Mathematical Methods CAS 3/4 Answers
January 1, 2018 | Author: iloveoj | Category: N/A
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Mathematical Methods CAS 3/4 Answers...
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CUAT018-EVANS
November 3, 2005
16:7
Answers Chapter 1
c –2 –1 0 1 2 3 4 5 6
Exercise 1A
d
1 a {7, 9} b {7, 9} c {2, 3, 5, 7, 9, 11, 15, 19, 23} d {2, 3, 5, 11) e {2} f {2, 7, 9} g {2, 3, 5, 7} h {7} i {7, 9, 15, 19, 23} j (3, ∞) 2 a {a, e} b {a, b, c, d, e, i, o, u} c {b, c, d} d {i, o, u} 3 a {6} b {2, 4, 8, 10} c {1, 3, 5, 7, 9} d {1, 2, 3, 4, 5, 7, 8, 9, 10} e {1, 2, 3, 4, 5, 7, 8, 9, 10} f {5, 7} g {5, 7} h {6} √ 4a [−3, 1) b(−4, 5] c (− 2, 0) 1 √ e (−∞, −3) d −√ , 3 2 f (0, ∞) g (−∞, 0) h [−2, ∞) 5 a (−2, 3) b [−4, 1) c [−1, 5] d (−3, 2] 6a –3 –2 –1 0 1
–8
–6
–4
–2
Exercise 1B 1 a Domain = R b Domain = (−∞, 2] c Domain = (−2, 3) d Domain = (−3, 1) e Domain = [−4, 0] f Domain = R
Range = [−2, ∞) Range = R Range = [0, 9) Range = (−6, 2) Range = [0, 4] Range = (−∞, 2)
3 1 0
x
Domain = R Range = [1, ∞)
0 1
2 3
y
c
0
–3
3
x
–3
Domain = [−3, 3] Range = [−3, 3]
b 2 3
y
b
y
2a
2
–4 –3 –2 –1 0 1
0
y
d
c (4, 4)
2
d
–4 –3 –2 –1 0
e
f 0
1 2 3
y
x
0
8
Domain = [0, ∞) Range = [0, ∞)
Domain = R + ∪ {0} Range = (−∞, 2]
e
–2 –1
(1, 2)
x
0
–4 –3 –2 –1 0 1
f
y (4, 18)
4 5
5
7a –3 –2 –1 0 1 2 3 4 5 6
0
b
x 5
Domain = [0, 5] Range = [0, 5]
–3 –2 –1 0 1 2 3 4 5 6
682
2 0
x 4
Domain = [0, 4] Range = [2, 18]
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683
Answers
4 0 –1 (–1, –5)
–
x –2 2
–2
2 0
–5
(6, 7) (2, 3)
x
Range = [3, ∞)
0
0
x (2, –1)
Range = (−∞, −1]
–2 3
x
(–4, –7)
y
(3, 11) 2 0
y
f
(6, 17)
(3, 4) –1 1 0
x
Range = (−∞, 11)
Range = [−7, ∞) e
y
d
1
1 2
x
Domain = R Domain = [−1, 2] Range = (−∞, 4] Range = [−5, 4] 3 a Not a function Domain = {−1, 1, 2, 3} Range = {1, 2, 3, 4} b A function Domain = {−2, −1, 0, 1, 2} Range = {−4, −1, 0, 3, 5} c Not a function Domain = {−2, −1, 2, 4} Range = {−2, 1, 2, 4, 6} d A function Domain = {−1, 0, 1, 2, 3} Range = {4} f Not a function e A function Domain = {2} Domain = R Range = Z Range = {4} g A function h Not a function Domain = R Domain = R Range = R Range = R i Not a function Domain = [−4, 4] Range = [−4, 4] 4 a g(−2) = 10, g(4) = 46 b [−2, ∞) 5 a f (−1) = −2, f (2) = 16, f (−3) = 6 b g(−1) = −10, g(2) = 14, g(3) = 54 c i f (−2x) = 8x 2 − 8x ii f (x − 2) = 2x 2 − 4x iii g(−2x) = −16x 3 − 4x − 6 iv g(x + 2) = 2x 3 + 12x 2 + 26x + 14 v g(x 2 ) = 2x 6 + 2x 2 − 6 3 c− 6a3 b7 2 2 7 a x = −3 b x > −3 cx= 3 8 a f : R → R, f (x) = 2x + 3 4 b f : R → R, f (x) = − x + 4 3 c f : [0, ∞) → R, f (x) = 2x − 3 d f : R → R, f (x) = x 2 − 9 e f : [0, 2] → R, f (x) = 5x − 3 y y b 9a
0
y
c
y
h
(2, 4)
1 3
x
x
–1 0 (–2, –7)
Range = [−7, 17]
Range = (−∞, 4] y
g
h
(–5, 14)
y (4, 19)
(–1, 2) 0
x
Range = [2, 14]
0 –1 0.2 (–2, –11)
x
Range = (−11, 19)
10 a f (2) = −3, f (−3) = 37, f (−2) = 21 b g(−2) = 7, g(1) = 1, g(−3) = 9 c i f (a) = 2a 2 − 6a + 1 ii f (a + 2) = 2a 2 + 2a − 3 iii g(−a) = 3 + 2a iv g(2a) = 3 − 2(2a) = 3 − 4a v f (5 − a) = 21 − 14a + 2a 2 vi f (2a) = 8a 2 − 12a + 1 vii g(a) + f (a) = 2a 2 − 8a + 4 viii g(a) − f (a)= 2 + 4a −2a 2 2 2 2 , −1 b − , 11 a 3 3 3 1 2 c 0, − d (−∞, −1) ∪ ,∞ 3
3 2 2 e −∞, − ∪ ,∞ 3 3 1 f − ,0 3 12 a f (−2) = 2 b f (2) = 6 c f (−a) = a 2 − a d f (a) + f (−a) = 2a 2 e f (a) − f (−a) = 2a f a 4 + a2 a+2 13 a {2} b {x : x > 2} c 3 8 13 d − e {1} f 3 18 4 7 bk =6 ck =− 14 a k = 3 3 2 1 fk =− dk =9 ek = 9 3 6 1 1 b d1 e −1, 2 c± 15 a 5 5 3
Answers
y
g
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684
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November 3, 2005
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Essential Mathematical Methods 3 & 4 CAS y
13 a
Exercise 1C
(–3, 8)
1 One-to-one functions are b, d and f. 2 a Functions are i, iii, iv, vi, vii and viii. b One-to-one functions are iii and vii. √ 3y= √ x + 2, x ≥ −2; range = R + ∪ {0} y = − x + 2, x ≥ −2; range = R − ∪ {0} y 4a
2 x
0
b g1 (x) = x 2 + 2, x ≥ 0 g2 (x) = x 2 + 2, x < 0 5 a Domain = R Range = R b Domain = [0, ∞) Range = [0, ∞) c Domain = R Range = [−2, ∞) d Domain = [−4, 4] Range = [0, 4] e Domain = R\{0} Range = R\{0} f Domain = R Range = (−∞, 4] g Domain = [3, ∞) Range = [0, ∞) 6 a Domain = R Range = R b Domain = R Range = [−2, ∞) c Domain = [−3, 3] Range = [0, 3] d Domain = R\{1} Range = √ √ R\{0} 7 a R\(3) b (−∞, − 3] ∪ [ 3, ∞) cR d [4, 11] e R\{−1} f (−∞, −1] ∪ [2, ∞) g R\{−1, 2} 1 i 0, h (−∞, −2) ∪ [1, ∞) 3 j [−5, 5] k [3, 12] 8 a Even b Odd c Neither d Even e Odd f Neither y 9a
–1
x
0
1
2
8 5 x
0
–3
b Range = [5, ∞) 14 a f (−4) = −8 b f (0) = 0 1 c f (4) = 4 ⎧ ⎨ 1 , a>0 d f (a + 3) = a + 3 ⎩ 2(a + 3), a ≤ 0 ⎧ 3 1 ⎪ ⎨ , a> 2 e f (2a) = 2a ⎪ ⎩4a, a ≤ 3 2 ⎧ ⎨ 1 , a>6 f f (a − 3) = a − 3 ⎩ 2(a − 3), a ≤ 6 √ 15 a f (0) = 4√ b f (3) = 2 c f (8) = 7 √ a, a ≥ 0 d f (a + 1) = 4, a 2 4 if 0 ≤ x ≤ 2 = 2x if x > 2
1 f (x) = x
0
y
f
−1
687
1 (x) = , dom = R + , ran = R + x
8
y
g
4 y = 12 x 1 √x
y=
2
4
2 V (x) = 4x(10 − x)(18 − x) Domain = (0, 10) 3 a A(x) = −x 2 + 92x − 720 b 12 < x < 60 c A
x
0
1 f −1 (x) = √ , dom = R + , ran = R + x h
x
0
y
(46, 1396)
y = 2x + 4 4
(60, 1200)
1 y = (x – 4) 2
0
–2
(12, 240)
x
4 –2
−1
7a
h (x) = 2x + 4, dom = R, ran = R y y b (1, 2) (3, 3)
(0, 1) x
0 (0, 0)
(3, 4) (4, 3) (2, 1) x
0 (1, 0) y
d
y
c
x
0
d Maximum area = 1396 m2 occurs when x = 46 and y = 34 4 a i S = 2x 2 + 6xh 3V ii S = 2x 2 + x b Maximal domain = (0, ∞) c Maximum value of S = 1508 m2 1 1 3 ,b = , c = 45, d = − , e = 75 5aa= 30 15 2 b S 7 75, 2
1
3 2
0
–4
0
2
e
x
3
y
1
x 3 45, 2
–4
f
0
y
3 3 –3
0
(1, 1) x (0, 0)
x
–3
(–1, –1)
h
y
g
y
t
7 2 6 a C = 1.20 for 0 < m ≤ 20 = 2.00 for 20 < m ≤ 50 = 3.00 for 50 < m ≤ 150 b C ($) c Range = 0,
3 0
x
0
x
2
y = –2
1
8aC bB cD dA 9 a A = (−∞, 3] √ b b = 0, g −1 (x) = 1 − x, x ∈ [−3, 1]
Domain = (0, 150] Range = {1.20, 2.00, 3.00}
m(g)
Answers
y
f
60 80 10 0 12 0 14 0 15 0
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688
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Essential Mathematical Methods 3 & 4 CAS
Multiple-choice questions 1E 6C
2B 7D
3E 8B
4C 9B
(5, 4)
5E 10 B
3 2
Short-answer questions (technology-free) y
1a
f(x) = x2 + 1 (0, 1) 0
x
Domain = R
Range = [1, ∞)
y
b
f(x) = 2x – 6
x
0
y
2a
y = g(x) x
0
–3
3 ,4 b Range of g = 2 3 −1 ,4 → R cg : 2 g −1 (x) = 2x − 3 Range = [0, 5] 3 Domain = ,4 2 7 d {5} e 2 1 3a b {11} 5 y
4
3
–6
(2, 3)
Domain = R
Range = R
y
c
0 5
x2 + y2 = 25
0
–5
x
5
–5
Domain = [−5, 5]
Range = [−5, 5]
y
d
y> – 2x + 1 1 x
0
–1 2
x 1 2 (0, –1)
√ √ 5 a R\{3} b R\[− 5, 5] d [−5, 5] e [5, 15]
Range = R
y y 100 km 2 a S = 6x 2 √ 2 3s 3a A= 4 √ 4 a d(x) = 9 − x 2 ,
2
b S = 6V 3 √ 2 3h bA= 3 b dom = [0, 3] ran = [0, 3]
d 3 x
3
0
160x x + 80 → R, 6 a V1 : (0, 12) 5 S(x) =
h2 h 4 √ b V2 : (0, 6) → R, V2 (r ) = 2r 2 36 − r 2 7a domain range f R R g R R V1 (h) = 36 −
ran f = dom g, ∴ g ◦ f exists g ◦ f (x) = 2 + (1 + x)3 b g ◦ f is one-to-one and therefore a function. ∴ (g ◦ f )−1 is defined, (g ◦ f )−1 (10) = 1 8a y
0
–2
2
→ R, f −1 (x) =
13 a i f (2) = 3 f ( f (2)) = 2 f ( f ( f (2))) = 3 ii f ( f (x)) = x −x − 3 b f ( f (x)) = , f ( f ( f (x))) = x, x −1 i.e. f ( f (x)) = f −1 (x)
Chapter 2 Exercise 2A
(3, 3)
2
a
b − xd c xc − a 2−x b i f −1 : R\{1} → R, f −1 (x) = 3x − 3 3 ii f −1 : R\ → R, 2 3x + 2 f −1 (x) = 2x − 3 1−x iii f −1 : R\{−1} → R, f −1 (x) = x +1 1 −x iv f −1 : R\{−1} → R, f −1 (x) = x +1 c If a, b, c, d ∈ R\{0}, f = f −1 when a = −d. 11 a i YB = r ii ZB = r iii AZ = x − r iv CY = 3 − r √ x + 3 − x2 + 9 br = 2 ci r =1 ii 1.25 q 12 b f (x) = x 3x + 8 c i f −1 (x) = = f (x) √x − 3 ii x = 3 ± 17 10 a f : R\
–4
b i −3 ii 3 c S = (−∞, 0] 2 4x − 4 if x d f ◦ h(x) = 2x if x 2 2x − 8 if x h ◦ f (x) = 2x if x ⎧ 2 3t ⎪ ⎨ ,0 ≤ t ≤ 1 2 9 A(t) = ⎪ ⎩ 3t − 3 , t > 1 2 Domain = [0, ∞] Range = [0, ∞]
1 2 7 7 17 g j f h 21 i2 5 2 9 2 a x = 12, y = 8 b x = 5, y = −8 c x = 3, y = 1 d x = 2, y = 1 e x = 17, y = −19 f x = 10, y = 6 3 80 km 4 96 km 5 Width = 6 cm, length = 10 cm 6 John scored 4, David 8 7 a w = 20n + 800 b $1400 c 41 units 8 a V = 15t + 250 b 1150 litres c 5 hours, 16 minutes and 40 seconds 9 a V = 10 000 − 10t b 9400 litres c 16 hours and 40 minutes 10 a C = 25t + 100 b i $150 ii $162.50 c i 11 hours ii 12 hours 1 a 10
x
0
dA bm =3 b k = −5
0 (0, –7)
y
f
Exercise 4C
(2, 0)
4 2 x b y = −x 2 25 2 c y = x + 2x d y = 2x − x 2 2 e y = x − 5x + 4 f y = x 2 − 4x − 5 2 g y = x − 2x − 1 h y = x 2 − 4x + 6 1 2 1 2 y = − x + x + 1, y = x 2 + x − 5 8 8 3 b = 2, B = 4, A = 1 1a y =4−
(0, –4)
(1, –2)
y
g
(0, 4) (1, 2)
y
0
(1, 2) 0
x
0
Exercise 4D 1a
(2, 0)
y
y
0
x
(2, –4) 2+
(–1, 2) 0
x
x
h b
x
x
(0, –28)
1 4 3, 0 3
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707
Answers y
c
(0.435, 0)
(0, 10)
(1, 2)
Answers
y
i
x
0
0
(3.565, 0)
x
(2, –6) y
j
y
d 1 –2 3 + 1, 0
x
0
(2.159, 0) (0, 161)
(0, –2) (1, –4)
x
0 (3, –1)
y
k
(3.841, 0) y
e x
0
–1
1
(–4, 1)
–31
(–1, –32)
l
x
0
(–5, 0)
(–3, 0)
y
(0, –255) 0
x
2
y
f
(1, –2) –4
x
0 (2, –3)
2 a = −3, h = 0, k = 4 (0, –51)
3 a = 16, h = −1, k = 7 4 a y = 3x
b y = (x + 1) + 1
3
3
c y = −(x − 2) − 3 3
d y = 2(x + 1)3 − 2 5a
ey=
y
x3 27
Exercise 4E 1a
(0.096, 0)
+
1
3
6
–6
–3
1
x
– (0, 1) 0
(1.904, 0)
x
b
(1, –2)
x
–
y
b
+
c (–2, 0)
d (0, –32)
3
–1
5
x
–
x
0
+
+ –
1 2
4
5
x
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708
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November 3, 2005
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Essential Mathematical Methods 3 & 4 CAS y
2a
y
y = f(x)
y = f (2x) 8.00
1
y = f (x) 6.00
x
–0.62 0
1
1.62
y=f
4.00
4 , –0.19 3
2.00 (2.73, 0.75)
(–0.73, 0.75)
y
b
0
–2.00 –1.00
y = f(x + 2) y = f(x) y = f(x – 2)
–3
–2
y
(2, 1)
1
– 9 + 1, 81
x
0
–1
1.00 2.00 3.00 4.00
Turning points for y = f (2x) are at (−0.18, 0.75) and (0.68, 0.75). 2a
(–2, 1)
2
3
2
9 + 1, 81 2 4
4
3.62 8
10 , –0.19 3
–2 , –0.19 3
For clarity the graph of y = 3 f (x) is shown on separate axes.
x 0 1
–2
4
y
b
y
9 81 , 2 2
– y = 3f(x)
9 81 , 2 2
3 y = f(x) 1
0
–3
x
3
x
0
y
c 4 , –0.56 3
– 9 – 1, 81 2
Exercise 4F
9 – 1, 81 2 2
2
16
y
1a
x
–1 0
–4
2
y = f(x) y
d
(–0.37, 0.75)
1
(1.37, 0.75)
– 9, 0
9, 0 2
2
x
0 y = f(x)
0, – 81
y = f(x – 2)
4
e
y – 9 , 85
9 , 85 2 4
2 4
0
x
0
y
b
x 2
x (1.63, 0.75)
(3.37, 0.75)
Graphs of dilations shown on separate axes for clarity.
(0, 1) 0
x
x
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Answers
709
11 (x + 5)(x + 2)(x − 6) 60 5 = (x + 1)(x − 3)2 9 y = x3 + x + 1 y = x3 − x + 1 y = 2x 3 − x 2 + x − 2 y = (2x + 1)(x − 1)(x − 2) 1 y = x(x 2 + 2) 4 y = x 2 (x + 1)
1y= 2y 3a b c 4a b c
d y = x 3 + 2x 2 − x − 2 e y = (x + 2)(x − 3)2 5 a y = −2x 3 − 25x 2 + 48x + 135 b y = 2x 3 − 30x 2 + 40x + 13 6 a y = −2x 4 + 22x 3 − 10x 2 − 37x + 40 b y = x 4 − x 3 + x 2 + 2x + 8 31 4 5 3 157 2 5 11 cy= x + x − x − x+ 36 4 36 4 2
Exercise 4H √ −1 1 1 − 4k 2 , k∈ , \{0} 2k 2 2 1 b x = 4a or 3a or 0 c x = 0 or x = a 3 √ k ± k 2 − 4k dx= , k > 4 or k < 0 2 √ f x = ±a e x = 0 or x = ± a g x = a or x = b √ 1 h x = a or x = a 3 or x = ± a if a > 0 1 b+c 3 7 2ax = b x = (a) 3 a c 13 dx= −b c x = (a − c)n a 3 b 1 ex= f x = (c + d) 3 a 1 1 b , , (0, 0) 3 a (0, 0), (1, 1)
2 2 √
√ √ 3 − 13 3 + 13 √ , 13 + 4 ; , 4 − 13 c 2 2 1ax =
−1 ±
4 a (13, 3), (3, 13) b (10, 5), (5, 10) c (11, 8), (−8, −11) d (9, 4), (4, 9) e (9, 5), (−5, −9) 5 a (17, 11), (11, 17) b (37, 14), (14, 37) c (14, 9), (−9, −14) 6 (2, 4), (0, 0) √
√ 5+5 5+5 7 , , 2 2
√ √ 5− 5 5− 5 , 2 2
√ √
−130 − 80 2 60 − 64 2 8 , , 41 41 √
√ 80 2 − 130 64 2 + 60 , 41 41 √
√ 1 + 21 −1 + 21 , , 9 2 2 √ √
1 − 21 −1 + 21 , 2 2 √ √
4 −6 5 3 5 10 ,2 11 , 9 5 5 1 12 −2, 13 (3, 2), (0, −1) 2 −8 −15 27 10 , b , , (5, 9) 14 a (3, 2), 4 2 3 5 −12 , −10 c (6, 4), 5 2 15 c − ac + b = 0 √ √ 16 (−1 √ − 161, √ 1 − 161), ( 161 − 1, 161 + 1) √ √ 4m + 25 + 5 4m + 25 + 5 17 a x = ,y= 2m 2√ √ 5 − 4m + 25 5 − 4m + 25 ,y= or x = 2 2m −25 2 5 bm = , − , 4 5 2 −25 cm < and m = 0 4
Multiple-choice questions 1E 6A
2D 7C
3E 8E
4C 9C
5E 10 C
Short-answer questions (technology-free) y
1a
h(x) = 3(x – 1)2 + 2 (0, 5) (1, 2)
x
0 y
b
y = (x – 1)2 – 9
(–2, 0)
0
(0, –8)
x (4, 0) (1, –9)
Answers
Exercise 4G
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710
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November 3, 2005
16:7
Essential Mathematical Methods 3 & 4 CAS y
c
y
c y = x2 – x + 6
(0, 6)
1 3 ,5 2 4
0
x
0
(–3, 0) (–2, –1) x
(0, –9)
d
y
y
d y = x2 – x – 6
(0, 26) (–2, 0) 0
x
(3, 0)
(–2, 0) x
0 (–3, –1)
(0, –6) 1 25 ,– 2 4
y
e
e y=
2x2
y
–x+5 1, 1 2
(0, 2)
(0, 5) 1 , 39 4 8
0
x
0
x (1, 0)
y
f
5a y=
2x2
+ x
–x–1
0
–
–2 –1 –1 2
x
0
1
b
+ x 0
1, – 9 4 8
(0, –1)
2
–1
4 1 4 1 2 y = x2 − ; a = , b = − 3 3 3 3 √ 1 3 (1 ± 31) 3 y 4a
– 1
3
c
+ x 0 –4
–
–2 –1
d
+
0
x –
–3
–1
(3, 0) x
0 (0, –18)
b
(1, –16)
y
2 3
6a8 b0 c0 7 y = (x − 7)(x + 3)(x + 2) y 8a y = f(x – 1)
(–1, 8)
(0, 7) 0 (1, 0)
x
0
3 2 –83 2 , 4 3 3
x
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Answers
y = f(x + 1)
x
0
–2
1
–8 2 –83 , 3 34 y
c
y = f(2x)
(6, 144π)
1 – ,0 2
0 –2
d
x
1
4,
224π 3
(3, 45π)
5 –83 , 6 34
64π 2, 3
y
0
y = f(x) + 2
(–1, 2)
6
d x = 5. The depth is 5 cm. h2 5 a r = 25 − 4 b V
x
0
x
3
(5.77, 302.3)
3
2 –8 1 , 4 +2 3
3
9 y = −x 3 + 7x 2 − 11x + 6
0
Extended-response questions 4 1ak = 3375 c i Rnew
b 11.852 mL/min
10
h
c V = 96cm √ d h = 2, r =√2 6, i.e. height = 2 cm and radius = 2 6 cm, or h = 8.85 and r = 2.33 6 a V = (84 − 2x)(40 − 2x)x b (0, 20) c y 3
(15, 40)
(8.54, 13 098.71) y = V(x)
0
20
t
ii 23.704 mL/min d i Rout 40 0
t
20 (35, –20)
ii 11.852 mL/minute out 2 a i 2916 m3 ii 0 m3 b V (m3) (0, 2916)
0
c 3.96 hours
9 t (hours)
0
20
x
d i x = 2, V = 5760 ii x = 6, V = 12 096 iii x = 8, V = 13 056 iv x = 10, V = 12 800 e x = 13.50 or x = 4.18 f 13 098.71 cm3 7 a i A = 2x(16 − x 2 ) ii (0, 4) b i 42 ii x = 0.82 or x = 3.53 c i V = 2x 2 (16 − x 2 ) ii x = 2.06 or x = 3.43 8 a A = x 2 + yx 2 b i y = 100 − x 100 ii A = 100x − x 2 iii 0, 2 c x = 12.43 2 x 100 d i V = 100 − x , x ∈ 0, 50 2 ii 248.5 m3 iii x = 18.84
Answers
3 a V = x(96 − 4x)(48 − 2x) = 8x(24 − x)2 b i 0 < x < 24 ii x ≈ 8, maximum volume ≈ 16 500 cm3 c When x = 10, V = 15 680 cm3 . d Max. volume = 14 440 cm3 e Min. volume = 9720 cm3 1 4 V = x 2 (18 − x) 3 a i 64 cm3 ii 45cm3 iii 224 cm3 3 3 b 144cm3 c V
y
b
711
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Answers
712
CUAT018-EVANS
November 3, 2005
16:7
Essential Mathematical Methods 3 & 4 CAS 1 2 17 1 x3 − x + x 12 000 200 120 b x = 20 29 2 1 1 3 x + x − x dy=− 6000 3000 20 y e i
9ay=
y
2a
y = 2 x +1 – 2
2 –1
0
–1
x 1
–2 (40, 3)
Range = (−2, ∞) b
0
40
y
x
80
ii Second section of graph is formed by a reflection of the graph of y = f (x), x ∈ [0, 40], in the line x = 40.
