Answers
CHAPTER 1 Coordinate geometry
c
Exercise 1A — Sketch graphs of
y = axm + bx–n + c where m = 1 or 2 and
n = 1 or 2
1 a
e
h
2 a
3 a
y = 2x b y = x − 2 c y = 2x
d y=3−x
y = x2 f y = x − 2 g y = 3 − 4x2
y = 2x + 1 The vertical asymptote is x = 0 in each case.
B
b A
c A
d E
b
y
y
–6 –4 –2 0
0 2 4 6 x
–6 –4 –2
–5 y = 2– + 2x
x
–10
d
y
5
–10
–3 –2 –1 0
y = 4–x + x – 2
y
5
2
1
0 1 2 3 x
y
–4 –3 –2 –1 0 1 2 3 4 x
x
2
7 b
A = 2π r2
A=
–6 –4 –2 1
02 4 6 x
–20 y = 8–x + 2x + 1
800
8 a A = 50x + --------- + 32, x > 0
x
b A (cm2)
A = 50x + 32
c B
b
Exercise 1B — Reciprocal graphs
1 a
0
5
10 x
–10
x-intercepts (–3, 0), (3, 0),
no turning points
1
y y = –––––
2
x –4
10
1
y = x––––
+2
y
20
10
0
–3 –2 –1
b
1
2 3 x
–10
Turning point (0, – 1–)
4
–3
–2 –1 0
–10
1 x
1A
➔
–6 –4 –2 0 2 4 6 x
No x-intercepts, turning
points (–2, 8), (2, 8)
10
–10 –5
x (cm)
y
y = 9–x – x
20
32
0
dC
x
10
c 432 cm2 when
x = y = 4 cm
(4, 432)
c Max (− 2 , − 4 2 ), min ( 2 , 4 2 )
d Max (1, 0) e Min (−1, 2), (1, 2) f Min (2, 1)
g Max (− 1--2- , 0)
h Max (−2, −7), min (2, 9)
b A
256π
––––
r
r (cm)
4 a Max (−1, −4), min (1, 4)
b Max (−2, −6), min (2, 2)
y y = 16
––2 + x2
Asymptotes A = 2π r 2
and r = 0; stationary
point is (4, 96π)
c 96π cm2
A (cm2) (4, 96π )
y
20
x-intercepts (1, 0),
turning points (–1, 4), (1, 0)
x-intercepts (–0.62, 0), (1, 0),
(1.62, 0), turning point
(1.26, –0.11)
0
0 1 2 3x
y = 2 – x – 1–x
2
4
y = 1–x – 4x2+ 3
4
y=x–2
y = x + ––
–2
x2
h
–10
5 aA
6 a
h
0 1 2 3 4x
y=x–2
1
y = ––
+ x2
x2
x
x-intercepts (–2, 0), (1, 0),
turning point (–2, 0)
3
y y = ––
1 +x–2
2
–3 –2 –1
2 4 6
–5
y=2–x
4
y = ––
x2
–2
10
–3 –2 –1
0
–10 –8 –6 –4 –2
3 , 3 2)
turning points ( 2–––
2
g
y
5
4
y = x – ––
x2 + 3
1 2 3 x
0
–2
y
f
1
y = –x + 8x2
–10
x-intercepts (– 1– , 0),
2
g
y
10
–2
f
4
–5
–10
x-intercepts (1, 0), (2, 0),
turning points ( 2 , 2 2 – 3),
(– 2 , –2 2 – 3)
5
0
–1.5 –1–0.5 0.5 1 1.5 x
–5
–5
y = 4–x + 2x
y
–3 –2 –1
0
x
–5–4–3–2–1 1 2 3 4 5
x
0 2 4 6x
–6 –4 –2
e
e
2 4 6 x
2
y = 3 – x2 – –x
2
No x-intercepts, turning
points (– 1– , –4), ( 1– , 4)
2
2
y
10
y
10 y = x – 3 + –x
–5–4 –3 –2–1 01 2 3 4 5 x
–10
10
5
d
y = 1–x + 4x
10
2
c
y
answers
599
Answers
1B
answers
600
Answers
y
20
c
10
d
1
y = ––––––
x 2 + 2x
i
1
