.2 ✓
Y − intercept ≈ 1.5 D ≈ 1.5 ✓ 10(ii) Intersection Cx 2 + D = 2 𝑌 =2 Intersection Pt (0.42, 2): X ≈ 0.42 x 2 ≈ 0.42 x ≈ ±0.65 ✓
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216
A math 360 sol (unofficial) 11(i)
Ex 8.2
Linearization y−x = kx n lg(y − x) = lg(kx n ) = lg k + lg x n = lg k + n lg x = n lg x + lg k lg x 0.30 0.48 0.60 0.70 0.78 lg(y − x) 1.00 1.26 1.45 1.59 1.71 Scale (optional) Y1 = mX1 + c c = Y1 − mX1 (1.00)−(1.71)
≈ (1.00) − [(0.30)−(0.78)] (0.30) ≈ 0.56 Domain = [0,0.78] X-interval =
0.78−0 10
= 0.078 ⇒ 0.05
X-scale: 1 cm to 0.05 units = [0.56,1.71]
Range
Y-interval =
(1.71)−(0.56) 12
≈ 0.096 ⇒
0.05 Y-scale: 1 cm to 0.05 units 11(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,0.55) (X 2 , Y2 ) = (0.68,1.56) Gradient = n
(0.55)−(1.56) (0)−(0.68)
≈ 1.5 ✓
Y − intercept ≈ 0.55 lg k ≈ 0.55 k ≈ 3.5 ✓ 11(iii) Substitution 𝑦 − 𝑥 = 3.5𝑥 1.5 −(1) 𝑦 = 𝑥 + 4.5 −(2) 𝑠𝑢𝑏 (2) 𝑖𝑛𝑡𝑜 (1): (𝑥 + 4.5) − 𝑥 = 3.5𝑥 1.5 4.5 = 3.5𝑥 1.5 9 7
𝑥
= 𝑥 1.5 ≈ 1.18
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217
A math 360 sol (unofficial) 12(i)
Ex 8.2
Linearization y = Ae−bx ln y = ln[Ae−bx ] = ln A + ln e−bx = ln A −bx = −bx + ln A x 1 2 3 4 5 6 ln y 2.59 2.29 1.99 1.69 1.39 1.10 Scale (optional) Y1 = mX1 + c c = Y1 − mX1 ≈ (2.59) − [
(2.59)−(1.10) (1)−(6)
] (1)
≈ 2.89 Domain = [0,6] X-interval =
(6)−(0) 10
= 0.6 ⇒ 0.5
X-scale: 1 cm to 0.5 units = [1.10,2.89]
Range
Y-interval =
(2.89)−(1.10) 12
≈ 0.15 ⇒ 0.1
Y-scale: 1 cm to 0.1 units 12(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,2.88) (X 2 , Y2 ) = (5.8,1.16) Gradient ≈ −b b
(2.88)−(1.16) (0.2)−(5.8)
≈ −0.307 ≈ 0.3 ✓
Y − intercept ≈ 2.88 ln A ≈ 2.88 A ≈ 18 ✓ 12(iii) Substitution y = (18)e−(0.3)x = 18e−0.3x At y = 6: 6 = 18e−0.3x 1
= e−0.3x
3 1
ln = −0.3x 3
x
=
ln
1 3
−0.3
≈ 3.66 ✓
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218
A math 360 sol (unofficial)
Ex 8.2
13(i) Linearization y=a+
b x
1
= b( ) + a 𝑥
Plot y against 1 x
1 x
2.50 0.59 0.50 0.39 0.31 0.25 0.20
Y -0.5 0.8 1.5 1.6 1.7 1.75 1.8 Scale (optional) Y1 = mX1 + c c = Y1 − mX1 (−0.5)−(1.8)
≈ (−0.5) − [(2.50)−(0.20)] (2.50) ≈2 Domain = [0,2.5] X-interval =
(2.5)−(0) (10)
= 0.25 ⇒ 0.25
X-scale: 1 cm to 0.25 units Range
= [−0.5,2]
Y-interval =
(2)−(−0.5) 12
≈ 0.21 ⇒ 0.2
Y-scale: 1 cm to 0.2 units 13 Correction (ii)(a) Abnormal reading: y = 0.8 (2nd data) ✓ Correct reading:y ≈ 1.4 ✓ 13 Gradient & Y-intercept (ii)(b) (X1 , Y1 ) = (0,2) (X 2 , Y2 ) = (2,0) (2)−(0)
gradient ≈ (0)−(2) b
≈ −1 ✓
Y − intercept = 2.0 a ≈2✓
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219
A math 360 sol (unofficial) 14(i)
Ex 8.2
Linearization T = kx n ln T = ln(kx n ) = ln k + ln x n = ln k +n ln x = n ln x + ln k ln x ln T
4.06 -1.43
4.68 -0.48
5.43 0.63
6.66 2.47
7.26 3.38
Scale (optional) Y1 = mX1 + c c = Y1 − mX1 (−1.43)−(3.38)
≈ (−1.43) − [ (4.06)−(7.26) ] (4.06) ≈ −7.53 Domain = [0,7.26] X-interval =
(7.26)−(0) 10
≈ 0.73 ⇒ 0.5
X-scale: 1 cm to 0.5 units Range
= [−7.53,3.38]
Y-interval =
(3.38)−(−7.53) 12
≈ 0.91 ⇒ 0.5
Y-scale: 1 cm to 0.5 units 14(ii) Gradient & Y-intercept (X1 , Y1 ) = (0, −7.5) (X 2 , Y2 ) = (5,0) Gradient ≈ n
(−7.5)−(0) (0)−(5)
≈ 1.5 ✓
Y − intercept ≈ −7.5 ln k ≈ −7.5 k ≈ 5.5 × 10−4 ✓ 14(iii) Substitution Put 𝑛 − 1.5, 𝑘 ≈ 5.5 × 10−4 𝑖𝑛𝑡𝑜 𝑇 = 𝑘𝑥 𝑛 , T = (5.5 × 10−4 )(x)1.5 149.6 × 106 km = 149.6 million km T|149.6 = (5.5 × 10−4 )(149.6)1.5 ≈1✓
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220
A math 360 sol (unofficial)
Ex 8.2
15(i) The graph shows an upward sloping curve instead of a line.
15(ii) Linearization Plot P against ln t P = a ln t + b [P] = a[ln t] + b ln t 0.00 0.69 1.61 2.30 2.71 3.00 3.40 P 30 42 50 65 73 77 85 Scale (optional) c = 30 Domain = [0,3.4] X-interval =
(3.4)−(0) 10
= 0.34 ⇒ 0.25
X-scale: 1 cm to 0.25 units Range
= [30,85]
Y-interval =
(85)−(30) 12
≈ 4.58 ⇒ 2.5
Y-scale: 1 cm to 2.5 units 15(ii) Gradient & Y-intercept (X1 , Y1 ) = (0.9, 44) (X 2 , Y2 ) = (3.2, 80) (44)−(80)
Gradient = (0.9)−(3.2) a
≈ 15.7 ✓
Y − intercept ≈ 30 b ≈ 30 ✓ 15(iii) Substitution P = 15.7 ln t + 30 P = 95: (95) = 15.7 ln t + 30 15.7 ln t = 65 ln t ≈ 4.15 t ≈ 63.6s ✓
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221
A math 360 sol (unofficial) 16(i)
Ex 8.2
Quadratic eqn y = ax 2 + bx y = ax + b x
Plot
y x
against x
x 2 3 4 5 6 7 y 9 11 13 15 17 19 x
Exponential eqn y = kh−x lg y = lg k + lg h−x = −x lg h + lg k = − lg h (x) + lg k Plot
y x
x lg y
against x 2 3 4 5 6 7 1.26 1.52 1.72 1.88 2.01 2.12
By inspection, Quadratic equation fits better ✓ 16(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,5) (X 2 , Y2 ) = (6.5,18) (5)−(18)
Gradient ≈ (0)−(6.5) a
≈2✓
Y − intercept = 5 b =5✓ 16(iii) Substitution y = 2x 2 + 5x y|x=4.5 = 2(4.5)2 + 5(4.5) = 63 ✓ 16(iv) No. x = 1 is outside the data range and thus the computed value is not reliable. ✓
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222
A math 360 sol (unofficial) 17
Ex 8.2
Quadratic Equation It is the only one among the 3 that has a turning point to fit the data given that increases then decreases. ✓
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223
A math 360 sol (unofficial)
Rev Ex 8 A2
Rev Ex 8 A1(a) Points:
m=
Y-intercept: Linear eqn: Non-linear eqn:
a
(0,5) (6,1)
Gradient:
5−1 0−6
=
4
=−
−6
2 3
b
=
y
=
a
2
b
b
=− X+
3
Gradient = 0.75 a − = 0.75
) =5
y
=− ( )+
Y
2xy
=5 3
b x
Linear eqn:
=5− 2x
2 b 2
3 2 3
y (1 +
a 1
b x a 1
= (− ) +
y
= − xy + 5
3
x
2
y
y+
a
=− +2
y 1
= mX + c =− X+5
2xy
y
y 1
Y
y
b
+ =2
x b
c=5 Y
Method 1 (linear) Non-linear eqn:
b
a
5 2 1+ x 3
15 3+2x
= −0.75b
Y − intercept = −0.5
✓
2
= −0.5
b
A1(b) Points: Gradient:
A(−1,1) or B(2,7) mAB =
Linear eqn:
Y − Y1 Y − (1) Y−1 Y Non-linear eqn: (lg y) lg y lg y ⇒y
1−7 (−1)−2
=
−6 (−3)
−(1)
= −4 ✓
b =2
−(2)
sub (2) into (1): a|b=−4 = −0.75(−4) =3✓
= m(X − X1 ) = 2[X − (−1)] = 2X + 2 = 2X + 3 = 2(lg x) + 3 = lg x 2 + lg 103 = lg 1000x 2 = 1000 x 2 ✓
Method 2 (non-linear) Gradient: m = 0.75 Y − intercept: c = −0.5 Linear eqn: Y = mX +c 0.75X −0.5 Non-linear eqn: 1
1
( )
= 0.75 ( ) −0.5
y
x
1
=
y
− 3 x a x
0.75
− +
x 4 y b y
+
1 y
0.75 x
−0.5
= −0.5 =2 =2
⇒ a=3✓ ⇒ b = −4 ✓ A3
y = √ax + b y 2 = ax + b Plot y 2 against x gradient = a ✓ y 2 − intercept = b ✓
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224
A math 360 sol (unofficial)
Rev Ex 8
A4(i) Linearization y = Ae−bt ln y = ln(Ae−bt ) = ln A + ln e−bt = ln A + [(−bt) ln e] = −bt + ln A = −bt + ln A t 1 2 3 4 5 ln y 2.50 1.95 1.39 0.83 0.26 A4(ii) Gradient & Y-intercept (X1 , Y1 ) = (0, 3.06) (X 2 , Y2 ) = (4.3, 0.65) Gradient = −b b
(3.06)−(0.65) (0)−(4.3)
≈ −0.56 ≈ 0.56 ✓
Y − intercept ≈ 3.06 ln A ≈ 3.06 A ≈ 21.3 ✓
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225
A math 360 sol (unofficial) A5(i)
1 u 1 v
Rev Ex 8
0.050 0.040 0.033 0.025 0.020 0.050 0.059 0.067 0.077 0.080
A5(ii) Linearization 1 u 1 u
1
1
v
f
+ =
=−
1 v
+
1 f
Scale (optional) Y1 = mX1 + c c = Y1 − mX1 (0.05)−(0.08)
≈ (0.05) − [(0.05)−(0.02)] (0.05) ≈ 0.1 Domain = [0,0.05] X-interval =
(0.05)−(0) 10
= 0.005 ⇒ 0.005
X-scale: 1 cm to 0.005 units Range
= [0.05,0.1]
Y-interval =
(0.1)−(0.05) 12
≈ 0.0042 ⇒
0.0025 Y-scale: 1 cm to 0.0025 units 2 cm to 0.005 units Gradient & Y-intercept Y − intercept = 0.10 1 f
f
= 0.10 ≈ 10 ✓
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226
A math 360 sol (unofficial) A6(i)
Rev Ex 8
Linearization xm yn lg(x m y n ) lg x m + lg y n m lg x + n lg y n lg y
= 200 = lg 200 = lg 200 = lg 200 = −m lg x + lg 200
lg y
=−
m n
lg x +
lg 200 n
lg x 0.00 0.48 0.70 0.86 lg y 1.15 0.43 0.10 -0.12 Scale (optional) c = 1.15 Domain = [0,0.86] X-interval =
(0.86)−(0) 10
= 0.086 ⇒ 0.05
X-scale: 1 cm to 0.05 units Range
= [−0.12,1.15]
Y-interval =
(1.15)−(−0.12) 12
≈ 0.11 ⇒ 0.1
Y-scale: 1 cm to 0.1 units A6(ii) Gradient & Y-intercept (X1 , Y1 ) = (0.1,1.0) (X 2 , Y2 ) = (0.78,0) (1.0)−(0)
Gradient = (0.1)−(0.78) −
m n
≈ −1.47
−(1)
Y − intercept = 1.15 lg 200 n
= 1.15 ≈2✓
n
−(2)
sub (2) into (1): m − ≈ −1.47 2
m ≈3✓ A6(iii) Substitution x 3 y 2 = 200 y = 10: x 3 (10)2 = 200 x3 =2 x 1.26 ✓
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227
A math 360 sol (unofficial) B1(i)
Rev Ex 8
Method 1 (linear) Non-linear eqn: y 1 y 1 y
= =
B1(i)
h
Method 2 (non-linear) Points: A(10,1) B(0, −4) C(r, −2) Gradient:
2x+k 2x+k
(1)−(−4) (10)−(0)
=
1 2
Y − intercept: c = −4
h 2
k
h
h
= ( )x +
Linear eqn: Y = mX +c 1
Linear eqn: 2
k
h
h
Y = X+
m =
= X −4 2
Non-linear eqn: 1
At A(10,1): 2
1 = (10) 1 =
h 20+k
+
k h
At B(0, −4): k
h k
h
−4
= (0) +
−4
=
y
=
y
=
y
=
2 x
2 x−8
−4
2 2 x−8 4 2x−16 h 2x+k
⇒ h=4✓ ⇒ k = −16 ✓
h
−4h = k
=
y
−(1)
2
=
y 1
h
h = 20 + k
1
= (x) −4
y 1
−(2)
sub (1) into (2): −4(20 + k) = k −80 − 4k =k −80 = 5k k = −16 ✓
B1(ii) C(r, −2) lies on Y = 1 X − 4.
Put k = −16 into (1): h = 20 + (−16) = 4 ✓
B2(a) y =
2
1
(−2) = (r) −4 2
1 2
r
=4✓
r
x y
=2
x px+q
= px + q ✓
B2(b) y = pq−x lg y = lg(pq−x ) = lg p + lg q−x = lg p +(−x) lg q = − lg q (x) + lg p ✓
B2(c) ey = px 2 − qx ey x
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= px − q ✓
228
A math 360 sol (unofficial) B3(i)
Rev Ex 8
Linearization axy − b = a(x 2 + bx) axy − b = ax 2 + abx axy − ax 2 = abx + b a(xy − x 2 ) = abx + b xy − x 2
= bx
+
x 1 2 3.00 xy − x Scale (optional) Y1 = mX1 + c c = Y1 − mX1
b a
2 1.00
3 -0.99
4 -3.00
(3)−(−3)
≈ (3) − [ (1)−(4) ] (1) ≈5 Domain = [0,4] (4)−(0)
X-intercept =
10
= 0.4 ⇒ 0.25
X-scale: 1 cm to 0.25 units 2 cm to 0.5 units Range
= [−3,5] (5)−(−3)
Y-intercept =
12
≈ 0.67 ⇒ 0.5
Y-scale: 1 cm to 0.5 units B3 Graphical Reading (ii)(a) x = 1.5 ⇒ X = 1.5 Pt (1.5,2.0) Y xy − x 2 (1.5)y − (1.5)2 y
=2 =2 =2 ≈ 2.83
B3(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,5) (b) (X 2 , Y2 ) = (3.5, −2) (5)−(−2)
Gradient = (0)−(3.5) ≈ −2 ✓
b
−(1)
Y − intercept = 5 b
=5
a
−(2)
sub (1) into (2): (−2) a
a
=5 = −0.4 ✓
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229
A math 360 sol (unofficial) B4(i)
Rev Ex 8
Linearization y = Ka−x lg y = lg(Ka−x ) = lg K + (lg a−x ) = lg K + (−x) lg a = − lg a (x) + lg K ✓ x 1 2 3 4 5 6 7 lg y 1.19 0.99 0.79 0.48 0.38 0.18 -0.05
B4 Graphical reading (ii)(a) Flawed reading ⇒ y = 3.0 (4th data) Correct reading Y ≈ 0.59 lg y ≈ 0.59 y ≈ 3.89 ✓ B4 Gradient & Y-intercept (ii)(b) (X1 , Y1 ) = (0,1.4) (X 2 , Y2 ) = (6.9,0) (0)−(1.4)
Gradient = (6.9)−(0) − lg a a
= −0.2 ≈ 1.58 ✓
Scale (optional) Y1 = mX1 + c c = Y1 − mX1 ≈ (1.19) − [
Y − intercept = 1.4 lg K = 1.4 K = 101.4 = 25.1 ✓
(1.19)−(−0.05) (1)−(7)
] (1)
≈ 1.38 Domain = [0,7] X-interval =
(7)−(0) 10
= 0.7 ⇒ 0.5
X-scale: 1 cm to 0.5 units B4 Graphical Reading (ii)(c) y = 10 lg y = 1 Y =1
Range
= [−0.05,1.38]
Y-interval =
(1.38)−(−0.05) 12
≈ 0.12 ⇒ 0.1
Y-scale: 1 cm to 0.1 unit
Point (2,1) ⇒X =2 x =2✓
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230
A math 360 sol (unofficial) B5(i)
Rev Ex 8
Linearization y = ln(ax 2 + b) ey = ax 2 + b x2 ey
0.04 0.16 0.36 0.64 1.00 1.38 1.62 2.02 2.58 3.30
B5(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,1.3) (X 2 , Y2 ) = (0.9,3.1) Gradient ≈ a
(1.3)−(3.1) (0)−(0.9)
≈2✓
Y − intercept ≈ 1.3 b ≈ 1.3 ✓ B5 Graphical Reading (iii)(a) y = ln 3 ey = 3 Y =3 Pt (0.85,3) X = 0.85 x ≈ ±0.92 ✓ B5 Graphical Reading (iii)(b) x = 0.1 x 2 = 0.01 X = 0.01 Pt (0.01,1.32) Y ≈ 1.32 y ≈ 0.28 ✓
Scale (optional) Y1 = mX1 + c c = Y1 − mX1 ≈ (1.38) − [ ≈ 1.3 Domain
(0.04)−(1)
] (0.04)
= [0,1]
X-interval =
(1)−(0) 10
= 0.1 ⇒ 0.1
X-scale:
1 cm to 0.1 units
Range
= [1.3,3.3]
Y-interval = Y-scale:
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(1.38)−(3.30)
(3.3)−(1.3) 12
≈ 0.17 ⇒ 0.1
1 cm to 0.1 units
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A math 360 sol (unofficial)
Rev Ex 8
B6(i) Linearization P = kc t lg p = lg(kc t ) = lg k + lg c t = lg k +t lg c = lg c (t) + lg k t 1 2 3 4 5 lg P 1.14 1.40 1.67 1.93 2.20 Scale (optional) Y1 = mX1 + c c = Y1 − mX1 ≈ (1.14) − [
(1.14)−(2.20)
] (1)
(1)−(5)
≈ 0.88 = [0,5]
Domain
X-interval =
(5)−(0) 10
= 0.5 ⇒ 0.5
X-scale:
1 cm to 0.5 units
Range
= [0.88,2.2]
Y-interval = Y-scale:
(2.2)−(0.88) 12
≈ 0.11 ⇒ 0.1
1 cm to 0.1 units
B6(ii) Gradient & Y-intercept (x1 , y1 ) = (0,0.87) (x2 , y2 ) = (3.5, 1.8) Gradient = lg c c
(0.87)−(1.8) (0)−(3.5)
= 0.27 ≈ 1.86 ✓
Y − intercept = 0.87 lg k = 0.87 k ≈ 7.41 ✓ B6(iii) Substitution P = (7.41)(1.86)t P|t=10 = (7.41)(1.86)10 ≈ 3.7 × 103 ✓
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A math 360 sol (unofficial)
Ex 9.1 1(c)
Ex 9.1 1(a)
1(b)
x -5 -4 -3 -2 -1 0 y ±3.16 ±2.83 ±2.45 ±2.00 ±1.41 0
y 2 = 4x for 0 ≤ x ≤ 5 x 0 1 2 3 y 0 ±2 ±2.83 ±3.46 Line of symmetry: y = 0
y 2 = −2x for − 5 ≤ x ≤ 0
4 ±4
5 ±4.47
Line of symmetry: y = 0
y 2 = 0.5x for 0 ≤ x ≤ 10 x y
0 1 2 3 4 0 ±0.71 ±1.00 ±1.22 ±1.41
5 6 7 8 9 10 ±1.58 ±1.73 ±1.87 ±2.00 ±2.12 ±2.24 Line of symmetry: y = 0
2(i)
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(2,4) lies on y 2 = 4ax, (4)2 = 4a(2) 16 = 8a a =2✓
233
A math 360 sol (unofficial) 2(ii)
Ex 9.1
y 2 = 4(2)x = 8x
4(i)
Points A & B At A & B, y = 2x + 4 meets y = x 2 − 4, 2x + 4 = x2 − 4 x 2 − 2x − 8 =0 (x + 2)(x − 4) = 0 x = −2 or x = 4 y|x=−2 = 2(−2) + 4 y|x=4 = 2(4) + 4 =0 = 12 ⇒ A(−2,0) ✓ ⇒ B(4,12) ✓
4(ii)
Length of AB
x 0 1 2 3 4 5 y 0 ±2.83 ±4 ±4.90 ±5.66 ±6.32
|AB| = √[(−2) − 4]2 + (0 − 12)2 = √36 + 144 = √180 = √36 × 5 = 6√5 ✓ 5(i)
Line & Curve y 2 = 2x − 1 y = mx
−(1) −(2)
sub (2) into (1): (mx)2 = 2x − 1 2 2 m x − 2x + 1 = 0
3
Discriminant For line & curve to meet at one point: b2 − 4ac =0 (−2)2 − 4m2 (1) = 0 4 − 4m2 =0 2 m −1 =0 (m + 1)(m − 1) = 0 m = −1 or m = 1 ✓
2y = x + 3 y
1
3
2
2
= x+
−(1)
y 2 = 2x + 3 −(2) sub (1) into (2): 1
3 2
2
2 3
9
2 1
4 3
2
4
( x+ ) 1 2 x 4 1 2 x 4 2
= 2x + 3
+ x+ − x−
= 2x + 3
5(ii)
=0
x − 2x − 3 = 0 (x + 1)(x − 3) = 0 x = −1 or x = 3 1
3
2
2
y|x=−1 = (−1) + =1 ⇒ (−1,1) ✓
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1
3
2
2
y|x=3 = (3) + =3 ⇒ (3,3) ✓
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Point (𝐚𝐭 𝐦 = −𝟏) y 2 = 2x − 1 −(1) y = −x −(2)
Point (𝐚𝐭 𝐦 = 𝟏) y 2 = 2x − 1 −(1) y=x −(2)
sub (2) into (1): (−x)2 = 2x − 1 2 x − 2x + 1 = 0 (x − 1)2 =0 x=1 y|x=1 = −1 ⇒ (1, −1)
sub (2) into (1): x2 = 2x − 1 2 x − 2x + 1 = 0 x 2 − 2x + 1 = 0 (x − 1)2 =0 x=1 y|x=1 = 1 ⇒ (1, −1)
234
A math 360 sol (unofficial) 6(i)
y2 = x + 4
Ex 9.1 7(i)
x -4 -3 -2 -1 0 y 0 ±1 ±1.41 ±1.73 ±2
x 5 6 7 8 y ±5.48 ±6.00 ±6.48 ±6.93
x 1 2 3 4 5 6 y ±2.24 ±2.45 ±2.65 ±2.83 ±3 ±3.16
6(ii)
6(iii)
4x 2 − 13x = −5 2 4x − 12x + 9 = x + 4 (2x − 3)2 =x+4
y 2 = 6x x 0 1 2 3 4 y 0 ±2.45 ±3.46 ±4.24 ±4.90
7(ii)
y>3 ⇒ x > 1.5 ✓
7(iii)
4x 2 − 10x + 1 = 0 4x 2 − 4x + 1 = 6x (2x − 1)2 = 6x 2 (2x − 1) = y2 ⇒y = ±(2x − 1) Draw y = 2x − 1:
2
4x − 13x = −5 (2x − 3)2 = x + 4 (2x − 3)2 = y 2 y = ±(2x − 3) Draw y = 2x − 3 or y = −2x + 3 ✓ x = 0.45 or 2.8
⇒ x = 0.1 or 2.4
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A math 360 sol (unofficial) 8(i)
Ex 9.1
Points A and B
10(i) 2
At A and B, y = x + meets y = 2x − 1: x
x+
2
= 2x − 1
x
x−1−
2
=0
x
x2 − x − 2 =0 (x + 1)(x − 2) = 0 x = −1 or y|x=−1 = 2(−1) − 1 = −3 ⇒ B(−1, −3) ✓ 8(ii)
x=2 y|x=2 = 2(2) − 1 =3 ⇒ A(2,3) ✓
Points A and B At A and B, y 2 = 3x meets y = 6 − x (6 − x)2 = 3x 2 x − 12x + 36 = 3x x 2 − 15x + 36 = 0 (x − 3)(x − 12) = 0 x=3 or x = 12 y|x=3 = 6 − (3) y|x=12 = 6 − (12) =3 = −6 ⇒ A(3,3) ✓ ⇒ B(12, −6) ✓
10(ii) Area of triangle OMB 1
Area of triangle PAB P(5,0) A(2,3) B(−1, −3)
Area of △ OMB = (Area of △ OAB) 2 1 1
0 12 3 0 | 0 −6 3 0 = [0 + 36 + 0 − 0 − (−18) − 0] = ⋅ | 2 2 1
Area of △ PAB 1 5 = | 2 0
2 3
−1 −3
5 | 0
4 1
= (54) 4
1
1
= [15 + (−6) + 0 − 0 − (−3) − (−15)]
= 13 unit 2 ✓
2
2
1
= [9 − (−18)]
11(i)
2
= 9(i)
27 2
unit 2 ✓
Points A & B 1
At A & B, y = meets y = 2x + 1: x
1
= 2x + 1
x
1 = 2x 2 + x 2x 2 + x − 1 =0 (x + 1)(2x − 1) = 0 x = −1
1
or x =
y|x=−1 =
1 (−1)
2
y|x=1 = 2
= −1 ⇒ B(−1, −1) ✓ 9(ii)
1
Points A & B At A & B, y 2 = 12 − 2x meets y = 2 − x (2 − x)2 = 12 − 2x 2 x − 4x + 4 = 12 − 2x 2 x − 2x − 8 =0 (x + 2)(x − 4) = 0 x = −2 or x=4 y|x=−2 = 2 − (−2) y|x=4 = 2 − (4) =4 = −2 ⇒ A(−2,4) ✓ ⇒ B(4, −2) ✓
11(ii) ⊥ bisector of AB
1 ( ) 2
AB⊥ ≡ ⊥ bisector of AB
=2 1
⇒ A ( , 2) ✓ 2
Point:
Point M At M, y = 2x + 1 cuts y-axis (x = 0): y|x=0 = 1 ⇒ M(0,1)
MAB = (
(−2)+4 4+(−2)
Gradient: mAB⊥ = − =−
,
2 1
mAB 1 6 ) −6
(
) = (1,1)
2
=−
1 (4)−(−2) ((−2)−(4))
=1
Ratio 𝐀𝐌: 𝐌𝐁 1
AB⊥ :
Recall A ( , 2) M(0,1) B(−1, −1) 2
By similar triangles (using x-coordinates) 1
AM: MB = ( − 0) : [0 − (−1)]
y − y1 = m (x − x1 ) y − 1 = (1)(x − 1) y−1 =x−1 y=x✓
2
=
1 2
=1
:1 :2✓
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A math 360 sol (unofficial) 12(i)
Ex 9.1
Points A & B At A & B, y = x 2 + x − 2 meets y = 2x x2 + x − 2 = 2x 2 x −x−2 =0 (x + 1)(x − 2) = 0 x = −1 or x = 2 y|x=−1 = 2(−1) y|x=2 = 2(2) = −2 =4 ⇒ B(−1, −2) ✓ ⇒ A(2,4) ✓
13(i)
A(2, −4) lies on y = x + k: −4 = 2 + k k = −6 Point B At B, y = x − 6 meets y 2 = 8x (x − 6)2 = 8x 2 x − 12x + 36 = 8x x 2 − 20x + 36 = 0 (x − 2)(x − 18) = 0 x = 2 or x = 18 (taken) y|x=18 = (18) − 6 = 12
12(ii) Length of AB |AB| = √[(−1) − 2]2 + [(−2) − 4]2 = √9 + 36 = √45
13(ii) ⊥ bisector of AB
= √9 × 5 = 3√5 units ✓
Recall A(2, −4) B(18,12)
12(iii) ⊥ 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐟𝐫𝐨𝐦 𝐂 𝐭𝐨 𝐀𝐁 Recall A(2,4) B(−1, −2) C(−3,4) |AB| = 3√5
AB⊥ ≡ ⊥ bisector of AB Point:
F ≡ Foot of ⊥ from C to AB |CF| ≡ ⊥ distance from C to AB
MAB = (
2+18 (−4)+12 2
Gradient: mAB⊥ = − Equating Area of △ ABC: 1 2 1 2
=−
(base)(height) = △ area by shoelace formula |AB||CF|
|AB||CF| 3√5 |CF| 3√5 |CF| |CF|
−3 −1 2 | 4 −2 4 2 −3 −1 2 | =| 4 4 −2 4 8 + 6 + (−4) ] =[ −(−12) − (−4) − (−4) 1
= | 2
2 4
AB⊥ :
,
1 mAB 1 −16 ( ) −16
2
=−
) = (10,4) 1
(−4)−(12) ( (2)−(18) )
= −1
y − y1 = mAB⊥ (x − x1 ) y − (4)= (−1) [x − (10)] y − 4 = −x + 10 y = −x + 14 ✓
= 30 = = =
30 3√5 10 √5 10√5 5
= 2√5 ✓
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A math 360 sol (unofficial) 14(i)
Ex 9.1
Points A & B At A & B, y 2 = 2x − 3 meets 2x + 3y = 7 y 2 = 2x − 3 −(1)
14(iii) ⊥ distance from P to AB F ≡ Foot of ⊥ from P to AB |PF| = ⊥ distance from P to AB
2x + 3y = 7 3y = −2x + 7 y
Area of △ PAB = 1
2
7
3
3
=− x+
2 1
−(2)
sub (2) into (1): (− x + ) 3 4 2 x − 9 4 2 x − 9 2
= 2x − 3
3
28 9 46 9
x+ x+
|AB||PF|
2 1 5√13
7 2
2
(base)(height) =
49
2
2
) |PF|
=
|PF|
= 2x − 3
9 76
(
=
= =
=0
9
2x − 23x + 38 = 0 (x − 2)(2x − 19) = 0 x=2
or 2
7
3
3
y|x=2 = − (2) +
x=
15 19 2 2 19
7
3
3
y|x=19 = − ( ) + 2
=1
= −4
⇒ A(2,1)
⇒ B ( , −4)
2
35 2 35 2 35 2 35 2 14 √13 14 13
units
√13 units ✓
Point B B(b1 , b2 ) lies on y 2 = 2x: (b2 )2 = 2b1 1
= (b2 )2
b1
2
1
⇒ B ( (b2 )2 , b2 ) 2
19 2
Midpoint of AB 1
Length of AB
A(2,6) B ( (b2 )2 , b2 ) 2
19 2
|AB| = √(2 −
2
) + [1 −
1
2+ (b2 )2 6+b2 2
(−4)]2
MAB = (
325
=√
=
4
5√13 2
,
2
)
= (x, y)
4
52 ×13
=√
2
⇒x=
1 2
2+ (b2 )2 2
[shown] ✓ ⇒y =
14(ii) Area of triangle PAB
6+b2 2
2y = 6 + b2 b2 = 2y − 6 −(2) sub (2) into (1):
19
P(6,3) A(2,1) B ( , −4) 2
19
1 6 2 6 2 | Area of △ PAB = | 2 3 1 −4 3 1 57 19 = [6 + (−8) + −6− − 2
2
2
(−24)]
1 2
2+ (2y−6)2
x
=
x
= 1 + (2y − 6)2
x
= 1 + (4y 2 − 24y + 36)
2 1 4 1 4
x = 1 + (y 2 − 6y + 9) x = y 2 − 6y + 10 0 = y 2 − 6y − x + 10 y 2 − 6y − x + 10 = 0 ✓
1
= (35) 2
1
= 17 unit 2 ✓ 2
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238
A math 360 sol (unofficial) 16(i)
Ex 9.1
Point P P(a, b) lies in y 2 = 8x: b2 = 8a b = √8a or b = −√8a (rej ∵ b > 0)
17(ii) Line AC Point:
⇒ P(a, √8a)
A(0,2at) or C(at 2 , −2at)
Gradient:
mAC =
AC:
y − y1
(2at)−(−2at)
=
(0)−(at2 )
4at −at2
=−
4 t
= m(x − x1 ) 4
y − (2at) = − [x − (0)] Length of PF P(a, √8a) F(2,0)
y 2
PF = √(a − 2)2
+ (√8a)
2
4
(− x + 2at) t 16 2 x − t2 16 2 x − t2 2
= √(a2 − 4a + 4) + 8a =
= − x + 2at t
Another point of intersection At point where AC meets y 2 = 4ax
+ (√8a − 0)
= √(a − 2)2 √a2
t 4
+ 4a + 4 2)2
= √(a + = a + 2 units ✓ 16(ii) Length of PQ PQ = (x − coordinate of P) −(x − coordinate of Q) =a −(−2) =a+2 ⇒ PF = PQ [shown] ✓ 16(iii) Area of triangle PQF P(a, √8a) PQ = a + 2
2
= 4ax
16ax + 4a2 t 2
= 4ax
20ax + 4a2 t 2
=0
16x − 20at 2 x + 4a2 t 4 = 0 4x 2 − 5at 2 x + a2 t 4 =0 (4x − at 2 )(x − at 2 ) =0 x= y|
at2
or
4
at2 x= 4
4 at2
=− ( t
4
x = at 2 (taken)
) + 2at
= −at + 2at = at Another point of intersection (
at2 4
, at) ✓
1
Area of △ PQF = (base)(height) 2 1
= (PQ) (y − coordinate of P) 2 1
= (a + 2)(√8a) 2 1
= (a + 2)(√4 × 2a) 2 1
= (a + 2)2√2a 2
= (a + 2)√2a ✓ 17(i)
𝑦
D(at 2 , 2at) 𝑦 2 = 4𝑎𝑥
A(0,2at) 𝑂 B(0, −2at)
M
𝑥
C(at 2 , −2at)
A(0,2at), B(0, −2at) & C(at 2 , −2at) ✓
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239
A math 360 sol (unofficial)
Ex 9.1
17(iii) A(0,2at)
18(i)
2
2
M
1
1
l2 : y = √x ∴ l1 ≠ l2 ✓
2 2 1 C(at , −2at) ⃗⃗⃗⃗⃗⃗ = OA ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ OM +AM 2 ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ = OA + AC
⃗⃗⃗⃗⃗ − OA ⃗⃗⃗⃗⃗ ) + (OC 3 2
⃗⃗⃗⃗⃗ = OA
3
x
3
⃗⃗⃗⃗⃗ + OC 3 2
𝑦2 = 𝑥
2
0 ) + ( at ) 3 −2at 2at
2
=(
𝑦 = √𝑥
2
3 2
⃗⃗⃗⃗⃗ = OA = (
3
at 2 2
− at
✓ 19(i)
)
3
2
2
3
= √(2.2)2 + 12
3
= √5.84 ≈ 2.416609195
Method 2 Let M be (xm , ym )
MC
=
Distance in km = (2.416609195)(93 × 106 )(1.609) ≈ 361 × 106 ≈ 3.61 × 108 ✓
2 1
⇒ x − coordinates xm −xa =2 xc −xm x1 −0
⇒ y − coordinates ym −ya =2 yc −ym y1 −2at
=2
at2 −x1
−2at−y1
=2
3x1
= 2at 2 − 2x1 y1 − 2at = −4at − 2y1 2 3y1 = −2at = 2at
x1
= at 2
x1
2
2
3 2
S(−0.1,0) C(−2.3,1) |SC| = √[(−0.1) − (−2.3)]2 + [(0) − (1)]2
2
⇒ M ( at , − at) ✓
AM
y
⃗⃗⃗⃗⃗ − OA ⃗⃗⃗⃗⃗ + OC
1 3 1
18(ii)
2+1 2
⃗⃗⃗⃗⃗ = OA
l1 : y 2 = x y = ±√x
2
y1
= − at 3
2
⇒ M ( at , − at) ✓ 3
3
17(iv) Line (through M and ⊥ 𝐭𝐨 𝐀𝐂) A(0,2at) B(0, −2at) C(at 2 , −2at) D(at 2 , 2at) 2
2
3
3
Point:
M ( at 2 , − at)
Gradient:
m =− =−
Line:
1
=−
mAC 1 (
4at ) −at2
1 (2at)−(−2at) ( ) (0)−(at2 )
=−
1 −
4 t
=
t 4
= m(x − x1 )
y − y1 2
t
3
4 t
2
1
4 t
6 1
2
4
6
3
y − (− at) = [x − ( at 2 )] 2
3
y + at
= x − at 3
y
= x − at 3 − at ✓
3
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240
A math 360 sol (unofficial)
Ex 9.1
19(ii) S(−0.1,0) C(c1 , c2 ) C(c1 , c2 ) lies on x = −2.3y 2 c1 = −2.3(c2 )2 ⇒ C(−2.3(c2 )2 , c2 ) |SC| = √[(−0.1) − (−2.3(c2 )2 )]2 + (c2 )2 = √[2.3(c2 )2 − 0.1]2 + (c2 )2 = √[5.29(c2 )4 − 0.46(c2 )2 + 0.01] + (c2 )2 = √5.29(c2 )4 + (0.54)(c2 )2 + 0.01 f(c2 ) = √5.29(c2 )4 + (0.54)(c2 )2 + 0.01 f′(c2 )=
21.16(c2 )3 +(1.08)(c2 ) 2√5.29(c2 )4 +(0.54)(c2 )2 +0.01
Minimum |SC| ⇒ f ′ (c2 ) =0 21.16(c2 )3 +(1.08)(c2 ) 2√5.29(c2 )4 +(0.54)(c2 )2 +0.01 21.16(c2 )3 + (1.08)(c2 ) c2 (21.16c 2 + 1.08)
=0 =0 =0
c2 = 0 Sign Test c2 0− 0 0+ f′(c2 ) sign − 0 + ∴ f(c2 ) is minimum 1 AU = 93 million miles = 93 × 106 miles = 93 × 106 [1.609] km = 93 × 106 [1.609][103 ]m = 93 × 106 [1609] m d = |SC||c
2 =0
≈ √f(0)
× 93 × 106 [1609]
× 93 × 106 [1609]
≈ √[0.01] × 93 × 106 [1609] ≈ 1.49637 × 1010 V=
k √d
=
1.17×1010 √1.49637×1010
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≈ 95645ms −1 ✓
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241
A math 360 sol (unofficial)
Ex 9.2 2(a)
Ex 9.2 1(a)
1(b)
Centre (0,1) Radius: r = 4 Circle: (x − a)2 +(y − b)2 = r 2 (x − 0)2 + (y − 1)2 = 42 ✓
Comparing coefficients: 2g = 2 2f = −10 g=1 f = −5
𝑦
Radius: r = √f 2 + g 2 − c = √(−5)2 + 12 − 1 = 4
2 (3, −2)
Method 2 (standard form) x 2 + y 2 + 2x − 10y + 1 x 2 + 2x +y 2 − 10y (x + 1)2 − 12 +(y − 5)2 − 52 (x + 1)2 +(y − 5)2 2 [x − (−1)] +(y − 5)2
Centre: (3, −2) Radius: r = 2 Circle: (x − a)2 + (y − b)2 = r2 (x − 3)2 + [y − (−2)]2 = 22 (x − 3)2 + (y + 2)2 = 22 ✓ 1(c)
2(b) 3
(−3,4) 𝑥
𝑂
Radius: √f 2 + g 2 − c = √(−3)2 + (−2)2 − 4 = 3 Method 2 (standard form) x 2 + y 2 − 4x − 6y + 4 x 2 − 4x +y 2 − 6y (x − 2)2 − 22 +(y − 3)2 − 32 (x − 2)2 +(y − 3)2 (x − 2)2 +(y − 3)2
(−2,2) 𝑥
Centre: (−2,2) Radius: r = 2 Circle: (x − a)2 +(y − b)2 = r 2 [x − (−2)]2 +[y − (2)]2 = 22 (x + 2)2 +(y − 2)2 = 22 ✓