(0, 4)
Chapter 5
Range = R +
Exercise 5A 1a
y=0 x
0
y
c
y y = 3x y = 2x y=0 (0, 1) x
0
x
0 –1
Range = (−1, ∞) y
d
y
b y = 3–x
5 (0, 1)
y = 2–x
y=0 x
0
2 1 0
y
c
x 1
2
Range = (1, ∞) e y = 5x
(0, 1) 0
0, 2
y=0 x
1 4
2
y = –5x
(0, –1)
y
x
0 y
d
Range = (2, ∞) y
f
(0, 1)
y=0 x
0 (0, –1)
(0, 6)
y = (1.5)x
y = –(1.5)x
2 0
Range = (2, ∞)
x
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November 3, 2005
16:7
Answers y
h
y
y = f(x – 2) (0, 1)
(2, 1)
y=0 x
0
0, 1 0
Range = R + b
y
y
c
1
1 0
x
y = 10 x –1
x
0
Range = (1, ∞)
Range = (−∞, 1)
y
y = –1
3 y=0 x
0
Range = (−1, ∞)
2 0
Range = R +
x
0
y
e 1
y
5a
(0, 2)
d
y=0 x
4
x
b
y x
y = 10 10 + 1
Range = (2, ∞)
(10, 11)
y
f
(0, 2)
x
0 –1
y
Range = (1, ∞)
(0, 2)
y=1 x
0 y
c
y=2 0
0
y
d
y = f(–x) + 2
(0, 3)
y = 2(10 x) – 20
y = f(x) + 1
(0, 2) y = 0 x
0
y
c
y
b y = f(x + 1)
(–1, 1)
x
0
Range = (−1, ∞) 4a
y=1
(0, –18)
0
x
x
y = –20
Range = (−20, ∞)
x y = –1
(0, –2)
(1, 0)
y
d
y = –f(x) – 1
e
y
y = f(3x) 0
f
y
y=1
x
y=f 2
(1, 8) (0, 1) y = 0 x
(0, 1) 0
0
(2, 2) y=0 x
y = 1 – 10 –x
Range = (−∞, 1)
y
g
e
y y = 10 x + 1 + 3
y = 2 f(x – 1) + 1
(0, 13) y=3
(–1, 4)
(0, 2)
y=1 0
x
x
0
Range = (3, ∞)
x
Answers
3a
713
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Answers
714
CUAT018-EVANS
November 3, 2005
16:7
Essential Mathematical Methods 3 & 4 CAS y
f
Exercise 5B
x
y = 2 10 10 + 4 (10, 24)
y
1a
y=4
(0, 2)
y=1
x
0
Range = (4, ∞)
y = 1 – ex
Range = (−∞, 1)
y
c
d
(100, 408.024)
(0, 1)
x
0
0
b i $408.02 c 239 days 7 36 days 8a i
y
y y = e x–1 – 2
ii $1274.70 d ii 302 days
(0, e –1 – 2) 0
y = 5 x y = 3x y = 2x
x y = –2
Range = (−2, ∞) f
y
(0,1)
y = 2ex x
0
ii x < 0 b i
iii x > 0 1 3
y=
x
0
x
y y=
x
1 2
(0, 2)
iv x = 0
1 5
x
Range = (0, ∞) y
g (0,1)
y = 2(1 + ex) x
0
ii x > 0 c i a>1
iii x < 0
(0, 4)
iv x = 0
y=2
ii a = 1
y
(0, 1) 0
x
0 y
y = ax
Range = (2, ∞)
y=1 x
1 0
x
h
y
iii 0 < a < 1 y=2
y 0 (0, 1) 0
y = ax x
y = 2(1 – e –x)
Range = (−∞, 2)
x
Range = (0, ∞)
Range = (−∞, 1) e
y y = e –2x
y=1
y = 1 – e –x
x
x
0
Range = (1, ∞)
0
y=1
x
0
6 a C1
y=
y
b y = ex + 1
(0, 6)
x
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November 3, 2005
16:7
715
Answers
1 d 4 5 g 2 3a1
y = 2e –x + 1
(0, 3)
y=1
x
0
e −2
Range = (1, ∞) j
y
i6 4 a 1, 2 e 1, 2 i −1, 2
(1, 2)
(0, 2e –1)
h6 b1 f4 3 j 5 b1 f 0, 1 j −1, 0
c4 f3
3 2 10 g− 3 1 k± 2 c2 g 2, 4 c−
i4 d3 h−
3 2
d1 h 0, 1
Exercise 5D
Range = (0, ∞) y
y
l
y=2 (0, 3e – 2) x
0
x
(0, –1) 0
y = –2
Range = (−2, ∞) Range = (−∞, 2) 2 a x = 1.146 or x = −1.841 b x = −0.443 c x = −0.703 d x = 1.857 or x = 4.536 y 3 a, b i y = f(x) 1
y = f (x – 2) e –2 x
0 y
ii, iii y = f (–x)
y = f (x) y=f x
3
1
x
0
Exercise 5C 1 a 6x 6 y 9
b
x
0
k
1 2 3 e 5
2a4
b 3x 6
d 18x 8 y 4
e 16
g 24x 5 y 10
h 2x y 2
6y 2 x2 5x 28 f 6 y
c
i x 2 y2
1a3 b −4 c −3 d6 e6 f −7 2 a loge 6 b loge 4 c loge 106 (= 6 loge 10) d loge 7 1 e loge (= − loge 60) 60 f loge u 3v 6 (= 3 loge uv 2 ) g 7 loge x = loge x 7
h loge 1 = 0
3 a x = 100 b x = 16 cx =6 d x = 64 e x = e3 − 5 ≈ 15.086 1 g x = −1 fx= 2 1 −3 i x = 36 h x = 10 = 1000 4 a x = 15 bx =5 cx =4 1 d x = 1(x = − is not an allowable solution) 2 3 ex= 2 5 a log10 27 b log2 4 (= 2) a 1 a 1 2 c log10 = log10 2 b b
10a d log10 1 b3 1 e log10 (= −3 log10 2) 8 1 6a1 b1 c2 d3 e0 2 c0 7 a −x b 2 log2 x 3e 8ax =4 ≈ 0.7814 bx= 5 + 2e √ −1 + 1 + 12e 9ax = , i.e. x ≈ 0.7997 6 b x = loge 2 ≈ 0.6931
Answers
y
i
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Answers
716
CUAT018-EVANS
November 3, 2005
16:7
Essential Mathematical Methods 3 & 4 CAS 10
1 ,2 4
y
f
x=2
3 2 8 11 N = = 3 27
(3, 0) 0
y
12 a
x
2
x=3
Domain = (2, ∞), range = R y
g 0
3
x
4
x = –1
Domain = (3, ∞), range = R
(0, 1)
y
b
x
0
–1
x = –3
(1.72, 0)
Domain = (−1, ∞), range = R
(4.39, 0)
y
h
x=2
0
x
–0.9
–3
(0, 0.69)
0
Domain = (−3, ∞), range = R
x
1
y
c
Domain = (−∞, 2), range = R
x = –1
y
i
x= 4 3
–1 0
(0.65, 0) (0, –1)
x (0, 0.39)
d
(0.43, 0)
y x= 2
Domain =
3
y = log2 2x
x
(0.79, 0)
4 , range = R −∞, 3
y
13 a
0 2
x
0
Domain = (−1, ∞), range = R
3
Domain =
0
2 , ∞ , range = R 3
y
e
y
b 0 (0, –1.4)
x
Domain = R +
x = –2
–2 –1
1, 0 2
x=5
x
y = log10 (x – 5) 0
(6, 0)
Domain = (−2, ∞), range = R Domain = (5, ∞)
x
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November 3, 2005
16:7
Answers y
i y = –log10 x
Answers
y
c
717
– 1, 0 3
(1, 0)
x
0
x
0 y = 4 log2 (–3x)
Domain = R −
Domain = R +
y
j
y
d
x=2
y = 2 log2 (2 – x) – 6
y = log10 (–x)
Domain = (−∞, 2) 14 a x = 1.557 b x = 1.189 y 15 b i, ii
Domain = R − y
e
x
(–6, 0) 0 (0, –4)
x
0
(–1, 0)
x=5 y = f (–x)
y = log10 (5 – x)
y = f (x)
(0, log10 5) x
0
(–1, 0)
(4, 0)
x
(1, 0)
y = –f(x)
Domain = (−∞, 5) f
0
y
iii, iv
y
y = f(3x) y = f (x)
1 3
1 ,0 4
0
x
0
(1, 0) 3 y=f
x
x 3
y = 2 log2 2x + 2
Domain = R +
Exercise 5E
y
g
y = –2 log2 (3x) 1,0 3
x
0
3 a = 200, b = 500
4 a = 250, b =
5 a = 3, b = 5
6 a = 2, b =
7 a = 2, b = 4
Domain = R + h
1 a = 2, b = 4 14 14 2a= ,b= (a ≈ 8.148, e−1 1−e b ≈ −8.148)
y
9 b = 1, a =
x = –5
10 a = 1 –5100, 0
0
y = log10 (–x – 5) + 2
Domain = (−∞, −5)
1 loge 5 3
1 loge 5 3 8 a = 2, b = 3
2 , c = 8 (a ≈ 2.885) loge 2
2 ,b = 4 loge 2
x
Exercise 5F 1 a 2.58 e −2.32 i 2.89 m −6.21
b −0.32 f −0.68 j −1.70 n 2.38
c 2.18 g −2.15 k −4.42 o 2.80
d 1.16 h −1.38 l 5.76
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Answers
718
CUAT018-EVANS
November 3, 2005
16:7
Essential Mathematical Methods 3 & 4 CAS 2 a x < 2.81 b x > 1.63 c x < −0.68 d x ≤ 3.89 e x ≥ 0.57 loge 5 3ax = ≈ 2.3219 loge 2 loge 7 bx= ≈ 1.7712 loge 3 loge 8 c x = 0 or x = ≈ 1.8928 loge 3 4 a 10 b3 c8 5 a 2.9656 b 5.8329 c 2.0850 d 2.8551 e −4.6674 f −0.8273 6 0.544 b 549.3 2 1 9 a 9u c bu+ u 2 2 10 625 11 p
g f −1 (x) = 10x − 1; domain = R, range = (−1, ∞) 1 h f −1 (x) = loge x + 1; 2 domain = (0, ∞), range = R y 4 a, c f –1
y=x
y=1
f x
0 x=1
b
f −1 (x) = − loge (1 − x), domain = (−∞, 1)
5 a, c
y
Exercise 5G
y=x
f
x = –3 (0, 2)
0 (2, 0) f –1
1
y y=x
y = –3
(0, 4) y=3
(4, 0) x=3
b
x
0
y = f –1(x)
−x
f (x) = e + 3 f −1 (x) = − loge (x − 3)
1 x +3 loge , 2 5 domain = (−3, ∞) f −1 (x) =
y
6 a, c
f –1
y
2
x
1 0, 2 loge 0.6
y = f (x)
f x=1 y = f –1(x) y=x
(0, 2)
0, e
–
1 2
0
y = f(x) y=x
y=1 0
(2, 0)
x
f (x) = loge (x − 1) f −1 (x) = e x + 1 1 3 a f −1 (x) = e x ; domain = R, 2 range = R + 1 x−1 b f −1 (x) = e 3 ; 2 domain = R, range = R + c f −1 (x) = loge (x − 2); domain = (2, ∞), range = R d f −1 (x) = loge (x) − 2; domain = R + , range = R 1 e f −1 (x) = (e x − 1); domain R, 2 1 range = − , ∞ 2 1 x f f −1 (x) = e 4 − 2 ; domain R, 3 2 range = − , ∞ 3
b
f −1 (x) = e
7ax =e cn=
–
1 2,
x 0
x−1 2 , range
y−5 2
loge
e
y
a loge x
= R+ 1 P b x = − loge 6 A d x = log10
y
1 5−y ex= e 3 2⎛ y ⎞ log e 1⎜ 6 ⎟ fn= ⎝ ⎠ 2 loge x 1 y (e + 1) 2 5 h x = loge 5 −y x +4 −1 8 a f (x) = loge 2 b (0.895, 0.895), (−3.962, −3.962) gx=
x−4
9 a f −1 (x) = e 2 − 3 b (8.964, 8.964), (−2.969, −2.969)
5
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November 3, 2005
16:7
Answers y = ex
y
y
b
y=x
y = loge x
1 0
f (x) = 10 –x + 1 (0, 2)
x
1
y=1
ii
y y=
x–3 e2
x
0
y=x
c
y
y = 2 loge (x) + 3
h(x) =
1 x (e – 1) 2
–3 e2
x 0
x
0
y=–
–3 e2
iii
y
y
d
y = 10 x
1 2
y=x y=2 (0, 1) x
0
1
y = log10 x
0
(–loge 2, 0)
x
1
y
e
f(x) = loge (2x + 1)
b f (x) and g(x) are inverse functions
x
0
Exercise 5H
1
1 m = 0.094, d0 = 41.9237 2 k = 0.349, N0 = 50.25 3 a i N0 = 20 000 ii −0.223 b 6.2 years 4 a M0 = 10, k = 4.95 × 10−3 b 7.07 grams c 325 days
x=–2
f
y x=1
x
0 (1 + e –1, 0)
Multiple-choice questions 1C 6C
2D 7B
3B 8A
4E 9C
h(x) = loge (x – 1) + 1
g
y
5A 10 D
g(x) = –loge (x – 1) (2, 0)
Short-answer questions (technology-free) y
1a
x
0
x=1 y
h
f (x) = –loge (1 – x)
f (x) = e x – 2 (loge 2, 0) x
0
0
x
(0, –1) y = –2 x=1
Answers
10 a i
719
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Answers
720
CUAT018-EVANS
November 3, 2005
9:27
Essential Mathematical Methods 3 & 4 CAS 1 loge (x + 1) 2 x b f −1 : R → R, f −1 (x) = e 3 + 2 c f −1 : R → R, f −1 (x) = 10x − 1 d f −1 : (2, ∞) → R, f −1 (x) = log2 (x − 1) 3 a y = e2 x b y = 10x c y = 16x 3 e3 x5 ey= dy= f y = e2x−3 x 10 loge 11 loge (0.8) 4ax = bx= loge 3 loge 2 loge 3 cx= loge ( 23 ) 2 1 bx= cx= 5ax =1 3 20 d x = log10 3 or x = log10 4 2 a f −1 : (−1, ∞) → R, f −1 (x) =
2
6 a = 2, b = 2−3 − 1 1 287 8 loge 9 2a 3 4 11 a = loge 5, b = 5, k = 2
6
7 10 5 − 1
8 k = −0.5, A0 = 100 9a x
x=8
b i 0 grams ii 2.64 grams iii 6.92 grams c 10.4 minutes 10 a k = 0.235 b 22.7◦ c 7.17 minutes 11 a N(t) 20e10
10 R
10e10
Extended-response questions 1 a 73.5366◦ C 2 a 770 3 a k = 22 497, = 0.22 4 a A = 65 000, p = 0.064 5a y
y=
00 10 0
b 59.5946 b 1840 b $11 627 b $47 200
00 t + 10
15
b i (12.210, 22 209.62) ii t = 12.21 iii 22 210 c ii (12.21, 12.21) d c = 0.52 1 6 a iii a = or a = 1 2 iv If a = 1, e−2B = 1 and B = 0, 1 1 A ∈ R + . If a = , B = loge 2. 2 2 v A = 20 000 n b
20 000
t
2 loge 10 loge (0.1) 1 = ≈ 6.644 1 loge 2 loge 2 2 After 6.65 hours the population is 18 000. 7 a 75 b 2.37 c 0.646 c
(0, 20) 0
t
50 70
b i N (10) = 147.78 ii N (40) = 59 619.16 iii N (60) = 20e10 = 440 529.32 iv N (80) = 220 274.66 c i 25 days ii 35 days
Exercise 6A t
0
y = 10e10
Chapter 6
y = 8000 + 3 × 2t 0
t
0
5 18 17 d 9 2 a 60◦ d 140◦ 3 a 45.84◦ d 226.89◦ 4 a 0.65 d 2.13 1a
34 45 7 e 3 b 150◦ e 630◦ b 93.97◦ e 239.50◦ b 1.29 e 5.93 b
25 18 49 f 18 c 240◦ f 252◦ c 143.24◦ f 340.91◦ c 2.01 f 2.31 c
Exercise 6B 1 a 0, 1 b −1, 0 d 1, 0 e 0, −1 2a0 b Not defined d0 e Not defined 3 a −1 b0 d0 e1 4 a 0.99 b 0.52 d 0.92 e −0.67 g −0.99 h 0.44 j −2.57 k 0.95
c −1, 0 f 0, 1 c Not defined c −1 f1 c −0.87 f −0.23 i −34.23 l 0.75
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November 3, 2005
9:27
Answers d
1 a 0.52 b −0.68 e −0.52 f 0.68 2 a 0.64, 2.5 c 3.61, 5.816 3 a 17.46, 162.54 c 233.13, 306.87
c 0.52 d 0.4 g −0.4 h −0.68 b 0.64, 5.64 d 1.77, 4.51 b 66.42, 293.58 d 120, 240
e f g h
Exercise 6D 1 0.6 3 61 7
i
2 0.6
3 −0.7
4 0.3
7 −0.3
8 0.6
9 −0.6 10 −0.3
5 −0.3
Exercise 6E √
1 1 3 ,− ,−√ 2 2 3 1 1 c √ , √ ,1 2 2 1 1 e √ , √ ,1 2 2
1a
1 1 b −√ ,−√ ,1 2 2 √ 1 √ 3 ,− , 3 d− 2 2 √ 3 1 1 ,−√ f− , 2 2 3 √ 1 3 d c− 2 2 1 1 h g −√ 2 2 √ 1 3 k l 2 2 1 1 o√ p− 2 2
1 1 b− 2a √ 2 2 1 1 f −√ e− 2 √ 2 1 3 i j 2 2 1 1 m√ n− 2 2 b 60◦ , 120◦ 3 a 60◦ , 300◦ d 120◦ , 240◦ e 60◦ , 120◦ 5 b − or 4 a − or − 6 6 6 6 5 5 b 5a 6 6 d +b e +b 2 2 4 7 b 6a 3 6 3 d −b eb+ 2 3 5 3 7a , b , 4 4 4 4 7 7 11 e , , d 4 4 6 6 5 7 13 15 , , , 8a 8 8 8 8 5 7 17 19 29 b , , , , , 18 18 18 18 18 5 13 17 c , , , 12 12 12 12
7 3 5 17 23 , , , , , 12 12 4 4 12 12 5 13 17 25 29 , , , , , 18 18 18 18 18 18 3 9 11 , , , 8 8 8 8 5 7 17 19 29 31 , , , , , 18 18 18 18 18 18 5 7 13 5 7 23 , , , , , 12 12 12 4 4 12 2 4 5 , , , 3 3 3 3
c 210◦ , 330◦ f 150◦ , 210◦ −5 5 c or 6 6 2 c 3
Exercise 6F 1 a i 2 2 bi 3 ci
ii 3 1 ii 2 ii 3
g i 4 y 2a 2
π 3
0
θ
2π 3
–2
Amplitude = 2 2 Period = 3 y
b 2
7 c 6
π
π
4
2
3π 4
π
θ
–2
Amplitude = 2 Period = y
c 3 0
31 18
ii 2 1 ii 2 ii 2
d i 6 ei 2 f i 2
0
fb+ 5 7 c , 6 6 2 4 f , 3 3
ii 3
–3
3π 3π 2
9π 2
Amplitude = 3 Period = 6
6π
θ
Answers
Exercise 6C
721
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CUAT018-EVANS
722
November 3, 2005
9:27
Answers
Essential Mathematical Methods 3 & 4 CAS y
d
Exercise 6G
1 3
y
1a
0 –1 3
π
π
4
2
θ
π
3π 4
2
Amplitude =
Period =
1 3
y
e
π 8
–3
π 4
θ
π 2
3π 8
θ
7π 3
y
b 1
Amplitude = 3 Period = 2
π 2
0
y
3
4π 3
–2 Period = 2 Amplitude = 2 Range = [−2, 2]
3
0
π 3
0
θ
π
5 –5π 6
–π 2
–π
–2π 3
π 2
0 –π 3
π 6
–π 6
π 3
–1