x – 4x + 3
y = –––––––––
2
y
20
0
–2 –1 0 1 2 3
–10
Turning point (–1, –1)
x
–1 0
–10
1
3 4 5 x
2
f
y
20
1
–––––
y = –2x
+5
10
–4 –3 –2 –1
–2
–6
k
–1
–10
0 1
l
y
1
––––––
y y = –x
2 + 2x
20
2
x
3
(–1, 3)
h
y
1
y = ––––––
4x2 – 9
2
1
2 3
x
–2 –1 0
1
g(x) = ––––
x–4
20
f(x) = x – 4
10
6
(3.95, 5.06)
4 (2, 4)
1
2 (2, – )
4
4
b
0 1 2 3 4 5 6 7 8
–10
c (x + 2)2 + y2 = 9
4
d
y
20
f(x) = x2 + 3x + 2
1
g(x) = ––––
3–x
10
0
–10
2 a (–3, 2) and 1
y
20
3
3
–3 –2 –1 0 1
–10
f(x) = 3x + x2
f
–2 –1
–4 –3
f(x) = 3x2 – 8x – 3
1
g(x) = ––––––
3x + x2
(– 3–2, – 9–4)
10
h
1
(x – 4)
g(x) = –––––––
2
0
−1
x
5
--2
x
( 4– , – 25
––)
3
3
2
1
0
−1
−2
−3
f(x) = (x + 3)2
y
10
5
1
g(x) = ––––––2
(x + 3)
–6 –5 –4 –3 –2 –1
–5
0 x
x
4
2
d (–5, 2) and 2
y
4
(−5, 2)
( –52, 0)
1 2 3 4 5 x
e (–2, 3--2- ) and
f(x) = (x – 4)2
0 2 4 6 8 10 x
–10
4 6 8
4
y
20
3x – 8x – 3
3
25 )
Turning point ( 4– , – ––
3
3
Turning point (– 3– , – 9– )
g
1
g(x) = –––––––––
2
–10
–5
c ( 5--2- , 0) and
(2, 1)
1
y
5
02
–6 –4 –2
–5
01 2 3 x
2
2
−4 −3 −2 −1 0
−1
y
10
10
5
(−3, 2)
x
2
2
y
b (2, 1) and 2
y
Turning point (– 3– , –4)
e
d x2 + (y – 1)2 = 3
1
f(x) = 3 – x
b (x + 4)2 + (y + 3)2 = 36
y
1
10 g(x) = x–––––––––
2 + 3x + 2
1 2 3 4 5 6 x
e Min. h = 0.25 cm when
length = width = 2 cm.
6 x(cm)
1 a (x – 3)2 + (y – 1)2 = 4
Turning point (2, – 1– )
c
c ∼ 5.06 cm
Exercise 1C — Graphs of circles and
ellipses
f(x) = x 2 – 4
0
x
–2–1 1 2 3 4 5 6 7
–5 (2, –4)
x
0 2 4
–4 –2
1
y g(x) = x–––––
2 – 4x
10
5
c E
h(cm)
Turning point (5– , 8)
9
y
d
2 3 4 x
1
3
1
b h = ---------------------( 4x – x 2 )
4 a A = 4x − x2
1
x + 2x + 4
g(x) = ––––––––––
2
–3 –2 –1 0 1 2 3 x
Turning point (–1, 1– )
b D
–10
Turning point (0, – 1–)
2 a
–2x + 5x – 3
10
1
0
–3 –2 –1
–1
3 a B
1
y = ––––––––––
2
y
1 2 3 x
f(x) = x2 + 2x + 4
3
(–1, 1–3 ) 1
2
g
1
x +x+4
g(x) = –––––––––
1–
2
2
1
g(x) = –––––
x2 + 2
–6 –4 –2 0 2 4 6 x
Turning point (0, 1– )
Turning point (1, 1)
y
4
f(x) = x2 + 2
2
1.5
1
0.5
10
0 0.5 1 1.5 2 2.5 3 3.5 4 x
–10
f(x) = x2 + x + 1–4
0
–4
Turning point (2, –1)
e
y
8
6
4
2
2 4 6 8 x
f(x) = –x2 + 4x – 4
–4 –2
–2
10
j
1
–x + 4x – 4
g(x) = ––––––––––
2
y
−7 −5
5
--2
−3
f (– 3--2- , 2) and
0
x
3
--2
y
y
4
3
(−2, –32) 2
1
−5 −4 −3 −2 −1 0 1
−1
2
4
( − —32 , 2 )
3
2
1
x