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c=4
Centre: (−g, −f) = (2,3)
𝑦
𝑂
Method 1 (general form) x 2 + y 2 − 4x −6y +4 = 0 x 2 + y 2 + 2gx +2fy +c = 0 Comparing coefficients: 2g = −4 2f = −6 g = −2 f = −3
Centre: (−3,4) Radius: r = 3 Circle: (x − a)2 +(y − b)2 = r 2 2 [x − (−3)] +(y − 4)2 = 32 (x + 3)2 +(y − 4)2 = 32 ✓
2
=0 = −1 = −1 = 25 = 52
Centre: (−1,5) Radius: 5
𝑦
1(d)
c=1
Centre: (−g, −f) = (−1,5)
𝑥
𝑂
Method 1 (general form) x 2 + y 2 + 2x −10y +1 = 0 x 2 + y 2 + 2gx +2fy +c = 0
=0 = −4 = −4 =9 = 32
Centre: (2,3) Radius: 3
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242
A math 360 sol (unofficial) 3
Method 1
Ex 9.2 𝑦 A(−8,0) 𝑂 𝑥 B(0, −4) 10
4
r
C(0, −16)
2
= −10 5(i)
centre (−8, −10) radius = 10
=5 = 25 = 20 = 20 ✓
Centre C C(0,0)
1
y
= x−3
−(1)
y − 2x = 0 sub (1) into (2):
−(2)
2
=0
At A(−8,0): (−8)2 + (0)2 + 2g(−8) + 2f(0) + c 64 − 16g + c −(1) At B(0, −4), (0)2 + (−4)2 + 2g(0) + 2f(−4) + c 16 − 8f + c −(2)
1
=0 =0
( x − 3) − 2x = 0 2
3
=0
− x
=3
x
= −2
2
=0 =0
1
y|x=−2 = (−2) − 3 = −4 2
⇒ P(−2, −4) Radius r r = |CP| = √[(−2) − 0]2 + [(−4) − 0]2 = √20 Circle (x − a)2 + (y − b)2
Solving (1), (2) & (3): g = 8, f = 10, c = 64 ∴ x + y + 2(8)x + 2(10)y + 64 x 2 + y 2 + 16x + 20y + 64
− x−3 2 3
At C(0, −16), (0)2 + (−16)2 + 2g(0) + 2f(−16) + c = 0 256 − 32f + c =0 −(3)
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=5
Point P on circle At point P on circle, where x − 2y = 6 meets y − 2x = 0, x − 2y = 6 2y =x−6
Circle: (x − a)2 +(y − b)2 = r2 [x − (−8)]2 +[y − (−10)]2 = 102 (x + 8)2 +(y + 10)2 = 102 ✓
2
−c
√5 − c 5−c −c c
(−4)+(−16)
2
+
g2
√12 + (−2)2 − c = 5
y − coordinate of centre
Method 2 x 2 + y 2 + 2gx + 2fy + c
=5
√f 2
x − coordinate of centre = −8
=
C(2, −1) = (−g, −f) Comparing coefficients: g = −2 f=1
(x − 0)2 + (y − 0)2 x + y2 =0 =0
5(ii)
sleightofmath.com
= r2 2
= (√20) = 20
Check point (2,6) lies on circle (2)2 + (6)2 = 20 40 = 20 [inconsistent] ∴ No, (2,6) does not lie in circle ✓
243
A math 360 sol (unofficial) 6(i)
Ex 9.2
Points A & B At A & B, 12x − 5y = −11 cuts x 2 + y 2 − 6x + 2y − 15 = 0 12x − 5y= −11 5y = 12x + 11 y
12x+11
=
−(1)
5
or
12x+11 2
2
5 144x2 +264x+121
)
5
y|x=1 = 2(1) y|x=3 = 2 ( ) 5
5
=2
=0
−6x + 2 (
) − 6x +
25
3
x=
3
−(2)
12x+11
5 24x+22
6
=
⇒ P(1,2) ✓
2
x +(
Points P & Q At P & Q, x 2 + y 2 + 4x − 6y + 3 = 0 & y = 2x meet x 2 + (2x)2 + 4x − 6(2x) + 3 = 0 x 2 + 4x 2 + 4x − 12x + 3 =0 2 5x − 8x + 3 =0 (x − 1)(5x − 3) =0 x=1
x 2 + y 2 − 6x + 2y − 15 = 0 sub (1) into (2): x 2 + y 2 − 6x + 2y − 15 x +(
7(i)
5
3 6
⇒ Q( , ) ✓ 5 5
) − 15 = 0 − 15 = 0
5
7(ii)
⊥ bisector of PQ 3 6
P(1,2) Q ( , ) 5 5
x2 +
144 2 x 25
264
+
25
x
+
PQ ⊥ ≡ ⊥ bisector of PQ
121 25
Point:
−6x +
24 5
x
+
+
234
x−
25
144
13
⇒ A (− 6(ii)
x=
24 13
12(− )+11
24 13
y−
6 13
,−
13
= ) ✓
⇒ B(
6
,
y
6 13
12( )+11 5
13
29 13 29
8(i)
43
13 43
A (−
13
,−
29
) B(
13
6
,
43
) 2
24 6 29 43 |AB| = √[(− ) − ( )] + [(− ) − ( )] 13
13
2
13
mPQ
,
6 5
2
4 8
)= ( , ) 5 5
1
= − (2)−(6 = − )
5 3 (1)−( ) 5
1 4 ( ) 5 2 ( ) 5
=−
1 2
(x − x1 )
=m
8
1
5
2
4
8 5
5
1
2
2 1
5
=− x+
=− x+2✓ 2
x 2 + y 2 − 2x + 8y − 23 = 0 x 2 + y 2 + 2gx +2fy +c =0
c = −23
Centre: (−g, −f) = (1, −4) Radius: √f 2 + g 2 − c = √(4)2 + (−1)2 − (−23)
= √36 =6✓
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1
Comparing coefficients 2g = −2 2f = 8 g = −1 f=4
13 13
13
2
2+
) ✓
13 13
Length of AB 24
y − y1
3 5
y − ( ) = (− ) [x − ( )]
y|x= 6 =
5
=−
PQ ⊥ :
=0 =0 or
13
=0 =0
25
24
y|x=−24 =
Gradient: m = −
5
169x +234x −144 (13x + 24)(13x − 6) x=−
MPQ = (
22
−15 169 2 x 25 2
1+
= √40 = √4 × 10 = 2√10
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244
A math 360 sol (unofficial) 8(ii)
Ex 9.2
Tangent AB
10
P(3, −10) 2√10 AB
⊥ bisector of AB AB⊥ ≡ ⊥ bisector of AB
C(1, −4)
Point:
Point:
P(3, −10)
Gradient: mAB = −
Points on circle A(2,3) B(−1,6)
1 mPC
=−
1 (−10)−(−4)
( (3)−(1) )
=−
1 −6 ) 2
(
1 AB⊥ :
3
y − y1
y−
3
9(i)
1
= x − 11 ✓
y
3 1
⊥ bisector of AB
9(ii)
9(iii)
(−2)+4 2
= 1✓
Centre C At C, y = 6 − 2x meets AB⊥ (x = 1) y|x=1 = 6 − 2(1) =4 ⇒ C(1,4) ✓
2 2
1 (3)−(6)
[(2)−(−1)]
1
= − −3 = 3
= m (x − x1 ) 1
9 2
2
=x−
1 2
=x+4
Radius r r = |AC| = √[2 − (−3)]2 + (3 − 1)2 = √29 Circle (x − a)2
Radius r r = |AC| = √[(−2) − 1]2 + (0 − 4)2 = √25 = 5 Circle (x − a)2 +(y − b)2 = r 2 (x − 1)2 +(y − 4)2 = (5)2 ✓
1 9
)=( , )
Centre C At C, AB⊥ (y = x + 4) meets 2x + 5y = −1 2x + 5(x + 4) = −1 2x + 5x + 20 = −1 7x = −21 x = −3 y|x=−3 = (−3) + 4 =1 ⇒ C(−3,1)
3
Points on circle A(−2,0) B(4,0)
x=
mAB
=−
2
y − (−10) = ( ) [x − (3)]
y
1
2
y − ( ) = (1) [x − ( )]
1
= x−1
y − y1
,
2
9
= mAB (x − x1 )
y + 10
(2)+(−1) (3)+(6)
Gradient: mAB⊥ = −
=
1
AB:
MAB = (
+(y − b)2
= r2
[x − (−3)]2 +(y − 1)2 = (√29) (x + 3)2 +(y − 1)2 = 29 ✓ 11(i)
2
Points A & B At A & B, x 2 + y 2 − 4x + 6y − 12 = 0 cuts x − axis (y = 0). x 2 + (0)2 − 4x + 6(0) − 12 = 0 x 2 − 4x − 12 =0 (x + 2)(x − 6) =0 x = −2 or x = 6 ⇒ A(−2,0) ⇒ B(6,0) Length of AB |AB| = 6 − (−2) = 8 ✓
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A math 360 sol (unofficial)
Ex 9.2
11(ii) Centre C x 2 + y 2 − 4x + 6y − 12 = 0 x 2 + y 2 + 2gx + 2fy + c = 0
12(ii)
𝑦 𝑦=𝑥
𝐶 6
Comparing coefficients: 2g = −4 2f = 6 g = −2 f=3
𝐴
𝑦 = −𝑥
= MAD
(2, −3) = ( 2 =
(−2)+(d1 ) (0)+(d2 ) 2
(−2)+(d1 ) 2
,
and
12(i)
2
−3 =
d1 = 6 ⇒ D(6, −6) ✓
Centre C of 𝐂𝟑 At C, AB⊥ (y = −x) cuts circle (x 2 + y 2 = 36) x 2 + (−x)2 = 36 2x 2 = 36 2 x = 18 x = ±√18 = ±√9 × 2 = ±3√2 x = 3√2 or x = −3√2 y|x=3√2 = −3√2 y|x=−3√2 = 3√2
) (0)+(d2 ) 2
d2 = −6
𝑦 𝑦=𝑥 5 5
O
5
⇒ C(3√2, −3√2)
⇒ C(−3√2, 3√2)
𝑥 Radius r of 𝐂𝟑 r = |BC|
5 Centre: A(−5, −5) or B(5,5) Radius: r = 5 C1 : C2 :
𝐶
Circle with centre (𝟎, 𝟎) & radius 6 (x − 0)2 + (y − 0)2 = 62 x2 + y2 = 36
Point D D(d1 , d2 ) C
6
𝑥
⊥ 𝐛𝐢𝐬𝐞𝐜𝐭𝐨𝐫 𝐨𝐟 𝐀𝐁 AB⊥ : y = −x
= (−g − f) = (2, −3)
Centre C
𝐵
O
2
2
= √(5 − 3√2) + [5 − (−3√2)]
[x − (−5)]2 + [y − (−5)]2 = 52 ✓ (x − 5)2 + (y − 5)2 = 52 ✓
= √(25 − 30√2 + 18) + (25 + 30√2 + 18) = √96 Circle 𝐂𝟑 C3 :
[x − (3√2)]
or [x − (3√2)]
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2 2
+[y − (−3√2)] +[y − (−3√2)]
2 2
2
= (√96) ✓ 2
= (√96) ✓
246
A math 360 sol (unofficial) 13(i)
Ex 9.2
Tangent x + 2y = −5 2y = −x − 5 y
1
5
2
2
=− x−
Normal Point:
14 C(−1,3) P
x + 2y = −5
Normal:
−1 mtan
=
−1 1 2
(− )
=2
y − y1 = mnorm (x − x1 ) [x − (−1)] y − (3)= (2) (x + 1) y−3 =2 y − 3 = 2x + 2 y = 2x + 5
Centre C x 2 + y 2 − 4x − 8y − 5 = 0 x 2 + y 2 + 2gx + 2fy + c = 0
Point P on circle 1
5
2
2
At P, tangent (y = − x − ) meets normal (y = 2x + 5) 1
5
2
2
Comparing coefficients 2g = −4 2f = −8 g = −2 f = −4
− x − = 2x + 5 5 2
x
x
=−
15 2
= −3
Centre C
1
5
2
2
y|x=−3 = − (−3) −
1
Area of △ ABC = | |
13(ii) Centre C(−1,3)
2
Radius r = |CP| = √[(−3) − (−1)]2 + [(−1) − 3]2 = √20
2 −1
5 8
2 4
2 || −1
=
1 16 + 20 + (−2) | | 2 −(−5) − 16 − 8
=
1 |15| 2
= 7.5 units 2 ✓ = r2
[x − (−1)]2 +(y − 3)2 = (√20) (x + 1)2 +(y − 3)2 = 20 ✓
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= (−g, −f) = (2,4)
Area of triangle ABC A(2, −1) B(5,8) C(2,4)
= −1 ⇒ P(−3, −1) ✓
Circle (x − a)2 +(y − b)2
=0 =0 =0
10x 2 −70x + 100 = 0 x 2 −7x +10 =0 (x − 2)(x − 5) =0 x=2 or x = 5 y|x=2 = 3(2) − 7 y|x=5 = 3(5) − 7 = −1 =8 ⇒ A(2, −1) ⇒ B(5,8)
C(−1,3) or P
Gradient: mnorm =
Point A & B At A & B, (y = 3x − 7) cuts (x 2 + y 2 − 4x − 8y − 5 = 0) x 2 + (3x − 7)2 −4x −8(3x − 7) −5 2 2 x + (9x − 42x + 49) −4x −24x + 56 −5 (10x 2 − 42x + 49) −28x + 51
2
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247
A math 360 sol (unofficial) 15
Ex 9.2
x 2 + y 2 + 4x +6y −12 = 0 x 2 + y 2 + 2gx +2fy +c = 0 Comparing coefficients 2g = 4 2f = 6 g=2 f=3
16
Line & Circle y = mx − 1 x 2 + y 2 − 4x + 3 = 0
−(1) −(2)
sub (1) into (2): x 2 + (mx − 1)2 −4x + 3 = 0 2 2 2 x +(m x − 2mx + 1) −4x + 3 = 0 (1 + m2 )x 2 + (−2m − 4)x + 4 =0
c = −12
Centre C = (−g, −f) = (−2, −3) Radius r = √f 2 + g 2 − c
Discriminant For two distinct points: b2 − 4ac (−2m − 4)2 − 4(1 + m2 )(4) (4m2 + 16m + 16) − 16(1 + m2 ) (4m2 + 16m + 16) − 16 − 16m2 −12m2 + 16m 3m2 − 4m m(3m − 4)
= √32 + 22 − (−12) = √25 = 5 𝑦
(−3, −1)
𝑂
𝑥
(−2, −3)
+
3
= √[(−2) − (−3)]2 + [(−3) − (−1)]2 =
+
+ 4
0
Distance from centre to point (−3, −1) √12
−
>0 >0 >0 >0 >0 0
h a h a
∴ two y − intercepts ✓
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A math 360 sol (unofficial) 22
Ex 9.2
Method 3 (discriminant) a(y − k)2 + h =0 a(y 2 − 2ky + k 2 ) +h = 0 ay 2 −2aky +ak 2 + h = 0 i.e. A = a, B = −2ak, C = ak 2 + h
23
9 4
Grad: mAB⊥ = AB⊥ :
−1 mAB
8
4
8
x
= =
31 8 31 18
31
y|x=31 = 2 ( ) 18
18
=
⇒ D(
31
9 31 31 18
,
9
)
31
2
31
= √[(3) − ( )] + [(1) − ( )] 18
=√
7
) = ( , 3)
2
2
9
2465 324
2
−1
=
(1)−(5)
[(3)−(4)]
=−
1
Circle (x − x1 )2 + (y − y1 )2
4
y − y1 = mAB⊥ (x − x1 ) 1
7
4
2
(x −
y − (3)= (− ) [x − ( )] 1
7
4 1
8
y−3 =− x+ y
7
Radius r r = |AD|
3+4 1+5 2
1
x
⊥ 𝐛𝐢𝐬𝐞𝐜𝐭𝐨𝐫 𝐨𝐟 𝐀𝐁 AB⊥ ≡ ⊥ bisector of AB ,
7
4
− x + 3 = 2x
Points A(3,1) B(4,5) C(−1,3)
Point: MAB = (
1
At D, AB⊥ (y = − x + 3 ) meets AC⊥ (y = 2x).
Discriminant B 2 − 4AC = (−2ak)2 − 4(a)(ak 2 + h) = 4a2 k 2 − 4a2 k 2 − 4ah = −4ah >0 ∵ a > 0, h < 0 ∴ two y-intercepts 23
Centre D
=− x+3 4
31 2
) + (y −
18
31 2 9
)
= r2 2
= (√
2465 324
) ✓
7 8
⊥ 𝐛𝐢𝐬𝐞𝐜𝐭𝐨𝐫 𝐨𝐟 𝐀𝐂 AC⊥ ≡ ⊥ bisector of AC Point: MAC = ( Grad: mAC⊥ = AC⊥ :
3+(−1) 1+3 2 −1 mAC
,
=
2
) = (1,2)
−1 (1)−(3) [(3)−(−1)]
=2
y − y2 = mAC⊥ (x − x2 ) y − (2)= (2) (x − 1) y − 2 = 2x − 2 y = 2x
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A math 360 sol (unofficial)
Rev Ex 9 A2
Rev Ex 9 A1(i)
Line AC A(4,4) C(6,0) Point: A(4,4) or C(6,0) (4)−(0)
Gradient: mAC = (4)−(6) =
4 −2
−(1) −(2)
sub (1) into (2): (mx + 1)2 =x 2 2 m x + 2mx + 1 =x 2 2 m x + (2m − 1)x + 1 = 0
= −2
y − y1 = m (x − x1 ) y − (0)= (−2)[x − (6)] y = −2x + 12
AC:
Line & Curve y = mx + 1, m > 0 y2 = x
Point B At B, AC meets y 2 = 4x (−2x + 12)2 = 4x 2 4x − 48x + 144 = 4x 4x 2 − 52x + 144 = 0 x 2 − 13x + 36 =0 (x − 4)(x − 9) =0 x=4 or x = 9 (taken) y|x=9 = −2(9) + 12 = −6 ⇒ B(9, −6) ✓
Discriminant For two distinct points: b2 − 4ac (2m − 1)2 − 4(m2 )(1) (4m2 − 4m + 1) − 4m2 −4m + 1 −4m
>0 >0 >0 >0 > −1
m
<
1 4
Combine inequalities: m > 0 and m <
1 4
1
⇒ 0 < m < [shown] ✓ 4
A1(ii) A(4,4) B(9, −6) C(6,0) |AC| = √(4 − 6)2 + (4 − 0)2 =
√(−2)2
+
A3(i)
𝑦 𝑃(10,18)
42
= √4 + 16
13
= √20 = √4 × 5 = 2√5 units ✓
13
𝑥
Radius: r = 13 Point: (10,18) y − coordinate of centre C = 13
|CB| = √(6 − 9)2 + [0 − (−6)]2 C1 : (x − a)2 + (y − b)2 = r2 2 2 (x − a) + (y − 13) = 132 [(10) − a]2 + [(18) − 13]2 = 169 (10 − a)2 + 52 = 169 2 (10 − a) = 144 a2 − 20a + 100 = 144 2 a − 20 − 44 =0 (a + 2)(a − 22) =0 a = −2 or a = 22 ⇒ C(−2,13) ⇒ C(22,13)
= √(−3)2 + (−6)2 = √9 + 36 = √45 = √9 × 5 = 3√5 ✓ A1(iii) AC: CB
= 2√5: 3√5 = 2: 3 ✓
∴ [x − (−2)]2 + (y − 13)2 = 132 ✓ or (x − 22)2 + (y − 13)2 = 132 ✓
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A math 360 sol (unofficial)
Rev Ex 9
A3(ii) Centre: (−2, −13) or (22, −13) Radius: r = 13 [x − (−2)]2 + [y − (−13)]2 = 132 ✓ C2 : or (x − 22)2 + [y − (−13)]2 = 132 ✓
A4(iii) Radius 1
r = |AB| 2
1
= √[(−2) − 6]2 + [4 − (−2)]2 2 1
A4(i)
= √100 = 5 2
Points on circle A(−2,4) B(6, −2)
Circle (x − x1 )2 + (y − y1 )2 = r 2 [x − (2)]2 + [y − (1)]2 = (5)2 (x − 2)2 + (y − 1)2 = 25
Centre C C = MAB = (
(−2)+6 4+(−2) 2
,
) = (2,1) ✓
2
A4(ii) Line DE Point: C(2,1) Gradient: ∵ DE ⊥ AB, mDE = DE:
−1 mAB
Points D & E =
−1 (4)−(−2)
[(−2)−(6)]
=
−1 [
6 ] −8
=
4 3
y − y1 = mDE (x − x1 ) y − (1)=
4 3 4
[x − (2)]
y−1 = x− y
3 4
3 5
3
3
5
3
3
+ [( x − ) − 1]
(x − 2)2
+( x− )
25 2 x 9 25 2 x 9 2
= x− ✓
4
(x − 2)2
− −
100 9 100 9
3 3 16 2 64 +( x − x 9 9 100
x+ x−
= 25 = 25
+
64 9
)= 25 = 25 =0
9
=0 =0 or x = 5
4
5
3
3
y|x=−1 = (−1) − = −3 ⇒ E(−1, −3) ✓
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3
9 125
x − 4x − 5 (x + 1)(x − 5) x = −1
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3 2
8 2
4
(x 2 − 4x + 4)
8
4
At D & E, circle meets DE (y = x − )
4
5
3
3
y|x=5 = (5) − =5 ⇒ D(5,5) ✓
253
A math 360 sol (unofficial) A5(i)
Rev Ex 9
Circle & Line x 2 + y 2 + 6x − 8y = 0 y = mx −
B1(i) −(1)
1
−(2)
3
sub (2) into (1): 1 2
1
x 2 + (mx − )
+6x − 8 (mx − ) = 0
3
2
x +
(m2 2
3
2
1
8
3 2
9 1
3 8
3
9
x − mx + ) +6x − 8mx +
(1 + m2 )x 2 − mx + (1 + m2 )x 2 + (6 −
26 3
+(6 − 8m)x +
m) x +
3
25
=0 =0 =0
9
Discriminant b2 − 4ac = (6 −
26 3
m)
B1(ii) Length of AB A(−9, −6) B(−4,4)
2
25
− 4(1 + m2 ) ( )
|AB| = √[(−9) − (−4)]2 + [(−6) − 4]2
9
= (36 − 104m + = (36 − 104m +
676 9 676 9
m2 ) −
100
2)
100
m
−
9 9
= √125
(1 + m2 ) −
100 9
= √25 × 5 = 5√5 units ✓
2
m
224
= 64m2 − 104m +
B1(iii) Area of triangle CAB A(−9, −6) B(−4,4) C(−4, −4)
9
For line to intersect circle at two distinct points: b2 − 4ac >0 224
Area of △ CAB
7
4
1 −4 −4 −9 −4 | | 2 −4 4 −6 −4 1 = [(−16) + 24 + 36 − 16 − (−36) − 24] 2 1 = (40) 2
24
3
= 20 unit 2 ✓
64m2 − 104m +
9
>0
=
576m2 − 936m + 224 > 0 72m2 − 117m + 28 >0 (24m − 7)(3m − 4) >0 +
m<
−
7
+
4
24
or m > ✓ 3
A5(ii) For line to be tangent to circle: b2 − 4ac =0 64m2 − 104m + m=
Points A & B At A & B, y 2 = −4x meets y = 2x + 12 (2x + 12)2 = −4x 2 4x + 48x + 144 = −4x 4x 2 + 52x + 144 = 0 x 2 + 13x + 36 =0 (x + 9)(x + 4) =0 x = −9 or x = −4 y|x=−9 = 2(−9) + 12 y|x=−4 = 2(−4) + 12 = −6 =4 ⇒ A(−9, −6) ✓ ⇒ B(−4,4) ✓
7 24
224 9
B1(iv) ⊥ distance from C to AB F ≡ Foot of ⊥ from C to AB |CF| = ⊥ distance from C to AB
=0
Equate area of △ CAB:
4
1
3
2 1
or m = ✓
2
A5(iii) For line to not meet circle: b2 − 4ac 5√2 ✓?
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A math 360 sol (unofficial) 21(ii) tan(2α + β)
= =
Ex 13.1
tan 2α+tan β
0° < 2α + β < 90° tan(2α + β)
1−tan 2α tan β 2 tan α ( )+tanβ 1−tan2 α 2 tan α 1−( ) tan β 1−tan2 α
5
5x
x
x2 +50
∵ tan α = , tan β =
>0
15x[x2 +25]
>0
x2 −25x2 +25
x
25+√(−25)2 −4(1)(25) 25−√(−25)2 −4(1)(25) [x− ][x− ] 2(1) 2(1)
,
tan(2α + β)
x
>0
25+√525 25−√525 [x− ][x− ] 2 2
5 2( ) x ] + ( 5x ) [ x 2 + 50 5 2 1−( ) x = 5 2( ) x ] ( 5x ) 1−[ 5 2 x 2 + 50 1−( ) x
−
+ 0
−
25−√525 2
22 A+B+C tan(A + B + C)
10 5x ) [ 2 x ]+( 2 x − 25 x + 50 x2 = 10 5x ) 1−[ 2 x ]( 2 x − 25 x + 50 x2
tan(A+B)+tan C 1−tan(A+B) tan C
⇒ tan(A + B) + tan C tan A+tan B 1−tan A tan B
+
25−√525 25+√525 2
00
+ tan C
2
or x >
25+√525 2
✓
= 180° = tan(180°) =0 =0 =0
tan A + tan B + tan C − tan A tan B tan C = 0 tan A + tan B + tan C = tan A tan B tan C [shown] ✓
10x 5x [ 2 ]+( 2 ) x − 25 x + 50 = 10x 5x ]( ) 1−[ 2 x − 25 x 2 + 50 10x 5x (x 2 − 25)(x 2 + 50) 2 − 25] + (x 2 + 50) x = × 2 10x 5x (x − 25)(x 2 + 50) ]( ) 1−[ 2 x − 25 x 2 + 50 [
=
10x(x 2 + 50) + 5x(x 2 − 25) (x 2 − 25)(x 2 + 50) − 50x 2
=
10x 3 + 500x + 5x 3 − 125x x 4 + 25x 2 + 25 − 50x 2
=
15x 3 + 375x x 4 − 25x 2 + 25
15x[x 2 + 25] = 2 x − 25x 2 + 25
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A math 360 sol (unofficial)
Ex 13.2 2(b)
Ex 13.2 1(a)
Coordinates 1
A = cos −1 (− ) for obtuse A 1
1
2
2
2
cos 2 22 ° − sin2 22 ° = cos(45°) =
√2 2
cos 𝐴 = −
1 2
2
y
✓
-1
y = √(2)2 − (−1)2 = √3 1(b)
π
π
8
8
2 sin cos
π
= sin ( ) =
1(c)
π 2 cos 2 12
√2 2
𝑦 𝑟
√3 , 2
=
cos 𝐴 =
𝑥 𝑟
=−
1 2
✓ Trigonometric ratio sin 2A = 2 sin A cos A
π
− 1 = cos ( ) 6
√3 2
= 1(d)
sin 𝐴 =
4
(cos 75° + sin 75°)2 = cos 2 75° + 2 sin 75° cos 75° + sin2 75° = (sin2 75° + cos 2 75°) +2 sin 75° cos 75° = (1) + sin 150° (1) = + sin 30° =1
−1
√3 2
= 2( )( )
✓
+
=−
2
√3 2
✓
−1 2
√3 2
cos 2A = cos 2 A − sin2 A = ( ) − ( ) 2
2
=−
1 2
✓
tan 2A =
1
2 tan A 1−tan2 A
2(
=
√3 ) −1
1−(
2
3
√3 ) −1
2
=
−2√3 1−3
= √3 ✓
= ✓ 2
1(e)
3(i)
sin2 67.5° = = = =
1(f)
2 tan 15° 1−tan2 15°
1−cos(135°)
3(
2 1−(− cos 45°)
3
2
2
1−(−
4
2
1 √3
3(ii)
✓
sin 2x
3
= ✓ 9
3(iii) 5
3
5
cos 8x = cos[2(4𝑥)] = 2 cos 2 4x −1 1 2
= 2( )
𝑥 x = √52 − 32 = 4 = , 5
=−
tan 𝐴 =
𝑦 𝑥
=
3 4
4(a)
Trigonometric ratio 3
4
24
5
5
25
sin 2A = 2 sin A cos A= 2 ( ) ( ) =
LHS = =
✓
3 2
7
5
5
25
✓
4(b)
LHS = =
2 tan A 1−tan2 A
=
3 4 3 2 1−( ) 4
2( )
=
24 7
✓
9 79
81
−1
✓
cos 2A cos A+sin A cos2 A−sin2 A cos A+sin A
=
(cos A+sin A)(cos A−sin A) cos A+sin A
=
cos A − sin A = RHS [proven] ✓
4 2
cos 2A = cos 2 A − sin2 A = ( ) − ( ) =
tan 2A =
3
1
sin 𝐴 = for acute A
𝑟
2
= ✓
2 2
3
cos 𝐴 =
=1
cos 4x = cos[2(2𝑥)] = 1 − 2 sin2 2x
✓
4
=1
= 1 − 2( )
Coordinates
𝑥
)
sin 2x
√2 ) 2
2 2+√2
sin 2x
= tan 30° =
2(a)
3 sin x cos x = 1
4(c)
LHS =
1−cos 2A sin 2A sin A cos A
=
1−(1−2 sin2 A) 2 sin A cos A
=
2 sin2 A 2 sin A cos A
= tan A = RHS [proven] ✓
1−cos 2A 1+cos 2A 2
=
1−(1−2 sin2 A) 1+(2 cos2 A−1)
=
2 sin2 A 2 cos2 A
= tan A = RHS [proven] ✓ © Daniel & Samuel A-math tuition 📞9133 9982
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356
A math 360 sol (unofficial) 4(d)
Ex 13.2
LHS = 4 sin A cos 3 A − 4 sin3 A cos A = 4 sin A cos A (cos 2 A − sin2 A) 1
6(i)
(cos 2 A − sin2 A)
= 4 ( sin 2𝐴) 2
sin A =
= 2 sin 2𝐴 cos 2𝐴 = sin 4A = RHS [proven] ✓ 5(a)
1 cos A
)(
LHS = =
1 sin A
)=
1 sin A cos A
=1 2
1 sin 2A
=
sin 2A
5(c)
LHS =
x
Trigonometric ratio cos 2A = cos 2 A − sin2 A
6(ii)
=
1
=− ✓ 2
cos 4A = 2 cos 2A − 1 1 2
1
=− ✓
= 2 (− ) − 1 2
)(
cos A sin A
)=
1 sin A cos A
=1 2
1 sin 2A
=
A sec 2 2
=
1 cos2
A 2
=
1 1+cos A 2
=
2 1+cos A
6(iii)
sin 4A = √1 −
1 2
6(iv)
=
cos A cos
21 2
= 2 cos
2
=
2 2 tan A ) 1−tan2 A
(
=
A =
1+cos A 1−cos A
=
1 2 1 1−[1−2 sin2 ( A)] 2
1+[2 cos2 ( A)−1]
1 cot 2 ( A) 2
=
2
√3 2
✓
A−1
2
1
1+cos A
2
2
1−tan2 A tan A
=
2 1 2
1+cos A
cos A = √ tan 2A
2
cos 2 4𝐴
= √1 − (− )
= 2 csc 2A = RHS [proven] ✓
LHS = 2 cot 2A =
LHS =
2
2
=√
1 2
1+(− ) 2
= cot A − tan A = RHS [proven] ✓ 5(e)
√3 2
2
RHS [proven] ✓ 5(d)
y
2
1 2
1 ) cos2 A sin A ( ) cos A
2
1
r
= (− ) − ( )
tan A
1
x
2 sin 2A
(
cos2 A
𝑥
cos A = = − , tan A = = −√3
sec2 A
=(
2 A
√3 2
= 2 csc 2A = RHS [proven] ✓ 5(b)
√3 2
x = −√(2)2 − (√3) = −1
LHS = sec A csc A =(
Coordinates obtuse A ⇒ 2𝑛𝑑 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡
1
1+cos A
2
2
or cos A = −√
1
(
rej ∵ cos A > 0 2 1
for 45° < A < 90°
)
2
=√
1 2 1 2 sin2 ( A) 2
2 cos2 ( A)
1 2
2
1
=√
4
= RHS [proven] ✓
1
= ✓ 2
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A math 360 sol (unofficial) 7(i)
Ex 13.2
Coordinates 2 tan A = 3 3π π 0 𝑓𝑜𝑟 67.5° < 𝑥 < 90° = RHS [shown] ✓
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A math 360 sol (unofficial)
Ex 13.3 4(a)
Ex 13.3 1(a)
R = √32 + 42 = 5 4 α = tan−1 ( ) ≈ 53.1° 3 ∴ y = 5 sin(θ − 53.1°)
cos θ + sin θ
R = √12 + 12 = √2 1 α = tan−1 ( ) = 45° 1
max = 5 ✓ ⇒ sin(θ − 53.1°) = 1 0° < θ < 360° −53.1 < θ − 53.1° < 306.9° θ − 53.1° = 90° θ ≈ 143.1° ✓
∴ cos θ + sin θ = √2 cos(θ − 45°) ✓ 1(b)
√3 cos θ − sin θ 2
R = √(√3) + 12 = √3 + 1 = 2
α = tan−1 ( ) = 30°
1
√3
∴ √3 cos θ − sin θ = 2 cos(θ + 30°) ✓ 1(c)
R = √32 + 42 = 5 (4) α = tan−1 [(3)] ≈ 53.1°
∴ 3 sin θ + 4 cos θ = 5 sin(θ + 53.1°) ✓ 1(d)
4(b)
2
R = √12 + (√2) = √1 + 2 = √3 α=
√2 tan−1 ( ) 1
3(i)
α≈
1
≈ 1.18 ✓
𝑦 1
90° 180° 270° 360°
𝑥
−1
I = 15 sin(120πt) − 8 cos(120πt)
Min = −√10 ✓ θ + 71.6° ≈ 180° θ ≈ 108.4° ✓
R = √152 + 82 = 17 8 α = tan−1 ( ) ≈ 0.489 96 15
∴ I = 17 sin(120πt − 0.489 96) ✓ 3(ii)
−1
R = √12 + 32 = √10 3 α = tan−1 ( ) ≈ 71.6°
Max = √10 ✓ ⇒ cos(θ + 71.6°) = 1 0° < θ < 360° 71.6° < θ + 71.6° < 431.6° θ + 71.6° ≈ 0°, 360° θ ≈ 288.4° ✓
R = √52 + 122 = 13 ✓ 12 tan−1 ( ) 5
𝑥
= 54.7°
5 sin θ − 12 cos θ = R sin(θ − α)
90° 180° 270° 360°
∴ y = √10 cos(θ + 71.6°)
∴ sin θ − √2 cos θ = √3 sin(θ − 54.7°) ✓ 2
𝑦 = sin 𝑥
y = cos θ − 3 sin θ
sin θ − √2 cos θ
𝑦 1
min = −5 ✓ ⇒ sin(θ − 53.1°) = −1 θ − 53.1° = 270° θ ≈ 323.1° ✓
3 sin θ + 4 cos θ
y = 3 sin θ − 4 cos θ
A = 17 ✓
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A math 360 sol (unofficial) 4(c)
Ex 13.3
y = 6 cos θ + 5 sin θ
√62
5(a)
3 cos x − 4 sin x = 1 R = √32 + 42 = 5 4 α = tan−1 ( ) ≈ 53.1° 3 ∴ 5 cos(x + 53.1°)= 1
52
R= + = √61 −1 (5) α = tan ≈ 39.8° 6
∴ y = √61 cos(θ − 39.8°)
cos(x + 53.1°) max = √61 ✓ ⇒ cos(θ − 39.8°) = 1 0° < θ < 360° −39.8° < θ − 39.8° < 320.1° θ − 39.8° ≈ 0° θ ≈ 39.8° ✓
𝑦
90° 180° 270° 360°
𝑥
T
A α α C
x + 53.1° ≈ α, 360 − α ≈ 78.5,281.5 x ≈ 25.3°, 228.4° ✓ 5(b)
√3 sin x − cos x = 1 2
R = √(√3) + 12 = √3 + 1 = 2
α = tan−1 ( ) = 30°
1
√3
∴ 2 sin(θ − 30°) = 1
3
sin(θ − 30°)
∴ y = 3√5 sin(θ + 63.4°)
min = −3√5 ✓ ⇒ sin(θ + 63.4°) = −1 x + 63.4° ≈ 270° x ≈ 206.6° ✓
S
0° < x < 360° 53.1° < x + 53.1° < 413.1°
−1
R = √32 + 62 = √45 = √9 × 5 = 3√5 6 α = tan−1 ( ) ≈ 63.4°
max = 3√5 ✓ ⇒ sin(θ + 63.4°) = 1 0° < θ < 360° 63.4° < θ + 63.4° < 423.4° x + 63.4° ≈ 90° x ≈ 26.6° ✓
5
1
y = 3 sin θ + 6 cos θ
1
α ≈ 78.5° ⇒ 1st or 4th quadrant
min = −√61 ✓ cos(θ − 39.8°) = −1 θ − 39.8° ≈ 180° θ ≈ 219.8° ✓ 4(d)
=
1 2
α = 30° ⇒ 1st or 2nd quadrant
𝑦 1
=
𝑦 = sin 𝑥 90° 180° 270° 360°
0° < x < 360° −30° < x − 30° < 330°
𝑥
−1
S
A
α T
α C
x − 30° = α, 180 − α = 30,150 x = 60°, 180° ✓ 5(c)
6 cos x − 2 sin x = 3.5
R = √62 + 22 = √40 = √4 × 10 = 2√10 2 α = tan−1 ( ) ≈ 18.4° 6
∴ √10 cos(x + 18.4°) = 3.5 cos(x + 18.4°)
=
3.5 2√10
α ≈ 56.4° ⇒ 1st or 4th quadrant 0° < x < 360° 18.4° < x + 18.4° < 378.4°
S T
A α α C
x + 18.4° ≈ α, 360 − α ≈ 56.4,303.6 x ≈ 38.0°, 285.2° ✓
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367
A math 360 sol (unofficial) 5(d)
Ex 13.3 6(i)
sin x + 2 cos x = √2
√12
R= + = √5 −1 (2) α = tan ≈ 63.4° 1
=√
R = √82 + 62 = 10 6 α = tan−1 ( ) ≈ 0.643 50 8
∴ G(t) = 10 cos(4t − 0.643 50) ✓
∴ √5 sin(x + 63.4°) = √2 sin(x + 63.4°)
G(t) = 8 cos 4t + 6 sin 4t
22
2
6(ii)
5
α ≈ 39.2° ⇒ 1st or 2nd quadrant 0° < x < 360° 63.4° < x + 63.4° < 423.4°
S
A
α T
α C
−1 ≤ cos(4t − 0.643 50) ≤1 −10 ≤ 10 cos(4t − 0.643 50) ≤ 10 min = −10 ✓ 10 cos(4t − 0.643 50) = −10 cos(4t − 0.643 50) = −1
x + 63.4° ≈ α, 180 − α, 360 + α ≈ 39.2,140.8, 399.2 x ≈ 77.3°, 335.8° ✓
𝑦
t >0 4t − 0.643 50 > −0.643 50
1
90° 180° 270° 360°
𝑥
−1
5(e)
4t − 0.643 50 = π t ≈ 0.946 ✓
2.1 cos x − sin x = 1.6
R = √(2.1)2 + (1)2 = 1
∴
α = tan−1 ( ) 2.1
1
1 10
√541
≈ 25.5°
7(i)
√541 cos(x + 25.5°) = 1.6 10
cos(x + 25.5°)
=
16
4
√541
α ≈ 46.5° ⇒ 1st or 4th quadrant
S
0° < x < 360° 25.5° < x + 25.5° < 385.5°
= 5(1 + 0) = 5 coulombs ✓
A α α C
T
7(ii)
16
R = √12 + ( ) = √
α = tan−1 ( 4 ) ≈ 0.245
=
√17 4
1
=
R= + ≈ 17.26 −1 ( e ) α = tan ≈ 40.9° π ∴ 17.26 cos(x − 40.9°) ≈ 2 ≈
17
4
∴ q=
e2
cos(x − 40.9°)
1 2
1
π cos x + e sin x = 2 √π2
1
q= 5e−10t (cos 60t + sin 60t) 4
x + 25.5° ≈ α, 360 − α ≈ 46.5,313.5 x ≈ 21.1°, 288.0° ✓ 5(f)
1 q = 5e−10t (cos 60t + sin 60t) 4 1 −10(0) (cos q|t=0 = 5e 0 + sin 0)
√17 5e−10t [ cos(60t 4 5√17 −10t
⇒A =
e
4 5√17 4
− 0.245)]
cos(60t − 0.245)
✓
⇒ B = 0.245 ✓ 2
7(iii)
17.26
α ≈ 61.2° ⇒ 1st or 4th quadrant 0° 0
2x + 2 > 0 x > −1 ✓
= 3x 2 + 3 ✓ > 0 for all x values ∵ x 2 ≥ 0
⇒ y increases for all real values of x
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418
A math 360 sol (unofficial) 6(b)
y = 2x 3 − 3x 2 + 6 dy dx
Ex 15.2 9(i)
dy
−2x3 +2x (x2 +1)2
4x +1)2
For increasing function: f ′ (x) > 0
y = 3x 2 + 4x − 3 = 6x + 4
4x (x2 +1)2
>0
4x x
> 0 ∵ (x 2 + 1)2 > 0 >0✓
9(ii)
For decreasing function: f ′ (x) < 0 ⇒x 0 x2 − 1 >0 (x + 1)(x − 1) > 0
dx
(x2 +1)⋅
f ′ (x) =
>0
dx
dy
x2 +1
= 6x 2 − 6x
For increasing function:
7(a)
x2 −1
f(x) =
6x + 4 < 0 2
90 x >9✓ 11(a)
𝑦
2
1 0 x+1 >0 x > −1 ✓ 8(ii)
√2x−1
= = = =
d √2x−1⋅dx(3x+4)
could be 0, not
𝑥<
−
7 3
1
7
2
3
1
dy
−(3𝑥+4)]
dx
2𝑥−1 (6𝑥−3 −3𝑥−4)
= = =
For increasing function:
=
>0 3
3
>0
=
=
2𝑥−1 (3𝑥−7)
3 (2x−1)2
dx 3x−7
∵ (2x − 1)2 > 0 for x >
(2x−1)2
3x − 7 > 0
2
x≠a
x−a
3x+4 √2x−1
2𝑥−1 3x−7
dy
1
13(i) y = x2 −2ax−2,
2x−1 [3(2𝑥−1)
and x >
⇒ 0 for x >
0
dx a2 +2
Combine inequalities: 𝑥>
7 3
⇒x>
and x >
1 2
7 3
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1 + (x−a)2 > 0 2+a2 (x−a)2
> −1
⇒ all ℝ values of a ∵ a2 + 2 > 0 and (x − a)2 > 0 for x ≠ a ✓
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420
A math 360 sol (unofficial)
Ex 15.3 4(i)
Ex 15.3 1(i)
h(t) = =
t(10−t)
r =3+
5 4
= t − t2
4(ii)
dr
−1
= 0 +2 [(1+t)2 ] ⋅ (1)
dt 5
5
2 5
4 5
2
2
−2
= (1+t)2
h′ (t) = − (2t) = − t
dr
|
5
2 3
2
42 1
h′ (2) = − (2)
= − cm s −1 ✓
−1
The balloon is deflating.