5π 6 2π 3
π
x
Period = Amplitude = 1 Range = [−1, 1]
–5
y
c
y
4 1 2
–π
–π 2
3
0 –
π 2
π
3π 2
2π
x
1 2
0
π 3
–2
x 6 y = 2 sin 3 x 1 7 y = cos 2 3 1 x 8 y = sin 2 2
θ
–3
2 5π 3 2π π 4π 2π 3 3
3π 4
Period = Amplitude = 3 Range = [−3, 3]
y
5
π 4
0
–π 4
x
y
d √3 0
π 2
5π 6
–√3
2 3 √ Amplitude = 3√ Range = [− 3, 3] Period =
7π 6
θ
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November 3, 2005
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Answers
5 4 3 2 1
3 2 1 0 –1
π 6
π 3
2 3 Amplitude = 2 Range = [−1, 3]
x
π 2
2π 3
y
2a
2 1
b
0 –1 –2 –3 –4
x
π
π 2
c 3a
b
Period = Amplitude = 3 Range = [−4, 2] y
g 4
c
3 2 1 0 –1
π 6
7π 6
θ
Period = √ Amplitude =√ 2 √ Range = [− 2 + 2, 2 + 2]
d
y
h
π 2
3π 2
0 –1
Period =
f
y
i
7
θ
π
Period = Amplitude = 3 Range = [−1, 5] 1 1 y = cos x− 2 3 4 y = 2 cos x − 4 1 y = − cos x − 3 3 Dilation of factor 3 from the x-axis 1 Dilation of factor from the y-axis 2 Reflection in the x-axis Dilation of factor 3 from the x-axis 1 Dilation of factor from the y-axis 2 Reflection in the x-axis Translation of units in the positive 3 direction of the x-axis Dilation of factor 3 from the x-axis 1 Dilation of factor from the y-axis 2 Translation of units in the positive 3 direction of the x-axis Translation of 2 units in the positive direction of the y-axis Dilation of factor 2 from the x-axis 1 Dilation of factor from the y-axis 2 Reflection in the x-axis Translation of units in the positive 3 direction of the x-axis Translation of 5 units in the positive direction of the y-axis
3
Exercise 6H 0 –1
π
π
x
1a
y
2
Period = Amplitude = 4 Range = [−1, 7]
3 2 1 0 –1
2π π 3
Intercepts:
4π 3
x 2π
4 2 , 0 , , 0 3 3
Answers
y
e
723
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Answers
724
CUAT018-EVANS
November 3, 2005
9:27
Essential Mathematical Methods 3 & 4 CAS y
b
y
d (0, 2 – √3 )
1 0 –1 –2 –3
π 2
π
3π 2
2π
(0, –2 – √3)
(0, √2 – 1) 0
(0, –√2 – 1)
Intercepts:
π 4
π
y 1 + √2 (0, 2)
(2π, 2)
x
3π 2
7 , 0 , , 0 4 4
0
π 2
1 – √2
π
Intercepts: (, 0),
y
2a –11π –2π 6
–7π 6
–π 2
π 6
–π 6
π 2
5π 6
3π 2
2π
x
0 (2π, –2)
–2 (–2π, –2) –4 y
b –23π 12 –2π
3 –15π 12
π 12
–7π 12 –π
9π 12
17π 12
0
2π
1 – √2
–1
(–2π, 1 – √2)
x
(2π, 1 – √2) y
c –2π
–π
π
0
2π
x
–1 (2π, –3)
–3
(–2π, –3)
–5 y
d 3 (–2π, 1) –2π –11π 6
x
Intercepts: (0, 0), (2, 0) e
y 7π 4 2π
2π
–4
11 , 0 , , 0 , Intercepts: 12 12 13 23 , 0 , , 0 12 12
c
π
0
x
1 –π –7π 6
0 –1
(2π, 1) π 6
π 5π 6
2π
x
x 2π
3 ,0 2
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November 3, 2005
9:27
Answers e
725
–23π 12
–19π –15π 12 12
–11π 12
–7π 12
–π 4
π 12
2
5π 12
9π 12
13π 12
17π 12
21π 12
0 π
–π
–2π
–√2 –2
(–2π, –√2)
Answers
y
2π (2π, –√2)
x
y
f (–2π, √2)
2
(2π, √2)
√2
0
–21π 12
–17π –13π 12 12
x
π
–π
–2π
–9π 12
–5π 12
–2
–π 12
3π 12
7π 12
2π
11π 12
15π 12
19π 12
23π 12
y
g 2
0
–π
–2π
x
π
2π
y
h –π
–2π (–2π, –1)
–π 2
x
π
0
2π (2π, –1)
3π 2
–1 –2
y
3
Exercise 6I y
1
1
y = sin θ + 2cos θ 2 –4
–2 –1
0
–2
2
1
–4 –3
4
2
3
0
x
x –1
1 y = cos2θ – sin θ 2
–2
y
4
y
2
4
y = 2cos2θ + 3sin2θ 2 –2
y = 3 cos θ + sin 2θ
2
–1
1 0 –2 –4
2
0 x
–4 –3 –2 –1 –2
x 1
2
3
4
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726
CUAT018-EVANS
November 3, 2005
9:27
Essential Mathematical Methods 3 & 4 CAS 5
c
y
y
4 (0, 2)
2 0
x
0 1
–4 –3 –2 –1
2
3
(2π, 2)
3π 4 π 4
x
5π 7π 4 4
4
–2
y
d
–4 y = 2sin θ – 4 cos θ
(2π, 1)
(0, 1) 0
π 2
x
3π 2
Exercise 6J y
e
− 3 4 A = −4, n = 6
4 3 A = 3, b = 5, n = 3
2 A = 0.5, ε =
1 A = 3, n =
− ,ε = 4 2 (Note: ε can take infinitely many values.) − 6 A = 2, n = , ε = 3 6 (Note: ε can take infinitely many values.) − and d = 2 7 A = 4, n = , ε = 4 2 (Note: ε can take infinitely many values.) − 8 A = 2, n = , ε = and d = 2 3 6 (Note: ε can take infinitely many values.) 5 A = 4, n =
π
0
2π
x
f y
π
π 2
0
3π 2
x
2π
Exercise 6K 1a
3
2a
b 2
c
2 3
d1
y
3
e2
y
2π 0 π 4
π 2
3π π 4
5π 3π 7π 4 2 4
–3 2
x
–2
2
y
b
0
π 6
π 3
π 2
2π 3
5π 6
–1 – 1
π
7π 6
4π 3
3π 2
5π 3
11π 6
2π
x
0
1 2
1
3 2
2
x
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November 3, 2005
9:27
Answers b
√ √
3 3 7 , , ,− 6 2 6 2
7 19 31 43 , , , 24 24 24 24 5 11 17 23 , , , b 12 12 12 12 11 23 35 47 59 71 , , , , , c 36 36 36 36 36 36 9 n = 3, A = 5 10 n = , A = 6 2 8a
–
π 2
–
π 4
0
π 4
x
π 2
5 a , 7 6 6 5 9 13 17 21 25 29 , , , , , , , 16 16 16 16 16 16 16 16 b 7 13 19 c , , , 12 12 12 12 5 11 17 23 , , , d 12 12 12 12 7 11 5 19 23 , , , , e , 4 12 12 4 12 12 f 0.4636, 3.6052 g 1.1071, 4.2487 3 7 11 15 h , , , 8 8 8 8 7 13 19 25 31 , , , , , i 18 18 18 18 18 18 4 7 10 13 16 , , , , j , 9 9 9 9 9 9 y 6 a, c y = cos 2x – sin 2x 1
–π
–3π –π 4 2
–π 0 4
y = cos 2x
π 4
π 2
3π π 4
y = –sin 2x x
–1
b
7 a, c
−5 −1 − 1 ,√ , ,√ , 8 8 2 2 3 −1 7 1 ,√ , ,√ 8 8 2 2 y
y = cos x + √3 sin x
√3
Exercise 6L 1 a i 2 ii 4 iii −4 −14 −10 14 10 4 8 , iii , ii bi , 3 3 3 3 3 3 2 a 2n ± , n ∈ Z 3 2n 2 2n + or + ,n ∈ Z b 3 9 3 9 c 3n + , n ∈ Z 3 5 11 5 c , , b 3a , 3 6 12 12 6 6 −11 −7 5 4a , , , 6 6 6 6 − 5 , , 5 3 3 3 6 a x = n − or x = n − , n ∈ Z 6 2 n bx= − ,n ∈ Z 2 12 5 or x = 2n − , n ∈ Z c x = 2n + 6 2 (4n − 1) or x = n, n ∈ Z , 7x= 4 −5 3 7 , −, , 0, , , 4 4 4 4 n − −2 8x= ;x = or or − or 0 3 3 3 3n + 2 6n − 1 or x = ; 9x= 12 6 −2 −7 −1 −1 1 5 5 11 , , , , , , , , x= 3 12 6 12 3 12 6 12
y = cos x
1
Exercise 6M 0
π 2
π
3π 2
2π
–1 –√3
y = √3sin x
x
4 4 1 − ,− 5 3 √ √ 2 6 , −2 6 3− 5
2−
12 5 ,− 13 12
Answers
y
4a
727
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Answers
728
CUAT018-EVANS
November 3, 2005
9:27
Essential Mathematical Methods 3 & 4 CAS b 5.89 metres d 6 times f 4.21 metres
Exercise 6N 1 a i Amplitude = 1
1 2
ii Period = 12 iii d(t) = 3.5 − 1.5 cos t 6 iv 1.5 m b [0, 3) ∪ (9, 15) ∪ (21, 24] y 2a
Multiple-choice questions 1C 6C
1 a c
2 t 2
8
14
20
24
d
b 2:00
c 8:00, 20:00 3 a A = 3, n = , ε = , b = 5 6 2 b i 2:21 a.m., 9:39 a.m., 2:21 p.m., 9:39 p.m. c y
e g h
y = h(t)
8 5
2E 7B
2a
y=5
2
5A 10 B
f (x) = sin 3x
t
0
6
12
18
24
4a5 b1 c t = 0.524 s, 2.618 s, 4.712 s d t = 0 s, 1.047 s, 2.094 s e Particle oscillates about the point x = 3 from x = 1 to x = 5. t ◦ 5 a 19.5 C b D = −1 + 2 cos 12 c D
0
6
12
18
24
π 3
0
2π 3
x
–1 y
b
f(x) = 2 sin 2x – 1 1
1
0 π 5π –1 12 12
t
π
x
–3
–3
d {t : 4 < t < 20}.
c
y
6a
h (metres) 31.5
f(x) = 2 sin 2x + 1 (120, 30.99)
16.5 1.5
4A 9E
5 −2 2 4 , b , , 6 6 3 3 3 − 11 , , 6 6 6 −3 − 5 7 , , , 4 4 4 4 − 3 7 − −5 7 11 f , , , , , 4 4 4 6 6 6 6 −3 −5 3 5 11 13 , , , , , 8 8 8 8 8 8 −7 −11 5 13 17 , , , , , , 18 18 18 18 18 18 25 29 , 18 18 y 1
–1 –2
3D 8C
Short-answer questions (technology-free)
10 (0, 8) 6
0
c 27.51 seconds e 20 times g 13.9 metres
3
1 0
(0, 5.89) 13.5 31.5 49.5 67.5 85.5 103.5 120 t (minutes)
–1
7π 12
11π 12
x
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November 3, 2005
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729
Answers c
y
y
f(x) = 2 sin x – π 4
2
0 π
5π 4
4
√3 + 2 y = √3
x
9π 4
0 √3 – 2
–2
e
y
d
19π 23π 9π 12 12 4
y = –3sin x
3
3
2
x
π 4
y
f(x) = 2 sin π x
0
x
π
0
3
x
6
2π
–3 y
e
–2 y
f
Answers
d
4
h(x) = 2 cos πx 4
2
2
x
0
4
2
y=3
x
0 π
6 8
7π 6
6
f
–2
13π 6
y
3 a 30, 150 b 45, 135, 225, 315 c 240, 300 d 90, 120, 270, 300 e 120, 240 y 4a
3 y=1 2π
0 π 3
–1
5
π
x
5π 3
y
y = 2 sin x + 3 + 2
4
y = cos x
1
(0, √3 + 2) y=2
x
0 –1 x
0
–π
2π 3
3
b
7π 6
5π 3
π
π 2
2π
3π 2
y = sin 2x
a 4 solutions c 2 solutions
b 4 solutions
y
y
6a 3 π 2
3
–π 6
y=1
–π 3
–1
0
x 2π 3
(0, 1 – √3 )
5π 3
0 –3
x 180
360
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Answers
730
CUAT018-EVANS
November 3, 2005
9:27
Essential Mathematical Methods 3 & 4 CAS b
y
2 , 4 a a = 20 000, b = 100 000, n = 365 ε ≈ 5.77, 2t i.e. y = 20 000 sin + 5.77 + 100 000 365 y b
1 90 0
270 180
x 360
120000
–1
80 000
y
c
0
1
360,
√3 2
c i t = 242.7, t = 364.3 ii t = 60.2, t = 181.8 d ≈ 117 219 m3 /day 5 a i 1.83 × 10−3 hours ii 11.79 hours b April 25 (t = 3.856), August 14 (t = 7.477) 6a D
√3 2
x
0 30
210
360
–1
13 10 7
− 3 −2 b or or 4 4 3 3 − 3 7 −5 or or or c 8 8 8 8 − 5 −2 or or or d 3 6 3 3
7a
0
Extended-response questions 1a
y y = cos π t + 4 6
5 4 3
3 6
5
t 3
6
9
12
15
18
21
24
y = –3 cos (2πt)
0.5
6
1 second 3 d t = 0.196 second
t
12
c A ship √ can enter 2 hours after low tide. 8 a i 25 3 ii 30 b 2.27, 0.53 d b = 8√ e = 0.927 or 1.837 fa=4 3 9c A (3, 5.196) π x
0
d i 2.055 ii 0.858 iii 0.0738 e nr tan n cos f i n sin n n ii A
1 A=π
–3
ci t =
t
24
3
t 0
18
7
b 9:00, 15:00 c 8:00, 16:00 2 a Maximum = 210 cm, minimum = 150 cm, mean = 180 cm t b y = 30 sin − + 180; 6 2 − A = 30, n = , b = 180, ε = 6 2 c i 165 cm √ ii 180 − 15 3 ≈ 154 cm d ≈ 4:24, ≈ 7.36 3 a a = −3, n = 2 b y 3
12
b {t: D(t) ≥ 8.5} = {t: 0 ≤ t ≤ 7} ∪ {t: 11 ≤ t ≤ 19} ∪ {t: 23 ≤ t ≤ 24} c 12.898 m 7 a p = 5, q = 2 b D
0 0
t 121 303.5 486
ii t =
1 second 6
(3, 1.3) 0
x
iv 0.0041
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Answers
Exercise 7A 1 a i f ◦ g(x) = 3 sin (2|x|) g ◦ f (x) = 3 |sin 2x| ii Range of f ◦ g = [−3, 3], domain of f ◦ g = R, range of g ◦ f = [0, 3], domain of g ◦ f = R b i f ◦ g(x) = −2 cos (2|x|) g ◦ f (x) = 2 | cos 2x| ii Range of f ◦ g = [−2, 2], domain of f ◦ g = R, range of g ◦ f = [0, 2], domain of g ◦ f = R c i f ◦ g(x) = e|x| g ◦ f (x) = e x ii Range of f ◦ g = [1, ∞), domain of f ◦ g = R, range of g ◦ f = (0, ∞), domain of g ◦ f = R d i f ◦ g(x) = e2|x| − 1, g ◦ f (x) = |e2x − 1| ii Range of f ◦ g = [0, ∞), domain of f ◦ g = R, range of g ◦ f = [0, ∞), domain of g ◦ f = R e i f ◦ g(x) = −2e|x| − 1, g ◦ f (x) = 2e x +1 ii Range of f ◦ g = (−∞, −3], domain of f ◦ g = R, range of g ◦ f = (1, ∞), domain of g ◦ f = R f i f ◦ g(x) = loge (2|x|), g ◦ f (x) = | loge 2x| ii Range of f ◦ g = R, domain of f ◦ g = R\{0}, range of g ◦ f = [0, ∞), domain of g ◦ f = R + g i f ◦ g(x) = loge (|x| − 1), g ◦ f (x) = | loge (x − 1)| ii Range of f ◦ g = R, domain of f ◦ g = R\[−1, 1], range of g ◦ f = [0, ∞), domain of g ◦ f = (1, ∞) h i f ◦ g(x) = − loge (|x|), g ◦ f (x) = | loge x| ii Range of f ◦ g = R, domain of f ◦ g = R\{0}, range of g ◦ f = [0, ∞), domain of g ◦ f = R + 2 a h(x) = f ◦ g(x) where f (x) = e x and g(x) = x 3 b h(x) = f ◦ g(x) where f (x) = cos x and g(x) = |2x| c h(x) = f ◦ g(x) where f (x) = x n and g(x) = x 2 − 2x
d h(x) = f ◦ g(x) where f (x) = cos x and g(x) = x 2 e h(x) = f ◦ g(x) where f (x) = x 2 and g(x) = cos x f h(x) = f ◦ g(x) where f (x) = x 4 and g(x) = x 2 − 1 g h(x) = f ◦ g(x) where f (x) = loge x and g(x) = x 2 h h(x) = f ◦ g(x) where f (x) = |x| and g(x) = cos 2x i h(x) = f ◦ g(x) where f (x) = x 3 − 2x and g(x) = x 2 − 2x x 1 2 c ex b loge 3 a 2e2x 2 2 1 1 4 a f −1 (x) = − loge x, g −1 (x) = (x − 1) 3 2 3 b f ◦ g(x) = e−2(x +1) , ran f ◦ g = R + , g ◦ f (x) = e−6x + 1, ran g ◦ f = (1, ∞) √ 1 5−1 −1 bx= 5 a f (x) = − 1 x 2 6 a f −1 (x) = √ e x − 1, dom f −1 = R, g −1 (x) = x + 1 − 1, dom g −1 = (−1, ∞) b loge (x 2 + 2x + 1) 1 f (x) + f ◦ g(x) = 0 7 f ◦ g(x) = loge x 8x 9 a f (g(x)) = (x 2 − 8)(x 2 − 10), g( f (x)) = (x − 4)2 (x − 6)2 − 4 bx =1 √ √ 10 x = ± 6 or x = ± 2 12 a = 6 b = 0 g(x) = e6x
Exercise 7B 1aa=1 c, e y
b [2, ∞) y = f(x)
(1, 2)
y = f –1(x)
(2, 1) 0
x
√ d f −1 : [2, ∞) → R, f −1 (x) = x − 2 + 1, domain of f −1 = [2, ∞), range = [1, ∞)
x 2a± = y, domain = R + ∪ {0}
2 x +1 b± = y, domain = [−1, ∞) 2√ c y = ±√x − 1, domain = R + ∪ {0} d y = ± x + 1 − 1, domain = [−1, ∞) x −1 + 1, domain = [1, ∞) ey=± √ 2 f y = ± 1 − x + 1, domain = (−∞, 1]
Answers
Chapter 7
731
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732