−3 −2 −1 0
x
6 a
3 a x2 + y2 = 4; [–2, 2], [–2, 2]
b x2 + y2 = 9; [–3, 3], [–3, 3]
2
2
c (x – 2) + (y – 1) = 16; [–2, 6], [1, 5]
d (x + 3)2 + (y – 2)2 = 2;
[–3 – 2 , –3 + 2], [2 – 2 , 2 + 2 ]
4 a D
b C
5 a
c A
2
2
(x – 2)
(y – 3)
y ––––––
+ ––––––
=1
9
5
4
(–1, 3) 3
2
1
–1
(2, 3)
(2, 1)
01 2 3 4 5
c
2
y ––––––
(x – 2)2 ––––––
+ (y + 4)
9
4
x
y
(y – 4)2
–– + ––––––
=1
9
4
(0, 6)
2
(x + 1)2 ––
3 ––––––
+ y =1
9
5
2
(2, 0)
(–4, 0) (–1, 0)1
–5–4 –3 –2–1–1 0 1 2 3 x
–2
–3
(–1, – 5 )
e
y
5
4
(–2, 3) 3
2
1
–2
g
(x – 3)2
(y – 3)2
f
h
5 (–1, 2 + 5 )
4
3
(–1, 2) 2 (3, 2)
(–5, 2)
(–1, 2 – 5 ) 1
–5 –4–3 –2–1–1 0 1 2 3 x
i
(x + 2)2 (y + 1)2
–––––– + –––––– = 1 y
9
7
–6–5 –4 –3 –2–1–1 0 1 2 x
(–5, –1) (–2, –1)–2 (1, –1)
–3
(–2, –1 – 7 ) –4
y
4
2
(4 – 12, 0)
–2
=1
y
10
8
6
4
2
g
(2, 4)
2
1
(y + 2)2
(x – 4)2
+ –––––– = 1
y ––––––
4
25
4
(4, 3)
2
y
6 (0, 6)
4
2
(5, 0)
(0, 0)
(–5, 0)
–6 –4 –2–2 0 2 4 6 x
i
2
k (x + 3)2
y2
–––––– + –– = 1 y
4
12
(–3, 2 3 ) 4
(4, –4)
y (x – 2)2
––––––
1
16
(y + 2)2
+ –––––– = 1
4
j
8 a
(x – 3)2
(y – 1)2
–––––– + –––––– = 1
9
36
y
8 (3, 7)
6
4
(0, 1) 2 (3, 1) (6, 1)
–1–20 2 4 6 8 10 x
–4
–6 (3, –5)
y
8
6
4
(0, 2) 2
0
–2
–4
–6
(x – 3)2
(y – 2)2
–––––– + –––––– = 1
9
36
(3, 8)
(3, 2)
(6, 2)
2 4 6 8 10
x
(3, –4)
l (x+2) 2
(y+3) 2
––––– + ––––– =1
4
16
y
(–2, 1) 2
0
–8 –6 –4 –2
2 x
–2 (0, –3)
(–4, –3)
–4
(–2, –3)
–6
(–2, –7) –8
b C
9(x – 5)2 + 16(y + 1)2 = 144
y
4
(5, 2)
2
–2 0 2 4 6 8 10 x
–2
(5, –1) (9, –1)
(1, –1)
–4
(5, –4)
b
16x2 + 25y2 = 400
y
4 (0, 4)
2
(–5, 0)
(0, 0)
(5, 0)
–6 –4 –2 0 2 4 6 x
–2
–4 (0, –4)
1C
➔
0 (2, 0)
x
–3–2–1
–1 1 2 3 4 5 6
(–2, –2) –2
(6, –2)
–3 (2, –2)
–4
–5 (2, –4)
h
(–3, –2 3 ) –4
7 a B
(x – 2)2 (y + 3)2
–––––– + –––––– = 1
4
9
(2, 0)
–1–1 0 1 2 3 4 5 x
–2 (0, –3)
(4, –3)
–3
(2, –3)
–4
–5
–6
(2, –6)
2
(–1, 0)
(–5, 0) (–3, 0)
0
–8 –6 –4 –2
2 x
–2
(4, 4)
(4 + 12, 0)
(4, 0)
01 2 3 4 5 6 7 8 x
2
–1–1 0 1 2 3 4 5 6 x
(2, –1)
–2
–3 (2 – 2 2 , –1)
(2 + 2 2 , –1)
–4
(2, –4)
2
x2 ––
––
+ y =1
(0, 2) 25 4
y
(0, –6)
(y + 1)
(x – 2)
+ –––––– = 1
y ––––––
8
9
2
(2,
2)
1
(2, 1)
f
x
y2
x2
–– + –– = 1
25 36
(8, 4)
(0, 0) (5, 0)
(–5, 0)
–6 –4 –2 0 2 4 6 x
–1
–2
(0, –2)
10 (0, 9)
8
6
(–3, 4) 4
(3, 4)
2 (0, 4)
0
–6 –4 –2–2 2 4 6 x
(0, –1)
(0, 0) (3, 0)
0 2 4 x