= km s 2
1(ii)
2
=−
dt t=3 5
1+(0)
= 5 cm ✓
4 2
2
r|t=0 = 3 +
4 10t−t2 5
2 1+t
5
8
✓ 5(i)
5
h′ (6) = − (6) 2
1
f(t) =
10 000 000
2
(6t 2 + 5)2
1
= − km s −1 ✓
11am: t = 60
2
2(i)
l = dl
t3 3
− 4t + 10
= t2 − 4
dt
dl
5(ii)
6(i)
v
s= + 8
ds
ds
v 40 1
|
= + 8
60 40
5
=1 h ✓ 8
= 2t 6(ii) |
dt t=2
= 2(2)
7(i)(a)
= 4cm s −1 ✓ 3(ii)
80
8
r =t +2
dr
v2
= +
dv
2
dt
10 000 000
1
dv v=60 dr
144(60)3 +120(60)
= 3.11 ≈ 3 people per min ✓
t − 4 = −4 t2 =0 t =0✓ 3(i)
[2(6t 2 + 5)][12t]
10 000 000
f ′ (60) =
= −4
dt 2
(6(60)2 + 5)2
10 000 000 144t3 +120t
=
For length is decreasing at 4 mm/s: dl
1
f ′ (t) =
=5
t −4 =5 2 t −9 =0 (t + 3)(t − 3) = 0 t = −3 (rej ∵ t ≥ 0) or t = 3 ✓ 2(ii)
10 000 000
≈ 46 ✓
For length is increasing at 5 mm/s: dt 2
1
f(60) =
C(x) =
=
t
The initial radius is 2 cm and is increasing at an increasing rate. © Daniel & Samuel A-math tuition 📞9133 9982
8
19 200 x
Annual gasoline costs if car gets 3km l−1 = C(3)
r r = t2 + 2 2 𝑂
5
The time taken to stop at v = 60km h−1 is 1 h ✓
sleightofmath.com
19 200 3
= $6400 ✓
421
A math 360 sol (unofficial) 7(i)(b) Annual gasoline costs if car gets 8km l−1 = C(8) =
Ex 15.3 7(ii)
19 200
Average rate of change in Ben’s annual gasoline costs =
8
= $2400 ✓
=
C(8)−C(3) 8 2400−6400 8−3
= −$800 per km/l ✓ 7(iii)
C ′ (x) = − C
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′ (5)
=−
19200 x2 19200 52
= −$768 per km/l ✓
422
A math 360 sol (unofficial)
Ex 15.4 1(c)
Ex 15.4 1(a)
dx dt
dx dt
= 2,
dy
|
dt x=1
=?
= 2,
dy dx dy dt
x 1
dy
x2
dt
dy
×
dx
= (4x −
1
x2 2
= 8x −
dx dt
) ×2
|
dt x=1
dx dt
dy
= 2,
|
dt x=2
dy
dx
×
dx
dt
At y = 10, x 3 + 2 = 10 x3 =8 x =2
= 8(1) −
2 12 −1
= 6 units s 1(b)
=
= 3x 2 × 2 = 6x 2
x2 dy
At x = 1,
=?
= 3x 2
dx
1
= 4x − =
|
dt y=10
y = x3 + 2 dy
y = 2x 2 +
dy
dy
✓
=?
= 6(2)2
|
dt x=2
= 24 units s −1 ✓ 1(d)
dx dt
= 2,
dy
|
dt y=2
=?
3
y = (2x−3)3 y=
= 3(2x − 3)−3
dy dy dx
= dy dt
dx
= 3[−3(2x − 3)−4 ] ⋅ 2
= = =
=
18 − (2x−3)4 dy 18 − (2x−3)4 36 − (2x−3)4
At x = 2,
= ×
dx
dy
|
dt x=2
x x+1 (x+1)(1) −x(1) = (x+1)2 x+1 −x (x+1)2 1 (x+1)2
dx dy
dt
=
dt
×2
dy
×
dx
dx dt
1
= (x+1)2 × 2 2
= (x+1)2 36
= − [2(2)−3]4
At y = 2,
= −36 units s −1 ✓
x x dy
|
dt x=−2
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x x+1
sleightofmath.com
=2 = 2x + 2 = −2 2
= (−2+1)2 = 2 units s −1 ✓
423
A math 360 sol (unofficial) 2(a)
dy dt
= 4,
dx
|
dt x=3
Ex 15.4 2(c)
=?
dt
y = x 3 − 2x 2 dy
= 4,
dy
= 3x − 4x =
dt
dy dx
=
dt
dt dx
dy
dt
4
dx
=
dt x=3
dt
= 4,
y= dy dx
=
= =
4 = dx dt
dt x=2
4
1
15
×
√2x+7
dx dt dx dt
= 4√2x + 7
dt
✓
At y = 3, √2x + 7 = 3 2x + 7 = 9 x =1
=?
1+x
=
dt
|
=
×
dx
3x2
=
dy
dx
4 3(3)2 −4(3)
⋅2
dy
4 =
3x2 −4x
At x = 3, | dy
=?
√2x+7
=
dt
dx
2(b)
2√2x+7 1
=
dx
×
4 = (3x 2 − 4x) × dx
|
dt y=3
1
=
dx
dy
dx
y = √2x + 7
2
dx
dy
=
dx
(1+x)⋅
d (3x2 ) dx
−3x2 ⋅
d (1+x) dx
|
dt x=1
= 4√2(1) + 7 = 12 unit s −1 ✓
(1+x)2
2(d)
−3x2 ⋅1
(1+x)⋅6x
dt
(1+x)2 6x+6x2
dy
dx
|
dt y=5
=?
−3x2 (1+x)2
y = x(x − 4) = x 2 − 4x
6x+3x2 (1+x)2 dy dx 6x+3x2 (1+x)2
× ×
dx
dy
dt dx
dx
dt
dy
4(1+x)2
dt
6x+3x2
4
dx
At x = 2, |
= 4,
dt x=2
=
4(1+2)2 6(2)+3(2)2
3 2
unit s −1 ✓
=
dy
×
dx
= (2x − 4) ×
dx
=
= 2x − 4
=
dt
=
dx dt dx dt
4 2x−4 2 x−2
At y = 5, x 2 − 4x =5 2 x − 4x − 5 = 0 (x − 5)(x + 1) = 0 x = 5 or x = −1 (rej) dx
|
dt x=5
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=
2 5−2
=
2 3
units s −1 ✓
424
A math 360 sol (unofficial) 3(i)
y= dy
2x−1
= = = 3(ii)
4(ii)
x+1
=
dx
Ex 15.4
dt dy
(x+1)⋅
d (2x−1) dx
−(2x−1)⋅
dx dy
d (x+1) dx
(x+1)2 (x+1)⋅2 −(2x−1)⋅1 (x+1)2 2x+2 −2x+1 (x+1)2 3 ✓ (x+1)2
y|x=0 =
−1 1
dy
|
dx x=0
=3
dt
dt dx dt
=3 =3
= −1 5(a) 3
=
⋅
dx
dx
(x + 1)2 =4 2 x + 2x + 1 =4 2 x + 2x − 3 =0 (x + 3)(x − 1) = 0 x = −3 or x = 1 ✓
⇒ A(0, −1) Gradient:
=3
dx 12 (x+1)2
Tangent Point:
dy
((0)+1)
2
=3
Let A ≡ area of circle & r ≡ radius dA dr = 2π, | =? dt dt r=6 A = πr 2 [area of circle]
Tangent: y − y1
=
dy
|
dx x=0
dA
(x − x1 )
dr
y − (−1) = 3 [x − (0)] y = 3x − 1 ✓ 3(iii)
dx dt
= 0.03,
dy dt
dA dt
=
dt
= =
dy
×
dx 3 (x+1)2 0.09 (x+1)2
dr
y= dy dx
dr
dr
×
dt dr
=
dt
1 r
dt
At r = 6,
× 0.03 5(b)
dr
|
dt r=6
=
1 6
cm s −1 ✓
Let A ≡ area of circle & r ≡ radius dA dr = 10π, | =? dt dt r=2
0.09
|
dt x=0
4(i)
dA
dx
At A(0, −1), dy
=
2π = 2πr ×
at A(0, −1) = ?
dt dy
= 2πr
= (0+1)2 = 0.09 units s −1 ✓
A = πr 2 [area of circle] dA
2x−10
dr
x+1
= = =
(x+1)⋅
d (2x−10) dx
(x+1)⋅2 2x+2
−(2x−10)⋅
dA
d (x+1) dx
dt
(x+1)2 −(2x−10)⋅1 (x+1)2 −2x+10 (x+1)2
= 2πr =
dA dr
×
10π = 2πr × dr dt
12
=
dt dr dt
5 r
At r = 2,
= (x+1)2
dr
dr
|
dt r=2
5
= = 2.5 m s −1 ✓ 2
2
> 0 [shown] ∵ (x + 1) > 0 for x ≠ −1 ✓
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425
A math 360 sol (unofficial) 6(i)
Ex 15.4
dV dx = 3, = ? dt dt
7(b)
Let A be Surface Area, V be Volume & x be length of side dA
2
dt
V = x + 3x [given] dV dx dV dx
= 2x + 3 ×
dx
3 = (2x + 3) × dx
=
dt
6(ii)
dx
3 2x+3
|
dt x=3
=
|
dt x=6
=
A=x dA dx dA dt
3
=
1 3
cm s −1
3
=
1 5
dx
5
=
× ×
cm s −1 ✓
dV
=
=
3 A 2 ( ) 2 6
⋅
dV
dA
×
dA
=
1
A
4
6
1
√
1
1
A
4
6
= √
6
dt
20
A 6 A
At x = 1, 1 = √
dV
dt dx
dt A=6
8(i)
x
At A = 4, x 2 = 4 ⇒ x = 2
dt x=2
−(2)
= √ × 0.2
dx
dt
6
1
dV
[area of square]
dA
A
A 2 ( ) 6
= 2x =
⇒x=√
3
A 6
2(6)+3
= 2x
|
= x2
6
V = (√ ) =
3
dx
dx
A
sub (2) into (1):
dt
✓
2(3)+3
10 dt
−(1)
dt dx
Let A be area & x be side dA dx = 10, | =? dt dt A=4 2
=?
A = 6x 2 ⇒
dt
7(a)
|
dt x=1
dx
dA dx
dV
V = x3
dV
=
= 0.2,
|
=
1 20
6
⇒1=
A 6
⇒A=6
6
√ = 0.05 cm3 s −1 ✓ 6
Let y ≡ Length & x ≡ Breadth dx dy = 4, =? dt dt y = 2x [given] dy =2 dx
5
= = 2.5 cm s −1 ✓ 2
dy dt
=
dy dx
⋅
dx dt
= (2) ⋅ (4) = 8 cm/s
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426
A math 360 sol (unofficial) 8(ii)
Ex 15.4
Let A ≡ Area of rectangle dx dt
= 4,
dA
|
10(ii)
dt
=?
dt x=8
dx
= 2,
da
|
dt x=4
=?
1
a = x(x + 1) A = xy −(1) y = 2x −(2) sub (2) into (1): A = x(2x) = 2x 2
= da dx
da dA
=
dt
dA
×
dx
dt
dx
At x = 8,
dA
11(i)
dp
dA
|
dt p=3
1
3
2
2
dt
=
dt
× 3
2
2
=( + p =
2)
dp dt
dA dx dA
|
dt p=3
10(i)
dt 3
4
4
dt
=? −(1)
Let a be area of triangle 1
a = (2x)(x + 1) sin 150° 2
1 2
= 8x + 1 =
dA
×
dx
dx dt
= (8x + 1) × 1.2 = 9.6x + 1.2
= + (3)2 = 7 units 2 s −1 ✓
= x (x + 1)
dA
× 0.5
1 3 2 + p 4 4
1
= 1.2,
sub (2) into (1): A = x(4x + 1) = 4x 2 + x
At p = 3, dA
= 2(4) + 1 = 9 cm2 s −1
Q(2x, y) lies on y = 2x + 1: y = 2(2x) + 1 = 4x + 1 −(2)
=?
dA dp 1
dt
×2
A = xy
= + p2
dA
dx
Let A ≡ Area of parallelogram OPQR dx
= p + p3 ✓ = 0.5,
×
dx
dt x=4
(1 + p2 ) 2
da
da
= (p) 1
2
At x = 4, |
= 16(8) = 128 cm2 s −1 ✓
|
2 1
2
2
1
= 2x + 1
dt x=8
2 1
=
1
2
1
dA
2
= (x + )
A = (height) (sum of bases)
dt
1
1
dt
= 4x × 4 = 16x
dp
2
= (2x) +
= 4x
dx
9(ii)
1
+ x
=x+
dA
9(i)
2 1 2 x 2
C
At x = 1.5,
dA
|
dt x=1.5
= 9.6(1.5) + 1.2 = 15.6 units/s
x+1 150° A 2x B
1
= x (x + 1) [shown] ✓ 2
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427
A math 360 sol (unofficial)
Ex 15.4
11(ii) Let z be length of diagonal OQ dx dz = 1.2, | =? dt dt x=1.5
13
dx dt
By distance formula,
= 0.6,
−(1)
dy
=
dx
= sub (2) into (1): = √4x 2 + (4x + 1)2
dy
=
dt
= √4x 2 + (16x 2 + 8x + 1)
=
= √20x 2 + 8x + 1 dz dx
= =
dz dt
= = =
1 2√20x2 +8x+1 20x+4
dz
√20x2 +8x+1 24x+4.8
dt
dy
× 1.2
dt x=1.5
≈ 5.36 units per sec ✓
14
dP dt
dt
z dz
dI
dx
×
dP
×
dx −x
dx dt
× 0.6
√36−x2 −0.6x √36−x2
=
= −50,
−0.6(3.6) √36−3.62
= −0.45
dz
|
dt x=144
=?
𝑥 𝑧
270
= √x 2 + 72900 =
dz dt
= = =
|
dy
1 2√x2 +72900 x
⋅ 2x
√x2 +72900
dI dt
= 800I × 2 = 2400I At I = 3,
⋅ (−2x)
√36−x2
|
= dP
2√36−x2 −x
By Pythagoras’ Theorem, z 2 = x 2 + 2702
= 800I =
𝑥
Let z be distance between helicopter & lorry dx
P = I2 R = 400I 2 ∵ R = 400
dI
𝑦
∴ The rate at which the ladder is sliding down is 0.45m s −1
dI dP = 2, | =? dt dt I=3
dP
1
dt x=3.6
√20x2 +8x+1 dz
6
dx
× 20x+4
=?
At y = 4.8, x 2 + 4.82 = 62 x2 = 12.96 x = 3.6
√20x2 +8x+1
At x = 1.5, | 12
=
⋅ (40x + 8)
dx
|
dt y=4.8
= √36 − x 2
y
Q(2x, y) lies on y = 2x + 1, y1 = 2(2x) + 1 = 4x + 1 −(2)
z
dy
By Pythagoras’ Theorem, x 2 + y 2 = 62 y2 = 36 − x 2
z = √(0 − 2x)2 + (0 − y)2 = √4x 2 + y 2
Let x be distance from wall & y be distance from ground
dt I=3
= 2400(2) = 4800 W s −1 ✓
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dz dx
× x
√x2 +72900 −50x
dt
× (−50)
√x2 +72900
At x = 144,
sleightofmath.com
dx
dz
|
dt x=144
= −23
9 17
m s −1 ✓
428
A math 360 sol (unofficial) 15(i)
Ex 15.4
Let x ≡ top of man′ s shadow from the lamp y ≡ man from the lamp By similar triangles, 2
=
7
|
1
1
1
2
3
6
≈ 11.1 kmh−1
6
y 17(a)
x
d dr
= x✓
(πr 2 ) = 2πr
𝛿𝐴 𝑟 𝛿𝑟
7
dy dt dy dx dy dt 5 3 dx dt
16(i)
h after B reaches P
dt t=−1
5
y
3
dx
2
x
1
⇒ t = (− ) + = −
7
x−y
2x = 7x − 7y 7y = 5x
15(ii)
16(ii)
5 dx
= ,
3 dt
=
LHS =?
=
=
dx 5 7 7
× ×
= ms
dx dt dx dt −1
3
= lim ✓
δr→0
= 2πr = RHS
B (0,30) at t = 0
17(b)
P A (0,0) at t = 0
+ (3600t 2 + 3600t + 900)
=
1 2√9225t2 +3600t+900 9225t+1800
d
4
( πr 3 ) = 4πr 2
dr 3
𝑟 𝛿𝑟 = =
⋅ (18450t + 3600)
√9225t2 +3600t+900
d
4
( πr 3 )
dr 3 d dr
dt
=
9225t+1800
δV δr→0 δr
= lim
= lim
√9225t2 +3600t+900
−1
≈ 87.2 kmh
(4πr2 )δr δr
𝛿𝑟
δV ≈ (Surface Area) (breadth) (δr) ≈ (4πr 2 )
∵ δV = (4πr 2 )δr
dx | ≈ 87.2 kmh−1 dt t=1
= lim 4πr 2 δr→0
3
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4𝜋𝑟 2 𝛿𝑉
(V)
δr⇒0
dx
𝛿𝑉
LHS
= √9225t 2 + 3600t + 900 =
δr
= lim 2πr
x = √[(−75t) − 0]2 + [0 − (30 + 60t)]2
dt
δA ≈ (length) (breadth) ≈ (2πr) (δr)
∵ δA = (2πr)δr
A(−75t, 0) B(0,30 + 60t)
= √5625t 2
𝛿𝑟
(2πr)δr
δr→0
t ≡ time after A reaches P x ≡ distance between the two motorists
dx
2𝜋𝑟 𝛿𝐴
=
7 dy
(πr 2 )
d (A) dr δA = lim δr⇒0 δr
5
=
d dr
= RHS
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429
A math 360 sol (unofficial)
Rev Ex 15 A2(ii) For ⊥ tangents: m1 ⋅ m2
Rev Ex 15 A1(i)
10
y= dy
x
( |
−x
=−
dx
dy
10 x2
dx x=−3 1
−1
10 3
1
−3=
3
⇒ (3, ) 3
−1 dy | dx x=3
=
=1
=
10
− 2 −1 3
9
A3(i)
19
dy
−1
y − y1
= dy
(x − x1 )
|
1
y−
=
3 1
y−
=
3
y
=
9 19 9 19 9 19
(x − 3) x− x−
Gradient:
27 19 62 57
2
−
x2
−1 =−
10
=
x2 2
7 2 7
5
dy
2
10 −2
dx
−2
y|x=2 =
10 2
−2
=3 ⇒ (2,3) ✓
8y = x 2 − kx + 17 1
1
17
8
8
8
= x 2 − kx +
y dy
1
1
4
8
= 9(1)2 − 5 = 4
|
dx x=1
=
dy
|
dx x=1
(x − x1 )
=
dy
|
dx x=1
9x 2 − 5 =4 2 x −1 =0 (x − 1)(x + 1) = 0 x=1 or x = −1 (taken) y|x=−1 = 3(−1)3 − 5(−1) + 4 =6 ⇒ (−1,6) ✓
x=2
= −3 ⇒ (−2, −3)✓ A2(i)
dy
A3(ii) tangent ∥ tangent at A: m1 = m2
2
x =4 x = −2 or y|x=−2 =
A(1,2)
y − (2) = (4)[x − (1)] y−2 = 4x − 4 y = 4x − 2 ✓
A1(ii) Tangent has gradient − : =−
= −64 = −64 =0 =0 =2✓
Tangent: y − y1
✓
7
10
8
= 9x 2 − 5
Tangent Point:
dx x=3
dx
= −1
y = 3x 3 − 5x + 4 dx
dy
)
(k + 6)(k − 10) k 2 − 4k − 60 k 2 − 4k + 4 (k − 2)2 k
1
Normal:
dx x=5 1
8
y|x=3 =
Gradient:
) ( |
[− (6 + k)] [ (10 − k)] = −1
Normal Point:
= −1
dy
= x− k
dx
Tangent gradient at x = −3, x = 5: dy
1
1
= (−3) − k
|
dx x=−3
4
=−
8 1
6
− k
8 1
8
= − (6 + k) ✓ 8
dy
|
dx x=5
1
1
4 5
8 1
= (5) − k =
4 1
− k 8
= (10 − k) ✓ 8
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430
A math 360 sol (unofficial) A4(i)
Rev Ex 15
Curve
A5
y = ax + dy dx
b
=a−
b
dy
x2
dx
−1
mnorm = dy
|
=
dx x=2
−1 b a− 2 2
=
−1 b a− 4
=
−4 4a−b
> 0 [shown] ✓ for x > 2 ∵ 2x > 0, (x − 2) > 0, (x − 1)2 > 0 A6(i)
V= dp
b (2)
A6(ii)
dp
b = 14 − 4a
−(1)
dV
=
= 3,
dV dp 60
=−
=1
p2
dV
|
dt p=20
× ×
=?
dp dt dp dt
At p = 20, dV
= 2x + 6 x
=−
60 202
×3
= −0.45 units 3 s −1 ✓ A7(i)
Let A ≡ area of circular patch r ≡ radius Equate area of circle, πr 2 =
for a = 2, b = 6
−2x + 11
|
dt p=20
A4(ii) Point Q At Q, normal (y = −2x + 11) meets y = 2x +
−4x + 11 −
✓
p2
= −2
2 = 4a − b −(2) (1) sub into (2): 2 = 4a − (14 − 4a) 2 = 4a − 14 + 4a 2 = 8a − 14 16 = 8a a =2✓ b|a=2 = 14 − 4(2) =6✓
x
60
|
Equate gradients,
6
p
=−
dt
4a−b
60
dt p=20
= 7 − 2a
4a−b 2
2x2 −4x (x−1)2 2x(x−2) (x−1)2
=
Curve at P(2,7):
−4
(x−1)2 4x2 −4x −2x2 (x−1)2
=
dV
(7) = a(2) +
(x−1)(4x) −(2x2 )(1)
= =
Normal y + 2x = 11 y = −2x + 11 mnorm = −2
2
x−1
x
Gradient of normal at P(2,7)
b
2x2
y=
6
dA dt
(t)
πr 2 = (4)(16) = 64 64
r2
=
r
=√
π 64 π
x
=0
4x 2 − 11x + 6 = 0 (x − 2)(4x − 3) = 0 x = 2 (taken) or x =
3 4 3
6
4
3 4
y|x=3 = 2 ( ) + 4
=
19
2 3 19
⇒ Q( , ) 4
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2
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431
A math 360 sol (unofficial) A7(ii)
dA dt
dr
= 4, |
dt t=16
Rev Ex 15 B1(i)
=?
y= dy
A = πr 2 dA dr
dx
x √x−2 d
=
dt
=
dA
×
dr
4 = 2πr × dr
=
dt
dr dt dr
=
dt
=
2 πr
= 64
At t = 16, r = √ ,
=
π
dr
|
dt t=16
A8(i)
dθ dt
2
=
π√
π ds
= ,
2 dt
= 64 π
2 π(
8 √π
)
=?
=
2 8√π
=
1
ds
(√x−2)
4√π
x−2
=
−
√x−2
dt
d (√x−2) dx
=
[2(x−2)
−x]
x−2 1 2√x−2 1 2√x−2
(2x−4
−x)
x−2 (x−4)
x−2 x−4 2(√x−2)
3
[proven] ✓
Point:
y|x=4 =
(4) √(4)−2
⋅
=
4 √2
= 2√2
⇒ (4,2√2) Gradient: − dy
dθ
x 2√x−2
B1(ii) Normal
=8 ds
1 ⋅1 2√x−2
x−2 1 2√x−2
1 |
dx x=4
ds
2
−x⋅
√x−2⋅1
cm s −1 ✓
s = rθ [arc length] = 8θ ∵ r = 8
dθ
−x⋅
= 2πr =
dA
√x−2⋅dx(x)
dθ
Normal:
=
−1
⇒ −∞
(4)−4 2(√(4)−2)
3
x=4✓
dt π
= (8) ⋅ ( ) 2
= 4π cm s −1 ✓ A8(ii)
dθ dt
π dA
= ,
2 dt
=?
1
A = r 2 θ[sector area] 2 1
= (8)2 θ
∵r=8
2
= 32θ dA dθ dA dt
= 32 =
dA dθ
⋅
dθ dt π
= (32) ⋅ ( ) 2
= 16π cm2 s −1 ✓
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432
A math 360 sol (unofficial)
Rev Ex 15
B2(a) 1st curve y = ax 2 + bx + 2 dy dx
B2(b) Tangent y = 2x − 16 mtan = 2
= 2ax + b
Curve y = ax 3 + bx
1
Tangent gradient at (1, ): 2
dy
= 2a + b
dy
1st curve at (1, ):
1
Gradient of tangent at x = 2
1
mtan =
mtan =
|
dx x=1
dx
2
2
= a(1)2 + b(1) + 2
= 3ax 2 + b
dy
|
dx x=2
= 3a(2)2 + b = 12a + b
3
a=− −b
At point of contact, y|x=2 = 2(2) − 16 = −12 ⇒ (2, −12)
2
2nd curve y = x 2 + 6x + 4 dy dx
= 2x + 6 (2, −12) lies on curve: −12 = a(2)3 + b(2) −12 = 8a + 2b −12 − 8a = 2b b = −6 − 4a
Normal gradient at (−2, −4): mnorm = − dy
1 |
=−
dx x=−2
1 2(−2)+6
=−
1 2
1
Tangent to 1st curve at (1, ) ∥ normal to 2nd curve 2
at (−2, −4): mtan
= mnorm
2a + b
=−
3
2 (− − b) + b
=−
−3 − 2b + b
=−
−3 − b
=−
2
=
b
=− 5
2
2
1 2 1 2 1 2 1 2
5
−b 3
Using gradients, 12a + b =2 12a + (−6 − 4a) = 2 8a − 6 =2 8a =8 a =1 b|a=1 = −6 − 4(1) = −10 ✓
2 5 2
a|b=−5 = − − (− ) = 1 ✓ 2
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433
A math 360 sol (unofficial) B3
Rev Ex 15
Curve y = (x + 1)2 dy
B4
Curve y=
= 2(x + 1)
dy
mtan = 2(x + 1)
dx
dx
1
3
4
4
= x+
mline =
P(a, b) dy
|
dx x=a
=− =
ab a2 dy
=−
b a
(x − x1 )
|
dx x=a b
y − (b) = (− ) [x − (a)] a
∵ Tangent ⊥ line, mtan ⋅ mline = −1 1
mtan ⋅ ( )
= −1
mtan
= −4
4
Tangent:
x2
Tangent: y − y1
4
Point:
ab
=−
Gradient:
1
Tangent Gradient:
x
Tangent Point:
Line 4y = x + 3 y
ab
dy
|
dx x=−3
(x − a)
y−b
=−
y−b
=− x+b
y
= − x + 2b
a b a b a
Point Q At Q, tangent meets x-axis (y = 0) y =0
2(x + 1) = −4 x+1 = −2 x = −3 (−3 y|x=−3 = + 1)2 =4 ⇒ (−3,4) y − y1 =
b
b
− x + 2b = 0 a b
− x
= −2b
a
x = 2a ⇒ Q(2a, 0)
(x − x1 )
[x − (−3)] y − (4)= (−4) (x + 3) y − 4 = −4 y − 4 = −4x − 12 y = −4x − 8 ✓
Point R At R, tangent meets y-axis (x = 0) b
y|x=0 = − (0) + 2b a
= 2b ⇒ R(0,2b) Distance PQ & RP P(a, b) Q(2a, 0)
R(0,2b)
PQ = √(a − 2a)2 + (b − 0)2 = √a2 + b 2 RP = √(0 − a)2 + (2b − b)2 = √a2 + b 2 ∴ PQ = RP [shown] ✓ B5(i)
v
v2
4
60
s= + ds
ds
1
v
4
30
= +
dv
|
dv v=45
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✓
1
(45)
4
30
= +
=
7 4
km2 h−1 ✓
434
A math 360 sol (unofficial)
Rev Ex 15
B5(ii) If one is travelling at 45km h−1 , for every 1km h−1 increase in speed, the stopping distance increases by 1.75 km ✓ B6(i)
dp dt dA dp
22
y=√ dy
x
22 2√ −x x
1
=
22 2√ −x x
(− (
22 x2
− 1)
x2
1 x
) =
= 2,
dt dy
=
dt
= =
dy
|
dt x=2
−22−x2 22 2x2 √ −x x
−22−x2
=
3
=
|
dt p=5
1
=? 1
= (2p) + (3p2 ) 2
=
−22−x2
=
22−x2 2x2 √ x
=
2 3 2 + p 2
dA
×
dp 3 (p + p2 ) 2 9 3p + p2 2
✓
2x2 √22−x2
2x2 √ √22−x2 dx
dA dt
−22−x2
−22−x2
=
dA
= 3,
=p
−x 1
=
dx
B6(ii)
B7(ii)
dA
At p = 5,
|
dt p=5
B8(i) ×
dx −22−x2 3
2x2 √22−x2
dt
×3
9
= 3(5) + (5)2 2
= 127.5 ✓
=?