CUAT018-EVANS
November 3, 2005
9:27
Essential Mathematical Methods 3 & 4 CAS 1 3 a g −1 : R + → R, g −1 (x) = 2 ± √ x b g −1 : (1, ∞) → R, g −1 (x) = 1 ± √
6a 1
+
y = x2 + 3 y = x2
(–√3, 6)
x −1
(√3, 6)
2 c g : R → R, g (x) = −1 ± √ x 1 + x, y ≥ 1 , domain = [0, ∞) dy= 1 − x, y < 1 x − 1, y ≥ 2 , domain = [3, ∞) ey= 5 − x, y < 2 1 + x, y ≥ −1 , fy= −3 − x, y < −1 domain = [−2, ∞) 5 x −6 3 , domain = R 4 a f −1 (x) = 2 5 x −6 2 by=± , domain = [6, ∞) 2 −1
y
−1
y=3
(–√3, 3)
(√3, 3) y
b
y = x2 + 2x + √x y = x2 + 2x y = √x x
0
c
y y = √x
(1, 1)
Exercise 7C
(1, 0)
−2x
1ai e − 2x ii b i e +1 ii x 2 a i sin − 2x 2 x ii −2x sin 2 b i −1 ii x ii 3 a i cos + ex 2 bi 2 ii y = x2 + 3x + 2 y 4
−2xe e
x
0
x
0 (1, –1)
−2x
y = –x2 + √x y = –x2 y
7 −2 x e x cos 2 1
(0, 1) 0
y = e –2x x y = e –2x – 2x
y = ( f + g)(x)
y = –2x
2
y = 3x + 2
5 a y = ( f + g)(x) y
0
y = x2 x –
8
y
y = 2e2x + x + 2 y = 2e2x
2 3
4
y=x+2
2
(2, 6) –2
(2, 3)
x
0
9 a i |x| + x
y
y = f (x) x (0, 0) y = g(x)
b
y
y = f(x)
ii x|x|
y
(1, 1)
(–1, 1) (–1, 0)
x
0
(1, 0) (–1, –1) (1, –1)
y = ( f + g)(x)
y = g(x)
x 0
x
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November 3, 2005
9:27
Answers y
2 x
0
ii −x|x|
3
y
4 x
0
5 c i x − |x|
y
1 e y =− x −4 m c ax= bc≤2 2 c−x , domain = R d f −1 (x) = 2 c c 1 f y = x +c e , 2 3 3 b −b b , a 0 and b 2 4 c i (0, 0) and (b − 1, 1 − b) ii b = 1 iii b ∈ R √\{1} √ a −1 ± 2 2 b ±2 2 c a = −8, b = 16 a (−∞, 2a]
√ √ −1 + 1 + 8a −1 + 1 + 8a b , 2 2
c2 + c ea= 2 a2 6 a (0, 0) and (a, 0) b (0, 0) c 4 d a = 3 or a = −5 c b 1 b ec −a 7 a loge b a a+1 loge (c) − b c d c a 8 ax =a b (a + 1, 0) 1 1 c (a + e c , 1) d c = loge (2 − a) ca=1
x
0
ii −x|x|
y
0
x
Exercise 7D 1 f (x − y) = f (x) − f (y) 2 f (x + y) = f (x) + f (y) − 3; a = −3 3 g(x) = 0 or g(x) = 1 4 f (x + y) = |x + y|. Let x = 2, y = −4. f (x + y) = 2 = |2| + | − 4| = f (x) + f (y) 5 5 f (x + y) = sin(x + y). Let x = 2 ,y = 2 . 5 f (x + y) = sin(3) = 0 = sin + sin 2 2 =2 2 7 a h(x + y) = (x + y) . Let x = 1, y = −1. h(x + y) = 0 = h(1) + h(−1) = 2 10 f (x y) = ax y. Let x = 2 and y = 3. f (x y) = 6a and f (x) = 2a and f (y) = 3a f (x) f (y) = 6a 2 6a 2 = 6a implies a = 0 or 1
Exercise 7E 4 b m ≥ 4 or m < 0 m x +4 , domain = R c f −1 (x) = m 4 4 d , , m ∈ R\{1} m−1 m−1
1 a
da=3
9 a y = −b
b (loge (b) + 1, 0) 1 1 ii 0 < b < c i b= e e 10 a = 5 − c, b = −1 where the rule 2 is y = ax + bx + c 3d + 4 11 a = , b = 2 − d and 6 (3d + 28) c=− where 6 3 2 y = ax + bx√+ cx + d √ 12 a c = 28 − 8 6 or c = 28 + 8 6 √ √ b (−∞, 8) ∪ (28 + 8 6, ∞) ∪ (8, 28 − 8 6) 5d − 9 41 − 10d ,b = , 30 30 −25d − 2 where y = ax 3 + bx 2 + cx + d c= ⎡ 30 ⎤ x −3 4k −3 ⎢ −4 ⎥ +2 a 14 ⎣ y − 2 ⎦ b y = c 3−x 2 k ⎤ ⎡ a − x ⎢ 4 ⎥ 15 a ⎣ y +2 ⎦ 2 x−a b y =2×2 4 −2 ca =0 13 a =
Multiple-choice questions 1B 6E
2E 7D
3E 8E
4D 9E
5B 10 B
Answers
b i |x| − x
733
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734
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November 3, 2005
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Essential Mathematical Methods 3 & 4 CAS
Short-answer questions (technology-free) 1 a i f ◦ g(x) = 3 cos (2|x|) g ◦ f (x) = |3 cos (2x)| ii maximal domain = R; range = [−3, 3] maximal domain = R; range = [0, 3] b i f ◦ g(x) = loge (3|x|) g ◦ f (x) = | loge (3x)| ii maximal domain = R\{0}; range = R maximal domain = R + ; range = [0, ∞) c i f ◦ g(x) = loge (2 − |x|) g ◦ f (x) = | loge (2 − x)| ii maximal domain = (−2, 2); range = (−∞, loge 2) maximal domain = (−∞, 2); range = [0, ∞) d i f ◦ g(x) = − loge (2|x|) g ◦ f (x) = | loge (2x)| ii maximal domain = R\{0}; range = R maximal domain = (0, ∞); range = [0, ∞) 2 a h(x) = f ◦ g(x) where g(x) = |2x| and f (x) = cos x (Note: answer is not unique.) b h(x) = f ◦ g(x) where g(x) = x 2 − x and f (x) = x n (Note: answer is not unique.) c h(x) = f ◦ g(x) where g(x) = sin x and f (x) = loge x (Note: answer is not unique.) d h(x) = f ◦ g(x) where g(x) = sin 2x and f (x) = −2|x| (Note: answer is not unique.) e h(x) = f ◦ g(x) where g(x) = x 2 − 3x and f (x) = x 4 − 2x 2 (Note: answer is not unique.) x + e−x 3 a i ( f + g)(x) = 2 cos 2 x ii ( f g)(x) = 2e−x cos 2 b i ( f + g)(0) = 3 ii ( f g)(0) = 2 2 4a b (−∞, 3] 3 c
√
2 3−x , range = ,∞ , 3 3 domain = (−∞, 3]
d f −1 (x) =
2+
e
y
2 3, 3
2 + √3 3
x
0
5ay=
2±
2 + √3 3
√
x −4 3
by=±
3−x 2
Extended-response questions 1 a D = 0.05t 2 − 0.25t + 1.8 b $3 000 000 −7 23 2a= ,b = , c = 7.5 48 24 35 At 12:00 noon, rate of rainfall = mm/h 6 23 Rainfall was greatest hours after 7 4:00 a.m., i.e. at ≈ 7:17 a.m. 3 a (0, 1), (−∞, 0) 1 b f −1 (x) = − loge x, g −1 (x) = + 1 x 1 ex = c i g ◦ f (x) = −x e −1 1 − ex y ii
0
x y = –1
d i (g ◦ f )−1 (x) = loge y
x x +1
x
0 x = –1
4a i
y
(5, √2 )
y 0
2 ,3 3
0
3, 2 3
x
2 + √3 3
x
√ ii [ 2, ∞) iii f −1 (x) = x 2 + 3, √ domain of f −1 = [ 2, ∞)
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Answers
Extended-response questions 1 a a = −0.09, b = 9 √b DE = 2.79 m c Length of bar = 2 × 30 ≈ 10.95 m 1 2 a a = −3 b x = −1, x = − , x = 2 2 3 7 c ii b = , c = 2 2 y 3a
y=x (1, 4) (4, 1)
x
0
5 a (0, 1]
b (0, 1]
1 c range of g ⊆ domain of f , f ◦ g(x) = sin x d Not defined as range of f is not contained in the domain of g 1 e g −1 (x) = , domain of g −1 = (0, 1], range x of g −1 = [1, ∞) f range of f = domain of g −1 1 g −1 ◦ f (x) = sin x domain = (0, ), range = [1, ∞) 6 a i n = 5790 ii 1158 n iii
x = – 4 sin π t
4
x
0
1
2
–4
bi x =0 ii x = −4 iii x = 0 7 ct = 6 2 = 2 seconds d Period = 4ai 0 ii 2.5 iii 0 b 1 second y c
5790
1158 0
t
0
179 −100 loge iv t = 3 1600 b i a = 2.518, b = 0.049, c = 5097.661 n ii n = 5097.661
1 t (s)
0.5
d t = 0.35 seconds 5 a k = 0.0292 b 150 × 106 c 6.4494 × 108 d 23.417 years 6 a 62.5 metres b x (m) 40
1449.08 0
t
0
45°
α
90°
c 24◦ 18
Chapter 8 Multiple-choice questions 1D 7C 13 A 19 B 25 D 31 E 37 C 43 C 49 C 55 B 61 A 67 E
2A 8A 14 E 20 D 26 E 32 D 38 D 44 D 50 D 56 A 62 A 68 C
3B 9B 15 D 21 C 27 A 33 E 39 C 45 B 51 B 57 B 63 D 69 C
4E 10 B 16 C 22 A 28 D 34 A 40 B 46 B 52 C 58 D 64 B 70 A
5A 11 C 17 E 23 C 29 D 35 A 41 E 47 E 53 C 59 D 65 D
6A 12 C 18 C 24 D 30 A 36 E 42 B 48 A 54 A 60 D 66 B
7 a A = 80, k = 0.3466 b 17.5◦ C c 6 hours 18 minutes and 14 seconds after 2:00 p.m., i.e. 8:18:14 d T (°C) 95°C 15°C 0
t (hours) n
8 a Carriage A (0.83) I, carriage B 0.66(0.89)n I b 6 stations x 9 a Area = 0.02(0.92) 10 b 0.0199 mm2 c Load = 0.02(0.92)10−2.9x d x < 2.59 m
Answers
ii h −1 (x) = x 2 + 3
b i p=3 y iii
735
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736
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November 3, 2005
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Essential Mathematical Methods 3 & 4 CAS
n=4
y
vi
1 2 1 n + n+1 2 2 b i f (4) = 8 + 2 + 1 = 11 25 5 ii f (5) = + + 1 = 16 2 2 c
10 a f (n) =
a=1
1 2
a=3 (2.5, 1)
8
x
0
7 9
(5, –1)
3
6 2
10
5
11
4
n=5
−4 3 c i a, a +a 27 4 4a iii A = √ 81 v 3 4 375
1
13
12
11 10
3 1
iv a = 3
Chapter 9
15 4 16
5
Exercise 9A
9
7
1ah+9 b 9.1 c9 2ax +1 b 2x 3 + 1 c 40 e5 f1 g 2x + 1 i 3x 3 + x j 6x 3 a 2 + 3h + h 2 b2 4 2x + h, 2x 5 h + 6, 6
8
11 a y = h − k cos
t 6
i 12 units ii OQ = h − k, OR = h + k b
4 3 2a a , 27 3
14
2 6
ii
T (°C)
d0 h 3x
Exercise 9B
16.5 12 7.5 0 1 2 3 4 5 6 7 8 9 10 11 12 t (months)
c h = 12, k = 4.5 1 1 12 a i 3 + √ , 0 3 − √ ,0 a a 1 ii √ a y bi a=1
a=2 a=3
0
x
(0, –1)
√ 3 3 ii a = 2 iv a = 3
√ 3 3 iii a > 2 1 v a=1 2
1 a 10x b3 c0 d 6x + 4 e 15x 2 f 10x − 6 2 a 5x 4 b 28x 6 c6 d 10x − 4 2 e 12x + 12x + 2 f 20x 3 + 9x 2 g −4x + 4 h 18x 2 − 4x + 4 3 a −2 b0 c 15x 2 − 6x + 2 6x 2 − 8 e 4x − 5 f 12x − 12 d 5 g 50x 4 h 27x 2 + 3 4 a 8x − 4 b 2x + 2 c 6x 2 − 12x + 18 d x 2 − 2x + 1 5 a (3, 16), gradient = 8 b (0, −1), gradient = −1 c (−1, 6), gradient = −8 d (4, 594), gradient = 393 e (1, −28), gradient = −92 1 f 2 , 0 , gradient = 0 2 6a1 b1 c (1, ∞) 2 d (−∞, 1) e2 f 4 or − 2 3 7 a {x: x < −1} ∪ {x: x > 1} b {x: −1 < x < 1} c {1, −1} 8 a (−1, 0.5) ∪ (2, ∞) b (−∞, −1) ∪ (0.5, 2) c {−1, 0.5, 2}
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Answers
Exercise 9C 1 a −6x −3 − 5x −2 15 8 c 4 − 3 x x 2 e− 2 x 4 2 a z2
15 x4 −4 d −18x − 6x −3 b 12x −
−18 − 2z z4 −2z 3 + z 2 − 4 −4 d c 3z z2 6 6 − 12z f −6x − 2 e x z4 1 1 3± 5 4 a = −1, b = 4 2 2 3 6 a = −9, b = 1 7 k = 0 or 2 1 3 c −1 d5 b 8 a 11 32 4 9a
b
f(x)
3√ 1 x−√ 2 x3 x 6a √ x2 + 2 2x + 2 c 5 5 (x 2 + 2x)4 e
5 3 x2 2 3 1 20 2 x3 d x− 2 − 2 3 1 1 5 f − x − 4 + 2x − 2 4 5 2 d c 2 9 −3 b √ 2 4 − 3x −1 d 3 (4 − 3x)2 √ 5x + 6 f3 x 2 3x 2 − 5 b 3 3 (x 3 − 5x)2 b
Exercise 9F 1
1 20x 4 + 36x 2 + 4x
2 9x 2 +
3 3(2x − 1)2 (8x − 1) 4 8x(2x 2 + 1)(6x 2 + 1)
3 −1 x 2 2
5x 2 − 8x + 1 √ 2x − 4 7 x 2 (3x 2 + 4x + 3)(3x 2 + 2x + 1)−2 1
5 5(3x + 1) 2 (3x + 4)
6
1
0
−2(2 + h) (1 + h)2
x
Exercise 9G c −4
Exercise 9D 1 a 8x(x 2 + 1)3 c 24(6x + 1)3
4
x− 5 5 5 3 3 1 c x2 − x2 2 2 6 − 13 e− x 7 7 1 1 b 2a 12 27 1 3a √ 2x + 1 x c√ 2 x +2 1a
8 2x 3 (5x 2 − 2)(2x 2 − 1)− 2 √ 2x 2 (x + 1) 9 2x 3 x 2 + 2x + 3 3 (x 2 + 2x)2
f(x) = 22 x
b PQ =
Exercise 9E
b 20x(2x 2 − 3)4 d an(ax + b)n−1 6x e 2anx(ax 2 + b)n−1 f (1 − x 2 )4 1 −4 2 2 g −3 x − 2 2x + 3 x x −2 h (1 − x) 2 a 6(x + 1)5 b 4x 3 (3x + 1)(x + 1)7 3 2 2 d −4(x + 1)−5 18x 2 − 2 c 4 6x 3 + x x − f (x) 3 a n[ f (x)]n−1 f (x) b [ f (x)]2
4 b (x + 4)2 1 1 − 1 x 2 −x2 d c 2 (1 + x)2 2 + 2x − x 2 e f (x 2 + 2)2 x 2 + 4x + 1 g 2 h (x + x + 1)2 2 a 81, gradient = 378 1a
4x (x 2 + 1)2 (x + 2)2 (x − 3)(x − 1) (x 2 + 1)2 −4x (x 2 − 1)2 −2(4x 3 + 3x 2 + 1) (2x 3 + 2x)2 b 0, gradient = 0 1 d , gradient = 0 2
c 0, gradient = 0 1 3 e , gradient = − 2 2 5 (7x 3 + 3x + 4) 2x 2 + x + 1 c bx √ 3a √ 2 3 (x + 3)2 2 x +1 x +1
Answers
1 1 9 a − , 2 ∪ (2, ∞) b −∞, − 4 4 1 c − ,2 4 7 1 , −8 c 10 a (1, −9) b (2, −8) 4 16 1 1 11 a − , 3 b (2, 32) or (−2, −32) 2 2 c (2, 6) d (0, 0) or (2, −4)
737
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738
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November 3, 2005
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Essential Mathematical Methods 3 & 4 CAS g
Exercise 9H
y dy
y = dx
y
1a
(0, 2)
x
0
dy =2 dx
2
x
0
h
y dy
y = dx
y
b
0 dy = –3 dx
(0, –3)
c
0
x
x 3
1
i
y
y
1 dy y = dx
dy y = dx
x
0
1
x
0
2
–1 y
j d
–3, 2 3
y
2 3
dy y = dx
–1
x
0
x
0
3, – 5 3
–5 3 y
k
y
e
x 0
x
0 dy
y = dx
f
y
l
y
(1, 2) dy y = dx
–1 0
x 2 –1
0
x 1
2
(–3, –0.25) (1, –2.5) (2, –2.5)
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Answers 2 a, b
y
y y = f '(x)
x
0
–1.5
0
1.5
ci 0 di 1
y
n
0
iv 96
y
y = f (x)
x 1
x
2
ii 0 iii 0 ii 0.423
3
o
1
y = f(x)
2
0
y
x
1
4 Gradient is 0 at 1, . 3 Gradient is positive for R\{1}. –2
y
4
x
0
1
y = f '(x) y = f (x)
y
p
x
0
x
0
Gradient is positive always; minimum gradient where x = 0. y 5a y = f '(x)
q
y
y = f(x)
2 1 0
x
x
0
b i x = −1.495 or x = 0.798 ii x = 0.630 r
6
y
y
y = f '(x) 4 0
x
0
x
Answers
m
739
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740
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November 3, 2005
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Essential Mathematical Methods 3 & 4 CAS a Minimum value of f (x) is 4. b x = −0.471 or x = 0.471 y 7a i
Exercise 9J 1a
y
(0, 12)
y = f'(x)
0
x
–1 x
0
–0.5
1
0.5
b
y
y
ii
0 (0, 16)
2
x
0
0.5
y = f'(x) x
0
y
iii
4
y
c –0.5
x
–3
y
d (0, 32)
y = f '(x) 0
–0.5
x 0.5
x
0
b Graphs and tables illustrate that as x → 0 from the left and from the right the value of g(x) approaches the gradient of y = f (x) when x = 2.
