Additional Mathematics Form 4 and 5 Notes
February 9, 2017 | Author: Iskandar Omar | Category: N/A
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SPM ZOOM–IN Form 4: Chapter 1 Functions 4. (a) gf : x → x2 + 6x + 2 gf (x) = x2 + 6x + 2 g(x + 4) = x2 + 6x + 2
Paper 1 1. The relation in the given graph can be represented using the following arrow diagram. A
Let x + 4 = u x=u–4
B
1
10
2
20
3
30
4
40
g(u) = (u – 4)2 + 6(u – 4) + 2 = u2 – 8u + 16 + 6u – 24 + 2 = u2 – 2u – 6 ∴ g(x) = x2 – 2x – 6
[
Based on the above arrow diagram, (a) the object of 40 is 3, (b) the type of the relation is many-to-many relation.
(b) fg(4) = f 42 – 2(4) – 6 = f(2) =2+4 =6
2.
5. Let g–1(x) = y g(y) = x 3y + k = x x–k y= 3 1 k y = x– 3 3 1 k ∴ g–1(x) = x – 3 3
–4 –3 –2 2 3 4
16 9 4
(a) The above relation is a many-to-one relation. (b) The function which represents the above relation is f(x) = x2.
It is given that g–1(x) = mx –
5 6
Hence, by comparison, 1 k 5 5 m = and – = – ⇒ k = 3 3 6 2
3. f 2 (x) = ff (x) = f (px + q) = p (px + q) + q = p2 x + pq + q It is given that f 2 (x) = 4x + 9 By comparison, p2 = 4 pq + q p =–2 –2q + q –q q
]
=9 =9 =9 = –9
The question requires p < 0.
1
Paper 2
2. (a) Let f –1(x) f(y) y –2 2 y 2 y y ∴ f –1(x) ∴ f –1(3)
hx x–3 hx f(x) = x–3
1. (a) f : x →
Let f –1(x) f(y) hy y–3 hy hy 3x 3x
=y =x =x
= 2(x + 2) = 2x + 4 = 2x + 4 = 2(3) + 4 = 10
Hence, by comparison, 2k + 4 = –4 2k = –8 k = –4
kx , x ≠ 2. x–2 Hence, by comparison, h = 2 and k = 3.
[
(c) hf(x) : x → 9x – 3 h[f(x)] = 9x – 3 x h – 2 = 9x – 3 2 x Let –2 =u 2 x =u+2 2 x = 2u + 4
]
gf –1(x) = g f –1(x) 3x =g x–2
1 3x x–2 x–2 = 3x gf –1(x) = –5x x–2 = –5x 3x x – 2 = –15x2 2 15x + x – 2 = 0 (3x – 1)(5x + 2) = 0 1 2 x = or – 3 5
(
=x+2
But it is given that f –1g : x → 6x – 4 f –1g (x) = 6x – 4
But it is given that f –1(x) =
=
=x
(b) f –1g(x) = f –1[g(x)] = f –1(3x + k) = 2(3x + k) + 4 = 6x + 2k + 4
= x (y – 3) = xy – 3x = xy – hy = y(x – h) 3x y= x–h 3x ∴ f –1 (x) = x–h
(b)
=y =x
)
h(u) = 9(2u + 4) – 3 = 18u + 33 ∴ h : x → 18x + 33
2
SPM Zoom-In Form 4: Chapter 2 Quadratic Equations Paper 1 1.
4. x2 + 2x – 1 + k(2x + k) = 0 x2 + 2x – 1 + 2kx + k2= 0 x2 + 2x + 2kx + k2 – 1= 0 x2 + (2 + 2k)x + k2 – 1 = 0
12x2 – 5x(2x – 1) = 2(3x + 2) 12x2 – 10x2 + 5x = 6x + 4 2 12x – 10x2 + 5x – 6x – 4 = 0 2x2 – x – 4 = 0 x=
–b
x=
–(–1)
a = 1, b = 2 + 2k, c = k2 – 1
b2 – 4ac 2a
If a quadratic equation has two real and distinct roots, then b2 – 4ac > 0.
(–1)2 – 4(2)(–4) 2(2)
b2 – 4ac > 0 (2 + 2k) – 4(1)(k2 – 1) > 0 4 + 8k + 4k2 – 4k2 + 4 > 0 8k + 8 > 0 8k > –8 k > –1 2
1 33 4 x = 1.6861 or –1.1861 x=
Product of roots = – 2 – 3 = 2 3 5 5
2. Sum of roots = – 2 + – 3 = – 19 3 5 15
5.
a = 3, b = –2m, c = 12
The required quadratic equation is x2 + 19 x + 2 = 0 15 5
If a quadratic equation has equal roots, then b2 – 4ac = 0. b2 – 4ac = 0 (–2m) – 4(3)(12) = 0 4m2 – 144 = 0 4m2 = 144 m2 = 36 m = ±6
x2 – (sum of roots)x + (product of roots) = 0
2
15x2 + 19x + 6 = 0 3. 3x2 + 4p + 2x = 0 3x2 + 2x + 4p = 0 a = 3, b = 2, c = 4p If a quadratic equation does not have real roots, then b2 – 4ac < 0. 0 k2 + 16k > 0 k(k + 16) > 0
(b) When m = –1 and n = 2, y = (x – 1)2 + 2 Hence, the minimum point is (1, 2). 4.
18
(2 + p)(6 – p) < 7 12 + 4p – p2 – 7 < 0 –p2 + 4p + 5 < 0 p2 – 4p – 5 > 0 (p + 1)(p – 5) > 0
–16
0
k
Hence, the required range of values of k is k < –16 or k > 0.