–4
–6
–6 –4 –2 0 2 4 6 8 10 x
y
–2
2
(x – 2)2 ––––––
––––––
+ (y –9 4) = 1
36
(2, 7)
y
(y – 4)2
–– + –––––– = 1
9
25
0
–2 2 (4,4–2) 6 8 10
(6, –2)
–4 (2, –2)
–6
(4, –7)
–8
y2
(x – 4)2
–––––– + –– = 1
12
16
–4
l
(–4, 4)
j
(–2, –1 + 7 ) 2
1
k
4
–2–1–1 0 1 2 3 4 5 6 x
–2 (2, –1)
(–1, –3) –3 (2, –3)
(5, –3)
–4
–5
–6 (2, –5)
y
(x + 1)2 (y – 2)2
–––––– + –––––– = 1
16
5
–4 –3 –2 –1–1 0 1 2 3 4 x
–2
–3 (0, –3) (3, –3)
–4
–5
–6
–7
(0, –7)
(–3, –3)
–4 (0, – 15)
(3, 4)
y (x – 2)2 (y + 3)2
–––––– + ––––––
9
(3, 1)
0 2 4 6 8 10 x
y
(y + 3)2
–– + –––––– = 1 1 (0, 1)
9
16
d x2
y2
x2
y
–– + –– = 1
9
15
4 (0, 15 )
(–3, 0)
–4 –2
–4 –3 –2 –1 0 1 2 3 4 x
(8, 3)
(3, 3)
6
5
4 (0, 4)
3
2
1 (0, 2)
(–3, 4)
–––––– + –––––– = 1
4
(3, 5) 25
b x2
2
e
d x2
y
(–1, 5 )
4
c
=1
–2–1–1 0 1 2 3 4 5 6 x
–2 (2, –2)
–3
(5, –4)
(–1, –4) –4
–5 (2, –4)
–6
(2, –6)
(5, 3)
2
d E
b
(2, 5)
2
(y + 3)
(x – 2)
+ –––––– = 1
y ––––––
4
9
(2, 0)
x
0
–1
–1 1 2 3 4 5 6
–2
(4, –3)
(0, –3) –3
(2, –3)
–4
–5
–6
(2, –6)
answers
601
Answers
1C
answers
602
c
Answers
16(x – 2)2 + 25y2 = 400
y
4 (2, 4)
d
4(x – 1)2 + 16(y + 3)2 = 64
y
(–3, 0)
(7, 0)
(2, 0)
–4 –2 0 2 4 6 8 x
–2
e
(–3, –3)
9 a
y
8
6
4
2
(5, –3)
–3
(1, –3)
–4
–5
(1, –5)
–6
(2, –4)
f
16(x – 2)2 + 9(y + 5)2 = 144
y
–2
–2
–4
(–1, –5)
–6
–8
–10
0 2 4 6 8 10 x
(2, –1)
(2, –5)
(5, –5)
(2, –9)
(4, 2) (6, 2)
0
2
4 6 8 10 x
–2
–4
(4, –4)
–6
36(x – 4)2 + 4(y – 2)2 = 144
x2 – ––
y2 = 1
––
9 9
5
y = – 4–3 x
10 x
f
y=x
y
y = – 1–2 x
2
dy
2
5
x2 – ––
y2 = 1
––
4
4
–4
–10 –5
10 x
–10 –5–2 0 2 5
h
y 3y + 2x = 13
10
9
y
9(x + 3) + 7y = 63
y
(–3, 3) 4
4(x – 4) + 9(y – 3) = 36
6
(1, 3)
(–3 – 7 , 0) (–3, 0)
–8 –6 –4 –2 0 2 x
–2
(–3, –3) –4
2
0
(7, 3)
(4, 3)
k
(4, 1)
2
4 6
x
8
2
2
2
2
( x + 2) ( y – 3)
c ------------------- + ------------------- = 1, top half of ellipse
1
4
[−3, −1] and [3, 5]
3 D
5 a
c
c x = 3 + 4 cos θ, y = −2 + 4 sin θ
d x = −3 + 5 cos θ, y = 1 + 3 sin θ
c E
y = – 2–3x y y = 2–3 x
6
4
x2 – ––
y2 = 1
2
––
9
–10 –5 –3 0 3 5
–2
–4
4
10 x
25x2 – 16y2 = 400
b
y = 5–4 x
(4, 0)
0 5 10 15 x
y 9x2 – 16y2 = 144
y = – 3–4 x
3
10 y = –4 x
5
(–4, 0)
–15 –10 –5
b
d C
y = – 3–2x y y =
6
4
2
–6 –4 –2 0 2
–2
–4
d
y=x
9x2 – 25y2 = 225
y = – 3–x y y = 3–x
5
5 10
5
(–5, 0)
(5, 0)
–15–10 –5 0 5 10 15 x