e
y
x
0
–1
1
y
f
Exercise 9I 1 a 17
b3
c −4
e3
f4
i −2
j 12
g2 11 k 9
1 8√ h2 3 1 l 4
d
2 a 3, 4 b7 3 a 0 as f (0) = 0, lim f (x) = 0, but lim f (x) = 2
x→0+
2 f (x) =
x
0
–1
1
−2x + 3 3 y
if x ≥ 0 if x < 0
x→0−
b 1 as f (1) = 3, lim f (x) = 3, but x→1+
lim f (x) = −1
x→1−
c 0 as f (0) = 1, lim f (x) = 1, but lim f (x) = 0
x→0−
4 R\{1}
x→0+
0
x
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741
Answers 2x + 2 −2
if x > 1 if x < 1
y
1B 6C
x
0
1
–2
4 Defined for R\{−1} −2x − 2 if x > −1 f (x) = −2 if x < −1 y
–1
x
0 –2
b c d 6a b
b (y + 1)2 26 3
Multiple-choice questions
4
5a
1 (1 − x)2 5 25 −3x(x 2 + 1)− 2 24 a
2 1 R\{1}; f (x) = (x − 1)− 3 3 4 1 R\{0}; f (x) = (x)− 5 5 2 1 R\{0}; f (x) = x 3 3 3 2 R\{−2}; f (x) = (x + 2)− 5 5 2x − 4 if x > 4 or x < 0 f (x) = 4 − 2x if 0 < x < 4 2x − 4 if x > 0 f (x) = 2x + 4 if x < 0
Exercise 9K 1a8 b −8 2aD bF cB dC eA fE −2x 5 + 4x 3 − 2x b {0} 3a (x 4 − 1)2 4 a −7 b −14 1 1 b 5a5 2 8 √ 6 2x 3x 2 + 1 7 a 4x − 3 b −3 c {1} 3 2 8 − 12 10 24 11 (0.28, 0.14) 25 3 8x 15 24x − 12 14 − 2 (x − 2)2 1 2 17 16 −63(5 − 7x)8 18 9 3 19 −70 20 0 21 −1 2 b −2 22 a − (2x + 1)2 23 a x = 0 or x = −2 b x > 0 or x < −2 c −2 < x < 0
2C 7D
3A 8E
4A 9A
5B 10 B
Short-answer questions (technology-free) 2 x 1a1− √ 1 − x2 3 c √ 2 1 + 3x
b d
3x − 15 e √ 2 x −3 4x g 2 (x + 1)2
f h
2 10x (2 + 5x 2 )− 3 3 1 k 4x(3x 2 + 2)− 3 2 a −6 b1 y 3a
j
i
−4x − 2x + 12 (x 2 + 3)2 −2 1 − 3 x2 2x 2 1 + 2x 2 √ 1 + x2 −x 2 + 1 (x 2 + 1)2 −2x 2 − 2x + 4 (x 2 + 2)2
c5
x
0
dy = –3 dx y
b
y=
0
c
dy dx x
1
y
y= 0
dy dx x
1.5
3 9 9 , x =± 42 4− 2 4x + x x 2
d
1 6
Answers
3 Defined for R\{1} f (x) =
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November 3, 2005
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Essential Mathematical Methods 3 & 4 CAS 3 3 , ∞ ∩ (−1, 4) = ,4 2 2 a x f (x) + f (x) 6 b f (x) if x ∈ [0, 4] and − f (x) if x ∈ (−∞, 0) ∪ (4, ∞) 2x f (x) − 2x 2 f (x) c [ f (x)]3
5 b
Extended-response questions 1 a i −4 iv −18
ii −6 v 6
iii −18 1 vi − 6
5 7 b a = , b = 1, c = − , d = 6 2 2 2 a i −1 and 3 ii x > 3 and x < −1 b i 3 and 7 ii (3, 6) and (7, 1) 1 5 5 1 c i and ii ,6 ,1 2 2 2 2 d i 2 and 10 ii (2, 6) and (10, 1) e i 2 and 10 ii (2, 18) and (10, 3) 3 a x = or x = b (x − )m−1 (x − )n−1 ((m + n)x − m − n) m + n c x = or x = or x = m+n m + n , x = d i x> m+n m + n ii x < or x > m+n nx n−1 d0 ex >0 4b n (x + 1)2
Chapter 10 Exercise 10A 1 1 1 y = 4x − 5, y = − x + 3 4 2 1 2 y =− x −1 3 3 y = x − 2 and y = −x + 3 1 4 y = 18x + 1, y = − x + 1 18 3 11 29 5 ,− ,c=− 2 4 4 1 1 6 a i y = 2x − 3 ii y = − x − 2 2 1 b i y = −3x − 1 ii y = x − 1 3 c i y = −x − 2 ii y = x 49 1 d i y = 8x + 2 ii y = − x − 8 8 2 3 ii y = − x + 1 e i y = x +1 3 2 1 1 ii y = −2x + 3 f i y= x+ 2 2
3 2 4 ii y = − x + x+ 2 3 3 h i y = 4x − 16 1 15 ii y = − x − 4 2 i i y = −2 ii x = 2 1 ii y = − x + j i y = 4x − 4 4 3 1 7 a y = −1 by= x+ 2 2 c y = −2x − 1 d y = −4x + 5 8ax =4 b x = −5 1 cx =− d x = −5 2 g i y=
7 2
1 4
Exercise 10B 1 11.31◦ or 11◦ 19 2 30.96◦ or 30◦ 58 3 161◦ 5 4 a tan−1 (0.5) = 26◦ 34 b tan−1 (0.33) = 18◦ 26 c tan−1 (2.83) = 70◦ 39 5 a (1, 1), (0, 0) b 8◦ 8 and 0◦ 6 71◦ 34 7 a (0, 0) = 45◦ , (1, 1) and (−1, −1) = 26◦ 34 b (0, 0) = 18◦ 26 , (1, 2) and (−1, −2) = 12◦ 32 c (1, 1) and (−1, 1) = 63◦ 26 d 14◦ 2
Exercise 10C dy = 12x 2 − 16x b 16 c 0.32 dx 1 1 3 2 a − x− 2 b− 2000 2 3 c− d 0.0985 2000 h −3 e i y ≈ − a 2 2 1 1 h ≈ √ − ii √ 3 a a+h 2a 2 3 −28q 23 p 230 p 4a % b 10 11 8 dy = 5 − 2 ; y = 3p 5 dx x p 6 a 3p b √ 2 a −6 p d 8(2a + 1)3 p c (3a + 1)2 (−2a 2 − 10a + 2) p f e 36a(6a 2 − 1)2 p (a 2 + 1)2 5 2a h p g p 2 (a + 1)2 3(a 2 + 10) 3 31 1 9 −0.48 8 cm 7 30 20 2 10 10 p cm 1a
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Answers
b
c x = −1, x =
+
ii 1
2(1 + 8 ; 2.5% 17 a 5
ii f (0) =
1 x) 2
1 2
–1
13 4%
1
15 a i f (x) =
√ 10 √ seconds 40 g
1 2
16 ; 3.75% b 5
0
–
max.
1 2 0
+
min.
d x = −3, x = 4 –3 –
0
4 +
0
–
Exercise 10D 1 a (2, −16), (−2, 16) b (1, −2) c (0, 0), (1, 1) d (4, 48) 2 16 −2 16 e (0, 0), √ , , √ , 3 3 3 3 1 14 1 2 g (3, 2), − , 20 ,4 f 3 27 3 3 h (0, −10), (2, 6) 2aa=6 bb=3 3 a = 2, b = −4, c = −1 1 1 2 4 a = , b = −2 , c = −3, d = 7 3 2 2 5 a a = 2 and b = 9 b (−1, −5) 1 1 − 4n 6 x = or x = 2 2n + 2 1 1 or −1, − 7 x = ±1 or x = 0 8 1, 2 2 √ √ √ √
2 3 16 3 2 3 16 3 9 a (2, 4) b , , − , 3 9 3 9 √ √ 8 256 c (0, 0), , d (0, 0), ( 2, 4), (− 2, 4) 3 27
min.
max.
e x = −3, x = 4 –3 +
0
4 –
max.
f x = 0, x =
+
min.
27 5 27 5
0 +
0
0
–
max.
0
+
min.
g x = 1, x = 3 1 +
0
3 –
max.
0
+
min.
h x = 1, x = 3
Exercise 10E
1
1ax =0 –
0
3 +
0
–
0 +
0
+
min.
inflexion
b x = 2, x = −5 –5 +
0 max.
2 –
0 min.
+
max.
2 a x = −2(max.), x = 2(min.) b x = 0(min.), x = 2(max.) 1 c x = (max.), x = 3(min.) 3 d x = 0(inflexion) e x = −2(inflexion), x = 0(min.) 1 1 f x = − √ (max.), x = √ (min.) 3 3
Answers
dT = √ dl lg 12 4% 1 14 a i (1 − x)2 11 a
743
P1: FXS/ABE
P2: FXS
0521665175ans-2.xml
Answers
744
CUAT018-EVANS
November 3, 2005
9:27
Essential Mathematical Methods 3 & 4 CAS 3 a i (0, 0) and y
4 ,0 3
e i (−1, 0), (0, −1), (1, 0) y
(1, 1)
(0, 0) 0
0
(–1, 0)
x
x
(1, 0)
4, 0 3
(0, –1)
ii x = 0 (inflexion), x = 1(max.)
ii (−1, 0)(inflexion), (0, −1)(min.), (1, 0)(inflexion) f i (−1, 0), (0, 1), (1, 0)
b i (0, 0) and (6, 0)
y
y
(0, 1)
(0, 0) 0
(6, 0)
x
0
(–1, 0)
x
(1, 0)
ii (−1, 0)(min.), (0, 1)(max.), (1, 0)(min.) 4 a (−2, 27)(max.), (1, 0)(min.) 7 b (1, 0) is a turning point. c − , 0 , (0, 7) 2 y d
(4, –32)
ii x = 0(max.), x = 4(min.) c i (0, 0) and (3, 0) y
(–2, 27)
(2, 4) (0, 7)
(3, 0) 0 (0, 0)
x
x
0 (1, 0)
– 7, 0 2
5 b a = 3, b = 2, (0, 2)(min.), (−2, 6)(max.) 1 6 a (0, −256), , 0 , (2, 0) 2 1 4 b , 0 (inflexion), , 40.6 (max.), 2 3 (2, 0)(min.)
ii x = 0(min.), x = 2(max.) d i (−4, 0), (−1, 0), (0, 4) y
y
(–3, 4)
(0, 4)
(–4, 0) (–1, 0) 0
4 , 40.6 3
x
0
(2, 0)
1, 0 2
ii x = −3(max.), x = −1(min.) (0, –256)
x
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CUAT018-EVANS
November 3, 2005
9:27
Answers
f (x) g(x)
1
0
–1 2
x
1 2
–1
1 1 1 x: x > √ ∪ ∪ x: − < x < 2 2 2 1 x: x < − √ 2 √ 66 ii x: x > ∪ 12 √ − 66 1 x: 0 for r ∈ 0, dr 6 dV iii is increasing for r ∈ (0, 5.21) dr 100 − 3r 2 28 a i h = 2r r (300 − 5r 2 ) ii V = 6 √ iii defined for 0 < r < 2 15 dV iv = (300 − 15r 2 ) dr 6 v V (cm3)
(4.47, 468.32)
2 a e x (x 2 + 2x + 1) b e2x (2x 3 + 3x 2 + 6x + 5) c 2e4x+1 (x + 1)(2x + 3) −8x − 7 d √ 2e4x x + 1 x −2e x 3e − 2e4x b 3 a 3x (e x − 1)2 (e + 3)2 2x −8e c 2x (e − 2)2 2x 3 (2 − x) b 2e2x+3 4a e2x 1 3 e x (x − 1) c (2e2x + 1)(e2x + x) 2 d 2 x2 1 2 x 2 e xe 2 f −x e−x x e ( f (x) − f (x)) 5 a e x ( f (x) + f (x)) b [ f (x)]2 c f (x)e f (x) d 2e x f (x) f (x) + [ f (x)]2 e x
Exercise 11B 1a d g j 2a c
2 x 3x − 1 x2 2 2x + 3 3 15 − x 2x x2 + 1 1 − loge x 2 x
e e x loge x +
0
2√15 r (cm)
4 2 dy 2 by= x = 225 dt 27 50 ds dx 5 c i = ii = dt 9 dt 54 100 3000 30 a i y = 2 ii s = + 60x 2 x x 3000 ds b i = − 2 + 120x ii 1538.978cm 2 dx x c 585 cm2 /s
29 a
c −12e−4x + e x − 2x 2
e (2x + 3)e x +3x+1 g 2e2x − 2e−2x
b −21e−3x e3x − e x d e2x 2 f (6x − 1)e3x −x
3 x
1 x +1 6 i 2x − 3
f
1 x
d 2x(2 loge x + 1)
f loge (−x) + 1 x 2 + 1 − 2x 2 loge x g1 h x(x 2 + 1)2 1 b (e, e), m = 2 3 a (e, 1), m = e 2e c (e, log (e2 + 1)), m = 2 e +1 1 e (1, 1), m = 2 d (−e, 1), m = − e f (1, 0), m = −1 g (0, 0), m = −2 h (0, 0), m = 0 1 3 1 + 2x 4 6 72 82 5 2 2 5 1+x +x
y
1
Exercise 11A
c 2x +
b loge x + 1
Exercise 11C
Chapter 11 1 a 5e5x
2 x 3+x e x −2 h −2x + 3 3 k4− x −6 b
1
0
x
Answers
√ √ 27 a i y = 100 − r 2 and h = 2 100 − r 2 √ ii V = 2r 2 100 − r 2 b i V
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752
CUAT018-EVANS
November 3, 2005
20:32
Essential Mathematical Methods 3 & 4 CAS 2 {x: −2 < x < 0}
100 3 x < 1, max. value = 4 ≈ 1.83 e 4 a Min. f (x) = f (0) = 0 5ay=2 by=x e c y = 4e2 x − 3e2 d y = (x + 1) 2 e y = 3xe − 2e f y = 4e−2 6 a (0, 1)(min) by=x y c
y = loge 5x y = loge x (0.2, 0) 0
24 0
x
0
y
x
(1, 0)
23 a p
y=x
f (x) = x + e–x
21 a y = 2x − 1 b y = kx − 1 c y = kx − 1 22 Tangents are parallel for any given value of x.
b 25
p 2
p a 26 0.01
dy 1 dy t 2 + 1 dy dx 2t = , = = 1, = 2 , dt t dx 2t 2 dx dt t +1 dV = ([loge (h + 1)]2 + h 28 a dh + 2 loge (h + 1) + h + 1). When h = 2, dV = ([loge (3)]2 + 2 loge (3) + 5) and dh dh dh =5× dt dv b When h = 10, dV = ([loge (11)]2 + 2 loge (11) + 21) and dh dh dh =5× dt dv 27
7 t = 0, v = 0.4 m/s; t = 1, v = 0.4e ≈ 1.087 m/s; t = 2, v = 0.4e2 ≈ 2.956 m/s y 8a
b 2.02
y = 600
t
0
9 Maximum population is 44 (44.107) at t = 5 10 a −2y b ky 11 a 0.18 kg b 3.47 hours c i 6.93 hours ii 10.4 hours d 0.2 m 12 2q 13 a f (x) = aeax b f (h) ≈ ha + 1 c f (b + h) ≈ eba (ha + 1)
29 a 2x loge x + x y d
cx =e
bx =1
−1 2
a
b y ≈ −2ea p 14 a y ≈ pe 2 c y ≈ ea (1 + a) p d y ≈ e−a (1 − a) p dx dy = 2e2t = et 15 a b2 dt dt 16 p = 1, q = −6, r = 9 5 5 b 5 17 a 4 e e 2e2 18 4 e −1 2 19 a (8x − 8)e4x −8x b (1, e−4 ) y c
1
x
0 (1, e – 4)
1 5 dy=− x+ 8 4 20 a Tangent y = x − 1, normal y = 1 − x b Tangent y = −x − 1, normal y = x + 1
0
x –1
e 2, –1 e –1 2
Exercise 11D 1 a 5 cos 5x b −5 sin 5x c 5 sec2 5x d 2 sin x cos x = sin 2x e 3 sec2 (3x + 1) f −2x sin (x 2 + 1) g 2 sin x − cos x − 4 4 = sin 2x − = cos 2x 2 h − sin x ◦ cos x ◦ i 90 60 sec2 (3x)◦ j 60 1 √ b 1; 0 c 2; 0 2a √ ; 2 2 d 0; 0 e 1; 0 f 1; 4
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November 3, 2005
20:32
753
Answers
Exercise 11E 1 a y = 2x
b y = −x +
2
+1 d y = 2x 2 f y = −x − 1 + e y = 3x 2 2 3 − 2 sin x, gradient always positive 3 Distance = − 1 5 18◦ 26 4 45◦ 2 6 a 4.197 b −0.4 p 7a i ii 2 p cos2 b +1 90 1 1 8 a √ ; −√ 2 2 1 ii 0.6967 b i √ (1 − h) 2 a 1 b cos p 9 a −2 sin (2a) p 2 2 1 2p a p d− c cos2 (2a) 2 cos2 2 a 1 −a p e sin f − cos p 4 2 2 5 , 10 a Max. at x = and 3 3 min. at x = 0, and 2 5 b Max. at x = , min. at x = , 6 6 3 point of inflexion at 2 3 c Max. at x = and , 2 2 7 11 min. at x = and 6 6 c y = 2x −
d Max. at x = min. at 11 a
, point of inflexion at x = , 3
5 3
12 m/s 5
b−
1 rad/s 5
Exercise 11F 1 a 0.69 b1 c −1.82 d 1.54 2 a Local max. (2.07, 1.61); local min. (4.49, 0.53) b Local max. (1.57, 0) c Local max. (0.56, 0.55), (3.22, 2.85); local min. (1.39, 0.087) (4.66, 1.24) d Local min. (0, 0) e Local max. (4.27, −3.56); local min. (5.15, 3.56) f Local max. (0.32, 0.20), , 1 , (2.82, 0.20), 2 (3.93, 0), (5.5, 0); local min. (0.79, 0), (2.36, 0), (3.46, −0.202), (4.71, 1), (5.96, −0.202) 5 a, c
y y = ex – 2
x = –2
0
x y = –2
b (−1.841, −1.841), (1.146, 1.146) c y = loge (x + 2) d R\[−1.841, 1.146] e i a = −1 ii y = x iii (0, 0) iv y = loge (x + 1) y v x = –1
y = ex – 1
y=x
y = loge (x + 1) 0
x y = –1
6b
dy = 2 for y = 2e x − 2 dx dy 1 x +2 = for y = loge dx 2 2
c y = 2x and y =
1 x 2
Answers
3 a −5 sin x − 6 cos 3x b −sin x + cos x c cos x + sec2 x d 2 tan x sec2 x 4 a 3x 2 cos x − x 3 sin x −(1 + x) sin x − cos x b (1 + x)2 c e−x (cos x − sin x) d 3 − 2 sin x e 3 cos 3x cos 4x − 4 sin 4x sin 3x f 2 cos 2x tan 2x + 2 sec2 2x sin 2x g 12 sin x + 12x cos x h esin x (x 2 cos x + 2x) i 2x cos2 x − 2x 2 cos x sin x j e x (sec2 x + tan x) 5 a −2 b −6 c −e 1 d −e e− f0 6 a tan x b tan x 2 2 = 7a sin 2x − 1 2 sin x cos x − 1 7 7 5 b sec2 xetan x c x 2 sin 3x + 3x 2 cos 3x 2
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Answers
754
CUAT018-EVANS
November 3, 2005
20:32
Essential Mathematical Methods 3 & 4 CAS d i f −1 : (−a, ∞) → R, x +a f −1 (x) = loge a 1 ii y = ax and y = x a iv Pairs of graphs of this form intersect at the origin.