6
Paper 2
(b) g(x) = –2x2 + 8x – 12 = –2(x – 2)2 – 4
2 5 = 4
(a) f(x) = 2x2 + 10x + k k = 2 x2 + 5x + 2 25 25 k = 2 x2 + 5x + – + 4 4 2 5 2 25 k =2 x+ – + 2 4 2 2 5 25 =2 x+ – +k 2 2
[
1
2
25
The maximum point is (2, –4). When x = 0, y = –12 ∴ (0, –12) The graph of the function g(x) is as shown below.
y
]
O (2, –4)
(b) (i) Minimum value = 32 25 – + k = 32 2 89 k = 2 (ii)
–12
3. y = h – 2x… 1 y2 + xy + 8 = 0 …
b2 – 4ac 2 10 – 4(2)(k) 100 – 8k – 8k
0 and q < 0)
7
SPM ZOOM–IN Form 4: Chapter 4 Simultaneous Equations Paper 2 1. 2x – 3y = 2 x2 – xy + y2 = 4
… …
From 3 , When x = 0.70156, y = 2 – 4(0.70156) = –0.80624
1 2
When x = –5.70156, y = 2 – 4(–5.70156) = 24.80624
1
From : 2 + 3y x= … 3 2 Substituting 3 into
2
Hence, the solutions are x = 0.70156, y = –0.80624 or x = –5.70156, y = 24.80624 (correct to five decimal places).
:
2 +23y – y 2 +23y + y – 4 = 0 2
2
(2 + 3y)2 – y(2 + 3y) + y2 – 4 4 2 (2 + 3y)2 – 2y(2 + 3y) + 4y2 – 16 4 + 12y + 9y2 – 4y – 6y2 + 4y2 – 16 7y2 + 8y – 12 (7y – 6)(y + 2)
3. (a) Since (16, m) is a point of intersection of 1 y = x – 2 and y2 + ky – x – 4 = 0, then 4 x = 16 and y = m satisfy both the equations.
=0
=0 =0 =0 =0 6 y= or –2 7
From
3
When y = –2, x =
2+3 2
( 76 ) = 16
(b) When k = 8, 1 y= x–2… 4
7
2 + 3(–2) = –2 2
2 6 Hence, the points of intersection are 2 , 7 7 and (–2, –2). 2. 4x + y = 2 … 1 x2 + x – y = 2 … 1
:
y = 2 – 4x… 2
=0 =0 = 16 =8
1
y2 + 8y – x – 4 = 0 …
From 1 : 4y = x – 8 x = 4y + 8 …
3
,
2
3
Substituting 3 into y2 + 8y – (4y + 8) – 4 y2 + 8y – 4y – 8 – 4 y2 + 4y – 12 (y – 2)(y + 6) y
2
Substituting 3 into x2 + x – (2 – 4x) = 2 x2 + 5x – 4 = 0 x = –5 ±
m2 + km – 16 – 4 22 + k(2) – 16 – 4 2k k
:
6 When y = , x = 7
From
Therefore, 1 m = (16) – 2 = 2 and 4
2 , =0 =0 =0 =0 = 2 or –6
From 3 : When y = 2, x = 4(2) + 8 = 16 When y = –6, x = 4(–6) + 8 = –16
52
– 4(1)(–4) 2(1) = –5 ± 41 2 = 0.70156 or –5.70156
Hence, the other point of intersection, other than (16, 2), is (–16, –6).
8
SPM ZOOM–IN Form 4: Chapter 5 Indices and Logarithms Paper 1
4.
5x lg 5x x lg 5 x lg 5 x lg 5 – 2x lg 3 x(lg 5 – 2lg 3)
5.
log10 (p + 3) log10 (p + 3) – log10 p log10 p + 3 p p+3 p p+3 9p
1. 2 x + 3 + 2x + 16 (2x – 1) 2x = 2x.23 + 2x + 16 2 = 8(2x) + 2x + 8(2x) = (8 + 1 + 8)( 2x) = 17(2x)
2.
3x + 3 – 3x + 2 3 (33) – 3x (32) 27(3x) – 9(3x) (27 – 9)(3x) 18(3x) x
3x 3x 3x x 3.
m = 3a log3 m = a
=6 =6 =6 =6 =6 = 6 18 = 1 3 = 3–1 = –1
= 32x – 1 = lg 32x –1 = (2x – 1) lg 3 = 2x lg 3 – lg 3 = – lg 3 = –lg 3 –lg 3 x = lg 5 – 2 lg 3 x = 1.87
= 1 + log10 p =1
=1 = 101
= 10p =3 1 p = 3
n = 3b log3 n = b
6.
mn4 log3 27 = log3 m + log3 n4 – log3 27 = log3 m + 4 log3 n – log3 33 = a + 4b – 3
log2 y – log8 x log2 x log2 y – log2 8 log2 x log2 y – 3 3 log2 y – log2 x log2 y3 – log2 x y3 log2 x y3 x y3
=1 =1 =1 =3 =3
=3 = 23
= 8x y3 x = 8
9
log2 8 = log2 23 = 3
SPM ZOOM–IN Form 4: Chapter 6 Coordinate Geometry Hence, the area of ∆PQR 1 4 0 2 4 = 2 0 –3 5 0 1 = |–12 – (–6 + 20)| 2 = 1 |–26| 2 = 1 26 2 = 13 units2
Paper 1 1. Let point A be (0, k). AB = 10 2 (0 – 8) + (k – 7)2 = 10 64 + k2 – 14k + 49 = 102 k2 – 14k + 13 = 0 (k – 1)(k – 13) = 0 k = 1 or 13 Based on the diagram, k < 7. ∴k=1 ∴ A(0, 1)
4. (a) 2y = 3x – 12 At point L (on the x-axis), y = 0 2(0) = 3x – 12 x =4 ∴ L (4, 0)
2. (a) x + 2y + 6 = 0 x + 2y = –6 x 2y –6 + = (–6) (–6) –6 x y + =1 (–6) (–3)
(b) mMN = – –3 = – 1 –6 2
At point N (on the y-axis), x = 0. 2y = 3(0) – 12 y = –6 ∴ N (0, –6)
Intercept form: x y + =1 a b
∴ M = 4 + 0 , 0 + (–6) = (2, –3) 2 2
m = – y-intercept x-intercept
(b) mLN = –6 – 0 = 3 0–4 2
Therefore, the gradient of the perpendicular line is 2.