–5
–10
3–x
2
x2
y
y = –x
10
(4, 0)
0 5 10 15 x
–5
5
x2 – y2 = 25
(–5, 0)
(5, 0)
–15 –10 –5 0 5 10 15 x
–5
Exercise 1D — Graphs of hyperbolas
2 a
(x – 1)2 (y + 1)2
–––––– – –––––– = 1
25
9
–5
b x = 2 cos θ, y = 3 sin θ
b B
5
(–4, 0)
–15 –10 –5
11 a x = 3 cos θ, y = 3 sin θ
1 a A
2
10
y = – 5–4x
y
–10 –5 0 5 10
(–4, –1)
(6, –1) x
–5
4 A
y
x
10
5y + 3x = –2 5y – 3x = –8
5
(x – 3) (y – 3)
– –––––– = 1
–10 ––––––
25
4
2
3y + 5x = 3 5
(–3, 1)
(3, 1)
–10 –5 0 5 10
3y – 5x = 3 –5
l
5y – 2x = 9
2
( x – 1) ( y – 2)
b ------------------- + ------------------- = 1, ellipse [−2, 4] and [−2, 6]
9
16
4
x2 (y – 1)2
y ––
– –––––– = 1
9
25
10
5
(–2, 3)
(8, 3)
15
–10
x
–5 0 5 10
–5
x
y
10 a ----- + ----- = 1, ellipse [−2, 2] and [−3, 3]
4 9
5y – 2x = 11
25
3y – 5x = –4
5
y
5y + 2x = 21
10
x2 – ––
y2 = 1
––
16 4
j
(–1, 2)
(5, 2)
–10 –5 0 5 10 15 x
–5
(4, 5)
4
2 (–3 + 7 , 0)
2
3y + 5x = 16
2
10 x
5
(7, 3)
(–3, 3)
–10
15
x
–5 0 5 10
–5 (x – 2)2 (y – 3)2
––––– – ––––– = 1
4
(x – 2)2 (y – 2)2
–––––– – –––––– = 1
9
25
5
y
5y + 2x = 19
10
5
(–1, 3)
(5, 3)
–5
0
5 10 15 x
–10
–5
3y – 2x =5 –10 ––––––
(x – 2)2 (y – 3)2
– –––––– = 1
i
4
0
–5
–5
(0, –4)
2
y = 1–2 x
5
10
c
10 x
–5
y = –x y
g
x2 – ––
y2 = 1
––
9
16
–10 –5 –3 0 3 5
–5
e
0
–4 –2
2 4 6 x
(–4, –1)
–2 (0, –1) (4, –1)
–4
4
y y = –3 x
5
9x2 + 16(y + 1)2 = 144
y
2 (0, 2)
(2, 2)
d
y=x
–10 –5 –3 0 3 5
25(x – 3)2 + 9(y + 2)2 = 225
y
4 (3, 3)
2
x
0
–2 2 4 6 8 10
(0, –2)–4 (3, –2) (6, –2)
–6
–8 (3, –7)
b
(4, 8)
y = –x y
–3 –2–1–1 0 1 2 3 4 5 x
–2 (1, –1)
2
–4
c
e
y2
–– – –– = 1
4
9
4 6x
3y + 4x = –7 y 3y – 4x = –23
10
5
–15–10 –5 0 5 10 15 x
(–1, –5)
(5, –5)
–10
–15
16(x – 2)2 – 9(y + 5)2 = 144
f
y
3y – 5x = –21
10
3y + 5x = 9 5
0 5 10 15 x
–10 –5
(0, –2) –5
(6, –2)
–10
25(x – 3)2 – 9(y + 2)2 = 225
g
y
h 4y + 3x = –4 y
10
y + 3x = 14
5
(2, 2)
–10 –5 0
–5
10
5
y – 3x = –10
(6, 2)
10 15 x
–15 –10 –5
j
y
(–3 – 7, 0)
–10 –5
(–3 + 7, 0)
0 5 10 x
–5
7y + 3x = –9
–10
2
2
5
(1, 3)
(7, 3)
0 5 10 15 x
–10 –5
–5
3y + 2x = 17
4(x – 4)2 – 9(y – 3)2 = 36
9(x + 3)2 – 7y2 = 63
2
x
y
6 a ------ – ----- = 1, hyperbola, (−∞, −4] ∪ [4, ∞), R
16 9
x y
b ----- – ----- = 1, hyperbola, (−∞, −2] ∪ [2, ∞), R
4 9
2
2
c ( x – 1 ) – ( y – 2 ) = 1 , hyperbola, [2, ∞), R
2
f
–2
1