dy
dp
dx
Let x ≡ man from lamp y ≡ shadow length
dt
dx dt
×2
dy
= 2,
dt
1.5
=? y
5
x
By similar triangles,
−22−x2
y
3
x2 √22−x2
1.5
=
x+y 5 3
3
2 3
2
5y = x + y At x = 2, dy
|
dt x=2
=
7 −22−22 3 22 √22−22
=
−26 2√2√18
=
−13 √36
=−
13 6
2
cm s −1
y = x
y
2 3
= x 7
✓ B7(i)
dy
y = x2 y|x=p = p2
dx dy dt
1
A = (RP)(PQ) 2 1
= [p − 2 1
= =
=
2 1 2 p 2
1
dy dx 3
⋅
dx dt
7
6
= ms −1 7
= (p + 1)p2 = (p2 + p3 )
7
= ( ) ⋅ (2)
(−1)](p2 )
2 1
3
B8(ii) z ≡ top of shadow from lamp z=x+y
+ p3 ✓ 2
Speed of the top of his shadow = = =
dz dt d dt dx dt
(x + y) +
=2 + =
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20 7
dy dt 6 7 −1
ms
✓
435
A math 360 sol (unofficial)
Ex 16.1 3(i)
Ex 16.1 1(a)
y = x 2 − 5x + 1 dy dx
=
At stationary point, =0
dx
=
2x − 5 = 0 x
=
2
2
5 2
5
2
2
= −5
=
1
1
2
4
4
(x 2 − 4) ⋅ 2
− (2x − 5) ⋅ 2 (x 2
−
4)2
2x 2 − 8
− 4x 2 + 10x (x 2 − 4)2
=
−2(x 2 − 5x + 4) (x 2 − 4)2
=
−2(x − 4)(x − 1) (x 2 − 4)2
y = 5 − 6x + x 2 dx
d d (2x − 5) − (2x − 5) ⋅ (x 2 − 4) dx dx (x 2 − 4)2
−2x 2 + 10x − 8 = (x 2 − 4)2
1
⇒ (2 , −5 ) ✓
dy
(x 2 − 4) ⋅
5
y|x=5 = ( ) − 5 ( ) + 1
1(b)
x2 −4
dy dx
= 2x − 5 ✓
dy
2x−5
y=
= −6 + 2x ✓
✓ At stationary point, dy
3(ii)
=0
dx
dy
−6 + 2x = 0 x =3 y|x=3 = −6 + 2(3) = −4 (3, ⇒ −4) ✓ 2(i)
y=
At stationary point, dx −2(x−4)(x−1) (x2 −4)2
dx
4(i)
y = 2x 3 − 9x 2 + 12x − 4
2x+1 dy
x−1
= = =
(x−1)⋅
d (2x+1) dx
−(2x+1)⋅
(x−1)2 −(2x+1)⋅(1) (x−1)2 −2x−1 (x−1)2
(x−1)⋅(2) 2x−2
4(ii)
At stationary point, dy
=0
dx
−3
3
= 6x 2 − 18x + 12 = 6(x 2 − 3x + 2) = 6(x − 1)(x − 2) [shown]✓
d (x−1) dx
6(x − 1)(x − 2) = 0 x = 1 or x = 2 y|x=1 = 1 y|x=2 = 0 ⇒ (1,1) ✓ ⇒ (2,0) ✓
= (x−1)2 ✓ 2(ii)
=0
x = 4 or x = 1 ✓
dx dy
=0
3
− (x−1)2 ≠ 0 ∵ − (x−1)2 < 0 for x ≠ 1 ⇒ no stationary pt ✓ 4(iii)
d2 y dx2 d2 y
= 12x − 18 |
= −6 < 0 ⇒ max pt (1,1) ✓
|
=6 >0
dx2 x=1 d2 y dx2 x=2
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⇒ min pt (2,0) ✓
436
A math 360 sol (unofficial) 5(i)
Ex 16.1 6(b)
y = (x − 5)√7 + x
y = x 4 − 8x 2 + 2 dy
dy
d
= (x − 5) ⋅
dx
dx
= = = = 5(ii)
d
√7 + x + dx (x − 5) ⋅ √7 + x 1
= (x − 5) ⋅
dx
2√7+x
(1) +(1)
At stationary point,
⋅ √7 + x
dy
dx 3x+9 2√7+x
x
+√7 + x
2√7+x x−5
4x 3 − 16x =0 x 3 − 4x =0 2 x(x − 4) =0 x(x + 2)(x − 2) = 0 x=0 or x = −2 or x = 2 y|x=0 = 2 y|x=−2 = −14 y|x=2 = −14 ⇒ (0,2) ⇒ (−2, −14) ⇒ (2, −24)
+2(7+x) 2√7+x
x−5
+14+2x 2√7+x
3x+9 2√7+x
✓
d2 y
=0 =0
d2 y
= −3
dy
−3
−3+
−
0
+
sign
dx
= 3x 2 − 24x + 36
dy
=0 2
3x − 24x + 36 = 0 x 2 − 8x + 12 =0 (x − 2)(x − 6) = 0 x=2 or x = 6 y|x=2 = 32 y|x=6 = 0 ⇒ (2,32) ⇒ (6,0)
At stationary point, =0 2
3x − 12 =0 2 x −4 =0 (x + 2)(x − 2) = 0 x = −2 or x = 2 y|x=−2 = 16 y|x=2 = −16 ⇒ (−2,16) ✓ ⇒ (2, −16) ✓ d2 y dx2
d2 y dx2 d2 y
= 6x
2|
dx x=−2 d2 y
|
dx2 x=2
= 6x − 24 |
= −12 < 0
⇒ max. pt (2,32) ✓
|
= 12 > 0
⇒ min. pt (6,0) ✓
dx2 x=2 d2 y
d2 y
⇒ min pt (2, −14) ✓
= 32 > 0
dx
dx
⇒ min pt (−2, −14) ✓
At stationary point,
= 3x 2 − 12
dy
⇒ max pt (0,2) ✓
= x(x − 6)2 = x(x 2 − 12x + 36) = x 3 − 12x 2 + 36x
y
dy
y = x 3 − 12x dx
= 32 > 0
|
dx
dy
|
dx2 x=2
⇒ min pt ✓ 6(a)
= −16 < 0
dx2 x=−2 d2 y
6(c) −3−
|
dx2 x=0 d2 y
Sign Test: x
= 12x 2 − 16
dx2
y|x=−3 = [(−3) − 5]√7 + (−3) = −16 ⇒ (−3, −16) ✓ 5(iii)
=0
dx
x−5
At stationary point, dy
= 4x 3 − 16x
= −12 < 0
= 12 > 0
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⇒ max pt (−2,16) ✓
dx2 x=6
⇒ min pt (2, −16) ✓
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A math 360 sol (unofficial) 7(a)
y = 2x +
Ex 16.1
18
7(b)
x
y = x2 +
= 2x + 18x −1 dy
dy
= 2x + 16(−x −2 )
dx
= 2 − 18x −2 =2−
x
= x 2 + 16x −1
= 2 + 18(−x −2 )
dx
16
= 2x − 16x −2
18
= 2x −
x2
16 x2
At turning point,
At stationary point,
dy
dy
=0
dx 18
2−
=0
x2
2
=
=0
dx
2x −
18
16
=0
x2
2x
x2
2
=
16 x2
3
x =9 x2 − 9 =0 (x + 3)(x − 3) = 0 x = −3 or x=3 y|x=−3 = −12 y|x=3 = 12 ⇒ (−3, −12) ✓ ⇒ (3,12) ✓
2x = 16 x3 − 8 =0 2 (x − 2)(x + 2x + 4) = 0 x=2 y|x=2 = 12 ⇒ (2,12) ✓
d2 y
d2 y
dx2
= −18(−2x −3 ) =
d2 y dx2 d2 y
| |
dx2
36
=2+
x3
x=−3
dx2 x=3
= 2 − 16(−2x −3 )
=
=
36 −27
36 26
0
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⇒ max pt (−3, −12) ✓
d2 y
|
dx2 x=2
32 x3
= 6 > 0 ⇒ min pt (2,12) ✓
⇒ min pt (3,12) ✓
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438
A math 360 sol (unofficial) 7(c)
Ex 16.1
4x2 +9
y=
7(d)
x
= 4x +
9
dy
x
dx
= 4x + 9x −1
x2
y=
x+1
= =
dy
= 4 + 9(−x −2 )
dx
=
= 4 − 9x −2 =4−
dx 9
4x 2
2
2
or
x=
y|x=−3 = −12
3
d2 y
2
3
3
2
2
dx2
⇒ (− , −12) ✓ ⇒ ( , 12) ✓
dx2
= −9(−2x =
d2 y
=
|
dx2 x=3 2
=0
(x+1)2 ⋅
d (x2 +2x) dx
−(x2 +2x)⋅
d (x+1)2 dx
(x+1)4 (x+1)2 ⋅(2x+2)
−(x2 +2x)⋅2(x+1) (x+1)4 −(x2 +2x)⋅(2)
(x+1)⋅(2x+2) (x+1)3 2x2 +4x+2
−2x2 −4x (x+1)3
2
=−
2
d2 y
=
−3 )
x3
|
= =
18
dx2 x=−3
=0
2
y|x=3 = 12
2
=0
x=0 or x = −2 y|x=0 = 0 y|x=−2 = −4 ⇒ (0,0) ⇒ (−2, −4)
3
(x + ) (x − ) = 0
d2 y
−x2
(x+1)2
dx x2 +2x (x+1)2 x(x+2) (x+1)2
=0
4 3
2
(x+1)2 2x2 +2x
dy
=9 9
x=−
(x+1)2 (x+1)(2x) −x2 (1)
At stationary point,
=0
x2
3
d (x+1) dx
x2 +2x
=0
x −
−x2 ⋅
x2
dy
2
d (x2 ) dx
= (x+1)2
9
At stationary point,
4−
(x+1)⋅
=
16 3
16 3
0
= (x+1)3
3
⇒ max pt (− , −12) ✓ 2
d2 y
3
⇒ min pt ( , 12) ✓ 2
|
dx2 x=0 d2 y
|
=2 >0
dx2 x=−2
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= −2 < 0
⇒ min pt (0,0) ✓ ⇒ max pt (−2, −4) ✓
439
A math 360 sol (unofficial) 7(d)
Ex 16.1
x −1 2 x +1 x +0x +0 −(x 2 +x) −x +0 −(−x −1) 1
8(i)
y = 8x +
= 8x + x 2
dy
1
= 8 + (−2x −3 )
dx
2
= 8 − x −3
x2
y=
=8−
x+1
= (x − 1) +
At stationary point, dy
=0
dx dx
1
= 1 + [−(x + 1)−2 ](1)
8−
= 1 − (x + 1)−2
8
=
x3
=
x
=
1
= 1 − (x+1)2 At stationary point, dy
=0
x3
1 x3 1 8 1 2
y|x=1 = 6
=0
dx
1 x3
1 x+1
= (x − 1) + (x + 1)−1 dy
1 2x2 1 −2
2
1
⇒ ( , 6) ✓
1
1 − (x+1)2 = 0
2
1 = (x + 1)2 1 = x 2 + 2x + 1 0 = x 2 + 2x x(x + 2) = 0 x=0 or x = −2 y|x=0 = 0 y|x=−2 = −4 ⇒ (0,0) ✓ ⇒ (−2, −4) ✓
8(ii)
d2 y
= −(−3x −4 )
dx2
= d2 y
3 x4
|
dx2 x=1
= 48 > 0
2
1
d2 y dx2
⇒ min pt ( , 6) ✓ 2
= 0 −[−2(x + 1)−3 ](1) 9(i)
2
= (x+1)3
y = x 3 − 6x 2 + 3 dy dx
d2 y
|
dx2 x=0 d2 y 2|
=2 >0
dx x=−2
= −2 < 0
⇒ min pt (0,0) ✓
9(ii)
= 3x 2 − 12x ✓
At stationary point, dy
⇒ max pt (−2, −4) ✓
=0
dx 2
3x − 12x = 0 x 2 − 4x = 0 x(x − 4) = 0 x = 0 or x = 4 ✓ 9(iii)
dy
0
⇒ min pt (−1, −9) ✓ 12
y = ax +
b x2
= ax + bx −2
−2(x − 2)(x + 1) (x 2 + 2)2 ✓
dy
=
= a + b(−2x −3 )
dx
=a−
2b x3
Curve at (3,5):
10(ii) At stationary point, dx −2(x−2)(x+1) (x2 +2)2
= 36x 2
dx2 x=−1
−2(x 2 − x − 2) (x 2 + 2)2
dy
= −1
= 12x 3 + 12
dx d2 y
−2x 2 + 2x + 4 = (x 2 + 2)2 =
12
y = 3x 4 + 12x y|x=−1 = −9 ⇒ stationary pt (−1, −9)
d d (x 2 + 2) ⋅ (2x − 1) − (2x − 1) ⋅ (x 2 + 2) dx dx = (x 2 + 2)2 =
12
b
5 = 3a +
=0
b 9
=0
9
= 5 − 3a
b = 45 − 27a
−(1)
x = 2 or x = −1 ✓ 10(iii)
At stationary pt (3,5):
dy dx −2(x−2)(x+1) (x2 +2)2 (x−2)(x+1) (x2 +2)2
>0
dy
>0 0
+ − + −1 2
27
=0 =0 =0
−(2)
sub (1) into (2): a−
⇒ −1 < x < 2 [shown] ✓ 11(i)
|
dx x=3 2b a − (3)3 2b
a−
2(45−27a) 27 2(9)(5−3a) 27
=0 =0
2
y = 3x 4 + kx
a − (5 − 3a) = 0
dy
3a − 10 + 6a = 0 9a = 10
dx
3
3
= 12x + k
a
At stationary pt: dy
10 9
✓
b|a=10 = 15 ✓
=0
dx
= 9
3
12x + k = 0 x3 x
=− 3
= √−
k 12 k 12
⇒ only 1 stationary pt ✓
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441
A math 360 sol (unofficial) 13
1st curve & its gradient y = 2x 2 − 4x + 5 dy dx
Ex 16.1 14(ii) At a = 9, dy
= 4x − 4
dy
d2 y
=0
dx
15(i)
dy
= a − 2b(2x − 1)−2 2b
= a − (2x−1)2 (2,7) lies on curve, (7) = a(2) + 7
At turning pt (1,3):
dy
= 21 − 6a
|
a−
2b
=0
9
−(2)
sub (1) into (2): a− a−
2(21−6a) 9 2(7−2a) 3
=0 =0
3a − 14 + 4a = 0 7a = 14 a =2✓
=0
6(−3) + 2a(−3) = 0 54 − 6a =0 a =9✓ b|a=9 = −8 ✓
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=0 2b
At stationary pt (−3,19): |
−(1)
a − (2(2)−1)2 = 0
= 6x 2 + 2ax
dx x=−3 2
b 3
= 7 − 2a
dx x=2
Curve at (−3,19): (19) = 2(−3)3 + a(−3)2 + b 19 = −54 + 9a + b b = 73 − 9a
dy
b 2(2)−1
At stationary pt (2,7):
y = 2x 3 + ax 2 + b dx
= 2a +
b 3 b
=0
3 + 2a + 1 = 0 2a = −4 a = −2 ✓ b|a=−2 = 3 ✓
dy
= a + b[−(2x − 1)−2 ⋅ (2)]
dx
= 3x + 2ax + 1
|
b 2x−1
= ax + b(2x − 1)−1
2nd curve at (1,3): 3=1+a+1+b b=1−a
14(i)
y = ax +
2
dx x=1
= −18 < 0
⇒ max pt (−3,19) ✓
2nd curve & its gradient y = x 3 + ax 2 + x + b
dy
|
dx2 x=−3
4x − 4 = 0 x =1 y|x=1 = 3 ⇒ turning pt (1,3)
dx
= 12x + 18
dx2
At turning point,
dy
= 6x 2 + 18x
dx d2 y
Put a = 2 into (1): b|a=2 = 9 ✓
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442
A math 360 sol (unofficial)
Ex 16.1 16(ii) Curve at (1,1): 1 = 12 (1 − k)2 1 = 1 − 2k + k 2 0 = k 2 − 2k k 2 − 2k = 0 k(k − 2) = 0 k=0 or k = 2 (rej ∵ k ≠ 0)
15(ii) At a = 2 & b = 9, y = 2x + dy dx
9 2x−1 18
= 2 − (2x−1)2
At stationary point, dy
=0
dx 18
2 − (2x−1)2
=0
2
= (2x−1)2
18
∴ y = x 2 (x − 2)2 dy
(2x − 1)2 =9 2 4x − 4x + 1 = 9 4x 2 − 4x − 8 = 0 x2 − x − 2 =0 (x − 2)(x + 1) = 0 x=2 or x = −1 (taken) y|x=−1 = −5 ⇒ (−1, −5) ✓
dx
At stationary point, dy
dx
=0
dx
2x(x − 2)(2x − 2) = 0 2x(x − 1)(x − 2) = 0 x=0 or x = 1 or y|x=0 = 0 y|x=1 = 1 ⇒ (0,0) ✓ ⇒ (1,1) ✓
15(iii) 1st derivative dy
16(iii) 1st derivative −2
dy
= 2x(x − 2)(2x − 2)
dx
= 2x(2x 2 − 6x + 4) = 4x 3 − 12x 2 + 8x
2nd derivative dx2
= −18[−2(2x − 1)−3 ] ⋅ = −18[−2(2x − 1) =
d2 y
|
d dx
(2x − 1)
−3 ]
. (2)
2nd derivative d2 y
72 (2x−1)3
dx2
8
dx2 x=2
d2 y
= >0 3
y=x dy dx
2 (x
2
− k)
=x ⋅
d dx
d2 y
[(x − k)
d2 y 2]
+
d dx
= x ⋅ 2(x − k) +2x = 2x(x − k)(2x − k) ✓
(x 2 )
=8 >0
|
= −4 < 0 ⇒ max pt (1,1) ✓
|
=8 >0
⋅ (x − k)
dx2 x=2
2
⋅ (x − k)2
⇒ min pt (0,0) ✓
|
dx2 x=1
2
2
= 12x 2 − 24x + 8
dx2 x=0
⇒ (2,7) is min pt ✓ 16(i)
x=2 y|x=2 = 0 ⇒ (2,0) ✓
18
= 2 − (2x−1)2 = 2 − 18(2x − 1)
d2 y
= 2x(x − 2)(2x − 2)
⇒ min pt (2,0) ✓
16(iv) 𝑦 𝑦 = 𝑥 2 (𝑥 − 2)2 (1,1) (0,0)
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(2,0)
𝑥 ✓
443
A math 360 sol (unofficial) 17
y= dy dx
1 20
=
Ex 16.1
7
18(ii) 1st derivative
x 5 − x 3 + 3x 2 6
1 4 x 4
−
7 2 x 2
dy dx
+ 6x
At stationary point,
At stationary point, dy dx 1 4 x 4 4
dy
−
+ 6x
3x 2 − 12x + 9 = 0 x 2 − 4x + 3 = 0 (x − 1)(x − 3) = 0 x=1 or x=3 y|x=1 = 0 y|x=3 = −4 ⇒ (1,0) ⇒ (3, −4)
=0
x − 14x + 24x = 0 x(x 3 − 14x + 24) = 0 x(x − 18(i)
Curve y = x 3 + ax 2 + bx + c
2nd derivative d2 y dx2
touch x − axis at x = 1 & cross x − axis at x = 4 y = k(x − 1)2 (x − 4) Compare x 3 : k = 1 y = (x 2 − 2x + 1)(x − 4) = x 3 −2x 2 +x = −4x 2 +8x −4 = x 3 − 6x 2 + 9x − 4 Compare x 2 : a = −6 ✓ Compare x1 : b = 9 ✓ Compare x 0 : c = −4 ✓
=0
dx
=0 7 2 x 2 2
= 3x 2 − 12x + 9
d2 y
= 6x − 12 |
= −6 < 0
⇒ max pt (1,0) ✓
|
=6 >0
⇒ min pt (3, −4) ✓
dx2 x=1 d2 y
dx2 x=3
18(iii) Tangent Recall
y = x 3 − 6x 2 + 9x − 4 dy dx
Point: Gradient: Tangent:
&
2
= 3x − 12x + 9
y|x=0 = −4 ⇒ (0,4) dy
|
dx x=0
y − y1
=9 =
dy
|
dx x=0
(x − x1 )
(x − 0) y − (−4) = 9 y = 9x − 4 ✓
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444
A math 360 sol (unofficial)
Ex 16.2 1(d)
Ex 16.2 1(a)
dy
dx
=
√1−2x
dy
=0
dx 1−3x
=4>0
=0 1
= ✓
x
3
⇒ min ✓
Sign test x
= x 2 + (2x − 1)2 = x 2 + (4x 2 − 4x + 1) = 5x 2 − 4x + 1
dy dx
dx
At stationary value, dy
y= dy
=0
dx
dx
10x − 4 = 0
dx2
= ✓ 5
−
(x2 +1)⋅
d d (x) −x⋅ (x2 +1) dx dx (x2 +1)2
(x2 +1)⋅1
−x⋅2x
(x2 +1)2 x2 +1
−2x2 (x2 +1)2
dy
= 4(x − 2) ⋅ (1)
4(x − 2)3 = 0 x =2✓
Sign test x dy
= 4[3(x − 2) = 12(x − 2) |
dx2 x=2
2]
=0
1−x =0 (x + 1)(x − 1) = 0 x = −1 or x = 1 ✓
=0
dx
=0
dx 1−x2 (x2 +1)2 2
3
dy
+1)2
At stationary value,
At stationary value,
d2 y
0
1−x2
= 4(x − 2)3
dx2
=
= (x2
= 10 > 0
y = (x − 2) + 3
d2 y
3
x
=
4
dx
1+
3
x2 +1
=
⇒ min ✓
dy
1
3
sign +
2
x
1−
⇒ max ✓
= 10x − 4 1(e)
d2 y
⋅ √1 − 2x
√1−2x
d2 y
dy
(x) ⋅ √1 − 2x
+ √1 − 2x
√1−2x 1−3x
At stationary value,
y
d dx
⋅ (−2) + 1
4x − 8 = 0 x =2✓
dx2
1(c)
2√1−2x x
+
=0
dx
1(b)
1
=−
At stationary value,
√1 − 2x
dx
=x⋅
= 4x − 8
dy
d
=x⋅
dx
y = 2x 2 − 8x + 3 dy
y = x√1 − 2x
dx
⋅ (1)
sign
−1− −
−1
−1+
0
+
⇒ min at x = −1 ✓
2
dy
=0
dx
⇒ not max/min ✓
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x
1−
1
1+
sign
+
0
−
⇒ max at x = 1 ✓
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445
A math 360 sol (unofficial) 2
Ex 16.2
2x + y = 10 y = 10 − 2x
4(iii)
dA
= 8 + 800(−x −2 )
dx
= 8 − 800x −2 =8−
At stationary value,
dA
dA
dx
= 10 − 4x
At stationary value, dA
800
dx
=
dx
d2 A
d2 A
Fence = 36 2x + y = 36 y = 36 − 2x [shown] ✓
3(ii)
A = xy = x(36 − 2x) = 36x − 2x 2 [shown] ✓
3(iii)
dA
|
5(i)
dx2
By Pythagoras Theorem, Height = √(5x)2 − [
=0
4(ii)
2
]
= √25x 2 − 9x 2 = 4x Area of trapezium
= −4 < 0
1
A = (height)(sum of bases) 2 1
= (4x)[(6x + y) + y] 2
Volume = 400 [given] x(8)h = 400 h
(6x+y)−y 2
= √(5x)2 − (3x)2
⇒ max 4(i)
= 1.6 > 0
Wire length = 104 y + 5x + 5x + (6x + y) = 104 16x + 2y = 104 2y = 104 − 16x y = 52 − 8x ✓
36 − 4x = 0 x =9✓ d2 A
x3
⇒ min ✓
At stationary value, dx
1600
dx2 x=10
= 36 − 4x
dA
= −800(−2x −3 ) =
0)
dx2 2 = −4
800
2
10 − 4x = 0 x = 2.5 A|x=2.5 = 12.5 d2 A
=0
x2
8
=0
x2
=0
dx
8−
3(i)
800
A = xy = x(10 − 2x) = 10x − 2x 2
=
50 x
= 2x(6x + 2y) ∵ y = 52 − 8x, A = 2x[6x + 2(52 − 8x)] = 2x(104 − 10x) = 208x − 20x 2 [shown]✓
✓
A = 8x + 2xh + 2(8)h 50
50
x
x
= 8x + 2x ( ) + 2(8) ( ) = 8x +
800 x
+ 10 ✓ [shown]
= 8x + 800x −1 + 10
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446
A math 360 sol (unofficial) 5(ii)
dA dx
= 208 − 40x
Ex 16.2 7
y = 2x 3 − 12x 2 + x + 9 dy dx
= 6x 2 − 24x + 1
At stationary value, dA
Normal gradient (n)
=0
dx
1
= − dy
208 − 40x = 0 x
= =
y|x=5.2 d2 A dx2
dx
208
=−
40 52
= −(6x − 24x + 1)−1
10
= 5.2 ✓ = 10.4 ✓
dn dx
= −40 < 0
1
⋅ (12x − 24)
−24x+1)2 12x−24
= (6x2
−24x+1)2
x 4 y = 32 y
=
32
At stationary value,
x4
12x−24 (6x2 −24x+1)2
= x2 +
32 x4
[shown] ✓
2
= x + 32x −4
dz
= 2x + 32(−4x −5 )
dx
8(i)
= 2x − 128x −5 = 2x −
128
dx
2x −
128 x5
128 x5
d2 z
ℎ
400 πr2
=0 =0
= 2πr 2 + 2πr (
= 2x
= 2πr 2 +
800 r
400 πr2
)
[shown] ✓
= 2πr 2 + 800r −1
= 2 − 128(−5x −6 ) =2+
d2 z
=
𝑟
Area A = 2πr 2 + 2πrh
2x 6 = 128 6 x = 64 x = −2 or x=2 (rej ∵ x > 0) ✓
dx2
Volume V = 400 2 πr h = 400 h
x5
At stationary value, dz
=0
12x − 24 =0 x =2 y|x=2 = −21 ⇒ P(2, −21)✓
z = x2 + y
6(ii)
= −[−(6x 2 − 24x + 1)−2 ⋅ (12x − 24)] = (6x2
⇒ max A ✓ 6(i)
1 6x2 −24x+1 2
|
dx2 x=2
640 x6
= 12 > 0
⇒ min z ✓
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447
A math 360 sol (unofficial) 8(ii)
Ex 16.2
1st derivative dA dr
9(ii)
= 4πr + 800(−r
1st derivative dA
−2 )
dx
= 4πr − 800r −2 800 = 4πr − 2 r
= 4x − 108x −2
4πr −
800
dA
4x −
800
4x
r2
3
4πr
=
πr 3
= 200 3
=√
r
=0
dx
=0
r2
x2
At stationary value,
=0
dr
108
= 4x −
At stationary value, dA
= 4x + 108(−x −2 )
108 x2
=0 =
108 x2
x = 27 x =3✓ h|x=3 = 2 ✓
200 π
≈ 3.99cm ✓ 2nd derivative d2 A
2nd derivative d2 A dr2
dx2
= 4π − 800(−2r −3 ) = 4π +
=4+
1600 r3
d2 A
> 0 for r > 0 ⇒ min A 9(i)
Volume x(2x)h = 36 h
= =
ℎ 2𝑥 𝑥
2x2 18
By Pythagoras Theorem, 25x2
4
16
5
= x 4
Perimeter PQRST = 30 2x + 2y +2RS = 30 x+y + RS = 15 5
x + y + ( x) = 15 4
= 2x + 6x ( 2 )
9
y
x
x
= 12 > 0
3x 2
18
108
x3
RS = √x 2 + ( ) = √
x2
2
216
⇒ min 10
36
|
dx2 x=3
Area A = 2[x(2x) + xh + 2xh] = 2(2x 2 + 3xh) = 2x 2 + 6xh = 2x 2 +
= 4 − 108(−2x −3 )
= 15 − x 4
[shown] ✓ A = triangle STR
= 2x 2 + 108x −1
1
3x
2 3 2 x 4 3 2 x 4 3 2 x 4
4
= (2x) ( )
+2xy
=
+2xy
= =
= 30x −
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+ rectangle PQRT
9
+2x (15 − x) +30x − 15x2 4
4 9 2 x 2
[shown]
448
A math 360 sol (unofficial) 10(i)
Ex 16.2
1st derivative dA dx
= 30 −
15
11(iv) At stationary value, dA
x
2
=0
dx
48x − At stationary value, dA
48x
=0
dx
30 −
15 2
=−
15 2
0)
Sign Test R
P|x=4 = 16 cm ✓
⇒ work done is greatest when R=r
dW dR
r− sign +
r 0
r+ −
2nd derivative d2 P dx2
= −32(−2x −3 ) =
64 x3
>0 ∵x>0
⇒ min P
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450
A math 360 sol (unofficial) 14
Ex 16.2
Perimeter
15 1
𝑦
x + 2y + ( ) πx = 6 x + 2y +
2 πx
𝑥
=6
2
2y
=6−
πx 2
−x
= 6 − x ( + 1)
y
= 3 − x ( + 1)
Area A = rectangle
+semicircle
x 𝑎
Rectangle area, A = xy
π
2y
x ≡ length y ≡ breadth A ≡ area of rectangle
y
−(1)
2
1
π
2
2
1
x 2
2 1
2 2
= xy
+ π( )
= xy
+ πx
−(1)
By Pythagoras Theorem, x 2 + y 2 = (2a)2 y 2 = 4a2 − x 2 y = √4a2 − x 2 or y = −√4a2 − x 2 (rej ∵ y > 0)
−(2)
8
−(2)
sub (2) into (1): sub (1) into (2): 1
π
1
2 1 π
2
8 1
A = x√4a2 − x 2
A = x [3 − x ( + 1)] + πx = 3x − ( + 1) x 2 2 π 1
= 3x − ( + ) x 4 1
2
+ πx
dx
2
8
1
= 3x + ( π − − ) x 4 1
=
2
2
= 3x + (− − ) x 2 = 3x − (
8 π+4 8
=
2
) x2
=
1st derivative dA
=3−(
dx
=3−(
π+4 8 π+4 4
dA
3−( (
4
)x
=
= 3(
π+4
= =
dx2
⋅ √4a2 − x 2
(−2x) +(1)
+√4a2 − x 2
−x2
+(4a2 −x2 ) √4a2 −x2
4a2 −2x2 √4a2 −x2
=0 =0
x = √2a or x = −√2a (rej ∵ x > 0)
12 π+4 12
)−(
π+4
36 π+4 18 π+4
−
y|x=√2a = √4a2 − (√2a)
✓ 12
) (
π+4
8
= −2 (
π+4 8
2
= √4a2 − 2a2 = √2a2 = √2a = x ∵ length = breadth, largest rectangle is square of side √2a
2 π+4
)
144 8(π+4)
Sign Test
[shown]
x dy
2nd derivative d2 A
(x) ⋅ √4a2 − x 2
=3
x
A|x= 12
d dx
⇒ 4a − 2x 2 = 0 −2x 2 = −4a2 x 2 = 2a2
)x = 0
π+4
1 2√4a2 −x2
−x2
√4a2 −x2 2
π+4 4
(√4a2 − x 2 )+
√4a2 −x2
dx 4a2 −2x2
)x
=0
dx
d dx
At stationary value,
) 2x
At stationary value, dA
=x⋅ =x⋅
+ πx 2
π
π
dA
8 1
2
2
8
2
dx
)0∵p>0
⇒ min z
=0
dx
8x −
27 x2
8x 8x
3 2 2
( )
= −54(−3p−4 ) =
dy
9
=4✓ 2nd derivative
= 8x − 27x −2 = 8x −
8 3 2
q|p=3 =
= 8x + 27(−x −2 )
dx
27
p3
2
dy
=0
p3
p3
27
2
54
54
=2✓ A2(i)
=0
=0 =
3
A4(i)
27 x2
= 27
x
=
3 2
y|x=3 = 27
Wire length = 2400 4(length + breadth + height) = 2400 4(3x + x + h) = 2400 4(4x + h) = 2400 4x + h = 600 h = 600 − 4x ✓
2
3
⇒ ( , 27) ✓ 2
A2(ii)
d2 y dx2
= 8 − 27(−2x −3 ) =8+
d2 y
|
dx2 x=3
54 x3
= 24 > 0
2
A4(ii) Volume V = 3x(x)h = 3x 2 h = 3x 2 (600 − 4x) = 12x 2 (150 − x) ✓ = 1800x 2 − 12x 3
3
⇒ min pt ( , 27) ✓ 2
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454
A math 360 sol (unofficial) A4(iii) 1st derivative dV dx
Rev Ex 16 A6(i)
= 3600x − 36x
2
Volume = 100 2x(x)h = 100 h
h
50
=
2x
x2
x
At turning value, dV
Total surface area S = 2[2x(x) + 2x(h) + x(h)] = 2(2x 2 + 2xh + xh) = 2(2x 2 + 3xh) = 4x 2 + 6xh
=0
dx
3600x − 36x 2 = 0 x 2 − 100x =0 x(x − 100) =0 x = 0 or x = 100 ✓ (rej ∵ x > 0) 2nd derivative d2 V
= 3600 − 72x
dx2
A6(ii)
= 4x
2
= 4x
2
dS dx
d2 V dx2
|
x=100
50
= 4x 2
+6x ( 2 ) x
+
300 x
[shown] ✓
+300x −1
= 8x + 300(−x −2 ) = 8x − 300x −2
= −3600 < 0
= 8x −
⇒ max V
300 x2
At stationary value,
A5(i)
dS
=0
dx
𝑦 𝑥 Wire length = 100 4x + 4y = 100 4y = 100 − 4x y = 25 − x✓
8x −
3
=
dx
3 75 2
x= √
2 75
d2 S
2
≈ 134 ✓
= 8 − 300(−2x −3 ) =8+
>0 ⇒ min ✓
= 2x −50 + 2x = 4x − 50
x2 75 3
dx2
= 2x +2(25 − x)(−1)
300
=√
x
A6(iii)
=0 =
x
A5(iii) 1st derivative dA
x2
8x
S| A5(ii) Total area A = x2 + y2 = x 2 + (25 − x)2 [shown] ✓
300
600 x3
∵x>0
At stationary value: dA
=0
dx
4x − 50 = 0 x
=
50 4
= 12.5 ✓ 2nd derivative d2 A dx2
=4>0
⇒ min
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455
A math 360 sol (unofficial) A7(i)
Rev Ex 16
Perimeter
B1(i)
1
y = x2 +
x + x + (2πr) = 80 = 80 = 80 − πr
dy dx
1
x
= 40 − πr
= 2x −
Area A = semicircle + triangle 1
1
2 1
2 1
1
2
2
dy
+ (40 − πr)
2 1
1 1 2
2 πr2
2 2
2x −
A7(ii)
dA
=0
x3
=
16 x3
4
x = 16 x=2 or y|x=2 = 8 ⇒ (2,8)
+ (80 − πr)2 [shown] ✓ 8 1
= πr + [2(80 − πr)] ⋅ (−π)
dr
32
x
1
2
x3
=0
dx
2
= πr 2 + ( ) (80 − πr)2 =
32
At stationary point,
= πr 2 + x 2 = πr
= 2x + 16(−2x −3 ) = 2x − 32x −3
2
2
x2
= 2x + 16x −2
2
2x + πr 2x
16
8 π
x = −2 y|x=−2 = 8 ⇒ (−2,8)
= πr − (80 − πr) 4
= πr −20π + = (π +
π2 4
π2 4
r
B1(ii)
dx2
(π +
4 π
d2 y
=0
(1 + ) r
= 20
4 π 4
(
4+π 4
)r
= 20
r
=
80 π+4
≈ 11.2 ✓ A7(iii)
d2 A dr2
=π+
π2 4
>0
⇒ min ✓
|
dx2 x=2 d2 y
) r − 20π = 0
(1 + ) r − 20
= 2x − 32(−3x −4 ) =2+
) r − 20π
At stationary value, π2
d2 y
|
96 x4
=8>0
dx2 x=−2
= −4 < 0
⇒ min pt (2,8) ✓ ⇒ max pt (−2,8) ✓
B2(i) y=
4 2−x
= 4(2 − x)−1
+
9 x−3
+9(x − 3)−1
dy dx
= 4 [−(2 − x)−2 ⋅
d dx
(2 − x)] +9 [−(x − 3)−2 ⋅
= 4[−(2 − x)−2 ⋅ (−1)] = 4(2 − x)−2 4
d dx
(x − 3)]
+9[−(x − 3)−2 ⋅ (1)] −9(x − 3)−2 9
− (x−3)2 ✓
= (2−x)2 d2 y dx2
= 4 [−2(2 − x)−3 ⋅
d dx
(2 − x)] −9 [−2(x − 3)−3 ⋅
= 4[−2(2 − x)−3 ⋅ (−1)] 8
= (2−x)3
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d dx
(x − 3)]
−9[−2(x − 3)−3 ⋅ (1)] 18
+ (x−3)3 ✓
456
A math 360 sol (unofficial)
Rev Ex 16
B2(ii) At stationary value, dy
B4(i)
AR =
=0
dx 4 (2−x)2 4 (x−2)2
9
− (x−3)2
=0 =
4(x − 3)2 4(x 2 − 6x + 9) 4x 2 − 24x + 36 5x 2 − 12x x(5x − 12) x=0
By Pythagoras’ Theorem,
=
9 (x−3)2
y|x=0 = −1
AS
− [ (10)]
S Q 13cm h
13cm P
2
2
B
C R 10cm
PQ BC x 10 6
12 − h = x 5
5
5
=
12
12
12
=
AR 12−h
6
h
5
⇒(
1
By similar triangles,
y|x=12 = −25
⇒ (0, −1)
−
√(13)2
A
BR2
= 12
= 9(x − 2)2 = 9(x 2 − 4x + 4) = 9x 2 − 36x + 36 =0 =0
or x =
√AB 2
= 12 − x 5
6
= (10 − x) ✓
h
, −25)
5
B4(ii) Area of △ PQR d2 y
1
|
dx2 x=0 d2 y dx2
B3
|
3
12 x= 5
=−
625 3
1
⇒ min pt (0, −1) ✓
= >0 0
1 2
3
⇒ 1st or 4th quadrant
S
0 0 cos x > sin x 𝑦 5π π 4
⇒0≤x<
dS dθ
π
or
4
5π 4
At stationary value, dS
64 cos 2θ − 64 sin θ =0 cos 2θ − sin θ =0 2 1 − 2 sin θ − sin θ =0 2 sin2 θ + sin θ − 1 =0 (2 sin θ − 1)(sin θ + 1) = 0
< x ≤ 2π ✓
𝑂
π 4
⇒ 17(i)
π 4
0)
2
= (BC)(AF) 2 1
2
θ = α, π − α π = ✓
E
Area S = area of triangle
π
1
⇒ 1st or 2nd quadrant π 00 ⇒ min at x = ln 2
>0
dx x
y = e−x dy dx
25m
= ln (
m
=
m
=
= e−x
2 +2x
= e−x
2 +2x
⋅
d dx
(∵ m =
2 +2x
✓
18(vi)
2 +2x
−
ln 199−ln 130 (9) 25
]
1 sign +
1 0
dP
> 18 000
mP
> 18 000
P
>
mt
1 −
)
25
650 000e
+
t ≡ t years after beginning of 2006
>
18 000 m 18 000 m 9
e
>
mt
> ln (
t
>
t (∵ m =
325m
1 m
9
)
325m 9
ln (
325m
)
> 28.5 ln 199−ln 130 25
)
⇒ t = 29 ✓ 19(i)
P = 650 000 emt
m = ae−kt Initial of 100g m|t=0 = 100 ae0 = 100 a = 100 ✓
2006 ⇒ t = 0 P|t=0 = 650 000 ✓ dt
(30)
ln 199−ln 130
mt
=0 =1 = e−1+2 = e [shown] ✓
Max ✓
dP
[shown] ✓
≈ 13 000 ✓
(−x 2 + 2x)
=0
−2(x − 1)e−x x y|x=1
18(ii)
25
ln 199−ln 130 (8) 25 ]
⋅ (−2x + 2)
dx
18(i)
)
2 +2x
dy
dx
ln (
25 130 ln 199−ln 130
−650 000 [e
17(ii) At turning point,
dy
)
ln 199−ln 130
dt
x
1
199
130 199
25 P|t=30 = 650 000e ≈ 1 083 000 ✓
= 650 000 [e
= −2(x − 1)e−x
17(ii)
130
P|t=9 − P|t=8
e −2 >0 ex >2 x > ln 2 ✓ 17(i)
=
18(v) 2014 ⇒ t = 8 2015 ⇒ t = 9
16(iii) For increasing function, dy
199
e25m
= 650 000memt = m(650 000emt ) = mP
⇒
dP dt
∝ P [shown] ✓
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475
A math 360 sol (unofficial)
Ex 17.2
19(ii) m = 100e−kt
21(iv)
𝑛 𝑛 = 20𝑒 0.1𝑡
th
At 40 h, mass reduces to 90g m|t=40 = 90 −k(40) 100e = 90 −40k e = 0.9p −40k = ln(0.9p) k 19(iii)
=− (
m = 100e
1
ln
40
9 10
21
= ab(e−bx ) >0 ∵ a > 0, b > 0, e−bx > 0 ⇒ increasing for all x ∈ ℝ ✓
1
= m|t=0
100e [
1
e 40 [
1 40
ln
2 1
1 9 ln ]t 40 10
= (100)
9 10
=
ln ]t 9 10
]t
2 1
22
y = ae2x + be4x dy
2
= ln
f: x ↦ a(1 − e−bx )
f ′ (x) = −a(e−bx )(−b)
Decay to half: [
dx
1
= 2ae2x +4be4x
2 d2 y dx2
1 9 ( ln )t 40 10
m = 100e dm dt
=(
1 40
ln
5
9
2
10
= ln 20(i)
= a(2e2x ) +b(4e4x )
≈ 263h ✓
t 19(iv)
✓
f(x) = a − ae−bx
✓
1 9 ln )t 40 10
m
𝑥
𝑂
9 10
e
= 4ae2x (
) 100e
[
1 9 ln ]t 40 10
= 2a(2e2x ) +4b(4e4x )
1 9 ln )t 40 10
Show:
d2 y dx2
+16be4x =6
dy dx
− 8y
✓ RHS
n = 20e0.1t n|t=5 = 20e0.5 ≈ 33 ✓
= 6(2ae2x + 4be4x ) −8(ae2x + be4x ) = 12ae2x + 24be4x −8ae2x − 8be4x = 4a2x + 16be4x = LHS [shown] ✓
20(ii) n > 200 0.1t 20e > 200 0.1t e > 10 0.1t > ln 10 t > 10 ln 10 t > 23.026 ⇒ least t = 24 ✓ 20(iii)
dn
= 20(e0.1t )(0.1)
dt
= 2e0.1t dn
|
dt t=15
= 2e0.1(15) ≈ 9 snails /week ✓
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476
A math 360 sol (unofficial) 23
Ex 17.2
y = (2 + 3x)e−x
24
dy
ex − x = 0 ex = x 𝑦
dx d
= (2 + 3x) ⋅
dx
(e−x ) +
= (2 + 3x) ⋅ (−e−x ) +3 = e−x [−(2 + 3x) +3] = e−x (−2 − 3x +3) = e−x (1 − 3x)
d dx
(2 + 3x) ⋅ e−x
𝑦=𝑥
⋅ e−x
y1 = ex y1 = ex > 0 ∴ y1 > y2
y2 = x y2 = x y2
y2 = 0
x > 0:
y1 = ex
y2 = x
dy1
dy2
x < 0: dx2 d
d (e−x )⋅ (1 dx −x ) (1
(1 − 3x) +
dx
−x
𝑥
𝑂
d2 y
= e−x ⋅
𝑦 = 𝑒𝑥
= e ⋅ (−3) +(−e ⋅ −x = e [−3 −(1 − 3x)] −x = e (−3 −1 + 3x) = e−x (3x − 4)
− 3x)
− 3x)
dx
At stationary point, dy dx −x (1
e
⇒
=0 − 3x) = 0
x
=
dy dx
=e
dy1 dx
>
dx
dx
∵ y1 > y2 at x = 0
3
&
dx
>
dy2 dx
for x >
0, y1 > y2 for x > 0
1−
1
1+
3
3
3
0
−
sign +
dy1
=1
dy2
1
⇒ stationary pt exist Sign Test: x
x
∴ ex − x = 0 does not have a real solution ∵ y1 > y2 for all x
⇒ max 1
2nd derivative at x = : 3
d2 y
1 3
−( )
|
dx2 x=1
=e
1
[3 ( ) − 4] 3
3
1 −e−3
= 0 y2 ≤ 0 ⇒ y1 > y2
dn y dxn
26
x > 0:
y1 > 0 y2 > 0 ⇒ cannot confirm y1 and y2 don′ t intersect dy1
x > 0:
dx
= ex > 1 [exponential function is an
increasing function] dy2 dx
⇒
= k n ekx [by inspection] ✓
Exponential function: (constant)variable (variable)constant Power function ex is an exponential function with e as a constant & x as a constant It is wrongly treated as a power function & power rule is wrongly applied.
=1
dy1 dx
>
dy2 dx
⇒ y1 > y2 x ∈ ℝ:
y1 > y2 y1 − y2 > 0 ex − x > 0 ex − x ≠ 0 ✓
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478
A math 360 sol (unofficial)
Ex 17.3 2(d)
Ex 17.3 1(a)
d dx
= = 1(b)
d dx
d dx 1
(ln x) +2
=1− (
d dx 1
=
(ln x)
+
x
dx
3(a)
) =
(1+ln x)⋅
d (x) dx
d (1+ln x) dx 2 (1+ln x)
−x⋅
−x⋅
1 x
(1+ln x)2 1+ln x −1 (1+ln x)2 ln x ✓ (1+ln x)2
d
d dx
ln(5x + 1) =
5 5x+1
[ln(4x − 3)2 ] =
=2
d dx
= d dx
ln(8 − x 3 ) = = =
8 4x−3 1
8−x3 1 8−x3 3x2 x3 −8
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⋅
=
dy
|
dx x=3
4
3(b)
⋅ (0
⋅ (−2)
⋅ ln(5 − 2x)
+ ln(5 − 2x) + ln(5 − 2x) ✓
1 2x−5 2
⋅ (2)
2x−5
=
2 2(3)−5
y = ln(3 − 2x)
dy
=
dx
= dy
)
|
dx x=1
1 3−2x 2
⋅ (−2)
2x−3
=
2 2(1)−3
= −2 ✓
✓ d
(5 − 2x) +1
(x) ⋅ ln(5 − 2x)
Curve crosses x − axis, y =0 ln(3 − 2x) = 0 3 − 2x = e0 =1 −2x = −2 x =1✓ (1,0) ⇒
4x−3
dx
d dx
=2✓
[ln(4x − 3)]
=2⋅(
d dx
+
y = ln(2x − 5)
=
[2 ln(4x − 3)]
dx
⋅
2x−5
dx
✓
d
5−2x 1
5−2x 2x
dy
d sin x ( ) dx ln x d d ln x ⋅ (sin x) − sin x ⋅ (ln x) dx dx = (ln x)2 1 ln x ⋅ cos x − sin x ⋅ x = (ln x)2 sin x cos x ln x − x = (ln x)2 ✓
dx
[ln(5 − 2x)]
Curve crosses x − axis, y =0 ln(2x − 5) = 0 2x − 5 = e0 =1 x =3 ⇒ (3,0) ✓
⋅ ln x
(1+ln x)⋅1
d
dx 1
(x − 1) ⋅ ln x
+ ln x ✓
=
2(c)
d
+1
x
=
2(b)
=x⋅
x
=
2(a)
(x)
1
dx 1+ln x
1(d)
dx
=x⋅
[(x − 1) ln x]
= (x − 1) ⋅
d
d
+2 ✓
x
x ln(5 − 2x)
=x⋅
(ln x + 2x)
= (x − 1) ⋅
1(c)
d dx
(8 − x 3 ) − 3x 2 )
✓
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479
A math 360 sol (unofficial) 3(c)
y = 3 ln(x − 3)
Ex 17.3 4(ii)
At stationary point, dy
dy
= 3(
dx
=
1
=0
dx ln x−1 = (ln x)2
Curve crosses x − axis, y =0 3 ln(x − 3) = 0 ln(x − 3) = 0 x−3 = e0 x−3 =1 x =4✓ (4,0) ⇒
0
ln x = 1 x =e y|x=e =
e =e ln e
⇒ (e, e) Sign Test x 3−
) ⋅ (1)
x−3
3
dy
x−3
dx
sign
−
e
e+
0
+
⇒ min dy
|
=
dx x=4
3 4−3
5
=3✓ 3(d)
y = ln(3x − 2)2 = 2 ln(3x − 2)
dy dx
= 2(
dx
|
y= dy dx
d [ln(x+1)] dx
−ln(x+1)⋅
d (x+1) dx
(x+1)2 1 (x+1)⋅ x+1
−ln(x+1)⋅1 (x+1)2
1
−ln(x+1) (x+1)2
>0
dx 1−ln(x+1) (x+1)2
>0
1 − ln(x + 1) ln(x + 1) x+1 x
) ⋅ (3)
6 3−2
>0 ∵ x > −1 0) x = ea ✓ 17(ii)
dy dx
= a − [x ⋅ = a − (x ⋅
d dx 1
(ln x) +
d dx
=x⋅
d dx 1
(ln x) +
d dx
(x) ⋅ ln x
+1 ⋅ ln x
x
=1
+ ln x
At stationary point, dy
=0
dx
1 + ln x = 0 (x) ⋅ ln x]
x
+1 ⋅ ln x)
x
=x⋅
dx
=
1 e
1
1
e 1
e
y|x=1 = − ln e
= a − (1 + ln x) = a − 1 − ln x
=−
e
1 1
⇒ M( , )✓ e e
At stationary point, dy
=0
dx
d2 y
a − 1 − ln x = 0 ln x =a−1 x = ea−1 ✓
dx2 d2 y
=
1 x
|
dx2 x=1
y|x=ea−1 = a(ea−1 ) −(ea−1 ) ln(ea−1 ) = a(ea−1 ) −(ea−1 )(a − 1) = ea−1 [a −(a − 1)] = ea−1 (a −a + 1) = ea−1 ⇒ (ea−1 , ea−1 ) d2 y dx2 d2 y dx2
=− |
e
𝑦 𝑂
1 1 e
( )
=e >0 ⇒ min without reflection ⇒ max at M 19
1
x x is (variable)variable x n is (variable)constant ⇒ Tom is wrong ax is (constant)variable ⇒ Doris is wrong
x
x=ea−1
=−
1 ea−1
0
⇒ max. ✓ 18(i)
=
𝑦 = |𝑥 ln 𝑥|
1 dy
𝑥 ✓
= x ( )+1(ln x)
( )
=1
y dx dy dx d dx
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1
( )
y dx 1 dy
(x x )
x
+ ln x
= y(1 + ln x) = x x (1 + ln x)
486
A math 360 sol (unofficial) 20
Clearly,
Ex 17.3
ln 3x
= ln 3 + ln x ≠ ln x The derivative might be the same because the derivative of the constant is ln 3 is 0.