∴ Gradient of perpendicular line = – 2 3
Hence, the equation of the straight line which passes through the point N and is perpendicular to the straight line MN is y = 2x – 3.
Hence, the equation of the perpendicular line is y – y1 = m(x – x1) y – (–3) = – 2 (x – 2) 3 3(y + 3) = –2(x – 2) 3y + 9 = –2x + 4 3y = –2x – 5
y 3. x – = 1 4 3 At point P (on the x-axis), y = 0. x – 0 =1⇒x=4 4 3 ∴ P is point (4, 0).
5.
PA (x – + (y – 2)2 2 (x – 1) + (y – 2)2 2 x – 2x + 1 + y2 – 4y + 4 –2x – 4y + 5 –2x + 2y – 4 –x + y – 2 y
At point Q (on the y-axis), x = 0. 0 – y = 1 ⇒ y = –3 4 3 ∴ Q is point (0, –3).
1)2
10
= PB = (x – 0)2 + (y – 3)2 = (x – 0)2 + (y – 3)2 = x2 + y2 – 6y + 9 = –6y + 9 =0 =0 =x+2
Paper 2
(c) A(–18, 0), B(2, 0), C(0, –6), D(–20, –6) Area of ABCD 1 –18 2 0 –20 –18 = 2 0 0 –6 –6 0 1 = |–12 – (120 + 108)| 2 = 1 240 2 = 120 units2
1. (a) y – 3x + 6 = 0 At point B (x-axis), y = 0. 0 – 3x + 6 = 0 ⇒ x = 2 ∴ B is point (2, 0). y – 3x + 6 = 0 At point C (y-axis), x = 0. y – 3(0) + 6 = 0 ⇒ y = –6 ∴ C is point (0, –6).
2. (a) (i) y – 3x + 6 = 0 At point P (on the y-axis), x = 0. y – 3(0) + 6 = 0 ⇒ y = –6 ∴ P is point (0, –6). (ii) The coordinates of point S are 4(0) + 3(7) , 4(–6) + 3(15) = (3, 3) 3+4 3+4
y = 3x – 6 mBC = 3 ∴mAC = – 1 3
Let A(k, 0). ∴ mAC = – 1 3 0 – (–6) = – 1 k–0 3 –k = 18 k = –18 ∴ A is point (–18, 0). (b) Let D (p, q). Midpoint of BD = Midpoint of AC 2 + p , 0 + q = –18 + 0 , 0 + (–6) 2 2 2 2
2 +2 p , q2 = (–9, –3)
(b) Area of ∆QRS = 48 units2 1 k 7 3 k = 48 2 0 15 3 0 15k + 21 – (45 + 3k) = 96 12k – 24 = 96 12k = 120 k = 10 (c) S(3, 3), Q(10, 0), T(x, y) TS : TQ = 2 : 3 TS = 2 TQ 3 3TS = 2TQ 9(TS)2 = 4(TQ)2 9[(x – 3)2 + (y – 3)2] = 4[(x – 10)2 + (y – 0)2] 9(x2 – 6x + 9 + y2 – 6y + 9) = 4(x2 – 20x + 100 + y2) 2 2 9x – 54x + 81 + 9y – 54y + 81 = 4x2 – 80x + 400 + 4y2 2 2 5x + 26x + 5y – 54y – 238 = 0
Equating the x-coordinates, 2 + p = –9 2 p = –20 Equating the y-coordinates, q = –3 2 q = –6 ∴ D is point (–20, –6).
11
SPM ZOOM–IN Form 4: Chapter 7 Statistics Paper 1
2 = 1. After the given score are arranged in ascending order, we have 6
6
6
k
k
=
9
Since the mode is 6, then k ≠ 9.
6
6
8
8
4.
9
After two new scores, 7 and 10, are added to the original scores, the mean of the eight scores = 6 + 6 + 6 + 8 + 8 + 9 + 7 + 10 8 = 7.5
(b)
10 132 318 – 10 10
2
Number
1
k
6
Frequency
2
2
1
(a) 1 < k < 6 k = 2, 3, 4, 5
Median = 7
2. (a)
2
= 1.96
For 7 to be the median, k = 8, as shown below. 6
∑fx 2 ∑fx – ∑f ∑f
k + 3 11 3
1
13 1
6 < k + 3 < 11 3 0) x3 dx 2 ∴ L is a minimum.
4.
y=
h = h(1 + 2x)–2 (1 + 2x)2
dy = –2h(1 + 2x)–3 (2) = – 4h dx (1 + 2x)3 δy = dy δx dx 8c – = – 4h 3 c 3 (1 + 2x) 8c 4h – =– c 3 [1 + 2(1)]3 – 8c = – 4hc 3 27 h = 8 27 3 4 h = 18
18
SPM ZOOM–IN Form 4: Chapter 10 Solution of Triangles 2. (a) In ∆PQS, using the sine rule, sin ∠QSP = sin 35º 8 7 sin ∠QSP = sin 35º 8 7 sin ∠QSP = 0.65552 ∠QSP = 40.96º
Paper 2 1. (a) ∠UST = 180º – 65º = 115º ∠SUT = 180º – 43º – 115º = 22º In ∆UST, using the sine rule, US = 9 sin 43º sin 22º US = 9 sin 43º sin 22º = 16.385 cm
∴ ∠PQS = 180º – 35º – 40.96º = 104.04º Hence, the area of ∆PQS = 1 8 7 sin 104.04º 2
U 22°
= 27.16 cm2 16
(b) This problem involves the ambiguous case of sine rule. The sketch of ∆QRS1 is as shown below.