g --------------- + --------------3 – 2x 3x + 1
3 – -----------------2 h ----------x + 2 ( x + 2 )2
2
( x – 2) ( y – 4)
d ------------------ – ------------------- = 1, hyperbola, [5, ∞), R
9
25
7 a x = 2 + 2 sec t and y = −1 + 3 tan t
b x = −3 + 5 sec t and y = 1 + 4 tan t
2
3
c ----------- + -----------x–2 x+5
3
2
d --------------- + ----------2x – 1 x – 4
4
2
e ----------- – ----------x+1 x–3
f
5
2
g --------------- + ----------3x + 2 3 – x
2 – -------------1 h -------------2x + 3 6 – 5x
i
1
--3
(3x + 7) + 1--3- (3x − 4) j
4 E
2
3
5 a ------------ + ------------------2x + 1 ( x + 1)
–2
3
----------- + --------------x – 4 2x + 5
1
--2
(2x − 1) + 1--2- (4x + 5)
2
5
3
2
c 2x + 1 – ----------- + --------------- d 4 + ------------ + ----------x – 2 2x + 5
x+3 x–4
1
2
2
3
e x – 5 – --------------- + --------------- f 3x + 2 + --------------- – -------------2x – 1 3 – 2x
4x + 3 3x – 1
–3
2
c ------------ + ------------------x + 3 ( x + 3 )2
5
4
d --------------- – ---------------------2x – 1 ( 2x – 1 ) 2
–2
3
e ----------- + ------------------4 – x ( 4 – x )2
f
8 a D
b A
Exercise 1F — Sketch graphs using
partial fractions
1 a
1
2
6 a ----------- + -----------x–3 x+2
3
2
b ----------- – ----------x+7 x–1
4
3
c ------------ + ----------x+2 x–4
2
3
d ----------- – ----------x+5 x–5
b
y
2
y
1
y = ––––
x–2
1
–4 –2
0 2
– 1– –1
2
4
c
6 x
d
y
2
y = – ––––
x–1
4
3x – 1
6
01 2
(0, 1–2)
x
1
f
y
4
2
–4 –2
–3 –2 –1
–2
g
y
2
(–31–2, 0)
2 x
4
6
x
x
(0, –2 1–3)
b
y
2
1
–2–1
–1 0 1 2
–2
0 234
(0, –4 2–3)
–8
–2x – 7
y = ––––––
x+3
–6 –4 –3 –2 0 2
–2
–4
–4
3 x
(3 1–2 , 0)
4
– 1–
2
2
–4x + 14
y = –––––––
x–3
8
6
(0, 5)
0 1
2–1
3
y
4x + 1
5
(– –4 , 0)
1
(–3, 0)
–2 –1 0
–1
e y = 4x
+5
–––––
y = –––––
3x – 2
2
–4
2 a
x
–8 –4 0 4 8
–1
2
–2
x+4
3–
4
–2
y
4
–4 –2
3
y = ––––
2
1
–2
6
7
--------------- – ---------------------5 – 2x ( 5 – 2x ) 2
1
5
1
1
g ----------------------- + -------------------------- h ----------------------- + -------------------------3 ( 3x + 2 ) 3 ( 3x + 2 ) 2 2 ( 2x + 1 ) 2 ( 2x + 1 ) 2
3
7
h 4x + ----------- + ------------------x – 3 ( x – 3 )2
5
8 g 6 + ----------- – -------------x – 5 2x + 5
3 B
4
1
b ----------- – ------------------x – 2 ( x – 2 )2
4
3
b 3 + ----------- – -----------x–3 x+4