21
No. d d ln|ax| = a ⋅ ln|x| dx dx For a = −1, LHS =
d dx
(ln 3x)
= =
d dx d dx
(ln 3 + ln x) (ln 3) +
=0 = d dx
(ln x)
=
+
d dx 1
d dx
ln|−x|
RHS = −1 ⋅ (ln x)
d dx
=
d dx
ln|x| = −
ln|x|
=
1 x
1 x
LHS ≠ RHS
x
1 x 1 x
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487
A math 360 sol (unofficial)
Rev Ex 17 A2(b)
Rev Ex 17 A1(a)
d dx
d dx
(x) −2
= 3(1)
d
(
d dx
(cos x)
A2(c)
x
d
d dx
(tan 3x)
+
= 2x ⋅ (3 sec 2 3x)
d dx
d dx
A2(d)
A2(e)
=3
d d dx
⋅ tan 3x
A3(i)
d dx
+
d dx
(3x 2 ) ⋅ ln x
+6x
x
⋅ ln x
+6x ln x ✓
=
ex [(2x + 1) (2x + 1)2
=
ex (2x−1) (2x+1)2
d
ln (
= d
(sin 8x)
d dx
d
✓ ) =
d dx
d
[ln(1 − 2x) − ln(1 + x)]
dx −2
2x−1
−
1 x+1
✓
(cot x) = − csc 2 x
sin2 x + cos 2 x sin2 x 1 =− 2 sin x
(x 2 + 2)
=−
= − csc 2 x [shown] ✓
[(x + 1)e−x ] dx
1+x
− 2]
d cos x ( ) dx sin x d d sin x ⋅ (cos x) − cos x ⋅ (sin x) dx dx = sin2 x sin x (− sin x) − cos x (cos x) = sin2 x
[sin(x 2 + 2)] = cos(x 2 + 2) ⋅
= (x + 1) ⋅
1−2x
Show:
(sin 8x)4 dx
d x d (e ) − ex ⋅ (2x + 1) dx dx (2x + 1)2
(2x + 1) ⋅ ex − ex ⋅ 2 (2x + 1)2
LHS
= cos(x 2 + 2) ⋅ 2x = 2x cos(x 2 + 2) A2(a)
(ln x)
)
(2x + 1) ⋅
(sin4 8x)
= 96 sin3 8x cos 8x ✓ d
ex
=
= 3 ⋅ 4(sin 8x)3 ⋅ (8 cos 8x)
dx
(
+2 tan 3x ✓
= 3 ⋅ 4(sin 8x)3 ⋅
A1(e)
+(2e2x ) ⋅ sin x +2 sin x) ✓
=
dx
(3 sin 8x) dx
d
=
4
=3
(e2x ) ⋅ sin x
dx 2x+1
(2x) ⋅ tan 3x
+2
= 6x sec 2 3x A1(d)
d dx 1
d dx
)
(2x tan 3x)
= 2x ⋅
(sin x) +
= 3x
d d (sin 2x) − sin 2x ⋅ (x) dx dx = x2 x ⋅ (2 cos 2x) − sin 2x ⋅ 1 = x2 2x cos 2x − sin 2x = 2 x ✓
dx
(3x 2 ln x) = 3x 2 ⋅ = 3x 2 ⋅
x⋅
A1(c)
d dx
+2 sin x ✓
sin 2x
d dx
= e2x ⋅ cos x = e2x (cos x
−2(− sin x)
=3 dx
(e2x sin x) = e2x ⋅
(3x − 2 cos x)
=3
A1(b)
d dx
(e−x )
= (x + 1) ⋅ (−e−x ) = e−x [−(x + 1)
+1]
= e−x (−x − 1
+1)
+
d dx
+1
(x + 1) ⋅ e−x ⋅ e−x
= −xe−x ✓
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488
A math 360 sol (unofficial)
Rev Ex 17 dy
A3(ii) t = 20 cot θ dt
2
= −20 csc θ
dθ dθ
=
dt
= −2(2 cos x) ⋅
dx
1
20
At turning points,
sin θ
dy
=0
dx dt
|
π θ= 4
=− =− =− =−
A4(i)
1 20 1
4 sin x cos x − sin x = 0 sin x (4 cos x − 1) = 0
π
sin2 ( ) 4
√2 20 2 1 2
2
sin x = 0
( )
40
rad s
or
0 < x < 2π:
( )
20 4 1
𝑦 −1
✓
1
90° 180° 270° 360°
y = 1 − 2 cos 2 x + cos x
𝑥
−1
x = 0, π ✓ Curve meets x − axis, y =0 1 − 2 cos 2 x + cos x =0 2 2 cos x − cos x − 1 =0 (2 cos x + 1)(cos x − 1) = 0 cos x = − α=
π
1 2
or
Sign Test x π− dy sign −
cos x = 1
1 4 α ≈ 1.32 ⇒ 1st or 4th quad. 0 < x < 2π: S A α α T C x = α, 2π − α ≈ 1.32,4.97 ✓ cos x =
π 0
π+ +
1.32 0
1.32+ −
4.97 0
4.97+ −
dx
0 < x < 2π:
3
⇒ min at x = π
𝑦
⇒ 2nd or 3rd quad. 1 0 < x < 2π: 𝑥 90° 180° 270° 360° S A −1 α α x = α, 2π − α T C = 0,2π (no x = π − α, π + α solution) 2π 4π = , ✓ 3
(cos x) − sin x
= −2(2 cos x) ⋅ (− sin x) − sin x = 4 sin x cos x − sin x
−20 csc2 θ 1 2
=− dθ
d dx
x dy dx
1.32− sign +
⇒ max at x = 1.32 x dy
3
dx
4.97− sign +
⇒ max at x = 4.97 A5(i)
y = xex−3
A5(ii)
dy
=x⋅
dx
d
(ex−3 ) +
dx x−3
d dx
(x) ⋅ ex−3
=x⋅e +1 x−3 = (x + 1)e ✓ dy
|
dx x=3
⋅ ex−3
= 4e0 =4✓
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489
A math 360 sol (unofficial) A6(i) A6(iii)
Rev Ex 17 A8(ii) n = 40: 18e0.2t = 40
𝑦 𝑦 = ln 𝑥
2
20 9
20
t = 5 ln ( )
2
9
✓ A6(ii) x 2 ex ln(x 2 ex ) ln x 2 + ln ex 2 ln x + x 2 ln x
= 12 = ln 12 = ln 12 = ln 12 = −x + ln 12
ln x ⇒a=−
1
1
2
2
9
0.2t = ln ( )
𝑥 1 1 𝑦 = − 𝑥 + ln 12
𝑂
20
e0.2t =
≈4✓ B1(a)
d dx
=
(sin 4x − 3 cos 2x) d
(sin 4x)
dx
= cos 4x ⋅
d dx
= − x + ln 12 ✓
= cos 4x ⋅ 4
= ax +b
= 4 cos 4x
1
−3
d dx
(cos 2x)
(4x) −3(− sin 2x) ⋅
d dx
(2x)
−3(− sin 2x) ⋅ 2 +6 sin 2x ✓
2
1
⇒ b = ln 12 2
B1(b)
≈ 1.24
d
1
(
dx 1+cos 4x
=
Draw 1
1
2
2
d
)
[(1 + cos 4x)−1 ]
dx
= −(1 + cos 4x)−2 ⋅
y = − x + ln 12 ✓
1
A7(i)
cos θ (
1
+
= − (1+cos
)
1+sin θ 1−sin θ (1−sin θ)+(1+sin θ)
= cos θ [
= cos θ ( =
1
1−sin2 θ 2 cos2 θ
)
B1(c)
d dx
2 cos θ
✓
(x cos 2 3x) d
=x⋅
= 2 sec θ [shown] ✓
4x)2
dx
(cos 2 3x)
= x ⋅ 2 cos 3x ⋅
d dx
=
[cos θ ( d dx
=2
1 1+sin θ
+
1 1−sin θ
)]
(2 sec θ) d
(
1
dx cos θ
=2⋅ =2⋅ =2⋅ = 2(
B1(d)
=
d cos θ⋅ (1) dx
d −1⋅ (cos θ) dx cos2 θ
−1⋅(− sin θ) cos2 θ
sin θ cos2 θ 1
sin θ
cos θ
cos θ
)(
)
B1(e)
+1
(x) ⋅ cos 2 3x ⋅ cos 2 3x
(x 2 − 3 tan2 4x) d dx
(x 2 )−3
d dx
(tan2 4x) d
(tan 4x)
−3 ⋅ 2 tan 4x ⋅
= 2x = 2x
−3 ⋅ 2 tan 4x ⋅ (4 sec 2 4x) −24 tan 4x sec 2 4x ✓
d
dx
[2 cos(1 − x 2 )]
=2
n = 18e0.2t
d dx
[cos(1 − x 2 )]
= 2[− sin(1 − x 2 )] ⋅
d dx
(1 − x 2 )
= 2[− sin(1 − x 2 )] ⋅ (−2x)
n|t=10 = 18e2 ≈ 133 ✓ © Daniel & Samuel A-math tuition 📞9133 9982
(cos 3x)
d dx
= 2x
dx
= 2 sec θ tan θ ✓ A8(i)
d dx
)
cos θ⋅0
d dx
+
= x ⋅ 2 cos 3x ⋅ (−3 sin 3x) + cos 2 3x = x ⋅ 2 cos 3x ⋅ (−3 sin 3x) + cos 2 3x = cos 2 3x −6x sin 6x ✓
k=2✓ A7(ii)
(1 + cos 4x)
⋅ (−4 sin 4x)
4x)2
4 sin 4x
= (1+cos
]
d dx
= 4x sin(1 − x 2 ) ✓ sleightofmath.com
490
A math 360 sol (unofficial) B2(a)
d
x2
(
dx ln 2x
Rev Ex 17 B3(i)
)
d 2 d (x ) − x 2 ⋅ (ln 2x) dx dx = (ln 2x)2 d ln 2x ⋅ 2x − x 2 ⋅ (ln 2 + ln x) dx = (ln 2x)2 1 ln 2x ⋅ 2x − x2 ⋅ x = (ln 2x)2 ln 2x ⋅
=
2x ln 2x (ln 2x)2
−x
y = sin 3x + cos 3 x dy
dy
dx
2(x2 −1)
6
1
√3 2
2
= 3(0) −3 ( ) ( )
✓
2
3 3
=− ( ) 8
d
d dx
[ln(2x + 1)] +
dx 2
=e
⋅
1
= = = d dx
=
dt
= 0.3,
d
dx −4
dt
= 0.3,
=?
(e2x ) ⋅ ln(5 − 4x) ln(5 − 4x)
=
|
dt x=π
=?
dy
×
dx
dx dt
π
At x = , 6
dy
9
|
dt x=π 6
+ ln(5 − 4x)] ✓
dy
6
1
d
dx 1 1 + ( e2x ) ⋅ 2 1
5−4x 4
|
dt x=π
⋅ ln(2x + 1) dx
[ln(5 − 4x)] +
dy
6
dt
4x−5
=
dx
dy
= e2x [
=
B3(ii)
[e ln(5 − 4x)]
1 x 2
=
⋅ ln(2x + 1)
1 x 2
1
d
(x 2 − 1)
+2x
2x+1
= e2x ⋅
dx
d dx
+2x ln(2x + 1) ✓
2x+1
B2(e)
π
6
=− ✓
= (x 2 − 1) ⋅
B2(d)
π
2
6
[(x 2 − 1) ln(2x + 1)]
B2(c)
π
= 3 cos −3 sin cos 2
|
dx x=π
(cos x)
2 4 9
= (x 2 − 1) ⋅
=
d dx
+3 cos 2 x ⋅ (− sin x) −3 sin x cos 2 x
= 3 cos 3x = 3 cos 3x
B2(b) d
+3 cos 2 x ⋅
= 3 cos 3x
dx
= (− ) ⋅ (0.3) 8
≈ −0.338 ✓
2
[ln(x + √x 2 + 1)] 1 x+√x2 +1 1 x+√x2 +1 1 x+√x2 +1 1 x+√x2 +1 1 x+√x2 +1 1 √x2 +1
⋅
d
(x +√x 2 + 1)
dx
⋅ [1
+
⋅ (1
+
⋅ (1
+
⋅(
1 2√x2 +1 1 2√x2 +1 x √x2 +1
√x2 +1 √x2 +1
⋅
d dx
(x 2 + 1)]
⋅ 2x)
)
+x
)
✓
[x 3 ln(cos 2 x)] d dx
[2x 3 ln(cos x)]
= 2x 3 ⋅ = 2x 3 ⋅ = 2x 3 ⋅
d
[ln(cos x )]
dx 1
cos x 1
⋅
(cos x)
(− sin x)
cos x 3 sin x
= −2x (
d dx
cos x
)
= −2x 3 tan x
© Daniel & Samuel A-math tuition 📞9133 9982
+
d dx
(2x 3 ) ⋅ ln(cos x)
+2(3x 2 ) ⋅ ln(cos x) +6x 2 ⋅ ln(cos x) +6x 2 ln(cos x) +6x 2 ln(cos x)✓
sleightofmath.com
491
A math 360 sol (unofficial) B4
Rev Ex 17
y = sin x cos 3 x
Sign Test x
dy dx
= sin x ⋅
d dx
(cos 3 x) d
2
= sin x ⋅ 3 cos x ⋅
dx
+
d dx
dy
(sin x) ⋅ cos 3 x
(cos x) + cos x
= sin x ⋅ 3 cos 2 x ⋅ (− sin x)
+ cos 4 x
= −3(1 − cos 2 x) cos 2 x
+ cos 4 x
= −3 cos 2 x + 3 cos 4 x
+ cos 4 x
2
2
0
−
sign −
π+
⇒ (0,0) is not max or min
⋅ cos x x dy
⇒( , 6
dy
dx 5π
=0
⇒(
4 cos 4 x − 3 cos 2 x =0 2 (4 2 cos x cos x − 3) = 0
π
π+
6
6
6
0
−
16
5π
5π+
) is max 5π−
x dy
π−
sign +
dx π 3√3
= 4 cos 4 x −3 cos 2 x At turning point, dx
π
2
3
+ cos 4 x
= −3 sin2 x cos 2 x
dx
π−
6
6
6
sign +
0
−
6
,−
3√3 16
) is min
√3
cos x = 0 or cos x = ± 2 π 0 < x < π: α= 𝑦 6 ⇒ 1st, 2nd, 3rd, 4th quadrant 1 𝑥 0 < x < π: 90° 180° 270° 360° x = α, π − α, π + α, 2π − α −1 π 5π π x= , 6 6 x= 2 π π y|x=π = sin cos 3 2
2
2
(0)3
= (1) =0 π
⇒ ( , 0) 2
π 3
π
y|x=π = sin (cos ) 6
6
6
1 √3 2 2 1 3√3
3
= ( ) = ( =
2 3
16 π 3√3
⇒( , 6
16
8
)
√3
)✓
y|x=5π = sin
5π
6
6
1
= (− 2
=− ⇒(
5π 6
,−
3
16 3√3 16
(cos
√3 ) 2
5π 3 6
)
3
√3
)✓
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492
A math 360 sol (unofficial)
Rev Ex 17
B5(a) y = 2e4x + 8e−4x 1st derivative dy
4x )
dx
B5(b) y = x ln x − 2x dy
= 2(−4e
+ 8(−4e
= 8e4x
−32e−4x
dx
−4x )
=0
dx 4x
−4x
8e − 32e e4x − 4e−4x e4x e8x 8x
= ln 22 8 2
= ln 2 8 1
B6
= ln 2 4
1 4
−4( ln 2)
+ 8e
= 2e
−2
=1 = ln x − 1
+ ln x
−2
=0
Curve y = (2x + c) ln x dy
4
ln 2
dx
− ln 2
+ 8e
= 2(2)
+8eln2
=4
+8 ( )
= (2x + c) ⋅ = (2x + c) ⋅
1
=2+
1
c
d dx 1
(ln x) +
x
⇒ ( ln 2 , 8) ✓
1
=−
|
dx x=1
2nd derivative
dx2
= 8(4e4x ) − 32(−4e−4x )
Line 2y = 5 − 3x
= 32e4x
y
|
+128e−4x 1 4
4( ln 2)
1 x= ln 2 4
⋅ ln x
+2 ln x
x
mnorm = − dy
4
d2 y
(2x + c) ⋅ ln x
Normal gradient at A(1,0)
1
dx2
+2
d dx
2
=8
d2 y
(x) ⋅ ln x −2
ln x − 1 = 0 ln x =1 x =e y|x=e = (e) ln(e) −2(e) = e −2e = −e ⇒ (e, −e) ✓
8 1
1 4
d dx
+(1) ⋅ ln x
dx
1
4( ln 2)
+
=x⋅( )
dy
= ln 4
y|x=1 ln 2 = 2e
(ln x)
At stationary point,
=0 =0 = 4e−4x =4 = ln 4
x
d dx 1 x
At turning point, dy
=x⋅
= 32e
+128 e
3
=− x+ 2
mline = −
1 4
−4( ln 2)
1 c 2+ +2 ln 1 1
2
−
+128 ( )
>0 1
⇒ min at ( ln 2 , 8) 4
B7(i)
1 2+c
1 2+c
2
2
= 32(2)
=−
3
1
+128eln2
1 2+c+0
5
Normal at A(1,0) ∥ line: mnorm = mline
1
= 32eln 2
=−
=−
3 2
2
2+c
=
c
=− ✓
3 4 3
t
m = 24e−8 7
m|t=7 = 24e−8 ≈ 10.0g ✓
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493
A math 360 sol (unofficial) 1
B7(ii) m
= m|t=0 t
2 1
24e−8 = (24)
Rev Ex 17 B8(ii) x|t=0 = 26 − 30e0 = −4 ✓
2
B8(iii) sub x = 25, 26 − 30e−0.2t = 25
= 12 −
t
= ln
8
t
1 2
= −8 ln
1 2
≈ 5.55h ✓ B7(iii)
dm
1
t − 8
=
−0.2t
= ln
t
= −5 ln
= 24 (− e )
dt
8
t − 8
= −3e |
dt t=8
B8(iv)
dx
1 30
= 6e−0.2t
3
dx
e
|
dt t=5
= 6e−1 ≈ 2.21° per minute ✓
x = 26 − 30e−0.2t
t →∞ −0.2t e →0 x → 26 ✓
x|t=9 = 26 − 30e−1.8 ≈ 21.0 ✓
© Daniel & Samuel A-math tuition 📞9133 9982
1 30
= −30(−0.2e−0.2t )
= −3e−1 = − g h−1 ✓
B8(i)
30
≈ 17.0 ✓ dt
dm
1
e−0.2t
sleightofmath.com
494
A math 360 sol (unofficial)
Ex 18.1 4(a)
Ex 18.1
∫(6x + 3) dx x2
= 6 ( ) +3(x) +c 1(i)
d dx
1(ii)
2
(x 2 + 5x) = 2x + 5 ✓
∫(2x + 5) dx= x 2 + 5 + c ✓
= 3x 2 4(b)
+3x +c ✓
∫(3 − √x) dx 1
2(i)
d
(
x
dx 1+2x
)=
= ∫ (3 −x 2 ) dx
d d (1+2x)⋅ x −x⋅ (1+2x) dx dx (1+2x)2
(1+2x)⋅1
=
3
x2
= 3x − ( 3 ) +c
−x⋅2
2
2
(1+2x)2 1+2x −2x (1+2x)2
=
1
= 3x − x√x +c ✓ 3 4(c)
= (1+2x)2 [shown] ✓ 2(ii)
2
x3
1
= 2(
1+2x 2x
= 3(a)
= ∫(3x 2 + 6x) dx 3
4(d)
x4
3(b) 3(c)
∫(x − 1)(x + 2) dx
=
= 2 ( ) +c =
+c ✓
4(e)
∫(−5) dx = −5x + c ✓
x3 3
1
2
=
= x +c ✓ 3(d)
1
∫ (− x2 ) dx= − ∫ x −2 dx = −( =
3(e)
1
1
1 x−2 2
=− 3(f)
∫
2 √x
−1
−2 1
dx = 2 ∫ x
+2x√x +c ✓
2
2
∫(2x − √x) dx
3
= ∫ (4x 2 − 4x 2 + x) dx 5
x3
x2
x2
= 4 ( ) −4 ( 5 ) + ( ) +c 3
) +c
4
= x3 3
+c ✓ 5(a) dx
∫
2x2 +3 x2
8
x2
5
2
− x 2 √x +
+c ✓
dx = ∫(2 + 3x −2 ) dx
= 2x
= 2 ( ) +c
2
2
= 2x
1 x2 1 2
= 4√x
2
x2
= ∫(4x 2 − 4x√x + x) dx
) +c
dx
4x2 1 − 2
4(f)
+c ✓
x
∫ 2x3 dx = 2 ∫ x = (
x−1
1
−3
x2
= ( ) +3 ( 3 ) +c
3 2
3
3
x2
+c
2
−2x +c ✓
2
∫ √x(√x + 3) dx
3 3 2
x2
= ∫ (x + 3x 2 ) dx
∫ √x dx = ∫ x dx =
+
= ∫(x + 3√x) dx
1 2
x2
+3x 2 +c ✓
= ∫(x 2 + x − 2) dx
∫ 2x 3 dx = 2 ∫ x 3 dx 4 1 4 x 2
2
= x3
) +c
+c ✓
1+2x
x2
= 3 ( ) +6 ( ) +c
∫ (1+2x)2 dx = 2 ∫ (1+2x)2 dx x
∫ 3x(x + 2) dx
+3 ( −
3 x
x−1 −1
) +c +c ✓
+c ✓
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495
A math 360 sol (unofficial) 5(b)
∫
x2 +1 2x2
dx = ∫ (
1
Ex 18.1
1
2 1 x−1
2 1
2 −1 1
2
2x
= x + ( = x − 5(c)
∫
x+1 √x
8(a)
1
+ x −2 ) dx
2
+c ✓
x2
y = 2 ( ) +3 ( 2
= x2
1
2 3
3
1
x2
x2
2 3 2
2
= =
6(b)
(3x+1)5 (3)(5) (3x+1)5 15
+c
x
x
+c ✓
8(b)
dy
⇒
(1−x)4
4
) +c
3
∫(1 − x)3 dx = (−1)(4) +c =−
−1
∴ y = x2 − + 1 ✓
+c
dx
(1−x)4
−
3
x−1
Curve at (−1,5): 5 = (−1)2 + 3 + c c=1
+c ✓
+2√x
∫(3x + 1)4 dx
3 x2
= 2x + 3x −2
) +c
dx = ∫ (√x + x −2 ) dx
= x
= 2x +
dx
= ( 3 ) + ( 1 ) +c
6(a)
dy
∝ (x 2 − 1) dy dx
= k(x 2 − 1) = kx 2 − k
+c ✓ x3
6(c)
∫(2x + 5)−3 dx =
(2x+5)−2
2
=− dy dx
3
+c
(2)(−2) 1 (2x+5)−2
= [
7
y = k ( ) −kx +c
−2 1
4(2x+5)2
1
= kx 3 3
] +c dy
+c ✓
dx dy
x3 3
=x
|
=9
k(2) − k = 9 3k =9 k =3 c|k=3 = 1
x2
y = 3( ) − ( ) + c 3
= 9 when x = 2,
dx x=2 2
= 3x 2 − x
2 1 2 − x 2
−kx +c
+c
y = 3 when x = 2, y|x=2 =3
Curve at (2,4), 1
(4) = (2)3 − (2)2 + c
1
2
3 2
4 =8−2+c c = −2
3
k(2)3 − k(2) + c = 3 k+c
=3 2
c
=3− k 3
1
∴ y = x2 − x2 − 2 ✓ 2
1
∴ y = (3)x 3 −3x +1 3
= x 3 − 3x + 1 y|x=3 = (3)3 − 3(3) + 1 = 19 ✓
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496
A math 360 sol (unofficial) 9
dV dt
Ex 18.1 12(i)
= 6(2t − 1)2 + 1
V = 6 [ (2)(3) ] + t + c
=x⋅
= (2t − 1)3 + t + c V|t=1 (2 − 1)3 + 1 + c 1+1+c c
=4 =4 =4 =2
=
(given)
= = =
3
∴ V = (2t − 1) + t + 2 ✓ dx dt
(x√x + 1)
=x⋅
(2t−1)3
10(a)
d dx
d dx
d
√x + 1 + dx (x) ⋅ √x + 1 1
2√x+1
⋅1
+1
x
⋅ √x + 1
+√x + 1
2√x+1 x
+2(x+1) 2√x+1
x
+2x+2 2√x+1
3x+2
[shown] ✓
2√x+2
12(ii) ∫ 3x+2 dx = 2 ∫ 3x+2 dx 2√x+1
= 3t 2 + 2
√x+1
= 2x√x + 1 + c ✓ t3
x = 3 ( ) + 2t + c
13(a) ∫ √6x − 1 dx
3
=t
3
+ 2t + c
1 2
= ∫(6x − 1) dx Initial radius is 1: x|t=0 =1 03 + 0 + c = 1 c =1
3
= =
x = t 3 + 2t + 1 ✓ 10(b)
dA dt
2
− t2
d
√6x + 5 1
+c ✓
=− = 11 = 11 = 11 = −3
(2x−7)−1 (2)(−1) (2x−7)−1 −2
+c +c
= −(2x − 7)−1 +c
13(c) ∫
1 2x−7
1 √3−2x
+c ✓ 1
dx = ∫(3 − 2x)−2 dx 1
=
(3−2x)2 1 2
(−2)( )
+c
= −√3 − 2x +c ✓
A = 2t 3 − t 2 + t − 3 ✓ dx
9
=2⋅
+t +c
Area is 11 when t = 2: A|t=2 2(2)3 − (2)2 + (2) + c 16 − 4 + 2 + c c
11(i)
(6x−1)√6x−1
=2⋅
t2
3
+c
= 2 ∫(2x − 7)−2 dx
A= 6( ) − 2( ) + t + c = 2t 3
3 2
(6)( )
13(b) ∫ 2 dx (2x−7)2
= 6t 2 − 2t + 1 t3
(6x−1)2
d (6x + 5) 2√6x + 5 dx 1 = ⋅6 =
=
⋅
2√6x+5 3 √6x+5
✓
11(ii) ∫ 1 dx = 1 ∫ 3 dx 3 √6x+5 √6x+5 1
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497
A math 360 sol (unofficial) 14
dy
Ex 18.1 15
= x 2 (x − k)
dx
dx
= x 3 − kx 2 x4
dy
⇒
∝ x(2x 2 − 3) dy dx
= kx(2x 2 − 3) = 2kx 3 − 3kx
x3
y = ( ) − k( ) +c =
4 1 4 x 4
−
3 k 3 x 3
1
= kx k
4
3 8
x2
4
2 3k 2 x 2
4
2
(2, −2) lies on curve, 1
x4
y = 2k ( ) − 3k ( ) + c
+c
−
(−2) = (2)4 − (2)3 + c
(1,3) lies on curve,
−2
=4
3 = k(1)4 −
c
= k−6
− k 3
+c
8
−(1)
3
1
3k
2 1
2
(1)2 +c
3
3= k
− k
2
+c
+c
2
c =k+3
−(1)
(4,2) lies on curve, 1
k
4
3 64
(2) = (4)4 − (4)3 +c
(2,9) lies on curve,
2
= 64
−
9 = k(2)4 −
0
= 62
−
3 64 3
k
+c
k
+c
0
= 62
−
0
= 56
−
56 3
k
3 56 3
k
3k
2
2
(2)2 + c
9 = 8k − 6k + c 9 = 2k + c
−(2)
sub (1) into (2): 64
1
−(2)
sub (1) into (2): 9 = 2k + (k + 3) 6 = 3k k =2
8
+ ( k − 6) 3
k
k= 56 c|k=2 = (2) + 3 = 5
=3
1
8
3
∴ y = (2)x 4 − (2)x 2 + 5
c|k=3 = (3) − 6 = 2
2
3
2
4
2
= x − 3x + 5✓ ∴y=
1 4 x 4
3
−x +2✓
16
dy dx
= x(2 − 3x)
= 2x − 3x 2 x2
x3
y = 2( ) − 3( ) + c 2
= x2
3
− x3
+c
Curve at (1,2), (2) = (1)2 − (1)3 + c c=2 ∴ y = x2 − x3 + 2 Curve at (−2, p), p = (−2)2 − (−2)3 + 2 = 14 ✓
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498
A math 360 sol (unofficial) 17
d2 y dx2 dy dx
Ex 18.1 18(ii) With c1 = 1 & c2 = 5,
= 6x − 16
d2 y
x2
= 6 ( ) − 16x + a
dx2 dy
= 3x − 16x + a
dx
2 2
= 6x − 4
= 3x 2 − 4x + 1
y = x 3 − 2x 2 + x + 6 x3
x2
3
2
y = 3 ( ) − 16 ( ) + ax + b
At max value,
= x 3 − 8x 2 + ax + b
dy
=0
dx
3x 2 − 4x + 1 = 0 (3x − 1)(x − 1) = 0
(0,3) lies on curve, 3 = 03 − 8(0)2 + a(0) + b 3=b
x=
1 3
or x = 1(min)
y|x=1 = 5
(−1,0) lies on curve, 0 = (−1)3 − 8(−1)2 − a + 3 0 = −1 − 8 − a + 3 a = −6
3
d2 y
|
dx2 x=1
4 27
✓
= −2 < 0
3
1
⇒ max at x =
3
∴ y = x 3 − 8x 2 − 6x + 3 ✓ 19(i) 18(i)
d2 y dx2 dy dx
d
(
d
2x
dx √x+1
= 6x − 4
)= =
x2
= 6 ( ) − 4x + c1 2 2
=
= 3x − 4x + c1 x3
x2
3
2
=
y = 3 ( ) − 4 ( ) + c1 x + c2 =
= x 3 − 2x 2 + c1 x + c2
dy
|
dx x=1 2
√x+1⋅(2)
−2x⋅
1 2√x+1
x+1 2√x+1
−
x √x+1
x+1 2(x+1) −x (x+1)√x+1 2x+2
−x
3 (x+1)2
x+2
=
At min (1,5),
d
√x+1⋅dx2x −2x⋅dx√x+1 x+1
3
[shown] ✓
(x+1)2
=0
3(1) − 4(1) + c1 = 0 c1 =1 (1,5) lies on curve, (5) = (1)3 − 2(1)2 + c1 (1) + c2 5 = −1 + c1 + c2 5 = −1 + (1) + c2 ∵ c1 = 1 c2 = 5
19(ii) ∫
x+2 3 4(x+1)2
1
x+2
1
(x+1)2 2x
dx = ∫ 4
3
= ( =
dx
) +c
4 √x+1 x 2√x+1
+c ✓
20(a) ∫ x2 +2x dx x2 +2x+1 =∫
∴ y = x 3 − 2x 2 + x + 5 ✓
(x2 +2x+1) −1 x2 +2x+1
= ∫ (1 −
dx
1
) dx
x2 +2x+1 1
= ∫ (1 − (x+1)2 ) dx = ∫[1 − (x + 1)−2 ] dx (x+1)−1
= x − (1)(−1) +c = x+
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1 x+1
+c ✓
499
A math 360 sol (unofficial)
Ex 18.1
20(b) ∫ 3x2 −4x dx 9x2 −12x+4 =∫
22
1 (9x2 −12x+4) 3 9x2 −12x+4
1
= ∫[
1
= ∫[
dx
dx
y=
] dx (3x−2)2
4
3
3
=
= ∫ [ − (3x − 2)−2 ] dx = =
1
4
(3x−1)−1
3
(3)(−1)
− ⋅
3 1
+
3
4
dy
+c
4 x3 4 3
) +c
2 3
3
3
4
4 3
= a2 (2x
9
2
3
−
3 4 x4 12
+ c1 x + c2
= −1 + c1 = −1 =−
1
1
2 5
12
4 = − 4 = 4 = c2 =
4 3
+ x ) +c ✓ 4
− ( ) + c1 x + c2
5 3
(1,4) lies on curve,
= 3a2 ( x 2 + x ) +c 3 2
+ c1
1 x4
2 x2
c1
1
+
+ c1
3 1 3 x 3
|
1−
= 3a2 ∫ (x 2 + x 3 ) dx = 3a2 (
x2
dx x=1 13
20(c) ∫ 3a2 (√x + 3√x) dx 1
x3
tangent at (1,4) is y = 5 − x,
+c ✓
9(3x−1)
3 x2 3 2
=x−
] dx
4 3
1
= 1 − x2
=x−
9x2 −12x+4
−
3
dx2 dy
4 3
4 3
−
3
−
d2 y
∴y=
12 5 12 21
+ c1 (1) + c2
+ c1 + c2 5
+ (− ) + c2 3
∵ c1 = −
5 3
4 x2 2
−
x4 12
5
21
3
4
− +
✓
21(i) 5
d 1
− (2t + 1)2 ]
dt 5
3
3 2
1 5
= ⋅ (2t + 1) ⋅ 5 2
d dt
3 2
d dt
(2t + 1)
1
1 3
= ⋅ (2t + 1)2 ⋅ 2
− ⋅ (2t + 1)2 ⋅ 2
5 2
= (2t + 1)
1
1 3
(2t + 1) − ⋅ (2t + 1)2 ⋅
3
1 5
3
1
[ (2t + 1)2
3 2
3 2
1
−(2t + 1)2
= (2t + 1)√2t + 1
−√2t + 1
= √2t + 1[(2t + 1)
−1]
= 2t√2t + 1 ✓ 21(ii) ∫ t(√2t + 1) dt 1 = ∫ 2t√2t + 1 dt 2 5
1 1
= [ (2t + 1)2 =
2 5 1 10
(2t + 1)
5 2
1
3
3 1
3
− (2t + 1)2 ] +c − (2t + 1)2 6
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500
A math 360 sol (unofficial) 23(i)
dy dx
Ex 18.1 24(i)
= kx + c
d2 y dx2
x2
dy
2
dx
y = k ( ) + cx + a
x2
= λ ( ) − 18x + a
k
= x 2 + cx + a At stationary pt (2, −3): |
2k + c c
18x + a
λ x3
x2
y = ( ) − 18 ( ) + ax + b 2 3 λ = x3 − 6
=0
dx x=2
2 λ 2 x − 2
=
2
dy
= λx − 18
=0 = −2k
−(1)
2
2
9x + ax + b
Origin (0,0) lies on curve, λ
0 = (0)3 − 9(0)2 + a(0) + b
(2, −3) lies on curve,
6
0=b
k
(−3) = (2)2 + c(2) + a 2
−3 −3 −3
= 2k + 2c + a = 2k 2(−2k) + a = −2k + a
(1,16) lies on curve, λ
16 = − 9 + a + b
−(2)
6 λ
16 = − 9 + a
∴b=0
6 λ
(0,1) lies on curve,
25 = + a
k
(1) = (0)2 + c(0) + a
−(1)
6
2
a
=1
Gradient at (1,16) is 9, dy
Put a = 1 into (2): −3 = −4k + 1 2k =4 k =2
|
=9
− 18 + a
=9
dx x=1 λ 2
a
= 27 −
λ 2
−(2)
sub (2) into (1): λ
Put k = 2 into (1): c = −2(2) = −4
λ
25 = + (27 − ) 6
2
1
25 = 27 − λ 3
∴y =
2 2 x 2 2
1
+ (−4)x + 1
3
λ
= x − 4x + 1 ✓
dx
6
a = 27 − = 24
= 2x − 4
Tangent Point: Gradient: Tangent:
= 6✓
24(ii) Put λ = 6 into (2):
23(ii) With k = 2, c = −4: dy
λ =2
2
∴ y == x 3 − 9x 2 + 24x ✓ (0,1) dy
|
= −4
y − y1
=
dx x=0
dy
|
dx x=0
(x − x1 )
y − (1)= −4(x − 0) y = −4x + 1 ✓
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501
A math 360 sol (unofficial)
Ex 18.1
24(iii) At a = 24, b = 0, λ = 6, y = x 3 − 9x 2 + 24x dy
26 (a) For f(x) = 1 g(x) = 2
= 3x 2 − 18x + 24
dx d2 y dx2
∫ f(x)g(x) dx = ∫(1)(2) dx = 2x + c
= 6x − 18
At turning pts, dy
[∫ f(x) dx][∫ g(x) dx] = [∫(1) dx][∫(2) dx] = x(2x) + c = 2x 2 + c ∴ ∫ f(x)g(x) dx ≠ [∫ f(x) dx][∫ g(x) dx]
=0
dx
3x 2 − 18x + 24 = 0 x 2 − 6x + 8 =0 (x − 2)(x − 4) = 0 x=2 or x = 4 y|x=2 = 20 y|x=4 = 16 ⇒ (2,20) ✓ ⇒ (4,16) ✓
26 (b) ∫ f(x) dx = ∫ 1 dx g(x) 2 1
= x+c 2
f(x) dx
d2 y dx2 d2 y
| |
x=2
dx2 x=4
= −6 < 0
⇒ max at (2,20)
=6>0
⇒ min at (4,16)
∫ g(x) dx [wrong notation ?] ∫ f(x)dx ∫ g(x)dx
=
Both are correct The arbitrary constant can factor in the difference in constant values. ∫(x + 1) dx ∫(x + 1) dx
= = = =
x2 2
2 2
x+a 2x+b 2
f(x) g(x)
dx ≠ ∫
f(x) dx g(x) dx
+ c1
x2 +2x+1 x2
2 dx
1
∴∫
(1)(2)
1 dx
= +d
+x+c
(x+1)2
=∫
+ c1 1
+ x + + c1 2
1 ⇒ c = + c1 2
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502
A math 360 sol (unofficial)
Ex 18.2 1(e)
Ex 18.2 1(a)
5
9 1
∫4
√x
9
1
5
∫2 3x dx = 3 ∫3 x dx
x2
=[1] 2
5