.3 85 cm
Q
115° 65° R
43° S
7 cm
9 cm
T 10 cm
(b) In ∆USR, using the cosine rule, UR2 = 72 + 16.3852 – 2(7)(16.385)cos 65º UR2 = 220.5238 UR = 14.85 cm
1 7 12 sin ∠RSV 2 sin ∠RSV Basic ∠ ∠RSV
R
7 cm
7 cm
43° R
S1
S
In ∆QRS, using the sine rule, sin ∠QSR = sin 43º 10 7 sin 43º 10 sin ∠QSR = 7
Area of ∆RSV = 41.36 cm2
(c)
7 cm
= 41.36 = 0.98476 = 79.98º = 180º– 79.98º = 100.02º
sin ∠QSR = 0.974283 Basic ∠ = 76.98º ∴ ∠QSR = 76.98º or ∠QS1R = 103.02º
S
100.02°
In ∆QS1R, ∠RQS1 = 180º – 43º – 103.02º = 33.98º
12 cm
In ∆QS1R, using the sine rule, RS1 10 = sin ∠RQS1 sin ∠RS1Q RS1 10 = sin 33.98º sin 103.02º 10 RS1 = sin 33.98º sin 103.02º = 5.737 cm
V
In ∆RSV, using the cosine rule, RV 2 = 72 + 122 – 2(7)(12)cos 100.02º RV 2 = 222.23064 RV = 14.91 cm 19
SPM ZOOM–IN Form 4: Chapter 11 Index Numbers Paper 2 1. (a)
(a) Supplement A x 100 = 120 400 x = 480
I2004 (based on 2002) = 115 P2004 100 = 115 P2002 69 100 = 115 P2002 P2002 = 69 100 115 P2002 = RM60.00
P2004 I= P 100 2002
y = 525 100 = 105 500 660 100 = 110 z z = 600
(b) Supplement B I2006 (based on 2002) P = 2006 100 P2002 P P = 2006 2004 100 P2004 P2002 = 130 120 100 100 100 = 156
– (b) I = 115 (120 20) + 130m + (105 80) + (110 40) = 115 20 + m + 80 + 40 15 200 + 130m = 115 140 + m 15 200 + 130m = 16 100 + 115m 15m = 900 m = 60
(c)
– (c) I 2006 (based on 2002) – = 100 + 25 I 2004 100 = 125 115 100 = 143.75
(115 3) + (120 2) + 105x 3+2+x 585 + 105x 5+x 585 + 105x 30 x – (d) I 2006 (based on 2004)
(d) Total yearly cost in 2006 = 143.75 5 500 000 100 = RM7 906 250
= 111 = 111 = 555 + 111x = 6x =5
= (150 3) + (130 2) + (120 5) 3+2+5 = 1310 10 = 131 P Thus, 2006 100 = 131 P2004 P2006 100 = 131 300 P2006 = 131 300 100 P2006 = RM393
2. Health I2004 (based supplement on 2002) A 115 B 120 C 105
– I 2004 (based on 2002) = 111
I2006 (based Weightage on 2004) 150 3 130 2 120 x
20
SPM ZOOM–IN Form 5: Chapter 1 Progressions 4.
Paper 1 1. (a)
T6 = 38 a + 5d = 38 a + 5(7) = 38 a =3
r = – 1 or 3 2 2
(b) S9 – S3 = 9 [2(3) + 8(7)] – 3 [2(3) + 2(7)] 2 2 = 279 – 30 = 249 2. (a)
T2 – T1 2h – 1 – (h – 2) h+1 h
5. 0.242424 … = 0.24 + 0.0024 + 0.000024 + … a = 0.24 S∞ = 1 – 0.01 1–r = 0.24 0.99 = 24 99 = 8 33
= T3 – T2 = 4h – 7 – (2h – 1) = 2h – 6 =7
(b) When h = 7, the arithmetic progression is 5, 13, 21, … with a = 5 and d = 8.
6. The numbers of bacteria form a geometric progression 3, 6, 12, …
S8 – S3 = 8 [2(5) + 7(8)] – 3 [2(5) + 2(8)] 2 2 = 264 – 39 = 225 3.
T3 – T2 = 3 ar 2 – ar = 3 4r 2 – 4r = 3 2 4r – 4r – 3 = 0 (2r + 1)(2r – 3) = 0
The number of bacteria after 50 seconds = T11 = ar10 = 3(210) = 3072 Paper 2 1. (a) The volumes of cylinders are πr 2h, πr 2 (h + 1), πr 2 (h + 2), …
T2 T3 = T1 T2 x+2 = x–4 9x + 4 x+2 (x = (x – 4)(9x + 4) x2 + 4x + 4 = 9x2 – 32x – 16 2 8x – 36x – 20 = 0 2x2 – 9x – 5 = 0 (2x + 1)(x – 5) = 0 x = – 1 or 5 2
T2 – T1 = πr 2 (h + 1) – πr 2h = πr 2h + πr 2 – πr 2h = πr2 T3 – T2 = πr 2 (h + 2) – πr 2 (h + 1) = πr 2h + 2πr 2 – πr 2h – πr 2 = πr2 Since T2 – T1 = T3 – T2 = πr 2, the volumes of cylinders form an arithmetic progression with a common difference of πr2.