3
1
7 a 2 + ------------ + ----------x+2 x–1
Exercise 1E — Partial fractions
1 a a = 1, b = 2 b a = 2, b = −1 c a = 3, b = 4
d a = −2, b = 3 e a = 4, b = −3 f a = −2, b = 2
g a = 3, b = 6 h a = −3, b = −2
1
2
4
3
b ----------- + -----------2 a ------------ + -----------x+2 x+1
x–3 x+3
6
7
----------- + -----------4–x x+2
–4
5
--------------- + ---------------------2x – 1 ( 2x – 1 ) 2
i
y 3y – 2x = 1
10
7y – 3x = 9
5
5
4
e --------------- – ----------2x + 1 x – 3
9x2 – 16(y + 1)2 = 144
10
2
–5
(4, –1)
x
5 10 15
–10
36(x – 4)2 – 4(y – 2)2 = 144
i
5
0
(–4, –1)
–10
4y – 3x = –4
answers
603
Answers
x2 –
x +1
y = ––––––––
x –1
x
y
2
1
3–
2
x
–2 –1 0 1 2
–2
2
+ 4x + 3
––––––––––
y = 2x
x+2
➔
1D
1F
answers
604
Answers
c
d
y
e
x2 – 7x + 11
y = –––––––––
x–3
y
5
2
4
2
x + 4x + 5
3 y = –––––––––
x+1
–3 –11 0
x
–1 1
e
–)
(0, –11
3
–2
0
y
–2x2 + 7x – 11
y = –––––––––––
x2 – 4x + 3
x
–32–3
–4
x
0 1 3
–2
(–5, –2)
2 34
–5
Chapter review
y
Multiple choice
2
– 4x – 3
–––––––––
y = 3x
x–2
4
2 0
1 2 3
–1
–2
3 a
x
b
y
c
2
x
d
5x – 2
y
x
–2 0 1
2x + 2
y = –––––––––––
(x – 2)(x + 4)
y = –––––
x2 – 4
1–
2
y
2 C
3 D
4 B
5 C
6 C
7 B
8 A
9 D
10 C
11 B
12 E
13 C
14 A
15 B
Short answer
–3
––––––––––
y = (x
+ 2)(x – 1)
1
1–2
10
–4 1
– –4 –1
y
1 A
1 x = 0 and y = 3 + 2x
2 As x → −∞, y → −2x from
above. As x → ∞,
y → −2x from above.
As x → 0, y → ∞.
y
20
3 – 2x
y = ––
2
10
x
–10 –5 0
–10
3x – 5
y = ––––––––
2
x – 2x – 3
10 x
5
2
1–3
x
0 2
–2
x
–1 0 3
3 a Turning point (1, −3) b x-intercept ( – 3 2 , 0)
c
y
y = –x2 – 2–x
5
e
3
– 2
–3 –2 –1 0
–5
y
–x + 2
y = ––––––––––
2x2 + 7x + 3
0
–3
(1, –3)
–10
x
– 1–
2
1 2 3 x
4
5
y
y
4
y = –x – ––
2
4 a C
b
5 C
x
3
( – 4, 0) 1 0
–2 –1–1 1 2
D
1
x
0
x
5
– 5
–3
6 D
(0,– –5)
1
y = –––––
2
–5
x –5
(2, –3)
7 a
b
y
y
x +5
y = ––––––––
x2 – x – 2
2
2
0 1
–1
(–7, 1)
x
2x2 + 2x – 2
y = ––––––––––
x2 – 1
–2 1–2
6 a
d
y
x
4
( –3 , –3)
–3x2 + 6x + 14
x –x–6
y = ––––––––––––
2
5
(0, 1– )
6
0
–2
y
5 (0, 5)
– 1–2 0 2–3
– 7x + 6
x
7 a
–2 0 –21–3 3
–3
x2
6
1
–1 0 2
–2
x
30x2 – 10
y = –––––––––
6x2 – x – 2
1
y y = –––––––––
2
x – 7x + 6
y=
12
–5
c
b
y
4
6
8 x
4
( 7–2 , – ––
)
25
0 12
–5
4
6
x
–– )
( 7–2 , – 25
4
y
(5, –1)
b
y (x + 1)2 y2
––––––
––
+
=
1
–2 0 2 4 6 8 10 12 x
(–1, 5) 6
36
25