x2
1
9
dx = ∫4 x −2 dx
4
= 3[ ]
1 9
2 3
= 2 [x 2 ]
3
= [x 2 ]53
4
2
=
9
= 2[√x]4
3 2 (5 − 32 ) 2
= 24 ✓ 1(b)
9 1 ∫1 x 2 dx
9
3
= 2[√9
− √4]
= 2(3
− 2)
=2✓
x2
=[3] 2
1
1(f)
3 9
2
4 3
4
∫1 x√x dx = ∫1 x 2 dx
= [x 2 ] 3
x2
=[5]
2 9 = [x√x]1 3 2 = (9√9 − 1√1) 3 2 = [9(3) − 1] 3
2
2 5 2
=
5
2
8
2(a)
1 x3
= [2] 3
4
2(b)
0
∫−1(3x 2 − 2x + 5) dx 0
x3
+5) dx
x2
= [3 ( )−2 ( ) +5x]0−1
3
]
3
3
2
+5x]0−1
= [x
1 1 3
= [(0)3 −(0)2 +5(0)] −[(−1)3 − (−1)2 + 5(−1)]
1 1
=− (
3 3
−x
2
1
3 x 2
1
−(4 + 4)
= ∫−1(3x 2 −2x
3
− [x −1 ]32 3
18
− 4x]1−1
= −8 ✓
=− [ ]
=
−1
= (4 − 4)
3 −1 2
=
2
= [4(1)2 − 4(1)] −[4(−1)2 − 4(−1)]
dx = ∫2 x −2 dx 3x2 3 = [
1
= [4x 2
2
− 13 )
2 3 = [(23 )3 − 1] 4 3 = (22 − 1) 4 1 =2 ✓
1 x−1
x2
= [8 ( ) − 4x]
4
1
1
∫−1(8x − 4) dx
1
2
3
3 1
5 2
= (83
∫2
62
= 12 ✓
3 2 8 = [x 3 ] 4 1
1(d)
− 1)
5
3
2
5
− 12 ]
= (25
1
8 1 −1 ∫1 2 x 3 dx
1 5 2
= [4
= 17 ✓ 1(c)
4
5
1
=7✓
1
− ) 2
✓
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503
A math 360 sol (unofficial) 2(c)
Ex 18.2
4
4(a)
∫1 (6x − 3√x) dx 1
4
= ∫1 (6x
− 3x 2 ) dx 4
3
x2
x2 2
3 2
= ∫ (1 + x −2 ) dx
1
1
4
= [3x 2
− 2x ]
= [3x 2
− 2x√x]1
1 4
= [x
− ]
= 32
=3
2
4(b)
2
x4
2
2
0
1
2
4
0
= [ x4 − x2 ]
= [( 1
4
4
2
3
−
2
(2)2 2
t2
2
2 2
1
− t2]
1
−
(2)2 ]
1
− [− (1) − (1)2 ] +2
2 1 2
4(c)
= ∫1 (x 2 − x − 2) dx −
− 2t) dt
= −2 ✓
2
3
− 2t) dt
− 2 ( )]
1
= −4
∫1 (x + 1)(x − 2) dx
=[
−1
)
t 1 [− (2) 1
=
=0✓
(2)3
t−1
= [−
−0
x2
dt
1
1
=0
=[
1
− (1)]
−0
4 3
t2 2 1 ∫1 (t2 2 ∫ (t −2
=
= [ (2)4 − (2)2 ] − [ (0)4 − (0)2 ]
x3
x 1 1 ] − [(1) (4)
3
2 1−2t3
∫1 =
x2
− 2 ( )]
4
−1 1 1 4
4
∫0 x(x 2 − 2) dx
=[
4
]
=3 ✓
= ∫0 (x 3 − 2x) dx
3(b)
+
= [(4) −
−1
x−1
= [x
= [3(1)2 − 2(4)√4] −[3(1)2 − 2(1)√1]
= 31 ✓ 3(a)
x +1 dx x2 1 4 1 = ∫ (1 + 2 ) dx x 1 4
= [6 ( ) − 3 ( 3 )] 2
4 2
∫
− 2x]
√x 1
4
− 2(2)] − [
3
2x − 1
(1)3 3
− (−
−
(1)2 2
− 2(1)]
1
3
1
x2
x2
= [2 ( 3 )
13 6
dx
= ∫1 (2x 2 − x −2 )
1
1
∫ 1
2
= (−3 )
4
)
2
4
1
= −1 ✓
=[ x 3
6
3 2
4
= [ x√x 3 4
dx 4
− ( 1 )] 2 1 2
1 4
− 2x ]
1
− 2√x]
4
1
= [ (4)√(4) − 2√(4)] 3
=
20 3
4
− [ (1)√(1) − 2√(1)] 3
+
2 3
1
=7 ✓ 3
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504
A math 360 sol (unofficial) 4(d)
9
∫ 1
1 3−2x2
x2
x−1 −1
= [−
3
−1.5 )
dx
1 − x 2 1 − 2
9
− 2x
−
3
= [− (9) +
)] 1
4 9
4 √(9)
− [− (1) +
4 √(1)
6(a)
]
2 −1 1 1 −1 2 x ] 2 1 1 2
− 2x −
]
2x 1 1
1
2(2)
2(1)
] − [2(1)4 − 2(1) −
3
+
4 1
]
1 2
0
∫−1 x(x − a)(x + a) dx 0
= ∫ x(x 2 − a2 ) dx = ∫ (x 3
−2⋅
3 2
1 2
=[ =[
]
1 2
1
x4
x4 (0)4
1 4
=0
1
a2
4
− 4x ]
2 3
2
+
3
=
2
− [ (1)√(1) − 4√(1)] 3 10
6(b)
3
2
0
2
−1
0 a2 x2
−
4
=[
x2
− a2 ( )]
4
4
= [ (4)√(4) − 4√(4)]
− a2 x) dx
−1
= [ x√x − 4√x] 3
−
2
a2 (0)2 2
]
−1
] −[
(−1)4 4
1
a2
4
2
− +
−
a2 (−1)2 2
]
1
− ✓
2
4
4
∫0 √x(a − √x) dx
= ✓
4
3
5(b)
4
)]
0
2
= −2
1 x−1
4
3
]
2
x4
−1
3 2
3
) dx
= 28 ✓
]
4 2 ∫1 (√x − ) dx √x 4 1 1 = ∫ (x 2 − 2x −2 ) dx 1 3 1 4 x2 x2
2
2x2 1 −2 2
= [2x 4
√x 1
−1
=[ x
+ x
− 2x −
=0✓
=[
−2
= [2x 4
= 27
1
=1
5(a)
) dx
+
= [2(2)4 − 2(2) −
1 9
+ 4x 2 ] +
x
2 ∫1 (8x 3
1
−2
= [8 ( ) − 2x + (
) − 2(
[−3x −1
2
∫1 (8x 3 =
dx
9 ∫1 (3x −2
= [3 ( =
5(c)
3 − 2√x dx x2
=∫ =
Ex 18.2
= ∫ (a√x − x) dx
2
0
4
∫1 (x 2 − x2 ) dx = ∫ (x 2
− 4x −2 ) dx
3
1
x3 =[ 3 x3 =[ 3
x2
= [a ( 3 ) −
2
−4(
x −1 )] −1 1
2
2
2
3
+ 4x −1 ] 1
2
1
− [ (1)3 + (1)]
=4
−4
3
3 1
=
3
1
=
= ✓ 3
© Daniel & Samuel A-math tuition 📞9133 9982
x2 2
4
]
x2
0 4
]
2 0 x2
4
]
2 0
2
(4)2
3
2
= [ a(4)√(4) −
4
= [ (2)3 + (2)] 3 2
−
= [ ax√x − 3
1 4 = [ x3 + ] 3 x1 4
3
= [ ax 2
2
1
1
4
= ∫0 (ax 2 − x) dx
2
sleightofmath.com
16 3 16 3
a−8
2
(0)2
3
2
] − [ a(0)√(0) −
]
−0
a−8✓
505
A math 360 sol (unofficial) 6(c)
2 2t2 +a
∫1
Ex 18.2 8
dt
t2 2
= ∫ (2 + at −2 ) dt
= [2t −
t−1
2
)]
=
−1 1 −1 ]2 at 1 a2 a
a
= [2(2) − (2)] − [2(1) − (1)] =4− a
=[
2
x2
a
2
1
=
− 3x
2)
1 12 1 12
x2
x3
1
2
3
2
[2x 2 ]1a
= [3x 2 −2(1)
2
= [3(1) −
2a2 −2
=2
2a2 −2
= −2
2a
9(i)
− x 3 ]12 2
(1)3 ]
(2x − 3)3
(2(1) − 3)3 +
a
=0
a
=0✓
dy 2
−[3(2) −
(2)3 ]
dx
+2
6
=x⋅
d dx
=x⋅[
−4
= = =
k
5
1 12
7
+
1
]
4(2x−3) 0 7
]
4(2(1)−3)
(2(0) − 3)3 +
7
]
4(2(0)−3)
5 6
y = x√1 + 2x 2
=0
2
−
1 7 (2x−3)−1 [ ]] 2 (2)(−1) 0
=1✓
dx
= [6 ( ) − 3 ( )]
2
7(b)
1 ∫2 (6x
[4 ( )]
2(a)
2
= −1
a ∫1 4x dx
2
7
2
−[
= +2✓ 7(a)
1 1
−2 + a
2
dx
1 (2x−3)3 [ [ (2)(3) ] 2
=[
= [2t − ] t 1 a
2(2x−3)2
= ∫0 [ (2x − 3)2 − (2x − 3)−2 ] dx
1
= [2t + a (
1 (2x−3)4 −7
∫0
√1 + 2x 2 1
2√1+2x2
(4x)]
2x2
+
d dx
+1
(x) ⋅ √1 + 2x 2 ⋅ √1 + 2x 2
+√1 + 2x 2
√1+2x2 2x2
+(1+2x2 ) √1+2x2
1+4x2 √1+2x2
[shown] ✓
∫−2 3(2 + x)2 dx
= 64
k 3 ∫−2(2
= 64
= [x√1 + 2x 2 ]0
= 64
= (2)√1 + 2(2)2 −(0)√1 + 2(0)2
+ x)2 dx
k
3 ∫−2(x 2 + 4x [ [ [
x3
3 (k)3 3
−[ 1 3 1 3
x2
k
2
−2
+ 4 ( ) + 4x]
3 x3
+ 4) dx
+ 2x
2
k
+ 4x]
−2
+ 2(k)2 + 4(k)] (−2)3 3
=
2
=
=
+ 2(−2) + 4(−2)]
k 3 + 2k 2 + 4k +
8
k 3 + 2k 2 + 4k −
56
3 3
k 3 + 6k 2 + 12k − 56 (k − 2)( + + ⬚) 2 (k − 2)(k + + ⬚) 2 (k − 2)(k + + 28) 2 (k − 2)(k + 8k + 28) ⇒ k=2✓ © Daniel & Samuel A-math tuition 📞9133 9982
=
9(ii)
2 1+4x2
∫0
√1+2x2
dx 2
=6
64 3
−0
=6✓
64 3
10(i)
64
Prove: (x − 4)(x 2 + 4x + 16) = x 3 − 64 LHS = x 3 +4x 2 +16x −4x 2 −16x −64 = x 2 − 64 [proven]✓
3 64 3
=0 =0
=0
sleightofmath.com
506
A math 360 sol (unofficial)
Ex 18.2
10(ii) ∫3 x3 −64 dx −1 x−4 =
12(ii) C| = 24 000 [10 + 4 (354 )] x=3 5
3 (x−4)(x2 +4x+16) dx ∫−1 x−4 3
= ∫−1(x 2 + 4x =[ =[ =[
x3
≈ $316 000 ✓
+ 16) dx
5
x2
3
2
−1
+ 4 ( ) + 16x]
3 x3
3
+ 2x 2
3 (3)3 3
12(iii) C| = 24 000 [10 + 4 (554 )] x=5
+ 16x]
≈ $384 000 ✓ 13
dV dt
−1
= 20 000 (t − 6)
= 20 000t − 120 000
2
+ 2(3) + 16(3)] −[
= 75
(−1)3 3
+14
Net change in Value
+ 2(−1)2 + 16(−1)]
4
= ∫0 [20 000t − 120 000] dt
1 3
= 89 ✓ 3
11(i)
2 ∫0 x dx
x2
t2
4
2
0
= [20 000 ( ) − 120 000t]
1
= [160 000 − 480 000 − (0)]
2
=[ ]
= −$320 000
2 0
1
= [x 2 ]20
Total loss = $320 000 ✓
2 1
= (22 −02 )
14(i)
2 1
3
8
Given: ∫0 f(x) dx = ∫3 f(x) dx = 12
= (4) 2
8
11(ii)
2
∫ |x| dx −2 2
2
= ∫−2|x| dx
+ ∫0 |x| dx
0
= [− ] 2
] − [−
(−2)2 2
= [x]83 =8−3 = 29 ✓
] +2
=2 =4✓
+2
C = 8000 (30 +
8
+2(12) +24
14(iii) ∫8 2f(x) dx 3 =
12(i)
+12
8
+2
2 −2
=[
= 12 = 24 ✓
= ∫3 1 dx +2 ∫3 f(x) dx
+ ∫0 x dx
0
−(0)2
8
14(ii) ∫8[1 + 2f(x)] dx 3
2
= ∫−2(−x) dx x2
3
∫0 f(x) dx = ∫0 f(x) dx + ∫3 f(x) dx
=2✓
x 1 3 ∫0 t 4 dt)
8 2 ∫3 f(x) dx
= 2(12) = 25 ✓
x 1
+1 +1 +1
= 24 000 (10 + ∫0 t 4 dt) x
5
t4
= 24 000 [10 + [ 5 ] ] 4
4
0 5 4
x
= 24 000 [10 + [t ] ] 5
0
4 5
= 24 000 [10 + x 4 ] 5
4
C|x=1 = 24 000 [10 + (1)] 5
= $259 200 ✓
© Daniel & Samuel A-math tuition 📞9133 9982
sleightofmath.com
507
A math 360 sol (unofficial)
Ex 18.2
8
∫3 [f(x) − mx] dx 8
8
∫3 f(x) dx −m ∫3 x dx x2
=0
8
−m [ ]
12
=0
2 3 m 2 8 − [x ]3 2 1 2
12
=0
2 1
12
− m(55)
=0
12
−
=0
55 2
2 55 2
m
m
−1
= −2 ∫0 x 2 dx
Given:
= −2m ✓ 16(ii) ∫−1(−x 2 ) dx = − ∫−1 x 2 dx 0 0 (c) = −m ✓
= −12
m 15(i)
0
= 2 ∫−1 x 2 dx
− m[(8) − (3)2 ] = 0
12
−
1 16(ii) ∫1 x 2 dx = ∫0 x 2 dx + ∫0 x 2 dx −1 −1 (b) 0 0 = ∫−1 x 2 dx + ∫−1 x 2 dx (equal area under curve)
=0
= 4 ∫−3 f(x) dx 4 ∫−1 f(x) dx
−1
24 55
17
✓
b
∫a f(x) dx ≥ 0 Not True
=a In theory
=b
b
4
∫ f(x) dx ≡ algebraic area ≠ geometric area
4
a
∫−3 f(x) dx = ∫−3 f(x) dx − ∫−1 f(x) dx =a−b✓
For example
15(ii) ∫−3 2f(x) dx = 2 ∫−3 f(x) dx −1 −1 𝑦
−1
= −2 ∫−3 f(x) dx
𝑦 = 𝑓(𝑥)
= −2(a − b) = 2b − 2a ✓
𝑃 𝑂
16(i)
𝑎
Q
𝑏
𝑥
𝑦 𝑦=𝑥
Algebraic area = P − Q Geometric area = P + Q
2
𝑥 Clearly f(x) can be negative between a and b 16(ii) ∫−1 x 2 dx = ∫0 x 2 dx (equal area under curve) 0 −1 (a) 1 ∫0 x 2 dx
= =
0 ∫−1 x 2 dx −1 − ∫0 x 2 dx
18
∵ symmetry in y − axis
In order to compute b
∫ f(x) dx a
f(x) must be continuous [no gaps] for a ≤ x ≤ b x ≠ 0 [gap]
= −m ✓
1
1 dx cannot be computed 2 −1 x
∴∫
© Daniel & Samuel A-math tuition 📞9133 9982
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508
A math 360 sol (unofficial)
Ex 18.3 2(e)
Ex 18.3 1(a)
1(b)
1(c)
2(a)
π
∫06 (3 cos x − 2) dx π
∫(sin x + 2) dx = (− cos x) + 2x + c = − cos x + 2x + c ✓
= [3(sin x) − 2x]06
∫(1 − 3 cos x) dx = x − 3(sin x) + c = x − 3 sin x + c ✓
= [3 sin ( ) − 2 ( )] −[3 sin(0) − 2(0)]
π
− 2x]06
= [3 sin x π
π
6
6
1
π
= [3 ( )
− ]
2
3
π
2
3
∫(5 sec x + 3) dx = 5 tan x + 3x + c ✓
1
3(a)
∫ cos 2x dx = 2 sin 2x +c ✓
3(b)
∫ sin 3x dx = − 3 cos 3x +c ✓
3(c)
∫ 2 cos 4x dx = 2 ∫ cos 4x dx
π
π 6
∫0 cos x dx = [sin x]06 π 6
1
=
−0
2 1
1
1
= 2 ( sin 4x) +c
= ✓
4
2
π 4
1
= sin 4x +c ✓ 2
π 4
2
∫0 sec x dx = [tan x]0
3(d)
π
= tan ( ) − tan(0)
1
1
∫ −3 sin 2 x dx = −3 ∫ sin 2 x dx
4
=1 = 1✓
1
1
= −3 (− 1 cos x) +c
−0
2
2
1
= 6 cos x 2
2(c)
− 0)
= − ✓
2
= sin ( ) − sin(0)
2(b)
−(0
3
+c ✓
π
π
∫02 sin x dx = [− cos x]02
3(e)
∫ 4 sec 2 5x dx = 4 ∫ sec 2 5x dx 1
π
= 4 ( tan 5x) +c
= −[cos x]02
5
4
= tan 5x +c ✓ 5
π
= −[cos ( ) − cos(0)] 2
= −(0
−1)
4(a)
=1✓
x
∫ 2 cos (2 + 1) dx x
= 2 ∫ cos ( + 1) dx 2
2(d)
π
1
∫02 (1 − 2 sin x) dx = [x − = [x + π
π 2(− cos x)]02 π 2 cos x]02 π
=( =
π 2
2
x
−(0
+c ✓
= 4 sin ( + 1) 2
4(b)
2
+ 0)
2
2
= [( ) + 2 cos ( )] −[(0) + 2 cos(0)] 2 π
x
= 2 [ 1 sin ( + 1)] +c
+ 2)
∫ 3 sin(2 − x) dx = 3 ∫ sin(2 − x) dx = 3 [−
−2 ✓
1 −1
cos(2 − x)] +c +c ✓
= 3 cos(2 − x) 4(c)
π
∫ 4 sec 2 (8x − 4 ) dx π
= 4 ∫ sec 2 (8x − ) dx 4 π
1
= 4 [ tan (8x − )] +c 8
4
1
π
2
4
= tan (8x − ) © Daniel & Samuel A-math tuition 📞9133 9982
sleightofmath.com
+c ✓ 509
A math 360 sol (unofficial) 5(a)
Ex 18.3
∫(sec 2 x − 4 sin x) dx = tan x − 4(− cos x) +c = tan x + 4 cos x +c ✓
9(a)
π 3 π 12
π
∫ cos (2x + 3 ) dx 1
π
2
3
π 3
= [ sin (2x + )] π 5(b)
5(c)
∫(3 cos x − 2 sin x) dx = 3 sin x − 2(− cos x) = 3 sin x + 2 cos x ∫(4 cos x = 4 sin x
+c +c ✓
12
1
π
2
3
= [sin (2x + )]
π 3 π 12
1
π
π
π
π
2
3
3
12
3
= [sin (2 ( ) + ) − sin (2 ( ) + )]
+ 3 sec 2 x) dx + 3 tan x +c ✓
1
π
= (sin π
− sin )
2
6
π 4
π 4
= [0
∫0 (1 + tan2 x) dx = ∫0 sec 2 x dx
1
π
=− ✓ 2
π
= tan − tan 0 4
9(b)
π
∫π 2 sin(π − x) dx 2
= 1 −0
π
= 2 ∫π sin(π − x) dx 2
=1✓
= 2 [−
π
∫04 [sec 2 x + 1] dx = [tan x + π
π
4
4
=1
+
π
−1
2
π 4
π
= 2[cos(0)
− cos ( )]
= 2(1
−0)
2
=2✓
∫[− sin(2x + 1)] dx
9(c)
= − ∫ sin(2x + 1) dx
1
π
=[
1
= − [− cos(2x + 1)] +c
−
2
2
1
+c ✓
= cos(2x + 1) 2
π
1
=[ π
=[
∫ cos (2x + 4 ) dx = 2 sin (2x + 4 ) + c ✓
π
∫0 [x − cos ( 3 − 6 x)] dx x2
8(c)
2
x)]ππ 2 π
= + 1 [shown] ✓
8(b)
π
cos(π − x)]π
= 2[cos(π − (π)) − cos (π − ( ))]
−0
4
1
= 2[cos(π −
π x]04
= [tan + ] −[tan 0 + 0]
8(a)
−1)
2
= [tan x]04
7
2
1
x2
1 π − 6
π
π
3
6
6
π
π
π
3
6
6
π
π
π
3
6
0 1
+ sin ( − x)]
2 (1)2 2
0
+ sin ( − (1))] −[
∫ 6 sin(3x + 2) dx = 6 ∫ sin(3x + 2) dx
=[
1
= 6 [− cos(3x + 2)] +c
=[
3
= −2 cos(3x + 2)
1
sin ( − x)]
+c ✓
1 1
π 6
6 1
+ ( )]
2
=(
6
π
+ sin ( )]
2
π 2
1
3
+ )
2
π
1
1
2
π
(0)2 2
6
π
π
π
3
6
6
π
π
3
+ sin ( − (0))]
− [0
+ sin ( )]
−[0
+ ( )]
−
6 √3 2
π
3√3 π
= + (3 − 3√3) ✓
© Daniel & Samuel A-math tuition 📞9133 9982
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510
A math 360 sol (unofficial) 9(d)
Ex 18.3
π
1
dx
= [− cos 5t] 5
d
12(a)
∫0 sin 5t dt π
sin(x 2 + 2x)
= cos(x 2 + 2x) ⋅
0
d dx
(x 2 + 2x)
= cos(x 2 + 2x) ⋅ (2x + 2)
1 = − [cos 5t]π0 5
= (2x + 2) cos(x 2 + 2x) ✓
1
= − (cos 5π − cos 0) 5
12(b) ∫(x + 1) cos(x 2 + 2x) dx
1
= − (−1
−1)
5
1
= ∫(2x + 2) cos(x 2 + 2x) dx 2
2
1
= ✓
= sin(x 2 + 2x) + c ✓
5
10(i)
d dx
2
(x cos x)
=x⋅
d dx
(cos x) +
d dx
(x) ⋅ cos x
= x ⋅ (− sin x) +1 cos x = cos x −x sin x ✓
13 Both are correct The arbitrary constant can factor in the differences in constant value ∫ sin x cos x dx 1
10(ii) d (x cos x) = cos x − x sin x dx d x sin x = cos x − x cos x ✓ dx
d
∫ x sin x dx = ∫ (cos x − dx x cos x) dx = sin x 11
− x cos x +c ✓
∫(sin x − cos x)2 dx = ∫(sin2 x −2 sin x cos x + cos 2 x) dx = ∫(sin2 x + cos 2 x −2 sin x cos x) dx = ∫(1 − sin 2x) dx 1
=x
− (− cos 2x)
=x
+ cos 2x
=x
+ (1 − 2 sin2 x) +c1
=x
+ − sin2 x
+c1
=x
− sin2 x
+ + c1
=x
− sin2 x
+c ✓
2
1 2 1 2 1 2
© Daniel & Samuel A-math tuition 📞9133 9982
= ∫ sin 2x dx 2
1
cos 2x
2
2
= (−
+ a)
1
1
4 1
2
= − cos 2x + a 1
1
2
4 cos2 x
1
= − [1 − 2 sin2 x] + a or = − [2 cos 2 x − 1] + a = =
4 sin2 x 2 sin2 x 2
1
1
4
2
− + a
=−
+ c1
=−
2 cos2 x 2
2
1
1
4
2
+ + a + c2
+c1 +c1
1 2
sleightofmath.com
511
A math 360 sol (unofficial)
Ex 18.4 2(c)
Ex 18.4 1(a) 1(b)
2x
∫e
dx =
3x
∫ 2e
1 2x e 2
1(c)
dx
1 2 ( e3x ) 3 2 3x e +c 3
0
= −2[e
= −2(
1
1
2(d)
−1)
e
2
1
∫1 e1−x dx = [−1 e1−x ]
2
1
2 1 x 2
= 2 (2e ) +c 1
= 4e2x
=
= −[e−1 −e0 ]
∫ 4e2x+1 dx = 4 ∫ e2x+1 dx 1 4 ( e2x+1 ) 2 2x+1
1
−1
= −[
+c
1−x )
e
1 e
2(e) +c
ln 5
∫0
5 (ex + 1) dx = [ex + x]ln 0
= [e(ln 5) + (ln 5)] −[e(0) + (0)] = 5 + ln 5 −1 = 4 + ln 5 ✓
= 3(−e1−x ) +c = −3e1−x +c ✓ 2
2(f)
∫0 ex dx = [ex ]20
−1]
e
=1− ✓
+c ✓
∫ 3e1−x dx = 3 ∫ e1−x dx 1
−[e1−x ]12
= −[e1−(2) −e1−(1) ]
+c ✓
= 2e
0
= e −e = e2 −1 ✓ 2(b)
1
e
∫ 2e dx = 2 ∫ e dx
2
]
1
= 2 ( 1 e2x ) +c
2(a)
(0)
= 2(1 − )✓
1 x 2
= 3(
1
−e−2
= −2(e−1 −1)
2
1(f)
2
1 − (2) 2
∫ e−2x dx= −2 e−2x +c
=
0
1 2 [−2e−2x ] 0
✓
1
1(e)
2
1
1
1 x 2
− x 1e 2 ]
= −2 [e−2x ]
+c
= − e−2x +c ✓ 1(d)
2
1
1
−
=
dx= 2 ∫ e
=
=[
+c✓ 3x
=
2 −1x ∫0 e 2 dx
ln 2
∫0
ln 2 3x
5e3x dx = 5 ∫0
e
1
ln 2
3
0
= 5 [ e3x ]
1 1 1 ∫0 e2x dx = [2 e2x ] 0
=
dx
5 3x ln 2 [e ]0 3 5
1 2x 1 [e ]0 2 1 = [e2 −e0 ]
= [e3(ln 2) −e3(0) ]
= [e2 −1] ✓
= [8
=
3 5
3
= [eln 2 3
2 1
5
2
3
−1] −1]
2
= 11 ✓ 3
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512
A math 360 sol (unofficial) 3
dy
Ex 18.4 5(c)
= e2x
dx
y=
1 2x e 2
1 2
∫0
2−x
dx = 2 ∫
= 2[
+c
−1
ln|2 − x|]
1 2x e 2
= 2 ln 2 ✓
2
5(d)
+2−
1 2 e 2
1
∫0
✓
1
3 2+3x
dx = 3 ∫0
1 2+3x
dx 1
1
= 3 [ ln|2 + 3x|] 3
4(a)
0
= −2(0 − ln 2)
e2
=2−
∴y=
1
1
= −2(ln 1 − ln 2)
+c =2
c
dx
= −2[ln|2 − x|]10
y = 2 when x = 1, y|x=1 = 2 1 2 e 2
1 2−x
2
1
∫ x dx = 2 ∫ x dx
= [ln|2 +
= 2 ln|x| + c ✓
= ln 5 − ln 2 5
1
= ln ✓
4(b)
∫ x+1 dx = ln|x + 1| + c ✓
4(c)
∫ 2x−1 dx = 2 ln |2x − 1| + c ✓
4(d)
∫ 2x+1 dx = 3 ∫ 2x+1 dx
1
2
1
3
0
3x|]10
6(a)
dy dx
1
=
1 2x+1
1
y = ln|2x + 1| + c 2
1
= 3 ⋅ ln|2x + 1| +c 2
3
= ln|2x + 1| 2
4(e)
2
1
1
∫ 1−2x dx = 2 ∫ 1−2x dx =2⋅
1 −2
2
1
1
∫ 4−3x dx = −3 ln|4 − 3x|
+c ✓ +c
1
= − ln|4 − 3x| +c ✓ 5(a)
=
=
1 2 1
1
2
2
∴ y = ln|2x + 1| + ✓ 6(b)
dy dx
3
44 ∫1 x dx
ln|1| + c = 0.5
c
ln|1 − 2x| +c
= − ln|1 − 2x| 4(f)
y = 0 when x = 0, y|x=0 = 0.5
+c ✓
=
1 3−2x 1
y = − ln|3 − 2x| + c 2
41 4 ∫1 dx x
= 4[ln|x|]14
y = 2 when x = 1, y|x=1 =2
= 4(ln 4 − ln 1)
− ln|1| + c = 2
= 4 ln 22
c
1 2
= 8 ln 2 ✓
=2 1
∴ y = 2 − ln|3 − 2x| ✓ 5(b)
1 2 ∫0 2x+1 dx
=
2
1 1 2 ∫0 dx 2x+1 1
1
= 2 [ ln|2x + 1|] 2
= [ln|2x +
0
1|]10
= ln 3 − ln 1 = ln 3 ✓
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513
A math 360 sol (unofficial) 7(a)
2
1
Ex 18.4
2
2
9(i)
x − x ∫0 (e3 + e 3 ) dx 1
4
4
= ∫ (e3x + 2 + e−3x ) dx 4
1 3
3 4
3
3
4
4 (0) 3
−[ e 4
3 4
= ( e3 3
−
4
7(b)
∫0
e−x
B 3−2x
⇒A=1
=B
1
]
9(ii)
3 −4(0) e 3 ] 4
⇒ B = −2
1 1−x
−
2 3−2x
✓
1
∫ (1−x)(3−2x) dx = ∫(
3 −4 e 3) 4 3
1 1−x
−
2 3−2x
) dx
= − ln|1 − x| + ln|3 − 2x| + c
4
= ln |
4 − 3
10
1 ex +e2x
)
∴ (1−x)(3−2x) =
0
= (e − e ) + 2 ✓ 4
+
=A
1 1 (− )( 2
2
− )
+0
4 3
3
+ 2(0) −
+2
4
−(
4 − (1) 3
+ 2(1) − e
A 1−x
1
4
3
=[ e 4
4
1 ( )(1)
3
x= :
0
− e−3x ]
4
4 (1) 3
e−3x ]
4 − 3
3
= [ e3x + 2x
x = 1:
1
4
1
=
Cover-up rule:
0
= [ 4 e3x + 2x +
1 (1−x)(3−2x)
dy dx
3−2x 1−x
|+c✓
1
= (x−1)(x−2) ≡
A x−1
+
B x−2
dx Cover-up rule:
1
= ∫ (e2x + e3x ) dx 0 1
= [ e2x
1
1
3
0
+ e3x ]
2 1
1
1
1
2
3
2
3
1 3 e ) 3
1
1 0 e ) 3
x = 1:
1 ( )(−1)
x = 2:
1 (1)( )
=A
=B
⇒ A = −1 ⇒B=1
= [ e2(1) + e3(1) ] − [ e2(0) + e3(0) ] =
8(i)
1 ( e2 2
+
1
1
5
2
3
6
−(
+
2
∴
dx
=−
1 x−1
+
1 x−2
= e2 + e3 − ✓
y = − ln|x − 1| + ln|x − 2| + c
2x−1 (x+1)(x+2)
y = 2 at x = 3, y|x=3 =2 − ln 2 + ln 1 + c = 2 c = 2 + ln 2
=
A x+1
+
B x+2
Cover-up rule: x = −1: x = −2:
−3 =A ( )(1) −5 =B (−1)( )
2x−1
∴ (x+1)(x+2) = − 8(ii)
dy
2
∫1
3 x+1
⇒ A = −3 y = − ln|x − 1| + ln|x − 2| + 2 + ln 2
⇒B=5
= ln | +
5 x+2
2x−4 x−1
|+2✓
✓
2x−1 dx (x+1)(x+2) 2
3 5 ) dx + x+1 x+2 1 = [−3 ln|x + 1| + 5 ln|x + 2|]12 = [−3 ln 3 + 5 ln 4] −[−3 ln 2 + 5 ln 3] 2 = −3 ln 3 + 5 ln 2 +3 ln 2 − 5 ln 3 = 13 ln 2 − 8 ln 3 ✓ = ∫ (−
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514
A math 360 sol (unofficial) 11(i)
1 (x−2)(x−1)2
A
=
x−2
+
Ex 18.4 B
1 12(ii) ∫5 dx 3 (x+3)(x−2)2
C
x−1
+ (x−1)2
5 1
= ∫3
Cover-up rule: 1 ( )(1)2
x = 2:
=A
=
1
[
+
25 x+3
1 5 1 ∫ [ 25 3 x+3
+
1
5
x−2 1 x−2
+ (x−2)2 ] dx + 5(x − 2)−2 ] dx
⇒A=1 = 1 (1)( )2
x = 1:
=C
⇒ C = −1
=
Substitution:
=
1 (−2)(−1)2
x = 0:
B ∴
B
+
−2
C
+ (−1)2
−1
=
= −1
1 (x−2)(x−1)2
= =
11(ii)
A
=
1 x−2 1 x−2
+ −
−1
+
x−1 1
−
x−1
=
−1 (x−1)2 1 (x−1)2
=
✓
=
5 1 ∫3 (x−2)(x−1)2 dx
13(i)
5
1 1 1 ] dx =∫ [ − − x − 1 (x − 1)2 3 x−2 =
5 1 ∫3 [x−2
−
1 x−1
− (x − 2)
−2 ]
= [ln|x − 2| − ln|x − 1| − [
1
2
4
1 (x+3)(x−2)2
1
1
4
2
2
(x−1)−1
=
x+3
+
x−2
1 25 1 25
−1
1 ( )(−5)2
x = 2:
1 (5)( )2
1
=
5
6
3
3
1
8
10
3
3
[ln − ln 6 + 4
10
9
3
[ln ( ) +
10
3
3
2
10
3
3
[2 ln +
3
]
]✓
y = ecos x = ecos x ⋅
d dx
(cos x)
= ∫ sin x ecos x dx
+
C (x−2)2
=A
3
1 25
x+3 1
[
+
+
= −(ecos x )
+(− cos x) +c
= −ecos x
− cos x
3
2 √e3x
∫0
ex−1
2 ex
dx = ∫0
ex−1
−2
−
1 25
x−2
1
© Daniel & Samuel A-math tuition 📞9133 9982
+
−
C 4
+ 1
x−2
+c ✓
dx
2
⇒C=
B
25 x+3
+ ∫ sin x dx
= ∫0 e dx
⇒A=
=C
A
]
]
2 2
[ln ( ) +
1
14(b)
25
1
∫ (e3
1
=
5
1 1 3
x−1
⇒B=−
− 3 cos 2x) dx
1
1 2
3
= 3e3x−1 − sin 2x 2
1
= 2e ✓
e3x−1 −3 ( sin 2x) +c 1
=
∴ (x+3)(x−2)2 =
8
13(ii) ∫ sin x (ecos x + 1) dx
Substitution: 3(4)
]
x−2 3
= e[x]20
x = −3:
1
5
|−
x−2
3
5
[[ln | | − ] − [ln | | − 5]]
Cover-up rule:
x = 0:
x+3
]]
5
]]
14(a) B
1 25
[ln |
−1
5
= − ∫ − sin x ecos x dx + ∫ sin x dx
1
A
1 25
dx
= ln + ✓ 12(i)
1 25
[ln|x + 3| + ln|x − 2| + 5 [
(x−2)−1
= ecos x ⋅ (− sin x) = − sin x ecos x ✓
= [ln | | + ] − [ln | | + ] 4
1 25
dx
x−2 1 |+ ] = [ln | x−1 x−1 3
3
25
dy
5
3
1
+c ✓
25
1 5
(x−2)2 5
+ (x−2)2 ] ✓
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515
A math 360 sol (unofficial) 15
dy dx
=
Ex 18.4 16(ii) ∫ sec x dx = ln(sec x + tan x) + c ✓
1
2x2 +1
− √e−2x
√x 3
1
17
1
= 2x 2 +x −2 −e−4x 5
1
x2
x2
2
2 1 2
x 1 − x 4 1 − 4 1 − x 4
y = 2( 5 ) + ( 1 ) −( 4 5
= x2 5
+2x
1
e
+c ∫
4 5
5
1
=
(0)2 + 2(0)2 + 4e−4(0) + c = 2
4e0 + c c ∴y=
4 2 x √x 5
=2 = −2 + 2√x + 4e
1 4
− x
x2 +a x+a
18
x2 2
+a −a2 ) a + a2
a+a2 x+a
) dx
−ax +(a + a2 ) ln|x + a| + c ✓
Recognize that the derivative of ln(−x) is also d
−2✓
+a
dx
= ∫ (x − a +
=2
1
+a
−a +0x +ax) −ax −(−ax
) +c
+4e
y = 2 when x = 0, y|x=0
x x2 −(x 2
dx
ln(−x) =
1 x 1
⋅
d
−x dx 1
(−x)
= − ⋅ (−1) 16(i)
d dx
= = = = = = = =
x
ln(sec x + tan x) 1 sec x+tan x 1 sec x+tan x 1 sec x+tan x 1 sec x+tan x 1 sec x+tan x 1 sec x+tan x 1 sec x+tan x
⋅
d dx
⋅[
=
dx
[(cos x)−1 ]
⋅ [−(cos x)−2 ⋅
d dx
1
possibility that the integral of can be either ln x + x
+ sec 2 x]
c or ln(−x) + c
(cos x) + sec 2 x]
19
⋅ [−(cos x)−2 ⋅ (− sin x) + sec 2 x] ⋅( ⋅(
sin x
+ sec 2 x)
cos2 x 1 cos x
⋅
x
The modulus sign is included to account for the
(sec x + tan x)
d
1
sin x cos x
+ sec 2 x)
⋅ (sec x tan x + sec 2 x)
d dx
2
(e−x ) = −2xe−x
∫ eax+b dx = but ∫ ef(x) ≠
eax+b a ef(x) f′ (x)
2
+c
+c
The student factored out the variable (−2x) outside the integral which makes it wrong.