21
2. (a)
(b) a = πr 2h, d = πr 2 T4 = 32π a + 3d = 32π πr 2h + 3πr 2 = 32π r 2h + 3r 2 = 32 r 2 (h + 3) = 32 …
2 1
1
2
: a (1 + r) = 150 45 ar (r – 1) 1 + r = 10 r (r – 1) 3
1 2
2
3 + 3r = 10r2 – 10r 10r – 13r – 3 = 0 (2r – 3)(5r + 1) = 0 r = 3 or – 1 2 5 2
2 : r 2(2h + 3) = 52 32 r (h + 3) 2h + 3 = 13 h+3 8 16h + 24 = 13h + 39 3h = 15 h =5
From 1 r 2 (5 + 3) r2 r
= 150 = 150 = 150 = 150 …
T3 – T2 = 45 ar 2 – ar = 45 ar (r – 1) = 45 …
1
S4 = 104π 4 (2a + 3d) = 104π 2 4a + 6d = 104π 4πr 2h + 6πr 2 = 104π 2r 2h + 3r 2 = 52 r 2 (2h + 3) = 52 …
S2 T1 + T2 a + ar a (1 + r)
(b) For the sum to infinity to exist, – 1 < r < 1. Thus, r = 3 is not accepted. 2 Therefore, r = – 1 5
: = 32 =4 =2
From
1
: 1 a 1– = 150 5
a = 187 1 2 1 187 a 2 ∴ S∞ = = 1–r 1– –1 5
22
= 156 1 4
SPM ZOOM–IN Form 5: Chapter 2 Linear Law Paper 2
Paper 1 y = 2 + qx x y = 2 +q x x2 y =2( 1 )+q x x2
1.
1 x2
1. (a)
y = hxk log10 y = log10 (hxk) 1
log10 y 2 = log10 h + log10 xk 1 log y = log h + klog x 10 10 2 10 log10 y = 2log10 h + 2klog10 x
y x
log10 y = 2k log10 x + 2 log10 h
5 = 2(1) + q q=3
(1, 5)
(b) 1 x2
q=3
y x (3, p)
2.
y k lg
x
=p
x
= lg p
y
k
lg y – lg k
x
x y log 10 x log 10 y
p = 2(3) + 3 p=9
1.5 142 0.18 2.15
2.0 338 0.30 2.53
2.5 660 0.40 2.82
3.0 1348 0.48 3.13
3.5 1995 0.54 3.30
The graph of log10 y against log10 x is as shown below. log10 y
= lg p
Graph of log10 y against log10 x
3.5
lg y – x lg k = lg p 3.0
lg y = x lg k + lg p ∴ Y = lg y, X = x, m = lg k, c = lg p
2.0
2 3. y – ax = b2 x x 2 3 xy – ax = b xy2 = ax3 + b
1.0 0.5
O
x3
1
xy2
–
2
:
12 = –6a a = –2
From
2
:
–2 = –2(5) + b b =8
0.55 – 0.06 = 0.49
1.55 1.5
10 = a(–1) + b … 1 –2 = a(5) + b … 2
(–1, 10): (5, –2):
3.35 – 1.75 = 1.6
2.5
23
0.1
0.2
0.3
0.4
0.5
0.6
log10 x
(c) 2k = Gradient 2k = 3.35 – 1.75 0.55 – 0.06 2k = 3.2653 k = 1.63 2 log10 h 2 log10 h log10 h h
(b) (i) 1 = y
1 = x+p y2 q 1 = 1x+ p y2 q q
= Y–intercept = 1.55 = 0.775 = 5.96
Gradient =
x y
0.1 0.3 0.78 0.60
0.4 0.54
0.5 0.50
0.7 0.44
0.8 0.42
1 y2
1.64 2.78
3.43
4.00
5.17
5.67
Graph of
Y-intercept p q p 0.17 p
1 against x y2
4.6 – 2 0.6 – 0.16
= 1.1 = 1.1 = 1.1 = 0.19
(ii) When x = 0.6, from the graph, 1 = 4.6 y2 y2 = 0.2174 y = 0.47
5.0 4.6 4.5 4.0 3.5 4.6 – 2 = 2.6
3.0 2.5 2.0 0.6 – 0.16 = 0.44 1.5 1.1 1.0 0.5
O
Squaring both sides.
1 = 5.91 q q = 0.17
2. (a)
1 y2 5.5
x+p q
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
x
24
SPM ZOOM–IN Form 5: Chapter 3 Integration 4. Area of the shaded region
Paper 1
=
k
1.
(y – 5) dy = 8 5 k
[
2
]
=
=
y2 –5y = 8 5 2 2
k – 5k – 5 – 5(5) = 8 2 2 2 k – 5k + 25 = 8 2 2 k2 – 10k + 25 = 16 k2 – 10k + 9 = 0 (k – 1)(k – 9) = 0 k = 1 or 9 2.