Exercise 11G 1 a = 0.1373, P0 = 30 b 9.625 hours c i 4.120 units/hour ii 1.373√units/hour 200 3 1 ◦ km/h b rad/s = 1.9099 /s 2a 3 30 3 a 8 cos b Area =16 (1 + cos )√sin ; maximum area = 12 3 square units 75 seconds 4a cos b 220 − 60 tan seconds dT 75 sin − 60 d = d cos2 −1 4 e = sin ≈ 53◦ 8 5 f T = 265 seconds; P is 400 metres from B for minimum time
Multiple-choice questions 1C 6C
2B 7A
3D 8E
4E 9B
Short-answer questions (technology-free) 2x b 3 cos (3x + 2) +2 1 x 2 d (2x − 2)e x −2x c − sin 2 2 1 e f 2 cos (2x) x −3 g 6 sin (3x + 1) cos (3x + 1) = 3 sin (6x + 2) 1 2 − 2 loge 2x h i 2x loge x x2
1a
5D 10 C
b aeax+b c −bea−bx 4 a aeax d abeax − baebx e (a − b)e(a−b)x 5 0.25 m/s; 0.25e m/s; 0.25e2 m/s; 0.25e4 m/s 6 a 25e100t ◦ C/second b 25e5 ◦ C/second 7 y = ex 8 b 20 cm/year 10 2 11 a = 2 or a = 1 √ x 1 by= √ − √ + 2 12 a y = x e 2 2 2 −1 3 dy= x cy=x− e 2
Extended-response questions 1a b
x dy −9 = sin dx 40 80 dy dx 0
x
80
–9π 40
c Magnitude of gradient is a maximum at the point (40, 12) 3 000 000e−0.3x 2 a f (x) = (1 + 100e−0.3x )2 b i 294 kangaroos/year ii 933 kangaroos/year 3 a a = 30 b (0, 8 loge 6)(25, 0) c f (20) = −0.8 x eR d f −1 (x) = 5 6 − e 8 y f (0, 8 loge 6)
x2
j 2x sin (2x) + 2x 2 sin (2x) a 2 e x sin 2x + 2e x cos 2x 1 − 3 loge x b 4x loge x + 2x c x4 d 2 cos 2x cos 3x − 3 sin 2x sin 3x 2 = 2 sec2 2x e cos2 2x f −9 cos2 (3x + 2) sin (3x + 2) g 2x sin2 3x + 6x 2 cos 3x sin 3x b0 3 a 2e2 ≈ 14.78 c 15e3 + 2 ≈ 303.28 d1
0
x (25, 0) x = 30
3 1 1 ,e , , c ,e 2 2 e e d 2 as g(x + 2) = g(x) 6 a i 30 g ii 12.28 g −300et dx = b dt (5et − 3)2 c ii
4b
dx dt
0
–6.837
x
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CUAT018-EVANS
November 3, 2005
20:32
755
Answers 2 tan
d OP + OQ =
c NQ = 8 tan 2 + 8 tan + 10 tan
dx = −2 cosec2 + 8 sec2 d f x = 18, = 26◦ 34 8 a f (x) = e x + e−x b {0} y d e
x
0
9 a x = 1 or x = e2 b Gradient of y = 2 loge x is 2, when x = 1; gradient of y = (loge x)2 = 0, when x = 1 y c
d When x = 6 bAebt dy = 13 b dt (1 + Aebt )2 e After 7 hours (to the nearest hour) 144 degrees per second = 1.833 degrees per 14 25 second 15 0.1 km/s 16 a y = ex b y = 2ex c y = kex 1 e i A unique real root k = or k ≤ 0 e 1 ii k > e x xe − e x 17 a f (x) = x2 bx =1 c (1, e); minimum x −1 f (x) d i = f (x) x f (x) ii lim = 1, i.e. f (x) → f (x) x→∞ f (x) as x → ∞ y e
y = (loge x)2 y = 2 loge x (e2, 4) 0
x
0
x (1, 0)
e2
1 ≈ 45.27 years, minimum occurred k in 1945 1 18 a A = 1000, k = loge 10 ≈ 0.46 5 dN b = k Aekt , where A = 1000 and dt 1 k = loge 10 5 dN c = kN dt dN d i ≈ 2905.7 dt dN ii ≈ 4.61 × 1012 dt 19 a t ≈ 34.66 years b t ≈ 9.12 years 20 a D(t) ft =
d {x: 2 loge x > (loge x)2 } = [x: 1 < x < e2 } 10 x = e 11 a h = a(1 + cos ) b r = a sin dV 1 3 3 d = a [−sin d 3 + 2sin cos (1 + cos )] 1 −1 ≈ 70◦ 32 = cos 3 32 3 eV = a cm3 81 12 a f (x) = loge x + 1 b x ≈ 0.37, i.e. during the fourth month of its life y c
13 10
1
7
1 e , 0.632 0
(1, e)
x 6
0
6
12
24
t
Answers
7 b MP =
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756
CUAT018-EVANS
November 3, 2005
20:32
Essential Mathematical Methods 3 & 4 CAS b {t: D(t) ≥ 8.5} = {t: 0 ≤ t ≤ 7} ∪ {t: 11 ≤ t ≤ 19} ∪ {t: 23 ≤ t ≤ 24} c i 0 metres/hour ii − metres/hour 2 iii metres/hour 2 d i t = 0, 12, 24 ii t = 6, 18 1 21 a The height is reached th of a second after 6 t = 0. 2 b 3 1 c When t = , speed = 0.6 m/s 3 (moving downwards). 2 When t = , speed = 0.6 m/s (moving 3 upwards). 1 When t = , speed = 0 m/s. 6 1 22 a p = 12, q = 8, r = 6 4 b T (3) = − ; hours of night decreasing by 3 4 4 hours/month T (9) = , hours of night 3 3 4 hours/month increasing by 3 8 c − hours/month 3 d t = 9 (after 9 months) 23 a A = 2x cos (3x) dA = 2 cos (3x) − 6x sin (3x) b i dx dA = 2. ii When x = 0; dx dA − . When x = , 6 dx c i A
0
π 6
x
ii x = 0.105 or x = 0.449 iii Maximum area = 0.374 when x = 0.287 1 t 24 a i N (t) = −1 + e 20 10 ii Minimum population is 974 and occurs when t = 20 loge 10. iii N (0) = 1002 iv N (100) = 900 + 2e5
v
(100, 900 + 2e5)
N 1002
y = N(t) (20 loge 10, 974) t
0
b i N2 (0) = 1002
1
ii N2 (100) = 990 + 2e 2 iv Minimum population is 974 and occurs when t = (20 loge 10)2 . c ii Minimum population is 297 and occurs when t = 100.24. 3 1 1 t d i N3 (t) = − t 2 + e 20 2 10 2 sec 25 c 180 26 a i BX = h tan h ii R = (1 + tan2 ) 2 h sin b i R = × 180 cos3 ii −0.1209 10 1 log a 27 e 3 3 5 b i x = 0 and x = 2√ −4 + 5a ± 25a 2 + 16 ii x = 4a
Chapter 12 Exercise 12A 1 a 3.81 square units b 1.34 square units 35 c square units 2 2 a 13.2 square units b 10.2 square units 3 a 10 square units b 10.64 square units 4 a 0.72 square units b 2.88 square units, decrease strip width 5 a 36.8 square units b 36.7 square units 6 a ≈ 48 square units b Distance travelled 7 11.9 square units 9 8 a square units b 9 square units 2 c 4 square units
Exercise 12B x4 +c 8 x4 − x3 + c c 5
1a
5 4 x − x2 + c 4 5 d 2z + z 2 − z 3 + c 2
b
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Answers
Exercise 12C −1 1 b (t − 2)4 + c (2x − 1)3 + c 4 6 3 2 1 d (3x + 6) 2 + c (5x − 2)4 + c c 9 20 9 1 1 2 f (2x − 4) 2 + c e (3x + 6) 2 + c 9 3 3 7 2 1 h − (2 − 3x) 2 + c g (3x + 11) 3 + c 9 7 1 1 j (3 − 4x)−4 + c i − (5 − 2x)5 + c 16 10 1 1 2 a loge |x| + c b loge |3x + 2| + c 2 3 −3 c loge |1 − 4x| + c d x − loge |x + 1| + c 4 −1 f +c e x + loge |x| + c x +1 2 x g + 2x + loge |x + 1| + c 2 h 2x − loge |x + 1| + c 3 i− +c 2(x − 1)2 b 3 loge |x − 4| + c 3 a 5 loge |x| + c c 5 loge |2x + 1| + c −3 e d −3 loge |2x − 5| + c 2x − 1 1 f − loge |3x − 4| + c 3 1 4 a y = loge |x| + 1 b y = 10 − loge |5 − 2x| 2 5 y = 10 loge |x − 5| |x − 2| 6 y = 3 loge + 10 2 5 5 7 y = loge + 10 4 |2x − 1| 1a
Exercise 12D x 1 2x e − 2e 2 + c b e x − e−x + c 2 x x 2 d 15e 3 − 15e 5 + c c e3x + e−x + c 3 1 2 a y = (e2x − x 2 + 9) 2 3 b y = − x − ex + 8 e 3 y = 9 − 2e−2 1 1 4ak =2 b y = e2x + e2 2 2
1a
Exercise 12E 7 3 1 e 2
1a
1 4 1 g 15 3 c−
b 20
d9
11 20 10 1 13 e 2 a 10 b1 d c 441 3 3 1 2 3 1 h 2 − 22 f 34 g 22 − 1 i 3 15 1 1 b y = (3 − e−2 ) 3 a y = (e2 − 1) 2 2 f
140 3
1
h 343
d y = e2 − e−2
c y = 6e 3 − 4
4 a 10 b17 c −5 1 1 b loge (3) 5 a loge 2 3
d9 3 c loge 2
e−3 19 7
Exercise 12F 1a3 b 44 c i 8 ii 10 1 1 1 4 d c 121 b 2a 6 2 6 3 y 3a
√ e4 3
y = 2x + 1
1 –1 0 2
x 1
4
y
b
3
0
y=3–x
x 3
f 108
Answers
−3 +c 2 a −3x −1 + c = x 2 b − 3 + 3x 2 + c 3x 3 5 √ √ 2x 2 4 x3 2 x5 4x 2 + +c = + +c c 3 5 3 5 2 3z 2 20 9 9 4 − +c e x4 +c d x3 − 2 z 4 9 14 3 12 7 x4 − x2 +c f 7 3 x4 by= +6 3 a y = x 2 − 3x + 3 4 3 2 1 22 c y = x 2 + x2 − 3 2 3 1 1 3 8 4 f (x) = x 3 + − 8 5 s = t2 + − 8 x 2 2 t 6 a k = −32 b 201
757
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758
CUAT018-EVANS
November 3, 2005
20:32
Essential Mathematical Methods 3 & 4 CAS y
c
y
10
y = x2
(4, 16)
x
0
x
0
1
4
d
2
y
4
–√2
11
1 2
√2
1
y
e
y
x
0
–1
Area = 0.5 square units
y = 4 – 2x2
–3 –2
x
2 3
0
y = √x
1 11 log e 3 8
12
y
x
0
2
4
y
f
y=2
y = (1 – x)(1 + x)2 1
log e 2 + 4
x
0
–1
x
–4 –2 0
–9 2
1
Exercise 12G
321 square units 4 10 y 5
1a
1 sin 3x 3
c sin 3x 1
0
x
0
–1
3 sq. units 4 1 5 b 8 sq. units 6 a sq. units 6 6 7 a A (0, 3), B (1, 0) b 2 sq. units 8 b Derivative = (loge a)e x loge a , e x loge a antiderivative = loge a y 9 −1
1 + x 3d x =
1 e − cos 2x − 2 3 1 1 g sin 4x + cos 4x 4 4 1 i − sin 2x + 4 3 1 1 b 2a1− √ 2 2 2 e1 f √ 3 1− 3 j −2 i √ 4 3 − 2 + 2 square units y 4a 1
7 y=3 0
3 0
x 1
Area = 2e2 + 1 ≈ 15.78 square units
4 0
π 4
π 2
x
1 cos x d x = √ 2
1 b −2 cos x 2 1 d −4 cos x 2 1 1 f sin 3x − cos 2x 3 2 1 1 h cos 2x + sin 3x 4 3 1 j − cos x 1 d2 c1+ √ 2 1 h4 g− 2
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November 3, 2005
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Answers y
f
Answers
y
b
759
1
π π 4 3
0
3
x
π 2
sin 2x d x =
0
3 4
− 4
√ 2 5a 2 6 y
y
c
4
π 4
0
–π 4
x
−1 2 √ 3 c− 3
1 − cos 2d = b−
1 3
1 d√ 2
1 3 2
π π 6 4
–π –π 0 4 6
1
x
− 6
2
x
π 2
cos + sin d = 2
0
y
e 2
1
0
0
2
2 + sin 3x d x =
x
2 2 + 3 3
Exercise 12H
√2
0
3
2π 3
0
y
d
π 3
0
√ 3 cos 2x d x = 2
6
π 2
x
sin 2 + 1 d = 1 +
√ 5 3 2 2 d −2 c 12 b2 1a4 4 3 3 2 2 e e4 g4 f + 4 loge 2 − e 3 2 2 1 51 5 2 j i 8 loge 2 + h +1 12 4 8 2 0.5 sq. units 3 a 139.68 b 18.50 c −0.66 d −23.76 e 2.06 f 0.43 4 b 5 loge 3 + 4 5 b 5 + 6 loge 2 1 7 dy 6a = −4 1 − x Hence dx 2 1 7 1 8 1 1 − x dx = − 1− x +c 2 4 2 dy b = − tan x. dx 3 Hence tan x d x = loge 2 0
2
1 7 f (x) = 1 − 2 cos x 2 1 8 a f (x) = sin 2x + 1 2 b f (x) = 3 loge x + 6 x
c f (x) = 2e 2 − 1
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760
CUAT018-EVANS
November 3, 2005
20:32
Essential Mathematical Methods 3 & 4 CAS 9 sin3x + 3x cos 3x. Hence 6 1 x cos 3x d x = − 18 9 0 10 a = 1, b = −2; area = 3 square units 11 a 1.450 square units b 1.716 square units 12 0.1345 1 13 f (x) = (x 2 − cos 2x + 3) 2 14 a b
(–1, 4)
f (x) − 4 d x = (x 2 + 1)3 − 4x + c
3h(x) d x = 3 sin x 2 + c
1 , tan x cos2 x 6x 19 1 b 2 , loge 3x + 7 6 7 √ x , loge (1 + 2) c1+ √ 1 + x2
15 a
y=
2 +4 x–1
0 (0, 2)
b 20
y
y = sin x
π 3
0
π
π 2
x
y = sin 2x
1 square units 2 8 P(loge 3, 3); area ≈ 2.197 square units
x 2 3 x=1
2 + 4 d x = (2 loge 2) + 4 x −1
y
Exercise 12J 2 2 2 c b d0 3 2 10(e5 − 1)e−5 ◦ C ≈ 9.93◦ C 3a v 1a
(5, 100)
(2, 1)
50 0
5 square units 6
7
y=4
17
1
1 c 4 square units d 4 square units 2 1 e 4 square units 2 4 a 2 square units b e + e−1 − 2 ≈ 1.086 square units 5 3.699 square units 1 6 square units 4
y
16
2
x
0
3 a 36 square units
− f (x) d x = −(x + 1)3 + c
3
(2, 1)
1
( f (x) + h(x)) d x = (x 2 + 1)3 + sin x 2 + c
f
f(x)
Area = 9 square units 2 36 square units
2
g(x)
5
h(x) d x = sin (x 2 ) + c
e
y
1
f (x) d x = (x 2 + 1)3 + c
c d
Exercise 12I
x 2
3
3
3 √ 1 2x − 4 + 1 d x = × 2 2 + 1 3 2 5 √ 1 22 4 2 2 c loge 4 b − 18 a 3 3 3 3 √ 2 1 e −2 d loge 3 + 3 f2 2−2 3 2
t
(0, 0)
b
v
24 48 π
0
t 4
e
1 2 (e − e−2 ) 2
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Answers
b 18 m2 a c i y − 3 + 3 cos 3 −1 a (x − a) = sin 3 ii 5.409 √ 3( 2 + 2) 18 a i 9 iii 12 ii 2 b Maximum value is 12, minimum value is 0.834 48( + 1) litres c 17 a 6 metres
v (5, 1 – e –5) (4e5 + 1)e –5 t
0
a2 147 m/s 5 6 10 2 6 a 3000(2 − 20.9 ) N/m 0.1 b 1000(4 − 1) N/m2 c0 7 a x = t 2 − 3t b x = 0 3 9 e m/s d metres 2 2 2t 3 8 a Displacement = − 4t 2 + 6t + 4 3 Acceleration = 4t − 8 20 b When t = 1, displacement = m 3 When t = 3, displacement = 4 m c When t = 1, acceleration = −4 m/s2 When t = 3, acceleration = 4 m/s2 9 Initial displacement −3 m 10 Velocity = 73 m/s 646 Position = m 3 11 a Velocity = −10t + 25 b Height = −5t 2 + 25t 5 c s 2 125 d m 4 e 5s y 1 1 12 a f −1 (x) = e x y = ex 2 2 4
c 4 loge (2) −
3 2
dt 7/4 1 200 60
120
3B 8C
4B 9C
5A 10 D
3
−5a 2 65 55 b0 c 1a d− 3 4 3 1 f1 g0 h0 e 2 23 33 44 5 820 2 2 5 5 85 8 7 6 3 3 4 c 9 d f (x) − g(x)d x + g(x) − f (x)d x + c b b f (x) − g(x)d x a
13 a H dH
1/4 0
2C 7E
Short-answer questions (technology-free)
x
0 1 2
3 2
1E 6D
y = loge 2x
1 2
b
Multiple-choice questions
180 240
b t ∈ [10, 50] ∪ [130, 170] c t = 30 or t = 150 d i 120 kilojoules ii 221.48 kilojoules 2 14 71 466 m3 3 15 a 46 5 m2 b 46 500 m3 16 1.26 m
10 a P(3, 9), Q(7.5, 0) b 29.25 square units 20 bp= 11 a 5 7 12 3.45 square units 1 b 15 square units 13 a A(0, 6), B(5, 5) 6 125 c square units 6 14 a 2 square units b e + e−1 − 2 ≈ 1.086 square units y 15 a
t
y = ex + 1
(0, 2)
0 b e2 + 1 ≈ 8.39
x
Answers
c
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762
CUAT018-EVANS
November 3, 2005
20:32
Essential Mathematical Methods 3 & 4 CAS y
16 a
c
y = e –x
y = ex
(0, 1)
–2
x
0
2
b 2 − 2e−2 17 a e − 1 ≈ 1.72 b 2(e − 1) ≈ 3.44 square units 18 2 + e2 ≈ 9.39 square units 1 19 3 square units 12 y
–1 0
20 a e2 + 1 2 +1 c 8
x
3 2 b loge − 3 2 1 1 1 5 d loge − 8 + 10 2 6 e e
1 a When t = 0, 1000 million litres/hour. When t = 2, 896 million litres/hour. b i t = 0 and t = 15 ii 1000 million litres/hour c dV dt 1000
10 15
t
d i 5000 ii 5000 million litres flowed out in the first 10 hours. 2 a When t = 5, ≈ 17.9 penguins per year; when t = 10, ≈ 23.97 penguins per year; when t = 100, ≈ 46.15 penguins per year. b R
R(t) = 10 loge (t + 1)
0
d i 3661 ii The growth in the size of the penguin population over 100 years (assuming zero death rate). 3 a 4y − 5x = −3 3 9 c (1, 0) b ,0 e 9 : 49 d 5 40 1 5 a square units 3 n−1 2 d1− = square units n+1 n+1 9 99 999 e , , 11 101 1001 f area between curves approaches 1 6 a 968.3◦ b θ (°C)
2
Extended-response questions
0
R
t = e 10 − 1; R ≥ 0 t R −1 (t) = e 10 − 1
t
(0, 30)
0
t (min)
c 2.7 min d 64.5◦ C/min 2 4 7 a 5 × 10 m/s b Magnitude of velocity becomes very small c 5 × 104 (1 − e−20 ) m d v(1 − e−t ) e x (m) v
0
t (s)
d 8 a (e−3x sin 2x) = −3e−3x sin 2x dx + 2e−3x cos 2x c e−3x sin 2x d x −1 −3x (3e sin 2x + 2e−3x cos 2x) 13 4 3 4 9 a i tan a = ii sin a = cos a = 3 5 5 b 2 square units e dy 10 a = loge x + 1, (loge x) d x = 1 dx 1 dy = (loge x)n + n(loge x)n−1 b dx e d (loge x)3 d x = 6 − 2e 1 √ 3 11 s = ba 2 √ 3 r = b2 a =
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November 3, 2005
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763
Answers
13 a i R(0) = 0
ii R(3) = 0 t 10 b R (t) = cos 3 3 t − sin 3 c i 1.41, 4.41, 7.41, 10.41 ii (1.41, 8.65) (7.41, 4.75) local maxima (4.41, −6.41) (10.41, −3.52) local minima d t = 0 or 3 or 6 or 9 or 12 e y
t e− 10
3
6
9
12
f i 16.47 litres iii 8.27 litres g 12.99 litres 14 a
1E 7D 13 B 19 E 25 B 31 A 37 D 43 B 49 C 55 C 61 B
2D 8A 14 E 20 D 26 A 32 C 38 A 44 B 50 A 56 D 62 B
3E 9D 15 D 21 B 27 D 33 D 39 A 45 C 51 C 57 C 63 D
4D 10 D 16 E 22 E 28 D 34 B 40 A 46 D 52 B 58 B 64 E
5B 11 A 17 B 23 B 29 A 35 A 41 B 47 B 53 E 59 D 65 B
6C 12 B 18 C 24 D 30 C 36 E 42 D 48 D 54 C 60 C 66 D
Extended-response questions 1 a 54.06 g b s 80
(10, 54.06)
50
y = R(t)
0
Chapter 13 Multiple-choice questions
t
0
t
1 ds c = −6e− 5 t dt e 0.8 g/litre 2 a 60◦ C b T
ii 12.20 litres
1 ds = − (s − 50) d dt 5 f 17 seconds
y area required in d π ,1 2
1 0 – π , –3 2 b 0
6
20
t 0
–1
f (x) d x = 2 −
60
area required in b √ 3− 6
c f −1: [−3, 1] → R, (x + 1) f −1 (x) = sin−1 2 1 2 −1 d f (x) d x = − f (x) d x 2 0 6 5 √ = − 3 6 15 b 1 − 4
t
dT c = −14.4e−0.36t dt dT d = −0.36(T − 20) dt 3 a 1.386 minutes b 2200, 5.38% c 66.4 spores/minute d 0.9116 minutes e 1200 1000 f g 0
0.91
t (minute)
Answers
dy x x = − e 10 and dx 10 1 dy = −x(100 − x 2 )− 2 dx dy b When x = 0, = 0 for both functions dx c −e d 6.71 square units e 8.55% f (25 − 50) square units g i 10(10e − 20) ii (25 − 100e + 200) square units
12 a
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764
CUAT018-EVANS
November 3, 2005
20:32
Essential Mathematical Methods 3 & 4 CAS 4a
60 60 b √ ≈ 9.61 km 13 a √ ≈ 9.61 km 39 39 12 m ≈ 2.7 m 14 4 − 7
V 100
15 a {x: x > 1}
0
t
b i 20e
−0.2t
V m/s2 ii 20 − 5
2
m/s
c 8.05 s 5 b 0.028 c 0.846◦ C/minute 6 a i 0.1155 ii 0.2 b 13.86 days 7 100 8 $600 9 a = ±3, b = ±2 y
b {x: 0 < x < 2} 3 c {x: x > 1}, x: 0 < x < 2
n+1 d {x: x > 1}, x: 0 < x < n √ 1 6 b ± 2, 2− 3 16 a (1, 1), (−1, −1) n √ − c ± 2n+2 n, n 2n+2
108 17 a A = 48 + 16x + x b A (m2)
y = 48 + 16x (2.6, 92.14)
–2 3
x
0
0
x = –2 3
10 a 5 × 104 m3 c −3500 m3/day e V
b −12 500 m3/day d After 222.61 days
(m3) (0, 5 × 104)
0
11 a
y
t (days)
C ($'000)
B
0
N 2 3
($)
(20, 12 000) V (km/h)
d V = 20, C = 12 000
2a 2 4 3 d2:3 c a 3 3 1 c a = 1 or a = −2 20 a −5 3 21 a i 50e−1 litres/minute ii t = 5 iii 2 minutes 18 seconds iv 3 minutes 48 seconds b 14.74 litres c 53 seconds h 2 sin 22 a S = cos2 100(1 + sin2 ) b i cos3 100(2 − cos2 ) ii S = cos3 2 − cos2 100 v 29.1◦ × iii cos sin 1 b
4(N 3 + 16) 4 c Rate of change of cost in $1000s with respect to the increase in the number of bottle tops produced 160 000 bC = + 10V 2 12 a $17 000 V c C
0
x
(0, – a2)
2
3N
A (a, 0)
0
(– a, 0)
(10, 5.65)
b
x (m)
6
√ √ 4 3 3 3 m, width = m c Height = 2 3 2 d 172 m 18 p = 4, number of items = 50 19 a (a, 0), (−a, 0)
e 12 560
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November 3, 2005
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Answers
Exercise 14A 2 b 3
9 250 b 0.35 b 0.047
c
b
6
5 5% 7 a 0.5
Second die
1 8 0.39 9 7 24 11 0.0479 10 59 12 a 0.486 b 0.012 2 5 2 14 a 5 5 15 a 14 13 a
16 a 0.735 3 17 44
d
1 15 7 b 40
d 0.4
c 0.138 c
d
7 15
3 5 24 b 53 b
Exercise 14B 1 a Discrete b Not discrete c Discrete d Discrete 2 a Not discrete b Discrete c Not discrete d Discrete 3 a {HHH, THH, HTH, HHT, HTT, THT, TTH, TTT } b Experimental outcome Value of x HHH THH HTH HHT HTT THT TTH TTT
2 3 4 5 6 7
First die 3 4
3 4 5 6 7 8
4 5 6 7 8 9
3 2 2 2 1 1 1 0
1 c 2 4 a {(1, 1), (1, 2), (1, 3), . . . , (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
5
1 6 = 36 6 1 d Pr(Y = 3|Y < 5) = 3 1 2 3 4 5 6 5a 1 1 1 1 1 1 1 2 1 2 2 2 2 2 3 1 2 3 3 3 3 4 1 2 3 4 4 4 5 1 2 3 4 5 5 6 1 2 3 4 5 6 b 1, 2, 3, 4, 5, 6 c 0.19 6 a 0.288 b 0.064 c 0.352 d 0.182 7 a See 4a. 1 1 5 b Pr(A) = , Pr(B) = , Pr(C) = , 6 6 12 1 Pr(D) = 6 1 1 c Pr(A|B) = , Pr(A|C) = , 6 5 1 Pr(A|D) = 6 d A and B are independent. A and D are independent.