sec x(tan x+sec x) sec x+tan x
= sec x ✓
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516
A math 360 sol (unofficial)
Rev Ex 18 A3(i)
Rev Ex 18
=
x3
−3 ( ) + c
4 x4
3
− x3
4
+c✓
=
A1(b) ∫ 2x2 −√x dx x
= 1
= ∫ (2x − x −2 ) dx x2
2
1 2
= 2( ) − = x2
+c
A2(a) ∫(2 + ex )2 dx = ∫(e2x + 4ex + 4) dx 1
= e2x +4ex + 4x + c ✓ 2
A2(b) ∫ e3x −2 dx = ∫(e2x − 2e−x ) dx ex
A2(c)
− sin2 x
1 2x 1 e −2 ( e−x ) 2 −1 1 2x e +2e−x +c 2
+c A3(ii)
✓
π 1 2 π 1−cos x 4
∫
dx
∫0 (cos x − 3 sin x) dx
= − ∫π2 − 4
π
=
π 4
4
π 1+cos 2 π sin 2
π
= [sin + 3 cos ] −[sin 0 + 3 cos 0] √2 2
= − [(
4
√2 2
−[0 + 3]
+ 3 ( )]
= − [(
= 2√2 − 3 ✓ π
dx
1+cos x 2 ] −[ sin x π
= [sin x + 3 cos x]0
=[
1 1−cos x π
= [sin x − 3(− cos x)]04
A2(d)
1 1 − cos x
π
4
π
∫π3 sin (2x − 3 ) dx
1+0 1
)
1
π
2
3
= [− cos (2x − )]
π 3 π 4
π 1 π 3 − [cos (2x − )]π 2 3 4 1 π π
−(
= − [1
−(
= − (1
−
= − (− = π
π
4
3
= − [cos (2 ( ) − ) − cos (2 ( ) − )] 2 1
= − (cos 2 1 1
=− (
2 2
3
π 3
3 π
2 √2
×
2 √2
π 4
1+cos
) −(
4
=
− cos x
[shown] ✓
π 4
π
− cos x − cos 2 x sin2 x
− sin2 x − cos 2 x sin2 x
=−
=
)
sin2 x + cos 2 x + cos x sin2 x 1 + cos x =− sin2 x 1 + cos x =− 1 − cos 2 x 1 + cos x =− (1 + cos x)(1 − cos x)
− 2√x + c ✓
=
sin x
=−
1
x2
1+cos x
sin x ⋅
= ∫(x 3 − 3x 2 ) dx x4
(
d d (1 + cos x) − (1 + cos x) ⋅ (sin x) dx dx = sin2 x sin x ⋅ (− sin x) − (1 + cos x) ⋅ cos x = sin2 x
A1(a) ∫ x 2 (x − 3) dx
=
d dx
π sin 4 √2 2 √2 2
1+
2+√2 √2
2 √2
)]
)]
)]
− 1)
)
√2 √2
= √2 ✓
− cos ) 6
√3 − ) 2
1
= (√3 − 1) ✓ 4
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517
A math 360 sol (unofficial) A4(i)
1 (x−3)(2x+1)
=
A x−3
+
Rev Ex 18 A6(a) ∫a(3 − 2x) dx 1
B 2x+1
1 ( )(7)
=A
1
1
2
(− )( )
1
1 7
x=− :
7 2
∴ (x−3)(2x+1) = =
7
1
=
⇒A=
=B
x−3 1
7(x−3)
7
⇒B=−
−
+
1
2 7
2 7
2x+1
−
2 7(2x+1)
✓
k 6 ∫2 (1 2
−
[x −
] dx
7(x−3) 7(2x+1) 7 1 2 ) dx − ∫ ( 7 4 x−3 2x+1
(k − k3
1 = [ln|x − 3| − ln|2x + 1|]74 7 1 = [(ln 4 − ln 15) − (ln 1 − ln 9)] 7 1 4 = (ln + ln 9) 7 1
15 12
7
5
A5(a)
d dx
3 3
[ln(cos x)] =
1 cos x
⋅
d dx
(cos x)
1 ⋅ (− sin x) cos x = − tan x [shown] ✓ =
π 3
1
2
2
= −100
− x 2 ) dx
= −100
k x3
]
=−
3 2 k3 3
8
) − (2 − )
−k−
3
52
=−
100 6 100 6
=0
3
k − 3k − 52 =0 2 (k − 4)(k + + ⬚) 2 (k − 4)(k + + 13) 2 (k − 4)(k + 4k + 13) =0 (k − 4)[(k + 2)2 − 4 + 13] = 0 (k − 4)[(k + 2)2 + 9] =0 k =4✓
✓
A7(i)
dy dx
= 3x 2 + k
At turning pt (−2,6):
π 3
dy
∫0 tan x dx = − ∫0 − tan x dx =
1
A6(b) ∫k 6(1 − x 2 ) dx 2
1
= ln
2
x2
= [4 ( )]
[3x − x 2 ]1a = 2[x 2 ]12 (3a − a2 ) − (3 − 1) = 2(1 − 4) 3a − a2 − 2 = −6 2 3a − a + 4 =0 2 a − 3a − 4 =0 (a + 1)(a − 4) =0 a = −1 or a = 4 ✓
1 A4(ii) ∫7 dx 4 (x−3)(2x+1)
= ∫4 [
a
[3x − 2 ( )]
Cover-up rule: x = 3:
x2
1
= ∫2 4x dx
|
dx x=−2 2
π −[ln(cos x)]03
=0
3(−2) + k = 0 k = −12 ✓
π = − [ln (cos ) − ln(cos 0)] 3 1 = − [ln − ln 1] 2
= ln 2 ✓ A5(b)
d dx
[ln(ex + 1)] =
1
= ln 2 ex
∫0
ex +1
⋅
d
(ex + 1)
ex +1 dx 1 = x ⋅ ex e +1 ex ex +1
[shown] ✓
2 dx = [ln(ex + 1)]ln 0
= ln(eln 2 + 1) − ln(e0 + 1) = ln(2 + 1) − ln(1 + 1) = ln 3 − ln 2 3
= ln ✓ 2
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518
A math 360 sol (unofficial) A7(ii)
dy dx
Rev Ex 18 B1(c) ∫1 (4x+1)4 −7 dx 0 2(4x+1)2
= 3x 2 − 12
1 1 2
y = 3 ( ) − 12x + a 3
=[
Curve at (−2,6): 6 = (−2)3 − 12(−2) + a a = −10
=[ =
3
∴ y = x − 12x − 10
4
(4x+1)3
2
(4)(3)
(4x+1)3 (4(1)+1)3
15
5 4
−[ =
632
−
15
= 41
9 10
3
1
2
2
0
1 2
6
π 1−sin2 x
= ∫π
3
6
1−sin x
dx
π (1+sin x)(1−sin x) 1−sin x
6
dx
π
= ∫ (1 + sin x) dx π 6
4
x2 2
+
2
= [x − cos x]ππ
5 2
(1)3
6
1
1
2 8
6
π 6
=
1
− (1)2 + (1)2 ] 5
π
6
= [(π) − (−1)] − ( −
2] 5
π
= (π − cos π) − ( − cos )
4 1 2 x ] 2 1
= [ (4) − (4) + (4) 3
12
B2(c) ∫ππ cos2 x dx 1−sin x
= ∫π
8 5
8
11
= − ln 8 ✓
2
5 x2 5 2
3
]
= −3 ln 2
= [4 ( ) − 4 ( ) + ( )]
4
7
8(4(0)+1)
✓
= 3 ln
4
5
+
= 3[ln 2 − ln 4]
= ∫1 (4x 2 − 4x√x + x) dx
− x
24
B2(b) ∫3 3 dx = [3 ln|x − 5|]13 1 x−5
B1(b) ∫4(2x − √x)2 dx 1
=
8(4(1)+1)
(4(0)+1)3
= 2 ln 5 ✓
=8 ✓
4 [ x3 3
−[
= 2(ln 5 − ln 1)
=
3
]
= 2[ln|2x + 1|]20
3 4 1 = [(2x + 1)2 ] 3 0
x3
]
= 4 [ ln|2x + 1|]
3 4
3 1 3 [92 − 12 ] 3 1 = (27 − 1) 3 26 = 3
7
8(4x+1) 0 7
B2(a) ∫2 4 dx = 4 ∫2 1 dx 0 2x+1 0 2x+1
∫0 √2x + 1 dx = ∫0 (2x + 1) dx (2x + 1)2 =[ ] 3 (2) ( ) 2 0
1
−
60 7
1
]
(4)(−1) 0
2
323
1 2
4
− ⋅
+
24
(4x+1)−1
7
+
24
=4
At point where curve meets y − axis (x = 0), y|x=0 = (10)3 − 12(10) − 10 = −10 ⇒ (0, −10) ✓
2
1
=[ ⋅
= x 3 − 12x + a
B1(a)
7
= ∫0 [ (4x + 1)2 − (4x + 1)−2 ] dx
x3
2
5π 6
+1+
√3 2
√3 ) 2
✓
7 30
✓
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519
A math 360 sol (unofficial) B2(d)
Rev Ex 18
π 2 π 4
B4(a)
∫ 2 csc 2 x dx
8x+13 (1+2x)(2+x)2
=
A
B
+
1+2x
C
2+x
+ (2+x)2
π
= 2[− cot x]π2
[not in syllabus]
Cover-up rule:
4
B3(i)
1
9 ( )(1.5)2
π π = 2 [(− cot ) − (− cot )] 2 4 π π = 2 (cot − cot ) 4 2 = 2(1 − 0) =2✓
x=− :
y = x 2 √2x − 1
Substitution:
dy dx
= x2 ⋅
d dx
= = = = =
2√2x−1
⋅2
x2
x = 0:
+2x ⋅ √2x − 1
x2
13 (1)(4)
=
B
= −2
8x+13
∴ (1+2x)(2+x)2 = =
+2x(2x−1) √2x−1
x2
2
∫1
+4x2 −2x √2x−1
2
√2x−1 √2x−1
4
1+2x
B
C
2
4
+ +
1
1+2x 4
4
−
1+2x
2
= ∫1 [4 ⋅
[shown] ✓
A
+ −
−2 2+x 2 2+x
1
+ (2+x)2 1
+ (2+x)2 ✓
8x+13 dx (1+2x)(2+x)2
= ∫1 [
5x2 −2x x(5x−2)
=C
⇒C=1
+2x√2x − 1
√2x−1
−3 (−3)( )2
x = −2:
d
1
=A
⇒A=4
√2x − 1 + dx x 2 ⋅ √2x − 1
2
=x ⋅
2
1
2
−2⋅
1+2x
1
+ (2+x)2 ] dx
2+x 1
+ (2 + x)−2 ] dx
2+x
1
= [4 ⋅ ln|1 + 2x| − 2 ⋅ ln|2 + x| + 2
B3(ii) ∫5 x(5x−2) = [x 2 √2x − 1]5 1 1
= [2 ln|1 + 2x| − 2 ln|2 + x| − = [2 ln | = [2 ln |
1+2x 2+x
|−
1+2(2) 2+(2)
5
1
4
4
6
1
4
12
= 2 ln +
sleightofmath.com
1
]
1
]
2
2+x 1
2
2+x 1
|−
= [2 ln − ]
© Daniel & Samuel A-math tuition 📞9133 9982
2
] (1)(−1)
1
√2x−1
= (5)2 √2(5) − 1 −(1)2 √2(1) − 1 = 75 −1 = 74 ✓
(2+x)−1
1 2+(2)
]
− [2 ln |
1+2(1) 2+(1)
|−
1 2+(1)
]
1
− [2 ln 1 − ] 3
✓
520
A math 360 sol (unofficial)
Rev Ex 18
B4(b) Partial fractions 1
B5(ii)
1
x2 +3x+2
A
= (x+2)(x+1) =
x+2
+
B
π
∫04 cos 3 2x dx π 4
x+1
= ∫ [(cos 2x)(cos 2 2x)] dx Cover-up rule:
0
1 ( )(−1)
x = −2:
π 4
=A
= ∫ [(cos 2x)(1 − sin2 2x)] dx
⇒ A = −1
0 π
1 (1)( )
x = −1:
=B
⇒B=1 1
∴
x2 +3x+2
=−
1 x+2
∫ x2 +3x+2 dx =
1
+
x+1
0
x+1
= ln |
π 4
dx
−
x+2
x+1
+c ✓
|
x+2
dx
1 6
1
− 0)
2
= 3(sin 2x) ⋅ cos 2x ⋅
dx
B6(a)
d dx
[ln(ln x)] =
∫2
1 x ln x
(2x)
= ln (
cos 2x
= ln (
⋅
ln x 1
d dx 1
(ln x)
x
x ln x
ln 4
)
ln 2 ln 22
)
ln 2 2 ln 2
)
ln 2
= ln 2 ✓
π
π
d
1 4 = ∫ 6 sin2 2x cos 2x dx 6 0 π 1 = [sin3 2x]04 6 1 π = (sin3 − sin3 0)
dx
(x n ln x)
= xn ⋅ = xn ⋅ =x
2
− 0)
⋅
= [ln(ln x)]42 = ln (
∫04 sin2 2x cos 2x dx
1 (1 6 1 = (1) 6 1 = ✓
1 ln x 1
= ln(ln 4) − ln(ln 2)
= 3(sin 2x) ⋅ cos 2x ⋅ 2
=
6
3
2
= 6 sin2 2x
1
= ✓
4
d
−
1
(sin 2x)
2
6
2
= d
1
2
3]
= 3(sin 2x)2 ⋅
6
π
= (1
dx
1
1
=
[(sin 2x)
6
−
2
(sin3 2x) d
−
= [sin 2x]0
= ln|x + 1| − ln|x + 2| +c
=
2
1
✓
1 ∫ (x+2)(x+1) dx 1 1
=∫
dx
π 4
= (sin − sin 0) −
1
d
1
= [ sin 2x]
Integral
B5(i)
π
= ∫04 cos 2x dx − ∫04 sin2 2x cos 2x dx
d dx 1
x n−1
(ln x) + +nx
d
dx n−1
+n x
(x n ) ⋅ ln x
⋅ ln x
n−1
ln x ✓
B6(b) n = 1:
6
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521
A math 360 sol (unofficial) d dx
Rev Ex 18 B7(a) ∫ 1 dN N
(x ln x) = 1 + ln x
⇒ ln x
=
d dx
ln|N| + a = kt ln|N| = kt − a N = ekt−a = bekt
(x ln x) − 1
∫ ln x dx = ∫[
d dx
(x ln x) − 1] dx
B7(b) N|t=0 = A be0 = A b=A ∴ N = ekt [shown] ✓
= x ln x −x + c ✓ n = 2: d dx
(x 2 ln x) = x + 2x ln x
⇒ 2x ln x =
d
dx 1 d
x ln x = (
= kt
x 2 ln x − x
2 dx
x 2 ln x − x)
∫ x ln x dx 1 d
=∫ [
2 dx 1 d
(x 2 ln x) − x] dx
= ∫ [ (x 2 ln x) − x] dx 2 dx 1
x2
2
2
= (x 2 ln x − ) + c ✓ n = 3: d dx
x 3 ln x = x 2 + 3x 2 ln x
⇒ 3x 2 ln x =
d
dx 1 d
x 2 ln x
= [
x 3 ln x − x 2
3 dx
(x 3 ln x) − x 2 ]
∫ x 2 ln x dx 1 d
=∫ [
3 dx 1 d
(x 3 ln x) − x 2 ] dx
= ∫ [ (x 3 ln x) − x 2 ] dx 3 dx 1
= (x 3 ln x
x3
−
3
3
)+c✓
n = m + 1: d dx
(x m+1 ln x)
= x m + (m + 1)x m ln x
⇒ (m + 1)x m ln x = x m ln x
=
d dx 1
(x m+1 ln x) − x m [
d
m+1 dx
(x m+1 ln x) − x m ]
∫ x m ln x dx =∫ = =
1
[
d
m+1 dx 1 d
m+1 1 m+1
(x m+1 ln x) − x m ] dx
∫ [dx (x m+1 ln x) − x m ] dx (x m+1 ln x −
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xm+1 m+1
)+c✓
sleightofmath.com
522
A math 360 sol (unofficial)
Ex 19.1 5
Ex 19.1 1
4 ∫0 x(4 4
= − ∫ sin(x − π) dx
𝑦
− x) dx
0 x2
x3
4
3
0
= [4 ( ) − ( )] 2
x3
= [2x 2 −
𝑥 𝑂
4
]
3 2
6 − [2(0)2 −
(0)3 3
= ∫ [(0) − (1 − ex )] dx
]
𝑦 = cos 2𝑥
𝑂 π 1 = [sin 2x]04 2 1 π = (sin − sin 0) 2 1
𝑥
𝜋
7
4
Shaded area =
2
= (1
−0)
2 1
4 ∫1 [(y
= [[
𝑦 2
− 3) + 2] dy
(y−3)3 3
4
4
] + 2y] 1
2
= unit ✓ Shaded area
((0)−3)
= [[
𝑦
1
= ∫ e2x dx =
0 1 2x 1 [ e ] 2 0
=8
𝑦 = 𝑒 2𝑥
1 2x 1 [e ]0 2 1 = (e2 − e0 ) =
𝑂
−[ =5
+ (3)]
𝑂
3
3
2
+ (0)] +
3
𝑥 = 𝑦(2 − 𝑦)
3
𝑂
2
y y = [2 ( ) − ( )] 2 3 0
𝑥
2 1 = [y 2 − y 3 ] 3 0
0
3
((0)−1)
2
0
𝑦 = (𝑥 − 1)2 + 1
3
3
𝑦
= ∫ [2y − y 2 ] dy 2
0
((3)−1)
3
0 2
= ∫ [(x − 1)2 + 1] dx
=[
2
= ∫ [y(2 − y) − (0)] dy
𝑦
3
+ x]
] + 2(0)]
2
Shaded area
3
+
3
3
3
Shaded area
2
=[
1
((0)−3)
8
= (e2 − 1) unit 2 ✓
(x−1)3
] + 2(0)] − [[
= 9 unit ✓
2 1
4
3
2
𝑥
1
3
𝑥 = (𝑦 − 3)2 + 2 𝑥
1
2
3
𝑦 = 1 − 𝑒𝑥
= −[x − ex ]20 = −[(2 − e2 ) − (0 − e0 )] = −[2 − e2 + 1] = (e2 − 3) unit 2 ✓
0
0
𝑥
= − ∫ (1 − ex ) dx
𝑦
= ∫ cos 2x dx
2
𝑂
2 0
π 4
π 4
2
0
−0
Shaded area
1
𝑦
2
3
= [ sin 2x]
𝑥
𝑦 = sin(𝑥 − 𝜋)
Shaded area
= 10 unit 2 ✓ 2
𝜋
4
3 0 (4)3 [2(4)2 − ] 3 2
= 10
𝑂
0
= −[− cos(x − π)]π0 = [cos(x − π)]π0 = cos 0 − cos(−π) =1 −(−1) 2 = 2 unit ✓
𝑦 = 𝑥(4 − 𝑥)
= ∫ (4x − x 2 ) dx
=
𝑦
π
Shaded area =
Shaded area
3
𝑥
=1 1 3
1
1
3
3
= [(2)2 − (2)3 ] − [(0)2 − (0)3 ] 1 3 1
−0
= 1 unit 2 ✓ 3
2
= 6 unit ✓
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523
A math 360 sol (unofficial) 9(i)
Ex 19.1 π
11(a) Area
At A, y = sin (2x + ) cuts x-axis (y = 0): 3
=0
=
sin (2x + ) = 0
=
y π 3
α π 2x +
= ✓
9(ii)
=[
3
Shaded Area π
π
= ∫03 sin (2x + ) dx 1
π
π 3
2
3
0
1
π
π 3
2 1
3
= [− cos (2x + )] = − [cos (2x + )]
1
𝐴
4
3
𝑂
+ (5)] − [ 1
4
√
Shaded area
𝑦
a 0 a
1
√
1
3 2
𝑦=𝑒
3 2
𝑂
𝑎
𝑥
1
3
√
x2
2
x
−
) dx
1
3 x2
2
2
3
√
x2
+ ∫ 3 (2 − 2
) dx ) dx
3 2 x √3 2
+4 +
1
3
2
2
2
3
3 2
3
−2 (2√ + 2
3 √
3 2
)
= 10 − 2 (2√ + 3√ )
2
ea
2
√
+ ∫ 3 (2 −
+ [2x + ]
3 √ 2
=2+3
unit ✓
1
3
2
2
2
3
= 10 − 4√ − 6√
10(ii) Shaded area = 0.5
e a
3 𝑥2
𝑥
3
= [2x + ]
= e−(0) −e−(a)
ea a
2
3 1
a 1 = [ e−x ] −1 0 = −[e−x ]a0 = [e−x ]0a
ea
𝑦 =2−
= ∫ 3 (2 −
−𝑥
0
1
+ (2)]
−6
4 1
= ∫ e−x dx
1−
4
∫1 − (2 − x2 ) dx
= ∫ e−x − (0) dx
1
(2)4
3
1 1 = − (−1 − ) 2 2 3 2 = unit ✓
=1
2
𝑦
𝑥
𝜋
𝑂
= − (cos π − cos )
10(i)
(5)4
3
π
2
4
4
𝑦 = sin (2𝑥 + ) 11(b)
0
+ x]
= 155 unit 2 ✓
𝜋
3
𝑥
𝑂 2 5
5
x4
= 161
𝑦
𝑦 = 𝑥3 + 1
+ 1] − (0) dx
x + 1 dx
=[
π
x
𝑦
2
=0 =π
3
5 ∫2 [x 3 4 ∫ 3
= 0.5 =
1
Note: Deal with the upper limits first then the lower limits.
2
=2 = ln 2 ≈ 0.693 ✓
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524
A math 360 sol (unofficial) 12
Ex 19.1 14(i)
𝑦 𝑦 = 𝑥2 + 1
y = x2 − x − 6 = (x − 3)(x + 2) 𝑦
A 𝑂
B
𝑥
1
4
Line PQ Point:
2−0
3
2 3 2
8
3
3
=− x+
=
=[
3
0
3
4
+ [−
+[
3 4
+
3 1
=[ 8
+ ] − (0)dx
=[
3
2 x2
8
3 2
3
4
=
+ [− [ ] + x]
= [( + 1) − (0)]
=
−2
4 2 + ∫1 [− x 3
1
=
3
+area of B
+ 1] − (0) dx
+ x]
−2
2) dx = − ∫−2(x 2 − x − 6) dx
1
x3
𝑦 = 𝑥2 − 𝑥 − 6 𝑂 𝑥 3
= ∫3 (x 2 − x − 6) dx
Shaded area = area of A 1 ∫0 [x 2
𝑦
3
= − ∫−2(x − 3)(x +
y − 2 = − (x − 1) y
✓
= ∫−2[(0) − (x − 3)(x + 2)] dx
2
y − y1 = m(x − x1 )
PQ:
𝑥
3
=−
1−4
3
14(ii) Area
P(1,2), Q(4,0)
Gradient: mPQ =
𝑂
−2
x2 3
(−
3
−
(−2)3 3
1
x2 2
−
−2
− 6x] (−2)2 2
3
− 6(−2)] − [
22
+
3 5
(3)3 3
−
(3)2 2
− 6(3)]
27 2
2
= 20 unit ✓ 6
4
8
x3
+ x] 3
16
+
3
1
32
)
1
3 8
3
3
− (− + )
15(i)
Curve: y = x2
−(1)
Line: x + y= 6 y =6−x
−(2)
]
9 3
2
= 4 unit ✓ 3
13
Shaded area
𝑦 𝑦 = (𝑥 + 1)(𝑥 − 2)
2
= ∫ 0 − (x + 1)(x − 2) dx
𝑂
−1
𝑥
2
−1
2
= − ∫ (x + 1)(x − 2) dx −1 2
= − ∫ (x 2 − x − 2) dx −1 −1
sub (1) into (2): x2 = 6 − x x2 + x − 6 = 0 (x + 3)(x − 2) = 0 x = −3 or (rej ∵ x > 0)
x=2 y|x=2 = (2)2 =4 ⇒ A(2,4) ✓
= ∫ (x 2 − x − 2) dx 2
=[ =[ =
x3
−
3
(−1)3 3
x2 2
−
−1
− 2x] (−1)2 2
2
− 2(−1)] − [
7
+
6 1
(2)3 3
−
(2)2 2
− 2(2)]
10 3
2
= 4 unit ✓ 2
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525
A math 360 sol (unofficial)
Ex 19.1
15(ii) Shaded region
16(ii) Method 2 (wrt x-axis) Line x=4−y y=4−x
𝑦
2
= ∫ [(6 − x) − x 2 ] dx
𝑦 = 𝑥2
0 2
6
𝐴
= ∫ [6 − x − x 2 ] dx 2
x2 x3 = [6x − − ] 2 3 0
=7
(2)2
−
2
2
𝑂
(2)3 3
] − [6(0) −
1
(0)2 2
𝑥
6 −
(0)3 3
Lower curve x = 4y − y 2 (y − 2)2 − 4 = −x (y − 2)2 =4−x y−2 = ±√4 − x y = 2 ± √4 − x ⇒y = 2 − √4 − x
]
−0
3 1
= 7 unit 2 ✓ 3
16(i)
x = 4y − y 2
−(1)
x=4−y
−(2)
𝐵
𝑥
𝑦 = 2 − √2 − 𝑥
Shaded Area 3
= ∫0 [(4 − x) − (2 − √4 − x)] dx 3
= ∫0 (2 − x + √4 − x) dx = [2x − = [2x −
16(ii) Method 1 (wrt y-axis)
x2 2 x2 2
𝑥 =4−𝑦
3
+
3 2
=
2 5
]
(−1)( )
0 3
2
3
− (4 − x)2 ] 3
9
2
2 2
3
3 2
0 3
2
− [0 − (4)2 ] 3
2
3
+ [ (22 )2 ]
=[ − ] 4
3
3
(4−x)2
= [6 − − (1) ]
𝑦
3
3 2 3 + 2 3
6 1
= 6 unit 2 ✓
𝑄
6
1 𝑃
17(i)
𝑥
2𝑂
d dx
1
=
+Area of Q
d
(√3x + 7) = ⋅ (3x + 7) 2√3x+7 dx =
Shaded Area = Area of P
𝑂 3
Line x=4−y y=4−x
sub (1) into (2): 4y − y 2 =4−y 2 y − 5y + 4 =0 (y − 1)(y − 4)= 0 y=1 or y = 4 x|y=1 = 4 − (1) x|y=4 = 4 − (4) =3 =0 ⇒ B(3,1) ✓ ⇒ A(0,4) ✓
𝑥 = 4𝑦 − 𝑦
𝐴
𝑦 =6−𝑥
0
= [6(2) −
𝑦 𝑦 =4−𝑥
1 2√3x+7 3 2√3x+7
⋅3 ✓
1
= ∫0 [(4y − y 2 ) − (0)] dy 4
+ ∫1 [(4 − y) − (0)] dy 1
= ∫0 (4y − y 2 ) dy y2
y3
1
3
0
= [4 ( ) − ( )] 2
= [2y 2 −
y3
1
]
3 0
1
= [(2 − ) − (0)] 3
=
5
4
+ ∫1 (4 − y) dy + [4y −
1
4
]
2 1 1
4
2
1
+ [4y − y 2 ]
1
+ [(16 − 8) − (4 − )] 2
+
3
y2
9 2
2
= 6 unit ✓ 6
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526
A math 360 sol (unofficial)
Ex 19.1
17(ii) Shaded area = = =
18(ii) Method 2 (without rectangle area) Horizontal line y|x=a = 1 + a2
𝑦
3 1 dx ∫−1 √3x+7 2 3 3 dx ∫ 3 −1 2√3x+7 3 2 [√3x + 7]−1 3
𝑦= −1
𝑂
1 √3𝑥+7
𝑥
3
a
B = ∫0 [(1 + a2 ) − (1 + x 2 )] dx a
= ∫ (a2 − x 2 ) dx
2
= (√3(3) + 7 −√3(−1) + 7) 3 2
= (√16 = (4 = 18(i)
3
= [a2 x −
−√4)
3 2 3 4
0
= (a3 −
−2)
a3
]
= (a +
3
(0)3 3
]
𝑥=𝑎 𝑎
= a(1 + a2 ) − (a + = a + a3
𝑥
−a −
a3 3
19
1 3 a 3 3
3
−a
=0
𝑦 𝑦=
2 k
𝑂
a3
Area of P
3
2 3 a 3
k 10
∫2 ( x2 ) dx k
∫2 10x −2 dx 2
= a3 3
10 [
=0
x−1
𝑥2
[ ]
a = 0 or a = −√3 or a = √3 ✓ (both rej ∵ a > 0)
2
k k
k
sleightofmath.com
−
= area of Q 5 10
= ∫k ( 2 ) dx x
5
= ∫k (10x −2 ) dx = 10 [
x−1
5
]
−1 k
1 5
=[ ]
x 2
1
𝑥
5
k
]
−1 2
1 k
a − 3a =0 2 a(a − 3) =0 a(a + √3)(a − √3) = 0
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10
𝑃
)
Region A = Region B a3
−a
a = 0 or a = −√3 or a = √3 ✓ (both rej ∵ a > 0)
18(ii) Method 1 (with rectangle area) Horizontal line y|x=a = 1 + a2 Region B = Rectangle − Region A
a+
3
a − 3a =0 a(a − 3) =0 a(a + √3)(a − √3) = 0
) unit 2 ✓
=
2
= a3
3
1 3 a 3 3
𝐴
1 𝑂
=B a3
a+
−0
3 a3
) − (0)
A 𝑦 = 1 + 𝑥2 𝐵
− [(0) + =a+
𝑦
]
3
3
3
a x3
= [(a) +
3 0
a3
= a3
Region A a = ∫0 (1 + x 2 ) dx 3 0 (a)3
a
]
2
unit 2 ✓
= [x +
x3
x k
1 2
1
1
5
k
= − = =
7 10 20 7
✓
527
A math 360 sol (unofficial) 20(i)
y = x 2 − ax − b
Ex 19.1 20(iii) Line PR Points:
P(−1,0) & R(0, −3)
(1, −4) lies on curve, −4 = (1)2 − a(1) − b −4 = 1 − a − b b =5−a
Gradient:
mPR = (−1)−(0) = −3
PR:
y − y1 = m(x − x1 ) y − 0 = −3[x − (−1)] y = −3x − 3
At min (1, −4),
Line RQ Points:
R(0, −3) & Q(3,0)
Gradient:
mRQ =
RQ:
y − y1 = m(x − x1 ) y − 0 = 1(x − 3) y =x−3
dy dx
dy
= 2x − a
|
dx x=1
=0
2(1) − a = 0 a =2✓ b|a=2 = 5 − (2) = 3 ✓ 20(ii) Curve with a = 2, b = 3: y = x 2 − 2x − 3 Points P & Q At P & Q, curve cuts x − axis (y = 0): y =0 2 x − 2x − 3 =0 (x + 1)(x − 3) = 0 x = −1 or x=3 ⇒ P(−1,0) ✓ ⇒ Q(3,0) ✓
(0)−(−3)
(0)−(−3) (3)−(0)
=1
𝑦 −1 𝑃
3
𝑂
𝑄
𝑥
𝑅 𝑦 = 𝑥 2 − 𝑎𝑥 − 𝑏 Shaded Area = left region
+right region
0
Point R At R, curve cuts y − axis y|x=0 = (0)2 − 2(0) − 3 = −3 ⇒ R(0, −3) ✓
= ∫−1[(−3x − 3) − (x 2 − 2x − 3)] dx 3
+ ∫0 [(x − 3) − (x 2 − 2x − 3)] dx 0
3
= ∫−1[−x 2 − x] dx + ∫0 [−x 2 + 3x] dx = [−
x3 3
−
x2
0
]
2 −1 1
1
3
2
= [(0) − ( − )]
+ [−
x3 3
+
+ [(−9 +
3x2 2
3
]
0
27 2
) − (0)]
2
= 4 unit 2 ✓ 3
21(i)
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P(2,0) lies on y = kx − x 2 : 0 = 2k − 22 k=2✓
528
A math 360 sol (unofficial)
Ex 19.1 22(a) ∫2π sin x dx = 0 0
21(ii) Point Q Q(3, h) lies on y = 2x − x 2 : h = 2(3) − 32 = −3 ⇒ Q(3, −3) Line OQ Points:
𝐴 𝑂
(−3)−(0) (3)−(0)
𝐵
𝑥
Algebraic area = A − B = 0 ∵ (symmetric) 22(b)
= −1
a
∫ cos x dx = 0 0
y − y1 = m(x − x1 ) y − (−3) = −1(x − 3) y = −x
OQ:
𝑦
𝑦
O(0,0) & Q(3, −3)
Gradient: mOQ =
[typo in book]
𝜋 2𝜋
𝑦 = 2𝑥 − 𝑥 2 3
By inspection a = kπ, k ∈ ℤ+
𝑥
𝑂
𝑄(3, ℎ) 3
Area = ∫0 [(2x − x 2 ) − (−x)] 3
= ∫ [3x − x 2 ] dx 0
=[ =[
3x2 2
−
3(3)2
=4
2 1 2 1
x3
3
]
3 0 (3)3
−
3
]− [
3(0)2 2
−
(0)3 3
]
−0
= 4 unit 2 ✓ 2
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A math 360 sol (unofficial)
Ex 19.2 3
Ex 19.2 1
Points P & Q At P & Q, y = x 2 − 8x + 20 meets y = 8 x 2 − 8x + 20 = 8 𝑦 x 2 − 8x + 12 = 0 𝑄 𝑃 (x − 2)(x − 6) = x = 2 or x = 6
Point P x=0 Point Q 1
At Q, y = cos x meets y = : 2
cos x =
𝑦0= 8
x
=
1 2 π 3
𝑦=7
2
𝑦 = 𝑥 − 8𝑥 + 20 2
𝑂
Shaded Area
Shaded Area
𝑥
6
1 = ∫ [(cos x) − ] dx 2 0
6
= ∫ [(8) − (x 2 − 8x + 20)] dx 2 6
= ∫ [8 − x 2 + 8x − 20] dx 2 6
=∫
(−x 2
+ 8x − 12) dx
= [−
=(
6
3
x + 4x 2 − 12x] 3 2 (6)3 3
− [−
4
+ 4(6)2 − 12(6)] (2)3
+ 4(2)2 − 12(2)]
3
2
= 10 unit 2 ✓ 3
2
π 3
2
0
π
π
3
6
Shaded Area 7
= ∫1 [(8x − x 2 ) − 7]
𝑃 𝑂 1
𝑄 7
√3 2
π
− ) unit 2 ✓ 6
Points P & Q At P & Q, y = x 2 − 12x + 42 meets y = x + 2 x 2 − 12x + 42 = x + 2 𝑦 x 2 − 13x + 40 = 0 𝑦 = 𝑥 2 − 12𝑥 + 42 𝑄 (x − 5)(x − 8) = 0 𝑦 =𝑥+2 x = 5 or x = 8 𝑂
Shaded Area (x + 2) 8 ] dx = ∫5 [ 2 −(x − 12x + 42)
= [−
3
𝑦=7 𝑥
= [− = [−
x3 3
+
512 3
13 2 x 2
3
98 3
= −74
7
1
+
13 2
2
(64) − 40(8)]
+79
3
5
125 3
+
13 2
(25) − 40(5)]
1 6
2
= 4 unit ✓
1
2
2
+ 4(7) − 7(7)] − [−
=
𝑥
8
− 40x]
− [−
+ 4x − 7x]
(7)3
8
8
7
2
5
= ∫5 [−x 2 + 13x − 40] dx
= ∫1 [−x 2 + 8x − 7] dx = [−
𝑄
𝑃
Points P & Q At P & Q, y = 8x − x 2 meets y = 7 8x − x 2 =7 2 x − 8x + 7 = 0 𝑦 (x − 1)(x − 7) = 0 𝑦 = 8𝑥 − 𝑥 2 x = 1 or x = 7
x3
𝑂
= (sin − ) − (sin 0 − 0)
2
= [−
1
= [sin x − x]
𝑦 = 8𝑥 − 𝑥 2 1 𝑦 7 𝑥
𝑃
π 3
+
(1)3 3
+ 4(1)2 − 7(1)]
10 3
= 36 unit 2 ✓
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A math 360 sol (unofficial) 5
Ex 19.2
Points P & Q At P & Q, y = x + 3 meets y = −x 2 + 8x − 7 x+3 = −x 2 + 8x − 7 x 2 − 7x + 10 = 0 (x − 2)(x − 5) = 0 x = 2 or x = 5 𝑦 Shaded Area 𝑦 =𝑥+3 5 (−x 2 + 8x − 7) ] dx =∫ [ 𝑄 −(x + 3) 2 𝑃 𝑥 5 5 𝑂 2 2 = ∫ [−x + 7x − 10] dx 𝑦 = −𝑥 2 + 8𝑥 − 7 2 =
x3
7 [− + x 2 3 2 125 7
= [−
= −4
3 1
− 10x]
7
Area of region A 1
= ∫0 [(ex ) − 1] dx
𝑦 = 𝑒𝑥 𝑥=1 1
= =
1 ∫0 [(1) 1 ∫ 2
− (1 − x
A B
Area of region B 2 )]
𝑂 dx
𝑦=1
𝑥 1 𝑦 = 1 − 𝑥2
x dx
0 x3
1
5
=[ ]
2
1 = [13 − 03 ] 3 1 = unit 2 ✓
3 0
8
7
3
2
+ (25) − 50] − [− + (4) − 20] 2
+8
6
𝑦
= [ex − x]10 = [e1 − 1] −[e0 − 0] = e − 1 −1 = (e − 2) unit 2 ✓
2
3
3
1
= 4 unit 2 ✓ 2
6
8(i)
𝑦 𝑦 = 2 sin 𝑥 𝑦=𝑥 𝑥
𝑂
By inspection, y = 2 sin x and y = x π π intersect at (0,0) and ( , ) 2 2 𝑦 𝑦 = 2 sin 𝑥 Area of region π
= ∫02 [(2 sin x) − x] dx = [−2 cos x −
=−
]
𝑂
2
2
2 π2 8
] − [−2 cos 0 −
] −[−2(1)
02 2
]
− 0]
+2
8
= (2 −
π 2 2
( )
−
π2
𝑦=𝑥 𝑥 𝜋
2 0
π
= [−2 cos − = [0
π x2 2
π2 8
) unit 2 ✓
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Points P & Q At P & Q, y = 7 + 6x − x 2 meets x-axis (y = 0). 7 + 6x − x 2 = 0 x 2 − 6x − 7 = 0 (x − 7)(x + 1) = 0 x = 7 or x = −1 ⇒ P(7,0) ✓ ⇒ Q(−1,0) ✓ Point R At R, y = 7 + 6x − x 2 meets y − axis (x = 0): y|x=0 = 7 ⇒ R(0,7) ✓ Point S At S, y = 7 + 6x − x 2 meets y = 7. y =7 2 7 + 6x − x = 7 6x − x 2 =0 2 x − 6x =0 x(x − 6) =0 x=0 or x=6 (taken at R) ⇒ S(6,7) ✓
531
A math 360 sol (unofficial) 8(ii)
Ex 19.2 9(i)
𝑦 𝑦 = 7 + 6𝑥 − 𝑥 2
𝑅
𝑆
𝑄 𝑂
𝑃
(−2,0)
y = x 2 − 7x + 15
−(1)
y=7−x
−(2)
sub (1) into (2): 7−x = x 2 − 7x + 15 x 2 − 6x + 8 =0 (x − 2)(x − 4) = 0 x = 2 or x = 4
𝑥
Left region = = =
𝑦
−1 ∫−2 0 − (7 + 6x − x 2 ) dx −1 − ∫−2 (7 + 6x − x 2 ) dx −2 ∫−1 (7 + 6x − x 2 ) dx −2 x3 2
= [7x + 3x −
− [7(−1) + 3(−1) − +3
3 1
(−2)3
3 (−1)3 3
𝑂
]
1 2 2
= ∫1 [(x 2 − 7x + 15) − (7 − x)] dx
3
2
= ∫1 (x 2 − 6x + 8) dx
2
= 4 unit ✓ 3
=[ Middle region 0
=[
= ∫ (7 + 6x − x 2 ) dx
x3
− 3x 2 + 8x]
3
= [7x + 3x −
x3
− 3(2)2 + 8(2)]
3
−[
0
]
3 −1 (0)3 2
= [7(0) + 3(0) −
3
− [7(−1) + 3(−1) − +3
=6 ]
(−1)3 3
2
2
(1)3
−5
3 1
3
− 3(1)2 + 8(1)]
1 3
2
= 1 unit ✓ 3
] 9(ii)
3
2
1
(2)3
−1
=0
𝑥
4
Region A
]
2
2
𝑦 =7−𝑥
𝐵
3 −1
= [7(−2) + 3(−2) −
=
𝐴
]
2
2
𝑦 = 𝑥 2 − 7𝑥 + 15
Region B 4
= ∫2 [(7 − x) − (x 2 − 7x + 15)] dx
2
= 3 unit 2 ✓ 3
4
= ∫ (−x 2 + 6x − 8) dx Right region =
6 ∫0 (7 6
2 4
x3 x2 = [− + 6 ( ) − 8x] 3 2 2
+ 6x − x 2 ) − 7 dx
= ∫ (6x − x 2 ) dx
= [−
0
= =
[3x 2
−
x3
6
]
3 0 (6)3 [3(6)2 − ] 3 (0)3 2
− [3(0) −
3
= [−
x3 3
+ 3x 2 − 8x]
(4)3 3
= −5
+ 3(4) − 8(4)]
3
+6
(2)3 3
+ 3(2)2 − 8(2)]
2 3
4
= 36 −0 = 36 unit 2 ✓
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1
2
2
− [− ]
4
= ✓ 3
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A math 360 sol (unofficial) 10(i)
Ex 19.2
y = x 2 + 4x − 5 dy dx
10(ii) Method 1 (complement)
= 2x + 4
𝑦
𝑦 = 𝑥 2 + 4𝑥 − 5
Gradient of tangent is 10, dy
𝑃
= 10
dx
2x + 4 = 10 x =3 y|x=3 = 32 + 4(3) − 5 = 16 ✓
𝑅 1
𝑂
Area of PQR = area of PR𝑇 =
10(ii) Point P P(3,16)
3 ∫1 (x 2
=[
x3
=( Point R At R, y = x 2 + 4x − 5 cuts x − axis (y = 0): x 2 + 4x − 5 = 0 (x + 5)(x − 1) = 0 x = −5 or x = 1 (rej ∵ x > 0) ⇒ R(1,0) Line PQ Points: Gradient: PQ:
−area of ⊿PQT
+ 4x − 5) − (0) dx
+ 2x 2 − 5x]
5
1
− (1.6)(16) 2
1
+ 18 − 15) − ( + 2 − 5)
15
7
2
−12.8
3
unit 2 ✓
𝑦
B A 𝑂
1
3
7
𝑥
5
Area of shaded region PQR = region A +region B 7
= ∫15(x 2 + 4x − 5) − 0 dx 3
+ ∫7 [(x 2 + 4x − 5) − (10x − 14)] dx 5 7 5
= ∫1 [x 2 + 4x − 5] dx 7 5
= ∫1 [x 2 + 4x − 5] dx
7 5
=[
7
⇒ Q ( , 0) 5
x3 3
+ 2x 2 − 5x]
3
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15
5
3
+ ∫7 [(x − 3)2 ] dx 5
+[
(x−3)3 3
3
]7 5
7 2 5
1
3 13
3
+ ∫7 [x 2 − 6x + 9] dx
+ 2 ( ) − 7]
− [ + 2 − 5] =1
7 5
1
7 3 ( ) 5
=[
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3
1
3 13
1
− (3 − ) (16)
Method 2 (break)
P(3,16) or Q mPQ = 10 y − y1 = mPQ (x − x1 ) y − 16 = 10(x − 3) 𝑦 − 16 = 10𝑥 − 30 y = 10x − 14
=
3 27
=1
Point Q At Q, PQ (𝑦 = 10𝑥 − 14) cuts x − axis (y = 0): y =0 10x − 14 = 0 10x = 14 x
𝑥
𝑄 3
+0
− (−
512
)
375
unit 2 ✓
533
A math 360 sol (unofficial) 11(i)
Ex 19.2
Point A At A, y = 7 + 6x − x 2 cuts y − axis (x = 0): y|x=0 = 7 ⇒ A(0,7)
12(ii) Area of region A k 1
= ∫1 =
=[
A(0,7) lies on y = mx + c, (7) = m(0) + c c =7✓
𝐴
=1− =
− m)x − x x3
2
3 0
[(6 − m) ( ) − [(6 − m) ( (6 −
(6−m)2 2
− (mx + 7)] dx = 20 = 20
6−m
x2
]
)−
1 m)3 [ ] 6 3
(6 − m) 6−m m
= 20 (6−m)3 3
] − [0]
= 20 =
k
k
𝑥
✓
−
=
−
=
= 20
dx
𝐴 1
k
1
=
2]
k−1
= (k − 1)(k + 1)
𝑥
+ 6x − x
𝐵
𝑄(1,1) 𝑂
𝑃(𝑘, 𝑘)
12(iii) By complement Area of region B = area of trapezium −area of region A
=
2)
1 𝑥2
k 1
𝑦 + 6𝑥 − 𝑥 2
Area of shaded region 6−m ∫0 [(7 6−m ∫0 [(6
−1 1 1 k
= − ( − 1)
𝑦 = 𝑚𝑥 + 𝑐
6−𝑚
𝑂
𝑦=
]
x 1 1
𝑦 𝐵
𝑦
= [− ]
11(ii) Point B At B, y = mx + 7 meets y = 7 + 6x − x 2 mx + 7 = 7 + 6x − x 2 x 2 + mx − 6x = 0 x 2 + (m − 6)x = 0 x[x − (m − 6)] = 0 x=0 or x = −(m − 6) (taken at A) =6−m✓ 11(iii)
dx
x2 k −2 ∫1 x dx k x−1
5 6 5
2 k2 −1 2
k−1 k k−1 k
k3 −k−2k+2 2k k3 −3k+2 2k (k−1)(k2 +k−2) 2k
13
[deduced] ✓
𝑦
6 5
𝑦 = √1 − 𝑥 2
6 5
𝑂
6
𝑥 𝑦 =1−𝑥
5 6
125
1
∫0 [√1 − x 2 − (1 − x)] dx
6
1
= 125 =5 =1✓
= ( area of circle) − (area of triangle) 4
1
= [π(1)2 ] 4 π
1
4
2
1
− (1)(1) 2
= − ✓ 12(i)
y=x✓ 14
∞ 1
∫1
x2
dx represents the convergent value of area as
→ ∞✓
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A math 360 sol (unofficial)
Rev Ex 19 A3
Rev Ex 19 A1(i)
Area of region A
𝑦
p 8
= ∫1
x
2 dx
1 p
𝑦=
= 8 [− ]
x 1 1 1
𝐴
= −8 ( − ) p 8
1
𝑂
= (8 − ) unit 2 ✓
−(1)
x=3 −(2) sub (2) into (1):