2 –1
4
3g (x) dx +
[ = 3[ =3
2 –1
4 –1
2
2 –1
2 –1
y dx (x2 – 2x + 1) dx 2
[ x3 – x + x] 3
2
–1
= 8 – 4 + 2 – – 1 –1 – 1 3 3 = 3 units2 Paper 2 y
1. y = –x3 – x
3g (x) dx
g(x) dx +
4 2
]
g(x) dx
1O
]
g(x) dx
x
2
P
Q
= 3(20) = 60 3. dy = x4 – 8x3 + 6x2 dx y=
4
3
Area P 0
x – 8x + 6x
5
2
–1
dx
=
+c 4
3
y = x –8 x +6 x 5 4 3 5 4 3 y = x – 2x + 2x + c 5
y dx
0 –1
(–x3 –x) dx 0
[
]
4 2 = – x – x 4 2 –1 4 2 (–1) =0– – – (–1) 4 2 = 1 + 1 4 2 = 3 4
[
Since the curve passes through the point
1, – 1 45 , – 9 = 1 –2 + 2 + c 5 5 c = –2
Area Q
Hence, the equation of the curve is 5 y = x – 2x4 + 2x3 – 2. 5
=
2 0
y dx 2 0
[
(–x3 – x) dx 2
]
4 2 = – x – x 4 2 0 4 2 2 2 =– – –0 4 2 = –4 – 2 = –6
25
]
Hence, the total area of the shaded region = Area P + |Area Q| = 3 + |–6| 4 = 6 3 units2 4 2. (a)
(b) When h = 1 and k = 4, y = x2 + 4 y y = x2 + 4 P
y = hx2 + k dy = 2hx dx
x =3
4
At the point (–2, 8), the gradient of the curve is – 4. ∴ dy = – 4 dx 2hx = – 4 2h (–2) = – 4 – 4h = – 4 h =1
x
O
2
3 Q
Volume generated, Vx = Volume generated by the curve – Volume generated by the straight line PQ (from x = 0 to x = 2) 3 = π 0 y2 dx – 1 πr2h 3 3 2 2 = π 0 (x + 4) dx – 1 π(4)2 (2) 3 3 4 2 = π 0 (x + 8x + 16) dx – 32 π 3 3 5 3 32 = π x + 8x + 16x – π 0 3 5 3 5 = π 3 + 8 (3)3 + 16(3) – 0] – 32 π 3 3 5 14 3 = 157 π units 15
The curve y = hx2 + k passes through the point (–2, 8). ∴ 8 = h(–2)2 + k 8 = 4h + k 8 = 4(1) + k k =4
[ [
26
]
SPM ZOOM–IN Form 5: Chapter 4 Vectors Paper 1
4. (a) If the vectors _a and _b are parallel, then
→ → 1. (a) EA = 3 DC = 3 (12p _ ) = 9p _ 4 4
a_ = hb _ (h is a constant). 2i_ – 5j _ = h(ki_ – 3j _) 2i_ – 5j _ = hki_ – 3hj _
→ → (b) EQ = 1 ED 2 → → → → = 1 EA + AB + BC + CD 2
Equating the coefficients of _j –3h = –5 h= 5 3
= 1 9p – 6r _ – 9q _ – 12p _ 2 _ = 1 – 3p _ – 9q_ _ – 6r 2
Equating the coefficients of _i hk = 2 5 k =2 3 k= 6 5
→ → → 2. XY = XD + DY → → = 1 BD + 2 DC 2 3 → → → = 1 BA + AD + 2 AB 2 3
(b)
2 = 1 –6b ) _ + 2a _ + (6b 2 3 _
|a| 5 |b| = 3 |a| : |b| = 5 : 3
= –3b _ + _a + 4b _ = _a + _b
→ → → 5. (a) AC = AB + BC = 9i_ – 4j _ + (–6i_ + mj_)
3. (a) _a + 1 _b + 2c _ 5 = 7j_ + 1 (10i_ – 5j_ ) + 2(–4i_ +_j ) 5 = 7j + 2i _ _ – _j – 8i_ + 2j _
= 3i_ + (m – 4)_j → (b) If AC is parallel to the x-axis, the coefficient of _j equals zero. m–4 =0 m =4
= – 6i_ + 8j _ 1 (b) |a _ + _b + 2c _ | = (–6)2 + 82 = 10 5 Hence, the unit vector in the direction of 1 _a + _b + 2c _ 5
= 1 –6i_ + 8j _ 10
_a = hb _ 5 _a = b_ 3
= – 3 _i + 4 _j 5 5
27
Paper 2
(c) Since the points O, T and S are collinear, → → then, OT = kOS , where k is a constant.
→ → → 1. (a) OT = OA + AT → = 4x _ + 1 AQ 3 → → = 4x _ + 1 (AO + OQ ) 3 = 4x _ + 1 (–4x _ + 6y _) 3 = 8 _x + 2y _ 3
→ → OT = kOS 8 _x + 2y = k [(6 – 6h)y + 16hx _] _ _ 3 8 _x + 2y = k (6 – 6h)y + 16hkx _ _ _ 3 8 _x + 2y = (6k – 6hk)y + 16hkx _ _ _ 3
→ → → (b) OS = OQ + QS → = 6y_ + hQP → → = 6y_ + h(QO + OP ) → → = 6y_ + h(QO + 4OA )
Equating the coefficients of _, x 8 = 16hk 3 1 = 6hk hk = 1 … 1 6
= 6y_ + h[–6y _)] _ + 4(4x
Equating the coefficients of _y, 6k – 6hk = 2 … 2
= (6 – 6h) _y + 16hx _
Substituting
1
into
6k – 6 1 = 2 6 6k = 3 k= 1 2 From
2
:
hk = 1 6
h 1 =1 2 6 h= 1 3
28
2
:
→ → 2. (a) (i) OM = 5 OB = 5 (14y _) = 10y _ 7 7
→ → → (c) AK = AL + LK
_ + 7 qy _ – 1 _x + 7 _y = –2px _ + 10py _ + 3 qx 2 2 2 2
→ → (ii) AK = 1 AB 4 → → = 1 AO + OB 4
–x _ + 7y _ + (20p + 7q)y _ = (–4p + 3q)x _ Equating the coefficients of _x , – 4p + 3q = –1 … 1
= – 1 _x + 7 _y 2 2
Equating the coefficients of _y , 20p + 7q = 7 … 2
→ → (b) (i) AL = pAM → → = p AO + OM
+
= p(–2x _ + 10y _) = –2px + 10py _ → → (ii) KL = qKO → → = q KA + AO
冧
– 1 _x + 7 _y = –2p + 3 q _x + 10p + 7 q _y 2 2 2 2
= 1 –2x _ + 14y _ 4
–20p + 15q = –5 … 20p + 7q= 7 … 22q = 2 q= 1 11 From
1 2
1
: – 4p + 3 1 = –1 11
→ → KA = –AK = 1_x – 7_y 2 2
– 4p = – 14 11 7 p= 22
= q – 3 _x – 7 _y 2 2 = q 1 _x – 7 _y – 2x _ 2 2
= – 3 qx _ – 7 qy 2 2 _
29
5
WebsiteZI F505_4th pp
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Page 30
SPM ZOOM–IN Form 5: Chapter 5 Trigonometric Functions Paper 1
3.