Exercise 14C 1ac 2a
b
x
1
2
3
p(x)
1 5
1 5
1 5
3 5 x
3
0
p(x) 4a
b
1
6
5 6 7 6 7 8 7 8 9 8 9 10 9 10 11 10 11 12
c Pr(Y < 5) =
200 = 0.519 385
8 15 7 c 16
b
1 2 3 4 5 6
41 500
6 7 c
b 0.385
41 125
1 2
4
5
1 5 1 c 3 2
1 5
0.36 0.48 0.16
x
0
1
2
3
p(x)
27 125
54 125
36 125
8 125
x
0
1
2
3
p(x)
5 30
15 30
9 30
1 30
Answers
b Y = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Chapter 14 1 1a 2 17 2a 500 3 a 0.65 4 a 0.067
765
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766
CUAT018-EVANS
November 8, 2005
10:32
Essential Mathematical Methods 3 & 4 CAS 5 a See 4a of Exercise 14B. b
x
2 3
p(x)
4
5
6
Exercise 14D 7
8
9
10 11 12
1 2 3 4 5 6 5 4 3 2 1 36 36 36 36 36 36 36 36 36 36 36
c p(x)
1 a No unique mode, median = 5, E(X ) = 4.6 1 b No unique mode, median = , 2 E(X ) = 0.5 c Mode = 2, median = 2, E(X ) = 2.38 d Mode = 0.6, median = 0.6, E(X ) = 0.569
6/36 5/36 4/36 3/36 2/36 1/36
e Mode = 7, median = 7, E(X ) = 7 f No unique mode, median = 0, E(X ) = 0 2 a c = 0.35 bX =1 c2 d E(X ) = 2.3 e Var(X ) = 1.61; sd(X ) = 1.27 f 0.95
0
2 3 4 5 6 7 8 9 10 11 12 x
5 7 d e 18 10 6 a {(1, 1), (1, 2), (1, 3), . . . , (6, 4), (6, 5), (6, 6)} 0 1 2 y b 5 2 11 p(y) 18 18 18 c p(y)
1 b5 15 d E(X ) = 3.667 f 0.9333
3ak =
c4 e Var(X ) = 1.556
4 Expected profit = $3000 5 A loss of 17c 6 1.54 7 a E(X ) = 7 b Var(X ) = 5.83 8 a E(X ) = 4.11 c E(5X − 4) = 16.55
11/18
c 0.944
b E(X ) = 78.57 1 dE = 0.255 X 3
9 a Var(2X ) = 64 c Var(1 − X ) = 16
b Var(X + 2) = 16 d sd(3X ) = 12
10 a 3
5/18 2/18 0 7a
b 8a
b
1
y
2
x
0
1
2
Pr(X = x)
1 3
8 15
2 15
b 1.5 c 0.9688 1 b E(X ) = 2 c Var(X ) = 3.5 11 a p = 16 20 91 1 c b 12 a k = 9 21 21 13 a x 1 2 3 4 6 8 9 12 16 Pr(X = x)
7 15
b i x
10
20
100
14 a
Pr(X = x)
3 4
6 25
1 100
15
y
20 30 40 110 120 200 9 9 36 3 3 1 Pr(Y = y) 16 25 625 200 625 10000
1 4 b {EENE, ENEE, ENNN, NEEE, NENN, 3 NNEN}, 8 3 c 8
1 4
1 1 1 3 1 1 1 1 1 16 8 8 16 8 8 16 8 16 ii
21 4
b
25 4
7 12
x 1 2 3 4 5 67
Pr(X = x) E(X ) =
275 16 497 c 48 iii
8
9
10 11 12
1 1 1 1 1 1 1 1 1 1 1 0 6 6 6 6 6 36 36 36 36 36 36 49 12
9a
Multiple-choice questions 1A 6E
2D 7C
3B 8C
4A 9B
5D 10 D
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Answers
A
40 81 4 0.4 5 a 0.1 3a
b b1
5 9
c1 d 1.3 53 b 256 √ 29 3 d 2 p x − 2 −2
6 a 21.5 c 630.75 7a
4 x −2 5
e 2.01
1 5
4 5
Pr(P = p) b
Pr (A′ ∩ B′) = Pr (A ∪ B)′
B
c x > $2.50
47 8 a 0.47 b 70 9 21.5% 1 17 10 a b 24 24
c
5 6
d
11 18
Extended-response questions 1 a 0.1 2a
b 0.2 0.6 J
0.3
c4 J
A
0.7
0.4
J
0.6
A
0.6
A
b i 0.396 c i
J
0.4
0.4
0.6
A J
0.4
A
ii 0.604 x 2 Pr(X = x)
3
0.6 0.4
ii 2.4 d 0.46 25 3 a 0.5 b 0.05 c 0.033 d 33 4 a i 1.21 ii Var(P) = 1.6659, sd(P) = 1.2907 iii 0.94 b i t 1 0.40 0 p(t)
0.39 0.27 0.34
ii E(T ) = 0.498 ≈ 0.50
iii 1
p(b)
0.677 0.270 0.053
ii E(B) = $37.60 7 a mean = 13.5%, sd = 16.2% b 0.95 c E(G) = 6.9%, sd(G) = 9.726% 8 Yes 9 $1.00 10 a i 0.65 ii 0.2275 iii 0.079625 iv 0.042875 b Expected cost = 8.439 375 million dollars c Expected profit is 10.703 125 million dollars. 1 2 c 27 11 a (304 − 2x) bx =2 3 9 1 49 12 b x = , 2 288 8 1 4 ii iii 13 a i 81 81 81 4 56 iv v 27 81 b 4.4197 cents. (As the lowest value coin is 5c/, he can settle for that.)
Chapter 15 Exercise 15A 1 a and b 2 a 0.0595 b 0.0512 c 0.0081 3 a 0.05358 b 0.0087 c 0.0623 55 54 4 a 5 ≈ 0.0804 b 5 ≈ 0.4019 6 6 5 a 0.1156 b 0.7986 c 0.3170 6 0.6791 7 a 0.1123 b 0.5561 c 0.000 01 d 0.000 01 8 0.6836 9 0.544 10 0.624 11 0.1356 12 a 0.0138 b 0.2765 c 0.8208 d 0.3368 6 1 13 a = 0.000 24 b 0.1694 4 14 0.9744 15 0.5432 b 0.001 23 16 a (0.8)8 ≈ 0.168 c 0.002 1 17 a (0.15)10 ≈ 0.000 000 006 b 1 − (0.85)10 ≈ 0.803 1 c 0.567 4 18 0.962 19 a 0.002 455 b 0.003 37 20 a 0.011 529 b 0.002 59 c 0.0392
Answers
5 $14 6 a E(Y ) = 2.002 b Var(X ) = 2.014; sd(X ) = 1.419 b 0 100 200 c i
Short-answer questions (technology-free) 1 Yes, as Pr(A ∩ B) = 0 2
767
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768
CUAT018-EVANS
November 8, 2005
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Essential Mathematical Methods 3 & 4 CAS
Exercise 15B 1a4 2
Exercise 15C
b 10
1 a Mean = 5, variance = 4 b Mean = 6, variance = 2.4 500 1000 c Mean = , variance = 3 9 d Mean = 8, variance = 6.4 2a1 b 0.2632 3 37.5 1 4 n = 48, p = , Pr(X = 7) = 0.0339 4 3 , Pr(X = 20) = 0.0076 5 n = 100, p = 10 √ 6 Mean = 10, sd = 5 The probability of obtaining between 6 and 14 heads is 0.95. √ 7 Mean = 120, sd = 4 3 The probability that between 107 and 133 students attend a state school is 0.95.
c 16
.40 .30 .20 .10 x 1
2
3
4
5
6
7
8
9 10
Legend: ◦ p = 0.3, skewed positively × p = 0.6, ‘slightly’ skewed positively • p = 0.9, skewed negatively 3 .40
Exercise 15D
.30
1 a i (0.8)5 ≈ 0.3277 ii 0.6723 b 14 2 a i 0.1937 ii 1 − (0.9)10 ≈ 0.6513 b 12 37 47 5 10 6 42 7 86
.20 .10 x 1
2
3
4
5
6
7
8
9 10
Multiple-choice questions Legend: ◦ p = 0.5; n = 6 × p = 0.5; n = 10 Both plots are symmetrical: ( p = 0.5; n = 6) is symmetrical about x = 3 ( p = 0.5; n = 10) is symmetrical about x = 5 4
5
1D 6B
16 81 54 2 125
7 a, b, c
c The distribution in part b is a reflection of the distribution in part a in the line X = 5. d When p = 0.5 the distribution is symmetric, when p = 0.2 it is positively skewed and when p = 0.8 negatively skewed.
3E 8C
4B 9E
5A 10 B
Short-answer questions (technology-free) 1a
6 a, b
2A 7C
b
32 81
3 0.40951
c
16 27
√ 3 5 b 5 b 4 p(1 − p)3 5 a (1 − p)4 c 1 − (1 − p)4 d p 4 e 1 − (1 − p)4 − 4 p(1 − p)3 4a2
6 120 5 p(1 − p)4 7 1 − (1 − p)5 5 8 16 32 9 625
d
65 81
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November 8, 2005
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Answers
1 a 0.0173 2
b 0.2131
p
Probability that a batch is accepted
0 0.01 0.02 0.05 0.1 0.2 0.5 1
1 0.9044 0.8171 0.5987 0.3487 0.1074 0.00098 0
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
0.1 0.2
p
1
0.5
3 a 0.0582 b Mean = 0.4, sd = 0.6197, ± 2 = 0.4 ± 1.2394 c Yes 4 0.0327 5 a i 0.0819 ii 0.9011 b i 15 p 2 (1 − p)4 dP = 30 p(1 − p)3 (1 − 3 p) dp 1 b n = 6, p = 6a2 3 c No. in group wearing glasses 0 1 2 3 4 ii
5
6
Theoretical 17.56 52.68 65.84 43.90 16.46 3.29 0.27 frequencies 7 a 0.9139 8 a 0.0735 1 9 ≤q≤1 3
Pr(L i ) 0.35 0.1 Pr(L i+1 ) × = b 0.8 0.65 0.9 Pr(Ti+1 ) Pr(Ti ) 5 a 0.214 b 0.096 i 0.57 0.47 Pr(A1 ) Pr(Ai+1 ) = × c 0.43 0.53 Pr(E i+1 ) Pr(E 1 ) i 0.536 ii 0.5236 6 a 0.007 b 0.398 i Pr(Ri+1 ) 0.43 0.16 Pr(R1 ) c = × 0.57 0.84 Pr(Fi+1 ) Pr(F1 ) i 0.781 ii 0.219
4a
Exercise 16B
y
b 0.04145 b 0.5015
c 10.702 c 27
Exercise 16A 1 a 0.36 b 0.26 0.65 0.44 2a b 0.5492 0.35 0.56 Pr(Wi ) Pr(Wi+1 ) 0.7 0.4 × = 3a 0.3 0.6 Pr(L i+1 ) Pr(L i ) b 0.517
769
67.7 59 1a i ii 32.3 41 0.7150 0.7126 b 0.2850 0.2875 70.31 67.7 c i ii 29.69 32.3 194.8 204.568 2a i ii 185.2 175.432 0.6223 0.5396 b 0.3777 0.4604 211.015 204.568 c i ii 168.985 175.432 0.2 714 0.23 3a i ii 0.7 286 0.77 0.3038 0.41 b i ii 0.6962 0.59 ⎡ 17 ⎤ ⎡2⎤
iii
70.31 29.69
71.4195 28.5805 211.015 iii 168.985
iii
218.078 161.922 0.280 435 iii 0.715 965 0.280 624 iii 0.719 376 iii
⎢ ⎥ ⎢ ⎥ ii ⎣ 36 ⎦ iii 4 a i ⎣3 1⎦ 19 36 ⎤ ⎡ 107 ⎡ 33 ⎤ ⎥ ⎢ ⎢ ⎥ b i ⎣8⎦ ii ⎣ 192 ⎦ iii 85 5 ⎡ 192 ⎤ ⎡8 ⎤ 0.636 0.28 5 a i ⎣0.04⎦ ii ⎣0.1596⎦ iii 0.2044 0.68 ⎤ ⎤ ⎡ ⎡ 0.4024 0.34 b i ⎣0.48⎦ ii ⎣0.2782⎦ iii 0.3194 0.18 ⎤ ⎤ ⎡ ⎡ 0.2966 0.80 c i ⎣0.19⎦ ii ⎣0.1251⎦ iii 0.5783 0.01 0.85 0.2 6a 0.15 0.8 b 186 179 at Surfside at Bayside, 0.6 0.3 b 57 200 7a 0.4 0.7 0.92 0.88 b 91.7% 8a 0.08 0.12
0.5162 0.4838 0.5160 0.4840 ⎤ 0.4718 ⎣0.1664⎦ 0.3618 ⎤ ⎡ 0.4761 ⎣0.1673⎦ 0.3566 ⎤ ⎡ 0.4808 ⎣0.1679⎦ 0.3513 ⎡
Answers
Extended-response questions
Probability that a batch is accepted
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P2: FXS
052161547Xans-4.xml
Answers
770
CUAT018-EVANS
November 8, 2005
10:32
Essential Mathematical Methods 3 & 4 CAS 0.91 0.13 0.09 0.87 b 53.6% at school A, 46.4% at school B 0.5 0.28 10 a 0.5 0.72 b i 0.359 ii 0.359 0.2 0.7 11 a 0.8 0.3 b i 0.6 ii 0.475 12 a 2000 young, 18 000 middle-aged b 6500 young, 3420 middle-aged, 10 080 old ⎤ ⎡ ⎤ ⎡ 60 0.80 0.40 0.35 13 a ⎣ 0.15 0.30 0.30 ⎦ b ⎣25⎦ 15 0.05 0.30 0.35 ⎡ ⎤ ⎤ ⎡ 63.3 65.4 c ⎣21.0⎦ d ⎣20.2⎦ 14.4 15.8 14 a 0.279 b 0.285
9a
Exercise 16C 1a
0.6 0.4
0.5 0.5
b Game (n) Pr(X n = 1|X 0 = 0) Pr(X n = 1|X 0 = 1) 1 0.4 0.5 2 0.44 0.45 3 0.444 0.445 4 0.4444 0.4445 5 0.4444 0.4445 6 0.4444 0.4444 4 5 d , c 0.4444, 0.5556 9 9 0.85 0.95 2a 0.15 0.05 b Day (n) Pr (X n = 0|X 0 = 0) Pr(X n = 0|X 0 = 1) 1 0.85 0.95 2 0.865 0.855 3 0.8635 0.8645 4 0.8637 0.8636 5 0.8636 0.8636 6 0.8636 0.8636 c 0.8636, 0.1364
d
19 3 , 22 22
3 64.3%, 35.7% 0.74 0.14 4a 0.26 0.86 b 338 at store A, 451 at store B c 276 at store A, 513 at store ⎡ 200B ⎤ 18.1818 ⎢ ⎥ 5a b ⎣ 11 ⎦ 31.8182 350 11
0.4167 5 b 0.5833 12 ⎤ ⎤ ⎡ ⎡ 0.2743 0.2743 b ⎣ 0.3521 ⎦ 7 a ⎣ 0.3521 ⎦ 0.3735 0.3735 ⎤ ⎡ 0.2743 c ⎣ 0.3521 ⎦ d All the same 0.3735 0.70 0.25 8a 0.30 0.75 b 105 in Melbourne, 125 in Ballarat 9 80% sad, 20% happy 10 5405 young, 4865 middle-aged, 9730 old 11 28.4% cool, 22.3% mild, 21.6% warm, 27.6% hot
6a
Exercise 16D 1 a 0.0034 3 a 0.0864 4 a 0.1029 c 0.0189 5 a 0.0625 6 a 0.0655 7 a 0.0696 8 a 0.0226
2 0.0154 b 0.0384 b 0.0353 d 0.0005 b 0.0366 b 0.0439 b 0.0700 b 0.0656
Multiple-choice questions 1C 6C
2A 7E
3C 8B
4B 9D
5B 10 D
Short-answer questions (technology-free) 0.2 0.8
1a
b
0.34 0.66
0.64 0.1 b 2a 0.36 0.3 ⎡4 2⎤ ⎢ ⎥ 3⎣7 5⎦ 4 0.34 5 0.36 3 3 7 5 6 59.3% club A, 40.7% club B 7 0.1029
Extended-response questions 1a
20 81
c i
1 9 7 ii 18
b 5 12
d
3 5
P1: FXS/ABE
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052161547Xans-4.xml
CUAT018-EVANS
November 8, 2005
10:32
Answers
0.1
−10
x
0
−5
Exercise 17A
10
5
b 0.190 11 a k = 1000 2 12 a 3 13 a 0.202 14 a 0.45 y 15 a
Chapter 17 11 2− 6 3 a, c
y
10 a
Answers
0.84 0.23 b 356 bus, 317 train 0.16 0.77 c 397 bus, 276 train d 0.0218 0.56 0.26 3a 0.44 0.74 b i 36.95% in Camberwell, 63.05% in Hawthorn ii 37.14% in Camberwell, 62.86% in Hawthorn
2a
771
b 0.5 17 b 30 b 0.449 b 0.711
1 y
x
0
2.5 2
1
1
iii e− 2
b i 1 − e− 2 ii e−1 f (x) 16 a
1.5 1
2a
0.5 x 0.5
1
–2
5 b Pr(X < 0.5) = 16 4a1 b 0.865 y 5a
ba=
–1
x
0
1
2
1 4
0.4
Exercise 17B
0.2
1 1 a F(x) = x if 0 < x ≤ 5 and 0 if x ≤ 0 and 5 x >5 3 b 5 2 2 a F(x) = 1 − e−x if x ≥ 0 and 0 if x < 0
x 1 2 3 4 5 6 7 8 9 10 11
b 0.259 6 b i 0.024 7 a 0.005 y 8a
ii 0.155 b 0.007
y
k
x
0
bk=1 9a y
b e−4
x
0
–1
1
c
3ak =
3 4
b 0.406
3 2
0
x 0.2 0.4 0.6 0.8
1
c 0.0182 1 48
b
Exercise 17C 1 1 2 c d Does not exist b 2 3 3 2a1 b 2.097 c 1.132 d 0.4444 3 a 0.567 b 0.458
1a
1
1 36
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P2: FXS
052161547Xans-4.xml
Answers
772
CUAT018-EVANS
November 8, 2005
10:32
Essential Mathematical Methods 3 & 4 CAS 2 ,B=3 9 6a2 b 1.858 7a1 b 0.5 8 a 0.632 b 0.233 c 0.693 9 0.1294 10 2.773 minutes 2 11 a f (x) = 12x 2 (1 − x), 0 ≤ x ≤ 1 b 3 12 2 13 a 1 b1 c1 14 a 0.714 b 0.736 15 a 12 b 12 √ 19 − 1 b 16 a 0.4 c1 6 (−kx + 1) 17 a ke−kx , e−kx k y c i,ii,iii 5A=
40
2
bk =
1 9
d 4.5
Exercise 17E 1 1300 2 a 0.708, 0.048 b $98.94, $0.33 3 a 0, 5.4 b 3, 0.6 x2 c g(x) = if −3 ≤ x ≤ 3 and 0 otherwise 18 4 a 7.5, 8.5 b (1.669, 13.331) 5 35, 10
Multiple-choice questions 1B 6B
2D 7C
3D 8E
4A 9A
5E 10 A
Short-answer questions (technology-free)
1.5 1 0.5 x
0
1
2
3
d The graph of e−x is dilated by a factor of from the x-axis and from the y-axis.