8 𝑥2
𝐵
1 P
Coordinates y 2 = 3x 𝑦 = ±√3𝑥
𝑦 = ±√3(3) = ±3
𝑥
6
p
Shaded region A1(ii) A
=B
8− 8− 8− 8− 8−
8
=
p 8 8 8
= −8 ( − )
p 8
6
8
3
p
=− +
p
= =
7
d
1
9
9
+6
✓ A4
(sin x) +
Area of region A =
d
(x) ⋅ sin x
− sin x
= x ⋅ cos x +1 ⋅ sin x = x cos x + sin x = x cos x [shown] ✓
− sin x − sin x
dx
𝑥
= 12 unit 2 ✓
(x sin x + cos x)
=x⋅
𝑥=3
1
=6
28 3 12
𝑂 −3
= [3(3) − (3)3 ] − [3(−3) − (−3)3 ]
p
4
𝑦 2 = 3𝑥
3
3 1 = [3y − y 3 ] 9 −3
x p 1 1
p
dx
3 1 = ∫ [3 − y 2 ] dy 3 −3
= −8 [ ]
p
𝑦
1 3
x p 1 6
p
d
6 8 ∫p (x2 ) dx 1 6
= 8 [− ]
p
16
A2(a)
3
= ∫−3 [(3) − ( y 2 )] dy
dx
1 ∫0 x(x
− 1)(x − 2) dx
𝑦 𝑦 = 𝑥(𝑥 − 1)(𝑥 − 2)
1 2
= ∫ x(x − 3x + 2) dx
𝐴 𝑂 1
𝐵
2
𝑥
0 1
= ∫0 (x 3 − 3x 2 + 2x) dx =[
A2 At A, y = x cos x cut x − axis (y = 0): (b)(i) x cos x = 0 x = 0 or cos x = 0 π (rej ∵ x > 0) x =
x4 4
1
− x3 + x2 ]
0
1
= ( − 1 + 1) − (0) 4
=
2
1 4
π
⇒ A ( , 0) ✓ 2
A2 Shaded Area π (b)(ii) = ∫02 x cos x − (0) dx
Area of region B
π
= ∫02 x cos x dx = [x sin x + π
π
2
2
𝑂
π cos x]02 π
2
= − ∫1 x(x − 1)(x − 2) dx
𝑦 𝑦 = 𝑥 cos 𝑥 𝐴 𝑥 𝜋 2
=
2
2
x4 4
1
− x3 + x2 ]
2
4
−1
≈ 0.57 ✓
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=[
1 = ( − 1 + 1) − (4 − 8 + 4) 4 1 =
= ( sin + cos ) −(0 + 1) π
1
= ∫2 (x 3 − 3x 2 + 2x) dx
∴A=B✓
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A math 360 sol (unofficial) A5(i)
Rev Ex 19
Point P At P, y = x 2 − 2x + 2 meets y − axis (x = 0): y|x=0 = 2 ⇒ P(0,2) ✓
A6(ii)
𝑦 = 2𝑥 𝐴 𝑂
Point Q At Q, y = x 2 − 2x + 2 has x-coordinate of 3 y|x=3 = 32 − 2(3) + 2 =5 ⇒ Q(3,5) ✓ A5(ii) Area of A =
3 ∫0 x 2
=[ =[
x3 3 3
−[
4
3
− x + 2x] −
Area of shaded region = left region +right region
𝑦 𝑦 = 𝑥 2 − 2𝑥 + 2
− 2x + 2 dx
(3)2
(0)3 3
+ 2(3)]
−
(0)2
𝑃 𝑂
𝐵 𝐴 (3,0)
+ ∫4 6x − x 2 dx
= [x 2 ]40
+ [3x 2 − x 3 ]
𝑥
B1(i)
2
4
+[108 − 72]− [48 −
= 16
+36
−48 +
64
3 64
]
3
2
Curve & its gradient y = x 2 − 4x + 5 dx
= 2x − 4
Normal PQ Point: y|x=3 = 32 − 4(3) + 5 = 2 ⇒ (3,2)
−6
Gradient:
1
= 4 unit 2 ✓
− 𝑑𝑦
1 |
𝑑𝑥 𝑥=3
2
PQ: Point A At A, y = 6x − x 2 meets y = 2x 6x − x 2 = 2x 4x − x 2 = 0 x 2 − 4x = 0 x(x − 4) = 0 x=0 or x = 4 y|x=0 = 2(0) y|x=4 = 2(4) =0 =8 ⇒ O(0,0) ⇒ A(4,8) ✓
3
=4 −0
dy
Area of B = area of trapezium −area of A 1
6
3
= 6 −0 = 6 unit 2 ✓
A6(i)
2
1
= 25 unit ✓
+ 2(0)]
= (3)(2 + 5)
6
= ∫0 2x dx
1
0
𝑥
4 6
2
2
(3)3
𝑦
y − y1
=−
1 2(3)−4
= − dy
1 |
=−
1 2
(x − x1 )
dx 𝑥=3
1
y − (2) = (− ) [x − (3)] 2
4 − 2y 2y + x
=x−3 = 7 [shown] ✓
B1(ii) Point Q At Q, PQ (2y + x = 7) cuts x − axis (y = 0): 2(0) + x = 7 x =7 ⇒ Q(7,0) ✓
Point B At B, y = 6 − x 2 meets x − axis (y = 0): 6x − x 2 = 0 x(x − 6) = 0 x=0 or x = 6 ⇒ O(0,0) ⇒ B(6,0) ✓
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536
A math 360 sol (unofficial) B1(iii)
Rev Ex 19 B2(b) Coordinates y = x(x − 3) −(1)
𝑦 𝑦 = 𝑥 2 − 4𝑥 + 5 𝑃 𝑄 𝑥 3 7
𝑂
Area of shaded region = left region 3
= ∫0 (x 2 − 4x + 5) dx =[
x3 3
− 2x 2 + 5x]
y = −2 −(2) (1) (2): sub into x(x − 3) = −2 2 x − 3x + 2 = 0 (x − 1)(x − 2) = 0 x = 1 or x = 2
+right region 7 7
1
2
2
+ ∫3 ( − x) dx
3
7
1
7
2 49
4 49
3
+ [ x − x2 ]
0
= [9 − 18 + 15] − [0]
+[
=6 = 10 unit 2 ✓
+4
2
−
4
]−[
𝑦
21
2
𝑦 = −2
x
3
2
= ∫ [(−2) − x(x − 3)] dx
4
1 2 ∫1 (−2 − 3x − x 2 ) dx 1 ∫2 (x 2 + 3x + 2) dx 1 x3 3 [ − x 2 + 2x] 3 2 2 1 3 8
=
=
= [ − + 2] − [ − 6 + 4]
𝑥
5
1 𝑂
2
9
− ]
𝑦 = 𝑥(5 − 𝑥)
𝑂
𝑦 = 𝑥(𝑥 − 3)
Area
= B2(a)
y
3 1
=
6
2
3
2
unit ✓
✓ B3(i)
Area of left region
𝑦
2
= ∫0 [x(5 − x)] dx = =
d dx
[(x − 1)ex ]
= (x − 1) ⋅
2 ∫0 (5x − x 2 ) dx 2 5 1 [ x2 − x3 ] 2 3 0 8
𝑂
2
5
d dx x
(ex )
= (x − 1) ⋅ e = ex [(x − 1) +1] = xex [shown] ✓
𝑥
+
d dx
(x − 1) ⋅ ex ⋅ ex
+1
= [10 − ] − (0) 3
=
B3(ii) Area of shaded region A
22
0
3
= ∫−2[(0) − (xex )] dx
Area of right region 5
=
5
= =
= ∫2 [x(5 − x)] − (0) dx = ∫2 [5x − x 2 ] dx 1
5
3 125
2
5
= [ x2 − x3 ] =[ =
2 125
2 27
−
3
𝑦
−2 ∫0 xex dx [(x − 1)ex ]−2 0 (−3)e−2
1 −2 A
−[−1(e0 )]
B 0.5
𝑥
3
8
=1
3
≈ 0.594 unit 2 ✓
] − [10 − ]
−
𝑂
𝑦 = 𝑥𝑒 𝑥
e2
2
Area of shaded region B Ratio =
22 3
=44
0.5
:
= ∫0 [(1) − (xex )] dx
27
= [x − (x − 1)ex ]0.5 0
2
: 81 [shown] ✓
1
1
= [0.5 − (− ) e2 ] −[0 − (−1)e0 ] 2
1
1
= ( + √e) 2 2
−1 2
≈ 0.324 unit ✓
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537
A math 360 sol (unofficial) B4(i)
Rev Ex 19
π 2 π 6
π
B5(ii) y-coordinate of P
∫ sin x dx = [− cos x]π2
4 π
y|x=π = 1 + ( ) = 2
6
=
π −[cos x]π2 6
Region B = area of trapezium −region A
π π = − (cos − cos ) 2 6
B4(ii)
√3 2
1 π
= ( ) (1 + 2) 2 4 3π
√3 ) 2
= − (0 − =
π 4
4
=(
unit 2 ✓
B6(i)
𝑦
dx
𝑦 = sin 𝑥
A B
= ex
Tangent Point:
1 2
Gradient: 𝑂
𝜋
𝜋
6
2
𝜋
𝑥
sin x =
⇒x=
2
sin x = 1
⇒x=
6 π
= = = = B5(i)
6 π 2 π 6 π 2 π 6
π 12 π 12 π 12 5 12
√3 2
π
+ ∫ 1 dx
− ∫π2 sin x dx
+[x]
√3 − 2
6
π
π
2
6
+ − π−
dy
(x − x1 )
|
dx x=2 2 (x
−
B6(ii) Area enclosed by curve & x − axis & lines x = 0, x = 2
√3 2
π 4
= [ex ]20 = e2 − e0 = e2 − 1 ✓
𝑦 𝑦 = sec 𝑥 𝑦 =1+
𝑃
𝐵
𝑥
2
Shaded region = (Bigger area) − △ ABC area
𝐵 𝐴 𝜋
𝑥
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1
= e2 − 1
− [2 − 1][e2 ]
= e2 − 1
− e2
=
4
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𝑂
4
1 𝑂
𝐶
𝜋
π
= [tan x]04 π = tan − tan 0 4 = 1 unit 2 ✓
𝐴
𝑦 = 𝑒𝑥
2
2
= ∫ (sec 2 x) dx
𝑦
= ∫0 [(ex ) − 0] dx
[shown] ✓
Region A
0
=
Point B At B, tangent cuts x − axis (y = 0): e2 x − e2 = 0 x =1 ⇒ B(1,0) ✓
+ [∫π2 1 − sin x dx]
2 6
= e2
|
dx x=2
2
π
1 π
dy
y − e2 = e − 2) 2 y = e x − 2e2 + e2 y = e2 x − e2
π
Area of the shaded region = rectangle A +region B = (1 − )
y|x=2 = e2 ⇒ A(2, e2 )
Tangent: y − y1
Coordinates 1
− 1) unit 2 ✓
Curve & its gradient y = ex dy
1
8
−1
1 ( e2 2
2 1 2
2
− 1) unit [shown] ✓
538
A math 360 sol (unofficial) 4(i)
Ex 20.1 1(i)
Displacement s = 2t 2 + t s|t=2 = 2(2)2 + 2 = 10 ✓
1(ii) 1
𝑠 = 2𝑡 2 + 𝑡 𝑂 𝑡
Velocity v =
ds dt
= 14t − 10
4(ii)
v|t=3 = 14(3) − 10 = 32 ms −1 ✓
5(i)
Displacement s = 7t 2 − 2t 3
2
Distance travelled in 1st 2s = s|t=2 −s|t=0 = 10 −0 = 10m ✓
Velocity v=
Velocity 1 v = 5t − t 2 2 v|t=0 = 0ms −1 ✓
2(ii)
Displacement s = 7t 2 − 10t
𝑠
−
2(i)
Ex 20.1
ds dt
= 14t − 6t 2 ✓ 5(ii)
Acceleration a =
At rest, v =0
dv dt
= 14 − 12t ✓
1
5t − t 2 = 0 2
t 2 − 10t = 0 t(t − 10) = 0 t = 0 or t = 10 ✓ 3(i)
5(iii)
a|t=2 = 14 − 12(2) = −10ms −2 ✓
6(i)
Acceleration a=t−9
Acceleration a = t − 5t 2
Velocity v = ∫ a dt = ∫ t − 9 dt
Accelerating at 0.05m/s 2 , a = 0.05 2 t − 5t = 0.05 2 20t − 100t =1 2 100t − 20t + 1 = 0 (10t − 1)2 =0 t = 0.1s ✓
=
Decelerating at 6 m/s 2 , a = −6 2 t − 5t = −6 5t 2 − t − 6 =0 (5t − 6)(t + 1) = 0 t = 1.2 or t = −1 (rej ∵ t > 0) ✓
2
− 9t + c
Initial velocity of −20 m/s, v|t=0 = −20 (0)2 2
3(ii)
t2
− 9(0) + c= −20
c
= −20 1
∴ v = t 2 − 9t − 20 ✓ 2
6(ii)
At rest, v 1 2 t 2 2
− 9t − 20
=0 =0
t − 18t − 40 = 0 (t − 20)(t + 2) = 0 t = 20s ✓ or t = −2 © Daniel & Samuel A-math tuition 📞9133 9982
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(rej ∵ t > 0)
539
A math 360 sol (unofficial) 7(i)
Velocity v=
Ex 20.1 9(i)
1 8e−2
Velocity
Displacement s = ∫ v dt = ∫ 8e
−
1 2
Displacement s = 12t − t 3
v =
ds dt
= 12 − 3t 2
dt
1 − 2
= 8e t + c
At rest, v =0 2 12 − 3t =0 2 t −4 =0 (t + 2)(t − 2) = 0 t = −2 or t = 2 (rej ∵ t > 0)
t ≡ time after leaving O: s|t=0 =0 1
8e−2 (0) + c = 0 c =0 1
∴ s = 8e−2 t ✓ 7(ii)
a=
s|t=5 = 8e (5) =
8(i)
Acceleration
1 − 2
40 √e
Acceleration at rest, a|t=2 = −6(2) = −12ms −2 ✓ 9(ii)
Velocity v =
ds
Recall
s = 12t − t 3 v = 12 − 3t 2 a = −6
dt
= 3t 2 − 24t + 36
Displacement Next at O, s =0 3 12t − t = 0 t 3 − 12t = 0 t(t 2 − 12) = 0 t = 0 or t = −√12 or t = √12 (rej both ∵ t > 0)
v|t=1 = 3 − 24 + 26 = 15 ms −1 ✓ 8(ii)
dt
= −6t
m✓
Displacement s = t(t − 6)2 = t(t 2 − 12t + 36) = t 3 − 12t 2 + 36t
dv
At rest, v =0 2 3t − 24t + 36 = 0 t 2 − 8t + 12 =0 (t − 2)(t − 6) = 0 t = 2 or t = 6 ✓
Velocity Velocity at t = √12: v|t=√12 = 12 − 3(√12)
8(iii)
= 12 − 36 = −24ms −1
Acceleration a =
2
dv dt
= 6t − 24 a|t=3 = 6(3) − 24 = −6ms −2 ✓
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540
A math 360 sol (unofficial) 9(iii)
Recall
s = 12t − t 3 v = 12 − 3t 2 a = −6
Ex 20.1 10(ii) s|t=2 = 18(2)2 − 24 = 56 Distance travelled in 3rd sec = s|t=3 − s|t=2 = 81 − 56 = 25m ✓
Distance 𝑦 12
𝑣 = 12 − 3𝑡 2 𝑥 −2 𝑂 2
10(iii) a = dv dt
Distance travelled in first 3 seconds
= 36 − 12t 2
3
= ∫ |v| dt = = =
0 2 ∫0 |v| dt 2 ∫0 v dt 2 ∫0 v dt [s]20
When a = −12, a = −12 36 − 12t 2 = −12 2 12t − 48 =0 2 t −4 =0 (t + 2)(t − 2) = 0 t = −2 or t = 2 (rej ∵ t > 0)
3 + ∫2 |v| dt 3 + ∫2 −v dt 2 + ∫3 v dt +[s]23
= = (s|t=2 − s|t=0 ) +(s|t=2 − s|t=3 ) = 2s|t=2 −s|t=0 −s|t=3 = 2(16) −0 −27 = 23m ✓ 10
Velocity Velocity when a = −12: v|t=2 = 36(2) − 4(2)3 = 72 − 32 = 40ms −1 ✓
Displacement s = 18t 2 − t 4 Velocity v =
ds
11(i)
dt
= 36t − 4t 3 At rest, v =0 36t − 4t 3 =0 3 t − 9t =0 2 t(t − 9) =0 t(t + 3)(t − 3)= 0 t = 0 or t = −3 or t = 3 (rej both ∵ t > 0) Displacement s|t=3 = 18(3)2 − 34 = 162 − 81 = 81 ✓
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Height h = 24t − 3t 2 Velocity v =
dh dt
= 24 − 6t v|t=3 = 24 − 6(3) = 6ms −1 ✓ 11(ii) At max height, v =0 24 − 6t = 0 t =4✓ Height h|t=4 = 24(4) − 3(4)2 = 96 − 48 = 48m
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541
A math 360 sol (unofficial)
Ex 20.1
11(iii) At h = 0, 24t − 3t 2 = 0 t 2 − 8t =0 t(t − 8) = 0 t = 0 or t = 8
13(i)
At rest: v =0 2t 2 − 3t − 2 =0 (2t + 1)(t − 2) = 0
Time of flight = 8 − 0 = 8s ✓ 12(i)
Velocity v = 2t 2 − 3t − 2
t=−
Displacement s = ∫ v dt 3t2
3
2
− 2t + c
dv dt
At t = 0, s = 3: s|t=0
= 3e0.4t (0.4) = 1.2e0.4t
2 3
2
e
2
(0)2
=3
− 2(0) + c = 3 =3
2
3
3
2
s = t 3 − t 2 − 2t + 3 Displacement at rest:
+ 6t + c
0.4 15 0.4t
−
3
Displacement with c = 3:
12(ii) Displacement s = ∫ v dt 3e0.4t
(0)3
c
Initial acceleration a|t=0 = 1.2e0 = 1.2ms −2 ✓
=
2
= t3 −
Acceleration
=
or t = 2 ✓
2
(rej ∵ t > 0)
Velocity v = 3(e0.4t + 2) = 3e0.4t + 6
a=
1
2
3
s|t=2 = (8) − (4) − 2(2) + 3
+ 6t + c
=
3 16 3
2
−6−4+3 5
=− m✓
t ≡ time after leaving O: s|t=0 =0 15 0 e 2
+ 6(0) + c
=0
c
=−
15 2
Displacement with c = − s=
15 0.4t e 2
3
+ 6t −
15 2
:
15 2
Displacement when t = 1: s|t=1 = =
15 0.4 e 2 15 0.4 e 2
+6−
15 2
3
− ✓ 2
12(iii) Velocity v = 3e0.4t + 6 > 0 ∵ e0.4t > 0 ∴ The particle will not return to O
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542
A math 360 sol (unofficial)
Ex 20.1
13(ii) Displacement when t = 3: s|t=3 2
14(i)
3
= (27) − (9) − 2(3) + 3 3
2
= 18 −
27 2
−6+3
27 2 1 =1 m✓
Velocity v = ∫ a dt = −t 2 + 6t + c
= 15 − 2
Initial Velocity of 7 m/s: v|t=0 =7 2 −(0) + 6(0) + c = 7 c =7
𝑣 𝑣 = 2𝑡 2 − 3𝑡 − 2 𝑂 −
Acceleration a = 2(3 − t) = 6 − 2t = −2t + 6
1
2
2
t Velocity with c = 7: v = 6t − t 2 + 7
Total distance travlled in first 3s At v = 7, v =7 2 −t + 6t + 7 = 7 −t 2 + 6t =0 t(t − 6) =0 t = 0 (rej) or t = 6 ✓
3
= ∫ |v| dt = = =
0 2 3 ∫0 |v| dt + ∫2 |v| dt 2 3 ∫0 −v dt + ∫2 v dt 0 3 ∫2 v dt + ∫2 v dt [s]02 +[s]32
= = s|t=0 − s|t=2 +s|t=3 − s|t=2 = s|t=0 −2s|t=2 +s|t=3 5
3
3
2
14(ii) Displacement s = ∫ v dt 1
= − t 3 + 3t 2 + 7t + c2
= (3) −2 (− ) + ( )
3
5
=7 ✓
t ≡ time after passing O: s|t=0 =0
6
1
3(0)2 − (0)3 + 7t + c2 = 0 3
c2
=0
Displacement with c2 = 0: 1
s = 3t 2 − t 3 + 7t 3
Position when v = 7: 1
s|t=6 = 3(36) − (216) + 42 3
= 78m ✓
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543
A math 360 sol (unofficial) 15(i)
Acceleration a = 8 − 4t = −4t + 8 Velocity v = ∫ a dt = −2t 2 + 8t + c
Ex 20.1 15(iii) Acceleration At max speed: a =0 8 − 4t = 0 t =2 da dt
0)
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544
A math 360 sol (unofficial) 16(iii) Displacement s = ∫ v dt = −t 3 +
13 2 t 2
Ex 20.1 17(iii) Distance Total distance travelled in first 2s 2
+ 30t + c2
= ∫ |v| dt =
t ≡ time after passing O: s|t=0 = 0 c2 = 0
=
= = s|t=1 − s|t=0 +s|t=1 − s|t=2 = 2s|t=1 − s|t=0 − s|t=2 = 2(47) − (40) − 42 = 12
Displacement with c2 = 0: s = −t 3 +
13 2 t 2
0 1 2 ∫0 v dt + ∫1 −v dt 1 1 ∫0 v dt + ∫2 v dt [s]10 +[s]12
+ 30t
Max Distance: s|t=6 =
13 2
average speed in first 2s 12 = 2 = 6ms −1 ✓
(36) − 216 + 30(6)
= 198m ✓ 17(i)
Displacement s = t 3 − 9t 2 + 15t + 40
s|t=2 = 8 − 36 + 30 + 40 = 42
Velocity v=
ds
s|t=6 = 216 − 324 + 90 + 40 = 22
dt
= 3t 2 − 18t + 15 At zero velocity, v =0 2 3t − 18t + 15 = 0 t 2 − 6t + 5 =0 (t − 1)(t − 5) = 0 t = 1 or t = 5
17(iv) 𝑣 𝑣 = 3𝑡 2 − 18𝑡 + 15 𝑂 1
Displacement s|t=1 = 1 − 9 + 15 + 40 = 47m ✓
Total distance travelled in first 6s 6
= ∫0 |v| dt 1
= ∫0 |v| dt
s|t=5 = 125 − 225 + 75 + 40 = 15m ✓
1
= ∫0 v dt 1
17(ii) Acceleration a=
dv dt
= 6t − 18 magnitude of 9: |a| = 9 a = −9 or a =9 6t − 18 = −9 6t − 18 = 9 t = 1.5 s t = 4.5s
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𝑡
5
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5
+ ∫1 |v| dt 5
+ ∫1 −v dt 1
6
+ ∫5 |v| dt 6
+ ∫5 v dt 6
= ∫0 v dt
+ ∫5 v dt
+ ∫5 v dt
= [s]10
+[s]15
+[s]65
= 2s|t=1
+s|t=6 −s|t=0
−2s|t=5
= 2(47) = 46 ✓
+22
−2(15)
−(40)
545
A math 360 sol (unofficial) 18(i)
Ex 20.1
Acceleration a = 6t − 8
18(iii) Acceleration At min velocity, a =0 6t − 8 = 0
Velocity v = ∫ a dt = 3t 2 − 8t + c
t
4 2
2 3
= − ms
✓
Velocity v = −3t 2 + 8t + 5 Acceleration a=
dv dt
= −6t + 8 ✓
or t = 2
19(ii) s = ∫ v dt t3
t2
3
2
= −3 ( ) + 8 ( ) + 5t + c
Displacement with c2 = 0: s = t 3 − 4t 2 + 4t Distance 2
= ∫03|v| dt 2
= ∫03 v dt
= −t 3 + 4t 2 + 5t + c ∵ t ≡ time after passing O, s|t=0 = 0 c =0 ∴ s = −t 3 + 4t 2 + 5t 19(iii) Velocity At speed of 2ms −1 , v = ±2 v =2 2 −3t + 8t + 5 = 2 3t 2 − 8t − 3 =0 (3t + 1)(t − 3) = 0 1
t = − or t = 3 ✓
= s|2 − s|t=0
3
3
2 3
2 2
2
3
3
3
(rej ∵ t > 0)
= [( ) − 4 ( ) + 4 ( )] − 0 27
3
−1
3
19(i)
t ≡ time after passing O: s|t=0 = 0 c2 = 0
32
3 4
3
Displacement s = ∫ v dt = t 3 − 4t 2 + 4t + c2
=
4
v|t=4 = 3 ( ) − 8 ( ) + 4
At rest: v =0 3t 2 − 8t + 4 = 0 (3t − 2)(t − 2) = 0 t=
3
Velocity Min velocity,
Initial velocity of 4m/s: v|t=0 =4 3(0)2 − 8(0) + c1 = 4 c1 =4 Velocity with c = 4: v = 3t 2 − 8t + 4
4
=
or v = −2 −3t 2 + 8t + 5 = −2 −3t 2 + 8t + 7 = 0 3t 2 − 8t − 7 = 0 t = = =
m✓
= 18(ii) Return to starting pt: s =0 3 2 t − 4t + 4t = 0 t(t 2 − 4t + 4) = 0 t(t − 2)2 =0 t = 0 (NA) or t = 2 ✓
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= t=
8±√64−4(3)(−7) 2(3) 8±√148 6 8±√4×37 6 8±2√37 6 4±√37 3 4−√37 3
or t =
4+√37 3
✓
(rej ∵ t > 0)
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546
A math 360 sol (unofficial) 19(iv) Displacement s = ∫ v dt = −t 3 + 4t 2 + 5t + c2
Ex 20.1 20(ii) Displacement s = ∫ v dt 1
= − t 3 + 9t + c 3
t ≡ time after passing O: s|t=0 = 0 c2 = 0
t ≡ time from start: s|t=0 = 0 c=0
Velocity with c2 = 0: s = −t 3 + 4t 2 + 5t
Displacement with c = 0: 1
s = − t 3 + 9t 3
At displacement of 20 m: s = 20 −t 3 + 4t 2 + 5t = 20 3 2 t − 4t − 5t + 20 =0 2 (t − 4)(t − 5) =0 (t − 4)(t − √5)(t + √5) = 0
At starting pt again: s = 0: 1
− t 3 + 9t = 0 3
t 3 − 27t = 0 t(t 2 − 27) = 0 t(t + √27)(t − √27) = 0 t(t + 3√3)(t − 3√3) = 0 t = 0 or t = −3√3 or t = 3√3 (rej both ∵ t > 0) ✓
t = 4 or t = √5 or t = −√5 (rej ∵ t > 0) 20(i)
Velocity v = 9 − t2 = −t 2 + 9 At rest, v =0 2 −t + 9 =0 2 t −9 =0 (t + 3)(t − 3) = 0 t = −3 or t = 3 ✓ (rej ∵ t > 0)
20(iii) 18m from starting pt, s = 18 or
s
1 − t3 3 1 − t3 3 3
1 − t3 3 1 − t3 3 3
+ 9t
= 18
+ 9t − 18
=0
t − 27t + 54 =0 2 (t + 6)(t − 6t + 9) = 0 (t + 6)(t − 3)2 =0 t = −6 or t = 3 ✓ (rej ∵ t > 0) 21
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= −18 + 9t
= −18
+ 9t − 18
=0
t − 27t − 54 =0 2 (t − 6)(t + 6t + 9) = 0 (t − 6)(t + 3)2 =0 t = 6 ✓ or t = −3 (rej ∵ t > 0)
max. speed ⇒ a = 0ms −2 ✓
547
A math 360 sol (unofficial)
Rev Ex 20 A1(iii) Displacement s|t=3 = 27(3) − 33 = 54 s|t=4 = 27(4) − 43 = 44
Rev Ex 20 A1(i)
Displacement s = 27t − t 3 = −t 3 + 27t
Distance travelled in fourth sec = |s|t=4 − s|t=3 | = |44 − 54| = 10cm ✓
Velocity v=
ds dt
= −3t 2 + 27 At rest: v 27 − 3t 2 t2 − 9 (t + 3)(t − 3) t = −3 or (rej ∵ t > 0)
A1(iv)
𝑣 = 27 − 3𝑡 2 =0 =0 =0 =0 t=3
−3
𝑂
3
t
Distance travelled in the first 4s 4
= ∫0 |v| dt 3
4
3
4
3
3
= ∫0 |v| dt + ∫3 |v| dt
Acceleration a=
𝑣
= ∫0 v dt + ∫3 −v dt
dv dt
= ∫0 v dt + ∫4 v dt
= −6t
= [s]30 Acceleration at rest: a|t=3 = −6(3) = −18ms −2 ✓ A1(ii) Displacement next at O: s −t 3 + 27t t 3 − 27t t(t 2 − 27) t(t + √27)(t − √27)
+[s]34
= 2s|t=3 −s|t=0
−s|t=4
= 2(54) −(0) = 64m ✓
−(44)
=0 =0 =0 =0 =0
t = 0 or t = −√27 or t = √27 (rej ∵ t > 0) Velocity Velocity when next at O, v|t=√27 = 27 − 3(√27)
2
= −54 cms −1 ✓
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548
A math 360 sol (unofficial) A2(i)
Rev Ex 20
Velocity
A3(i)
v = 12 sin
t
Displacement 3t
s=
2
t2 +9
Velocity Speed of 8 m/s: |v| = 8 v = 8 or 12 sin sin
t
v
=8
2
t
sin
3
α ≈ 0.729 t 2
= −8
12 sin
2
=
2
ds
v=
t 2
t 2
t 2
≈ 0.729 t ≈ 1.46 ✓
(t2 +9)2 (t2 +9)⋅3
=
2 3
3t2 +27 −6t2 (t2 +9)2 27−3t2
= (t2
=π−α
+9)2 −3(t2 −9) (t2 +9)2
=
≈ 2.41 t ≈ 4.82 (NA)
v|t=1 = Acceleration a=
=
dv dt
6 25
−3(12 −9) (12 +9)2
ms −1 ✓
−3(e2 −9) A3(ii) v| t=3 = (32 2 = 0✓
t 1
= 12 cos ( ) 2 2
= 6 cos
−3t(2t)
(t2 +9)2
=
α ≈ 0.729
=α
d dt
=
= −8 =−
dt (t2 +9)⋅ (3t) −3t⋅(t2 +9)
+9)
t
∴ Particle comes to instantaneous rest when t = 3 ✓
2
a|t=1.46 = 6 cos (
1.46 2
≈ 4.47 ms
)
−2
✓
A3(iii) For positive velocity, v >0 −3(t2 −9) (t2 +9)2 2
A2(ii) Displacement s = ∫ v dt
>0
= −24 cos + c
−3(t − 9) > 0 t2 − 9 0
=0 = 24
−3 < t < 3 and t ≥ 0 ∴0≤ t 0)
−(2)
3 2
|v| = |24√1 − ( ) |
Displacement s = ∫ v dt
4
7
= |24√ |
1
= − t 3 + t 2 + 6t + c2
16
= |24
6
√7 | 4
∵ t ≡ time after passing O, s|t=0 = 0 ⇒ c2 = 0
= 6√7 ms −1 ✓
1
∴ s = − t 3 + t 2 + 6t 6
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A math 360 sol (unofficial) A5(ii)
𝑣 𝑣=
Rev Ex 20
1 − 𝑡2 2
−2 𝑂
6
+ 2𝑡 + 6
A6(ii) Displacement s = ∫ v dt = −2e−t + 3t 2 − 2t + c
𝑡
Start from rest: s|t=0 = 0 −2e0 + c = 0 c =2
Distance travelled 6
= ∫0 |v| dt 6
= ∫0 v dt
Displacement with c = 2: s = −2e−t + 3t 2 − 2t + 2
= s|t=6 −s|t=0 = [36] −[0] = 36m ✓ A6(i)
Acceleration a = 2(3 − e−t ) = 6 − 2e−t = −2e−t + 6
s|t=1 = −2e−(1) + 3(1)2 − 2(1) + 2 2 =− +3−2+2 e ≈ 2.26m ✓ B1(i)
Velocity 144
v = (2t+3)2 − 4k
Velocity v = ∫ a dt = 2e−t + 6t + c
Initial velocity of 12 m/s: v|t=0 = 12 144 (2(0)+3)2
Start from rest: v|t=0 = 0 2 + c= 0 c = −2 Velocity with c = −2: v = 2e−t + 6t − 2
− 4k = 12
16 − 4k −4k k
= 12 = −4 =1✓
B1(ii) Velocity with k = 1: 144
v = (2t+3)2 − 4
Velocity when t = 2: v|t=2 = 2e−(2) + 6(2) − 2 ≈ 10.3 ms −1 ✓
At rest, v 144 (2t+3)2 144 (2t+3)2 (2t+3)2
=0
−4
=0 =4 =
144
1 4
(2t + 3)2 = 36 2 4t + 12t + 9 = 36 2 4t + 12t − 27 = 0 (2t − 3)(2t + 9)= 0 t=
3 2
or
t=−
9 2
(rej ∵ t > 0)
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A math 360 sol (unofficial)
Rev Ex 20
B1(iii) Velocity v = 144(2t + 3)−2 − 4
B2(ii) At rest, v =0 4e−t −
Acceleration a=
5
=0 2
dv
4e−t
=
dt
−t
e
=
−t
= ln
t
= − ln
= 144[−2(2t + 3)−3 ](2) 576 − (2t+3)3
=
2
5 1 10 1 10 1 10
≈ 2.30 ✓
1
Acceleration at t = : 2
a|t=1 = −
B2(iii) Acceleration
576 43
2
a=
= −9ms −2 ✓
=−
(2t+3)−1 (2)(−1)
72 2t+3
a|t=2.30 = −4e−(2.30) = −0.4 ms −2 ✓
] − 4t + c B3(i)
− 4t + c
∵ t ≡ time after leaving O, s|t=0 =0 −
72 3
v=
72
− 4t + 24
Acceleration
1
Displacement at t = :
a=
2
∴ s|t=1 = − 2
=−
72 1 2( )+3 2
72 4
1
− 4 ( ) + 24 2
− 2 + 24
Displacement 2
s = 4 − 4e−t − t 5
Velocity v=
ds dt
= 4e−t −
2 5
Initial velocity: v|t=0 = 4 −
dv dt
= 42 − 12t a|t=1 = 42 − 12(1) = 30ms −2 ✓
= 4m ✓ B2(i)
dt
v|t=1 = 42(1) − 6(1)2 = 36 ms −1 ✓
Displacement with c = 24: 2t+3
ds
= 42t − 6t 2
= 24
s=−
Displacement s = 21t 2 − 2t 3 Velocity
+c =0
c
dt
= −4e−t
Displacement s = ∫ v dt = 144 [
dv
B3(ii) Velocity For change in motion, v =0 2 42t − 6t = 0 t 2 − 7t =0 t(t − 7) =0 t=0 or t = 7 ✓ (rej ∵ t > 0)
2 5
= 3.6ms −1 ✓
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A math 360 sol (unofficial)
Rev Ex 20
B3(iii) s|t=7 = 21(49) − 2(343) = 343 s|t=0 = 0 s|t=10 = 2100 − 2000 = 100
B4(ii) At rest, v =0 2 t − 8t + 7 =0 (t − 1)(t − 7) = 0 t = 1 or t = 7 𝑣 𝑂1
Distance 𝑣
𝑡
7
Point A:
𝑣 = −6𝑡 2 + 42
1
s|t=1 = − 4 + 7 3
7
𝑂
=3
𝑡
1 3
Point B: Total distance travelled for 1st 10s =
10 ∫0 |v| dt
=
7 ∫0 |v| dt 7
= ∫0 v dt 7
s|t=7 =
3
=− 10 + ∫7 |v| dt
− 4(49) + 49
98 3
Distance between A & B = s|t=1 − s|t=7 = 36m ✓
10
+ ∫7 −v dt 7
= ∫0 v dt
+ ∫10 v dt
= [s]70
7 +[s]10
B4(iii) Distance Distance travelled in 1st 9s 9
= 2s|t=7 − s|t=0 −s|t=10 = 2(343) − 0 = 586 ✓ B4(i)
343
= ∫ |v| dt
−100 = =
Velocity v = t 2 − 8t + 7 Displacement
=
0 1 7 9 ∫0 |v| dt + ∫1 |v| dt + ∫7 |v| dt 1 7 9 ∫0 v dt + ∫1 −v dt + ∫7 v dt 1 1 9 ∫0 v dt + ∫7 v dt + ∫7 v dt [s]10 +36 +[s]97
= = s|t=1 − s|t=0 +36
s = ∫ v dt
=
1
= t 3 − 4t 2 + 7t + c
10 3
− 0 +36
+s|t=9 − s|t=7 98
+ [−18 − (− )] 3
= 54 ✓
3
∵ t ≡ time after passing O, s|t=0 = 0 c=0 Displacement with c = 0: 1
s = t 3 − 4t 2 + 7t ✓ 3
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553
A math 360 sol (unofficial)
Rev Ex 20
B4(iv) Displacement s=
1 3 t 3
B5(i)
2
− 4t + 7t
Velocity v = t 2 − 8t + 7 Acceleration a=
Velocity v = at 2 + b Initial Velocity of 3 m/s: v|t=0 = 3 ⇒b =3
dv dt
= 2t − 8
Velocity with b = 3: v = at 2 + 3
At zero acceleration: a =0 2t − 8 = 0 2t =8 t =4✓
Displacement s = ∫ v dt
Point O: s|t=0 = 0 Point C:
t ≡ time after passing O: s|t=0 = 0 ⇒c =0
1
= at 3 + 3t + c 3
1
s|t=4 = (4)3 − 4(4)2 + 7(4) 3
= −14
2
Displacement with c = 0:
3
Point B: s|t=7 = −
1
s = at 3 + 3t
98
3
3
Back at O after 3s: s|t=3 = 0 9a + 9 = 0 9a = −9 a = −1
OC = |s|t=4 | 2
= |−14 | 3
= 14
2 3
Velocity Velocity with a = −1: v = −t 2 + 3
BC = |s|t=4 − s|t=7 | = |(−
44 3
98
) − (− )| 3
= 18
Speed = |v|t=3 | = 6 ms −1 ✓
OC < BC ∴ OC is nearer to O ✓
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A math 360 sol (unofficial)
Rev Ex 20
B5(ii) At turning point, v =0 2 −t + 3 =0 (t + √3)(t − √3) = 0 t = −√3 or (rej ∵ t > 0)
B6(i)
t = √3
Acceleration 18 − 3t for 0 ≤ t ≤ 4 a={ t + 2 for t > 4 Velocity 0 ≤ t ≤ 4: V = ∫ a dt 3
= 18t − t 2 + c 2
Displacement Initial velocity of v m/s: V|t=0 = v c =v
1
s = − t 3 + 3t 3
3
1
s|t=√3 = − (√3) + 3√3 3
Velocity with c = v:
= 2 √3
3
V = 18t − t 2 + v
𝑣
−√3
𝑂
2
𝑣 = −𝑡 2 + 3 𝑡 √3
Velocity of 50 m/s when t = 4: V|t=4 = 50 3
18(4) − (4)2 + v = 50 2
Total Distance travelled between t = 0 and t = 3
48 + v v
3
= ∫0 |v| dt √3
= ∫0 |v| dt =
√3 ∫0 v dt √3
3
+ ∫√3|v| dt
B6(ii) 0 ≤ t ≤ 4: 3
V = 18t − t 2 + 2
3 + ∫√3 −v dt
2
√3
= ∫0 v dt
+ ∫3 v dt
= [s]√3 0
+[s]√3 3
= 2s|t=√3 −s|t=0
−s|t=3
= 2(2√3) −0
−0
= 50 =2✓
t > 4: 1 V = t 2 + 2t + c2 2 V|t=4 = 50: 1 2
= 4√3 ✓
(4)2 + 2(4) + c2 = 50
c2
= 34
t > 4: 1
V = t 2 + 2t + 34 2
Velocity at t = 5 = V|t=5 1 = (5)2 + 2(5) + 34 2 = 56.5 ms −1 ✓
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A math 360 sol (unofficial)
Rev Ex 20
B6(iii) Check for turning point At turning pt, V = 0 0 ≤ t ≤ 4: V 18t −
3 2 t 2
=0
+2 =0
3t 2 − 36t − 4 = 0 36±√(−36)2 −4(3)(−4)
t =
2(3)
36 ± 8 √21 6
=
4
= 6 ± √21 3 4 = 6 − √21 < 0 (rej ∵ t > 0) 3 4 = 6 + √21 > 4 (rej ∵ t ≤ 4) 3 ∴ no turning pt t > 4: V 1 2 t 2 2
=0
+ 2t + 34
=0
t + 4t + 68 =0 [(t + 2)2 − (2)2 ] + 68 = 0 (t + 2)2 + 64 =0 ∴ no turning pt Distance 𝑉
𝑉=
1 2 𝑡 + 2𝑡 + 34 2
3 𝑉 = − 𝑡 2 + 18𝑡 + 2 2 𝑡 4
𝑂
Total distance travelled 5
= ∫0 |V| dt 4
5
= ∫0 |V| dt
+ ∫4 |V| dt
4
5
= ∫0 V dt 4
+ ∫4 V dt 5 1
3
= ∫0 (18t − t 2 + 2) dt + ∫4 ( t 2 + 2t + 34) dt 2
=
[9t 2
−
1 3 t 2
+ 2t]
2
4 0
= [144 − 32 + 8] − 0
1 + [ t3 + t2 6 125
+[
6
+ 34t]
5 4
+ 25 + 170]
−[
64 6
+ 16 + 136]
1
= 173 m ✓ 6
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