1.
3 tan θ = 2 tan (45º – θ) 3 tan θ = 2
1tan+ tan45º45º– tantanθθ
3 tan θ = 2
tan θ 11 +– tan θ
1 + p2
θ
1 O
–p
3 tan θ + 3 tan2 θ = 2 – 2 tan θ 3 tan2 θ + 5 tan θ – 2 = 0 (3 tan θ – 1)(tan θ + 2) = 0 tan θ = 1 or tan θ = –2 3
sin (90º – θ) = cos θ =–
p
When tan θ = 1 , 3 Basic ∠ = 18.43º θ = 18.43º, 198.43º
1 + p2 2.
3 – 10 tan x = 0 cos2 x 3 sec2 x – 10 tan x = 0 2 3(tan x + 1) – 10 tan x = 0 3 tan2 x + 3 – 10 tan x = 0 3 tan2 x – 10 tan x + 3 = 0 (3 tan x – 1)(tan x – 3) = 0 tan x = 1 or tan x = 3 3
When tan θ = –2, Basic ∠ = 63.43º θ = 116.57º, 296.57º ∴ θ = 18.43º, 116.57º, 198.43º, 296.57º
When tan x = 1 , 3 x = 18.43º, 198.43º When tan x = 3, x = 71.57º, 251.57º ∴ x = 18.43º, 71.57º, 198.43º, 251.57º
30
WebsiteZI F505_4th pp
10/15/08
9:40 AM
Page 31
Paper 2
2. (a), (b)
1. (a) LHS = = =
y
1 – cos 2x sin 2x
y = 3 sin x 2
3 2
1 – (1 – 2 sin2 x) 2 sin x cos x
O
2
x
–2
2 sin2 x 2 sin x cos x
y = 2 – 2x 2
sin x cos x = tan x = RHS =
x 2x + =2 2 π x 2x 3 sin =2– 2 π
3 sin
(b) (i), (ii) The graph of y = |tan x| is as shown below.
Sketch the straight line y = 2 – x y
y y=
π
0 2
2π –2
(2π, 1) O
π
π
3π 2
π
Number of solution = Number of intersection point =1
x
1 – cos 2x x – =0 sin 2x 2π 1 – cos 2x x = sin 2x 2π x |tan x| = 2π
Sketch the straight line y= x . 2π
Number of solutions = Number of points of intersection =4
31
2x π
WebsiteZI F506_4th pp
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9:40 AM
Page 32
SPM ZOOM–IN Form 5: Chapter 6 Permutations and Combinations Paper 1
3. Number of different committees that can be formed
1.
Number of arrangements 2
1
3P 2
2
3
3P 2
2
5
3P 2
Choosing a female secretary and a female treasurer from 7 females
= 4C1 7C2 8C3
Hence, the number of 4-digit odd numbers greater than 2000 but less than 3000 that can be formed = 3 3P2 = 18
Choosing a male president from 4 males
= 4704 2. Each group of boys and girls is counted as one item. B1, B2 and B3 √ This gives 2!.
G1, G2 and G3 √
√ √ √ At the same time, B1, B2, and B3 can be arranged among themselves in their group. This gives 3!. √ √ √ In the same way, G1, G2, and G3 can also be arranged among themselves in their group. This gives another 3!. Using the multiplication principle, the total number of arrangements = 2! 3! 3! = 72
32
Choosing 3 subcommittee members from 8 (males or females).
SPM ZOOM–IN Form 5: Chapter 7 Probability Paper 1 1. P(Not a green ball) = 3 5 h+5 = 3 h+k+5 5 5h + 25 = 3h + 3k + 15 2h = 3k – 10 h = 3k – 10 2 2. P (not getting any post) =2 3 4 3 5 7 8 = 35 – 3. There are 3 ‘E’ and 4 ‘E ’ in the bag (a) P(EE) = 3 2 = 1 7 6 7 – (b) P(EE ) = 3 4 = 2 7 6 7
33
SPM ZOOM–IN Form 5: Chapter 8 Probability Distributions Paper 1
3. (a) X – Mass of a crab, in g X ∼ N(175, 15)
1. X – Number of penalty goals scored X ∼ B n, 3 5 P(X = 0) = 16 625
Co 3 5
n
Z = X–µ σ = 190 – 175 15 =1
0
2 5
n
= 16 625
(b) P(175 < X < 190) = P 175 – 175 < Z < 190 – 175 15 15 = P (0 < Z < 1) = 0.5 – 0.1587 = 0.3413
16 = 625 25 = 25
(1)(1) 2 5
n
n
4
0.1587
∴n =4 O
2. X ∼ N(55, 12 ) Area of the shaded region P (X < 37) = P Z < 37 – 55 12 = P (Z < –1.5) 0.0668 = 0.0668 2
4. 0.1841
–0.9
P(Z > – 0.9) = 0.8159 ∴ k = –0.9
–1.5
34
0.8159
1
(ii) P(10 < X < 13) = P 10 – 12 < Z < 13 – 12 3.1201 3.1201 = P(–0.641 < Z < 0.321) = 1 – 0.2608 – 0.3741 0.2608 = 0.3651
Paper 2
1. (a) X – Number of blue beads drawn
X ∼ B 10, 1 3
X ∼ B 10, 6 18
0.3741
–0.641
(i) P(X ≥ 3) = 1 – P(X = 0) – P(X = 1) – P(X = 2)
23 C1 2 3 3
= 1 – 10C0 1 3 – 10
0
10
2
2. (a) X – Number of customers requiring a supplementary card
23
– 10C1 1 3
0.321
1
X ∼ B 7, 14 25
9
X ∼ B 7, 280 500
8