1
1a2 b 0.21 1 2 a = ,b = 2 3 3 mean = median = mode = 2 1 1 b 4a 2 2 y 5a
c
1 2
2 , 16 3 9
Exercise 17D 1 a 0.630
c 0.44
b 0.909
c 0.279 √ 2 1 , sd(X ) = b Var(X ) = 18 6
2 a 0.366
y
3a
b 1.386
x
0
1 loge 9 5 a 0.366 6 0.641 7 a 0.732
b E(X ) = 3.641 Var(X ) = 4.948 b E(X ) = 0.333 Var(X ) = 0.056 4 2 b E(X ) = Var(X ) = 3 9 16 b c 2.21 3√ b2 5
8 a 0.0004 3 9a 3 4a y 10 a
1 2
1
x
5 16 2 2 16 6 a k = 12 b x = c Pr X < = 3 3 27 1 2 3 d Pr X < |X < = 3 3 16 2 − 1 e b 7 a 1 − e−2 e23 1 1 b e4 − e4 8 a e2 1 c f (x) = , 1 < x < e x 9 (320, 340) 10 (246, 254) b Pr(X < 0.5) =
0.5
4a
0
Extended-response questions
3k
−2 81 1 2a 4 1a
x 1
2
3
4
5
6
x 2 − 200x + 10 000 b 700 hours c 810 000 √ 5− 5 1 8 b c , 5 3 15
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CUAT018-EVANS
November 8, 2005
10:32
Answers
5 c 32 2 2 d b 3 3 2 b E(X ) = 2, Var(X ) = √ 3 4 5 d 5
7 a 25 1 8ak = 4 3 c 4
f (x)
–6 –5 –4 –3 –2 –1 0
x
7 a = 0, = 3 f (x) b
–9 –8 –7 – 6 –5 – 4 –3 –2 –1 0
1 2 3 4 5 6 7 8 9
x
Chapter 18 Exercise 18A 1c 2
μ1
3a1
μ2
∞
1 − 1 b i E(X ) = x √ e 2 −∞ 3 2 ii 2
c i E(X 2 ) =
∞
x−2 3
1 − 1 x2 √ e 2 3 2 −∞ iii 3
ii 13 4a1 2 ∞ 1 x+4 − 1 2 5 b i x √ e dx −∞ 5 2 ii −4 2 ∞ 1 x+4 − 1 2 2 5 c i x √ e dx 5 2 −∞ ii 41 iii 5 5 a = 3, = 10 b
f (x)
–30 –20 –10 0
x 10 20 30 40
6 a = −3, = 1
2
x−2 3
dx
2
dx
8 a Translation of 3 units in the positive direction of the x-axis; dilation of factor 2 from the y-axis; dilation of factor 12 from the x-axis b Translation of 3 units in the positive direction of the x-axis; dilation of factor 12 from the y-axis; dilation of factor 2 from the x-axis c Translation of 3 units in the negative direction of the x-axis; dilation of factor 2 from the y-axis; dilation of factor 12 from the x-axis 9 a Translation of 3 units in the negative direction of the x-axis; dilation of factor 13 from the y-axis; dilation of factor 3 from the x-axis b Translation of 3 units in the negative direction of the x-axis; dilation of factor 2 from the y-axis; dilation of factor 12 from the x-axis c Translation of 3 units in the positive direction of the x-axis; dilation of factor 12 from the y-axis; dilation of factor 2 from the x-axis
Exercise 18B 1 Mean = 135, sd = 5 4 2 Mean = 10, sd = 3 3 16% 4 a 68% b 16% c 0.15% 5 21.1 and 33.5 6 68% of the values lie within one standard deviation of the mean; 95% of the values lie within two standard deviations of the mean; 99.7% of the values lie within three standard deviations of the mean. 7 2.5% 8 a 16% b 16% 9 a 68% b 16% c 2.5% 10 a 95% b 16% c 50% d 99.7%
Answers
b
10 3 b E(X ) = 6, Var(X ) = 4.736 7 b $22.13 4a 25 8 5c= ,4 3 6 b E(X ) = 2, Var(X ) = 0.2 3 a Median = 6, IQR =
773
P1: FXS/ABE
P2: FXS
052161547Xans-4.xml
Answers
774
CUAT018-EVANS
November 8, 2005
10:32
Essential Mathematical Methods 3 & 4 CAS 5 b− c 1.5 4 12 a −1.4 b 1.1 c 3.5 13 Michael 1.4, Cheryl 1.5; Cheryl 14 Biology 1.73, History 0.90; Biology 15 a Mary: French 1, English 0.875, Mathematics 0 Steve: French −0.5, English −1, Mathematics 1.25 Sue: French 0, English 0.7, Mathematics −0.2 b i Mary ii Mary iii Steve c Mary
11 a 0
Exercise 18C 1 a 0.9772 b 0.9938 c 0.9938 d 0.9943 e 0.0228 f 0.0668 g 0.3669 h 0.1562 2 a 0.9772 b 0.6915 c 0.9938 d 0.9003 e 0.0228 f 0.0099 g 0.0359 h 0.1711 3 a 0.6827 b 0.9545 c 0.9973 4 a 0.0214 b 0.9270 c 0.0441 d 0.1311 5 c = 1.2816 6 c = 0.6745 7 c = 1.96 8 −1.6449 9 −0.8416 10 −1.2816 11 −1.9600 12 a 0.9522 b 0.7977 c 0.0478 d 0.1547 13 a 0.9452 b 0.2119 c 0.9452 d 0.1571 14 a 9.2897 b 8.5631 15 a c = 10 b k = 15.88 16 a 0.7161 b 0.0966 c 0.5204 d c = 33.5143 e k = 13.02913 f c1 = 8.28; c2 = 35.72 17 a 0.9772 b 0.9772 c 10.822 d 9.5792 e c2 = 10.98; c1 = 9.02
Exercise 18D 1 a i 0.2525 ii 0.0478 iii 0.0901 b 124.7 2 a i 0.7340 ii 0.8944 iii 0.5530 b 170.25 cm c 153.267 3 a i 0.0766 ii 0.9998 iii 0.153 b 57.3 4 a 10.56% b 78.51% 5 mean = 1.55 kg; sd = 0.194 kg 6 a 36.9% b 69 7 a 0.0228 b 0.0005 c 0.0206 8 1004 ml 9 a small 0.1587 medium 0.7745 large 0.0668 b $348.92
10 a i 0.1169 ii 17.7 b 0.0284 34 b 11 a 0.0228, 0.1587 3
Multiple-choice questions 1A 6E
2C 7D
3B 8C
4B 9A
5E 10 D
Short-answer questions (technology-free) 1 a 1 − p b 1 − p c 2p − 1 2 a a = −1 b b = −1 c 0.5 x −8 3 (x, y) → , 3y 3 q 1− p 4a b1−q c p 1−q
5 a Pr Z < 12 b Pr Z < − 12 c Pr Z > 12
e Pr − 12 < Z < 1 d Pr − 12 < Z < 12 d 0.68 c 0.16 b 0.5 6 a 0.84 d 0.02 c 0.32 b 0.34 7 a 0.16 d 0.68 c 0.15 b 0.19 8 a 0.69 9 Best C, worst B
Extended-response questions → high 1 > 63 [56, 62] → moderate [45, 55] → average [37, 44] → little < 37 → low 2 3.92 3 a i 0.1587 ii 0.9747 iii b 53 592 c 3.7 × 10−11 4 a 3.17 × 10−5 b False c c1 = 13.53, c2 = 16.47 5 0.0802 6 0.92% 7 a 0.9044 b 5.88 d 0.2651 e $17.61 8 a = 0, = 2.658 9 a = 60.058; = 0.2 10 a 0.1056 b 0.0803
0.0164
c 9.044 b 0.882 b 10% c 0.5944
Chapter 19 Multiple-choice questions 1E 6D 11 B 16 E 21 A 26 C 31 D 36 E 41 B
2D 7D 12 D 17 A 22 D 27 E 32 C 37 C
3A 8C 13 C 18 A 23 C 28 B 33 E 38 A
4B 9E 14 B 19 D 24 A 29 B 34 A 39 B
5E 10 C 15 E 20 B 25 D 30 C 35 B 40 B
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CUAT018-EVANS
November 8, 2005
10:32
Answers
9 16 3 a 0.27 c i 0.6 d 0.4995
2 $0.76
1b
4a P =
b 0.3025 ii 0.27 0.75x − 0.5s, 0.5s − 0.25x,
b $5.95 c E(P) =
s
x ≤s x >s
(0.75x − 0.5s) p(x)
x = 24
+
30
(0.5s − 0.25x) p(x)
x=s+1
d 27
1 1 ii 36 6 41 4 ii b i 100 25 121 c 600 ⎡3 1⎤ 11 ⎥ ⎢ i 7 a ⎣4 2⎦ 1 1 16 4 2 1 b 3 0.85 0.06 8a 0.15 0.94
iii
6a i
1 6
20 120 b E(X ) = 49 49 6180 c Var(X ) = 2401 3 18 a i 0.2 ii 0.7 iii 0.125 iv 160 128 ii b i 0.360 15 625 c 0.163 08 19 a i 0.0105 ii 0.0455 1149 b 0.4396 c 1909 20 a i = 4.25 ii = 0.9421 iii 0.94 iv 0.9 b i Binomial ii 18 iii 1.342 iv 0.3917 17 a C =
Chapter 20
1a i
1 , 8 2
ii Minimum
x (60 − 5x) 12 iii Maximum area is 15 cm2 . 2 a p = 1, q = 3, k = 2 b i m = −2 ii y = −2x 3 + 10x 2 − 14x + 6 y iii b ii A =
ii
85 128
6 4 2
b i 51% Dr Laslett, 49% Dr Kildare
7 , 64 3 27 x
0
c Yes, Dr Laslett during the 9th year 9 a 0.6915 b 0.1365 10 a 0.0436 b 0.2667 c 183 d 59 271 11 a i 0.1587 ii 511.63 b 0.1809 12 a −1 b 28.847 c 0.5276 1 13 a i ii X : 0.6915, Y : 0.5625 8 iii E(X ) = 10, E(Y ) = 10.67, machine 1 3 b 4 16 1 ii iii 0.8281 iv 0.7677 14 a i 3 2500 b 0.9971 15 a i Pr(X > 80) = 0.98 –
Pr(X > – 104) = 0.04 80
104
ii = 92.956, = 6.3084 b i 16.73% of sensors ii 81◦ C 16 a 0.1056 b Machine should be set at 1027.92.
3a
1 2 3 (1, 0) Local minimum at (1, 0) 7 64 Local maximum at , 3 27
y a , a2 2 4
0
a
x
a3 square units 6 a3 2a 2 2a 2 c i y= iii square units ,y = 162 9 9 3 1 4ay=− x+ 2 2 dy b i = cos − 2 sin d 1 ii = tan−1 = 26.57◦ 2 iii (26.57, 2.2361), exact values √ 1 , 5 tan−1 2 b
Answers
Extended-response questions
775
P1: FXS/ABE
P2: FXS
052161547Xans-4.xml
Answers
776
CUAT018-EVANS
November 8, 2005
10:32
Essential Mathematical Methods 3 & 4 CAS iv r = v
√
5, = 63.435◦ y
iii
y
1 2
√5, Tan–1
y = f (–x) y = f (x)
y = sin θ + 2 cos θ
2 y = 2cos θ
0 –2
(90, 1)
1
2
–1 A reflection in the y-axis
θ
90
0
x 1
–1
iv
y = sin θ
y
c i Q (2 sin , 2 cos ) iii = 74.4346◦
y = f (x)
1
t
5 a i f (t) = −100e− 10 (t 2 − 30t + 144)
x
0
t 10e− 10 (t 2
− 50t + 444) ii f
(t) = b i t ∈ (6, 24)√ √ ii t ∈ (25 − 181, 25 + 181) ≈ (11.546, 35) iii t ∈ (11.546, 24) 6 a i A = x 2 − 5x + 50 ii (0, 10) A iii
1
2 y = –f (x)
–1
A reflection in the x-axis y
v
y = f (x+ 2) y = f (x) 0
x
–2
(10, 100)
1
(0, 50)
2
A translation of 2 to the left
b f does not have an inverse function as it is not one-to-one. y c
(2.5, 43.75) x
0
y=x
iv Minimum area = 43.75 cm2 1 b i f (x) = (10 − x)x 2 ii y
y = g–1(x) 2 y = g(x) 0
x
2
5 , 25 2 f (x) =
0
1 (10 – x)x 2
x
10
c AYX : OXYZ : ABY : CBYZ = 1 : 2 : 2 : 3 y 7a i y = f(x) 0
x 1
2
d i Gradient = 15 1 ii Gradient = 15 8 a i 0.995 ii x = 0.2 1 b i h(x) = (x − )2 − 1 2 ≈ −0.989 98 9 a S = (60x − 6x 2 ) cx =5 d
–1
ii
S
y = 2f(x)
–2
b 0 < x < 10
(5, 150)
A dilation of factor 2 from the x-axis y
0
y = f(2x)
y = f(x) x 0 –1
1
2
A dilation of factor 1 2 from the y-axis
5
10
1 sin 1 − cos ii BQ = sin
10 a i OP =
x
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052161547Xans-4.xml
CUAT018-EVANS
November 8, 2005
10:32
Answers
6
19 b i h = c i
3 y = 2πa r
y = 400 cosθ
c
6
, 200
6
ii Coordinates of local minimum 2 √ √ 2 5 3 3 0.6 a, a 2 √ 0.6 + 3 0.6 3
√ + 3
T
20 a 0.0023 b
π, a π + √3 6 6 2a
r
0 θ
π 2
Maximum at
Q s − 1 −1
π, πa 2 2
3 4
Pr(Q = q) θ
0
a . Minimum value for T is 2 12 a iii x = 1 or x = k − 2 b i b = 3 − 2a; c = a − 2 ii h = a − 2 iii a = 0, b = 3, c = −2 iv a = −1, b = 5, c = −3 1 13 a Z = (7t − 2t 2 ) 2 7 , 49 b Z 4 16
7 2
c Maximum value of Z = 4 3 ii 15 8 8 27 ii b i 125 125 b 15 a k = 2 a b b x+ b i y= 2a 2 d S1 : S2 = 27 : 37
(37.3333, 289.0798)
40, 284 4 9
1, 11 18
x
ii The maximum value of P is 289.0798 tonnes. x (56 − x) cA= 90 A i
t
28, 8 32 45
7 49 , when t = 16 4
14 a i
16 a i 0.9332
1 4 √
3 3 s c E(Q) = s − 1 sd(Q) = 4 4 21 a 0.091 21 b 0.2611 c 0.275 dP 1 22 a = (112x − 3x 2 ) dx 90 P b i
0
Z = 1 (7t – 2t2) 2
0
3 2 y = π 2a + 5r r 3
π, 100π 2
y = 200θ
5r 2 2a 3 + r 3
2 y = 5πr 3
y = 200(θ + 2 cos θ)
0
ii S =
y
6
400
3a 3 − 2r 3 3r 2
1, 11 18
38 iii 125 ii
−a b , 2 4
40, 7 1 9 x
0
32 ii Amax . = 8 tonnes/man, x = 28 45 y 23 a i
ii 0.0668 iii 0.1151 iv 0.1151
b i 33.3% ii 866.4 iii 199.4 √ 17 90 − 8 3 metres from A towards E 18 a i y =−e−n x+ e−n n + e−n 1 1 b i n 1− e e
(0, 2)
y = 2(x – 1)2
ii x = n + 1 ii e : e − 2
0
x (1, 0)
Answers
√ 3 d Minimum value of S = 2 √ 2 3−3 AP = units 3 11 b y π, 200 π + √3
777
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Answers
778
CUAT018-EVANS
November 8, 2005
10:32
Essential Mathematical Methods 3 & 4 CAS ii
7 d 2 e
y
2
y
y = –16x + 2
3 +2 (x + 1)2
y= (0, 5)
y=2
0
x
1 8
iii
y
–1
(0, 2) x (–7 + 5√2, 0)
(–7 – 5√2, 0)
2 − 3k −(2 − 3k)2 + 2(k + 2) , k+2 k+2 2 2 i k: −2 < k < ii 3 3 14 ∪ {k: k < −2} iii k: 0 < k < 9 14 iv {k: −2 < k < 0} ∪ k: k > 9 c k < −2 14 14 ii 0 < k < d i k = 0 or k = 9 9 24 a x = 2 − loge 2 dy b i = −2e2−2x + 2e−x dx 1 iii 2, − 2 ii x = 2 e y iv b
(0, e2 – 2)
c i 1250 square units 14 375 ii square units 18 36 875 square units iii 18 27 c = 6 d A π , 12√3 6
16
0
y=
–
θ
T = 21 + 79e–0.02t
2e–x x
0
T = 21
2, – 1 e2
1 c − 2,0 e 25 a
d Average rate of change = −1.6◦ C/minute. e i 2.0479◦ C/minute ii −0.8826◦ C/minute 3 bb=4 30 a 16A c
(0, ba2 + c) (0, ba2)
y = b(x + a)2
(0, 1)
y = x2
(–a, c) y = (x + a)2 –a
t
0
y
y = b(x + a)2 + c
c
π 2
√ The maximum value of A is 12 3. 28 a = 5.0290, = 0.0909 b $409.28 1 79 loge ≈ 0.02, A = 79 29 a k = 10 63 b Approx 2 : 44 p.m. T c (0, 100)
e2–2x
1
26 a i y = 50 ii y = x − 25 25 1 c=− ba=− 15 3
(–7, 100) y = –2x2 – 28x + 2
0
0
y= 1 x2 x
0
x
A dilation of factor 3 from the x-axis A translation of 1 unit in the negative direction of the x-axis A translation of 2 units in the positive direction of the y-axis
0
1
p
d i 0.076 ii 0.657 e i A ( p) = −20 p(1 − p)3
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052161547Xans-4.xml
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November 8, 2005
10:32
Answers ii
A'
0
1
y
P
Answers
ii
(50, 11.5)
11.5
(37.5, 4) x
0
1 , – 135 4 64
1 4 iv Most rapid rate of change of probabilities 1 is occurring when p = . 4 31 a 91.125 cm b [0, 15] c V = 0.64 (4.5 − 0.3t)3 d h is a one-to-one function. 1 10t 3 h −1 (t) = 15 − 3 dom of h −1 = [0, 91.125] y e iii p =
b (2.704, 10), (22.296, 10) y c (62.5, 30)
(0, 15)
(125, 15)
0
(250, 15)
x
(187.5, 0)
(x − 10) d i h(x) = 15 + 15 sin 125 ii y
(0, 91.125) (72.5, 30) y = h(t)
(0, 15)
(260, 15) y = h–1(t)
0
(15, 0)
(10, 15) t
(91.125, 0)
32 a 0.065 36 b i 0.6595 c i 23.3% 33 a i 0.32 b 0.64 c i 0.043 95 34 a, b y
0
ii 0.198 14 ii c = 0.1075 ii 0.18 iii 0.5 ii 0.999
iii
y = ex + 1 x y = ex
y= 1 x x
0
1 x c y = +e x 1 dy = − 2 + ex dx x d ii 2 loge x < 0, ∴ x ∈ (0, 1) y iii y = 2 loge x
7 128
x
(197.5, 0)
36 a k = 4
√ √ 13 10 − 2 2 ii b i E(X ) = iii 6 4 12 c 0.1857 2 37 a k = 2 a a a2 b E(X ) = , Var(X ) = 18 √3 √ 6−4 2 c d a = 1000( 2 + 2) 9 1 cy= x − loge 10 38 a 7 10 d ii 36.852 n+1 39 a k = n + 1 b E(X ) = n+2 n+1 c (n + 2)2 (n +3) n+1 1 d Median = e Mode = 1 2 √ dy x 40 a b(0, 2 6) = √ dx x 2 + 24 c Even d y
10 0
x (0.7, –0.7)
y=–x
y=x
2√6 x
y = –x
y = – 5x + 10
iv (0.7, 3.4) 35 a i e = 12.5, g = 15, d = 37.5, a = 7.5, b = 7.5
e y = −5x + 10
779
f 14 units/second
P1: FXS/ABE
P2: FXS
052161547Xans-4.xml
Answers
780
CUAT018-EVANS
November 8, 2005
10:32
Essential Mathematical Methods 3 & 4 CAS √ √ 35 h 12 loge ( 7 − 1) − 2 7 + √ 2 3 x 3x b V = a r = 41 9 3 60 dx 1 2 dV = d = x c dt x 2 dx 3 1 8 1 53 23 3 180t 3 f ex = 3 42 a i 0 ii −0.6745 iii 0.6745 iv 1.3490 v 99.3% vi 0.7% b i ii − 0.6745 iii + 0.6745 iv 1.3490 v 0.9930 vi 0.7%
Appendix A
d
n+1
i4
i=1
3a b
n i=1 5
i=0
d
4
x i · 25−i = 32 + 16x + 8x 2 + 4x 3 + 2x 4 + x 5 x i × 2i × 36−i = 36 + 2 × 35 × x + 22 × 34 × x 2 + 23 × 33 × x 3 + 24 × 32 × x 4 + 25 × 3 × x 5 + 26 × x 6 (x − xi )i = (x − x1 ) + (x − x2 )2
i=0
Exercise A1 1 63 4 a 5040 5 a 120 6 18 7 a 5 852 925 8 100 386 9 a 792 10 a 200 d 462
2 26 b 210 b 120
3 336
c
b
4 i=1 5
c
i 3 = 1 + 8 + 27 + 64 = 100 k 3 = 1 + 8 + 27 + 64 + 125 = 225 (−1)i i = −1 + 2 − 3 + 4 − 5 = −3
i=1
(k − 1)2 = 0 + 1 + 4 + 9 = 14
k=1
4 1 1 (i − 2)2 = (1 + 0 + 1 + 4) = 2 3 i=1 3 6 h i 2 = 1 + 4 + 9 + 16 + 25 + 36 = 91
g
i=1
2a
n i=1
i
x 2−i . 22−i
d
(2x)3−i . 3i
i=0
c6
i=1
f
i=0
3
Exercise A3
b 336 b 75 e 81
5 1 1 i = (1 + 2 + 3 + 4 + 5) = 3 d 5 i=1 5 6 i = 1 + 2 + 3 + 4 + 5 + 6 = 21 e 4
x 5−i . 3i
+ (x − x3 )3 + (x − x4 )4 5 b x 5−i . (−3)i
b 1 744 200
k=1
5
i=0 2 i=0
Exercise A2 1a
4a
5
5 1 i i=1
xi = x + x2 + x3 + · · · + xn
i=0
6 c
e
b
11 i=1
xi
c
10 1 xi 10 i=1
1 a x 6 + 36x 5 + 540x 4 + 4320x 3 + 19 440x 2 + 46 656x + 46 656 b 32x 5 + 80x 4 + 80x 3 + 40x 2 + 10x + 1 c 32x 5 − 80x 4 + 80x 3 − 40x 2 + 10x − 1 d 64x 6 + 576x 5 + 2160x 4 + 4320x 3 + 4860x 2 + 2916x + 729 e 64x 6 − 1152x 5 + 8640x 4 − 34 560x 3 + 77 760x 2 − 93 312x + 46 656 f 16x 4 − 96x 3 + 216x 2 − 216x + 81 g x 6 − 12x 5 + 60x 4 − 160x 3 + 240x 2 − 192x + 64 h x 10 + 10x 9 + 45x 8 + 120x 7 + 210x 6 + 252x 5 + 210x 4 + 120x 3 + 45x 2 + 10x + 1 2 a −960x 3 b 960x 3 c −960x 3 d 192 456x 5 e 1 732 104x 5 f −25 344b7 x 5 16 7 x 3− 4 −336 798x 6 243 5 (−x + 1)11 = −x 11 + 11x 10 − 55x 9 + 165x 8 − 330x 7 + 462x 6 − 462x 5 + 330x 4 − 165x 3 + 55x 2 − 11x + 1 6 a 40 b −160 c −80 d 181 440 e 432 f 1080 7 83 026 944 8 −768
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