2
= 0.7009 (i) P(X = 3)
11 25
(ii) Mean = np = 10 1 = 3 1 3 3
= 7C3 14 25 = 0.2304
Standard deviation = npq
C 14 11 25 25
5
2
= 0.1402
P(X > 8) = 90%
2
7
(b) X – Lifespan of a species of dog X ∼ N(12, σ2)
(b) X – Time taken to settle invoices X ∼ N(30, 52)
P Z > –4 = 0.9 σ
P Z > 8 – 12 σ
4
(ii) P(X = 3) = P(X = 0) + P(X = 1) + P(X = 2) 0 11 7 +7C 14 1 11 6 + = 7C0 14 1 25 25 25 25
= 10 1 2 3 3 = 1.49
(i)
3
= 0.9
(i) P(28 ≤ X ≤ 36)
= P 28 – 30 ≤ Z ≤ 36 – 30 5 5 = P(–0.4 ≤ Z ≤ 1.2) = 1– 0.3446 – 0.1151 = 0.5403
0.9 0.1
–1.282 0.3446
0.1151
– 4 = –1.282 σ –0.4
σ = 3.1201 years
35
1.2
(ii) P(X < 22)
= P Z < 22 – 30 5 = P(Z < –1.6) = 0.0548
Hence, the expected number of invoices which are given discounts = 0.0548 220 = 12
36
SPM ZOOM–IN Form 5: Chapter 9 Motion Along a Straight Line Paper 2
(c) When particle A reverses its direction, vA = 0 12 + t – t2 = 0 t2 – t – 12 = 0 (t + 3)(t – 4) = 0 t = –3 or 4 t = –3 is not accepted ∴t=4
1. (a) For particle A, at maximum velocity, dvA = 0 dt 1 – 2t = 0 t= 1 2 d 2vA = –2 (negative) dt 2
vB = dsB dt vB = 6t2 – 14t – 15
Hence, vmax = 12 + 1 – 1 2 2 = 12 1 m s–1 4
2
aB = dvB dt aB =12t – 14
(b) sB = 2t 3 – 7t 2 – 15t When particle B returns to O, sB = 0 2t 3 – 7t 2 – 15t = 0 t(2t 2 – 7t – 15) = 0 t(2t + 3)(t – 5) = 0 t = 0, – 3 or 5 2 t = 0 and t = – 3 are not accepted 2 ∴t=5
When t = 4, aB =12(4) – 14 = 34 m s–2 2 (a) a = 12 – 6t v = a dt v = (12 – 6t) dt v = 12t – 3t2 + c When t = 0, v = 15. Thus, c = 15 ∴ v = 12t – 3t2 + 15 At maximum velocity, dv = 0 dt 12 – 6t = 0 t=2
sA = vA dt sA = (12 + t – t2) dt 2 3 sA = 12t + t – t + c 2 3 When t = 0, sA = 0. ∴ c = 0 2 3 ∴ sA = 12t + t – t 2 3
When t = 2, v = 12(2) – 3(2)2 + 15 = 27 m s–1 d2v = –6 (< 0) dt2 Therefore, v is a maximum.
When t = 5, 2 3 sA = 12(5) + 5 – 5 = 30 5 m 6 2 3
37
(b) s = v dt s = (12t – 3t2 + 15) dt s = 6t 2 – t 3 + 15t + c When t = 0, s = 0. Thus, c = 0. ∴ s = 6t 2 – t 3 + 15t
(c) When the particle travels to the right, v >0 12t – 3t 2 + 15 > 0 3t 2 – 12t – 15 < 0 t 2 – 4t – 5 < 0 (t + 1)(t – 5) < 0
At maximum displacement, ds = 0 dt 12t – 3t2 + 15 = 0 3t2 – 12t – 15 = 0 t2 – 4t – 5 = 0 (t – 5)(t + 1) = 0 t = 5 or –1 t = –1 is not accepted ∴t=5
–1
5
t
–1 < t < 5 Since the values of t cannot be negative, therefore 0 ≤ t < 5.
When t = 5, s = 6(5)2 – 53 + 15(5) = 100 m d 2s = 12 – 6t dt 2 2 When t = 5, d s2 = 12 – 6(5) = –18 dt Therefore, s is a maximum.
38
SPM ZOOM–IN Form 5: Chapter 10 Motion Along a Straight Line (c) (i) x = 4 y 3 3x = 4y y = 3x 4
Paper 2 1. (a) I 180x + 90y ≤ 5400 2x + y ≤ 60 x y
0 60
30 0
The furthest point on the straight line y = 3 x 4 inside the feasible region R is (20,15). ∴ xmax = 20, ymax = 15
II 3x + 4y ≤ 120 x y III
0 30
40 0
0 0
30 60
(ii) Profits = 200x + 150y Draw the straight line 200x + 150y = 3000
y ≤ 2x x y
200 150 0.1 = 3000
The optimal point is (24, 12).
(b)
Hence, the maximum profit = 200(24) + 150(12) = RM6600
y 60 2x + y = 60
2. (a) Mixing: 30x + 10y ≤ 15 60 3x + y ≤ 90
y = 2x 50
Baking: 40x + 40y ≤ 26 2 60 3 x + y ≤ 40
40
30 y=
3 x 4
Decorating: 10x + 30y ≤ 15 60 x + 3y ≤ 90
(20, 15)
20
Max (24, 12) 10
R 3x + 4y = 120
O
10 15 20 200x + 150y = 3000
30
40
x
39
x y
0 90
30 0
x y
0 40
40 0
x y
0 30
90 0
(b) y 90
80
70 3x + y = 90
60
50
40
Max (15, 25)
30
20
17
R
x + 3y = 90
10 5
O
x + y = 40
5x + 10y = 50 10
2023
30
40
50
(c) (i) When y = 17, xmax = 23 (ii) Profits = 5x + 10y Draw the straight line 5x + 10y = 50. From the graph, the optimal point is (15, 25). Hence, the maximum profit = 5(15) + 10(25) = RM325
40
60
70
80
90
x
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