–2
4
–4
2
(11, –6)
(–1, –6) –6
(5, 0)
(–7, 0) (–1, 0)
(5, –6)
–8
–8 –6 –4 –2–2 0 2 4 6 x
(5, –11)
–10
–4
–12 (x – 5)2 (y + 6)2
––––– + ––––– = 1
(–1, –5) –6
25
36
c
d 3y + 5x = 3 y
y
1 (1, 7 – 2)
3y – 5x = 3
10 a
–1 –1 0 1 2 3 4 x
(3, –2)
(–1, –2) –2 (1, –2)
–3
–4
–5 (1, – 7 – 2)
2x + 5x – 3
2
–10
x2
––
9
f 3y + 2x = –6 y
3y + 4x = –4 y 3y – 4x = 4
10
(x + 1)2
–––––
9
g
–
c
x – 4x + 4
(−5, 4)
i
−11
x
4
2
−5
y
2 × 10
4
10
4
(1, 1)
j
y
1
(x + 2) y −2 =
y −2 = −—
2
−10 −8 −6
01 x
4
b y = 2x + -------------------x(1 – 4)
8
c S = ------------------- + 4x – 4x 2
x(1 – x)
120 2 + x 2
b t w = --------------------------2
90 – x
c S b = 90 – x and t b = -------------4
(1, −5)
(−4, 2)
−2
c r ≈ 31.7 cm, min. S = 9466.1 cm2
2 a Length = y − 2x, width = 2 − 2x
3 a Sw =
x
−1 01
3
(3, −2)
(−1, −2) −2
(1, −2)
−5
r (cm)
0 10 20 30 40 50 60 70 80 90
4
d The minimum surface area is 33 m2.
e The length of roll to cut off is 17 m.
y
1
x
2
Analysis
1 b
S (cm 2)
10
1
0
y = 2x + 1
1
1–4
0
4(x + 3)2 – 9y2 = 36
(2, 3)
1–
2
y
3y – 2x = 6
5
3
x
1
– 5–3
2
2x – 7x + 4x + 5
y = ––––––––––––––
2
h
y
–3
2
(y – 1)
=1
– ––––––
25
(–6, 0)
(0, 0)
–20 –15–10 –5 0 5 10 15 x
–5
y2
–– = 1
16
0
9x – 33
y = ––––––––––
x2 – 7x + 12
5
(–4, 0)
(2, 0)
–15–10–5 0 5 10 15 x
–5
x
3 4
11
– ––
4
10
5
( 5–3 , 0)
0
3
e
–3x + 5
y y = ––––––––––
2
(3–3, 0)
5
(–3, 1)
(3, 1)
–8 –6 –4 –2 0 2 4 6 8 x
–5
7(x – 1)2 + 4(y + 2)2 = 28
b
y
10
answers
605
Answers
1
—
(x
2
+ 2)
120 2 + x 2 90 – x
d t = --------------------------- + -------------- , x = 69.28 m
2
4
e 74.46 s
4 a
y
x
6
4
2
3
2 (0, 2)
(−2, 2)1
−4
−2
0
−1
−2
2
120 2 + x 2
y1 = x + 2
4
−2−20
−4
y2 = x4_
x
2
8 a (x + 1)2 + (y − 2)2 = 9
2
2
( x – 4) ( y – 3)
b ------------------- + ------------------- = 1
1
4
2
2
( x + 2) ( y – 1)
c ------------------- – ------------------- = 1
4
9
6
3
9 a ----------- + ----------x–3 x–4
1
2
b --------------- – ----------2x – 1 x + 3
x
y
(2, 6)
y1 = x + 2
6
4
y2 =
2
−2 0
(−2, −2) −2
−4
2
4_
x
x
1F
➔
1
c 2x + 1 + ------------------( x – 2 )2
b No x-intercept; no y-intercept; stationary points at
(−2, −2) and (2, 6); asymptotes at x = 0 and y = 0
(see graph).
y = x + 2 + 4_
1F
answers
606
Answers
c
2
y = x − 2x + 4
x −2 y=x
y
6
3
b
d
2
e 2
1
f − ------
g 1
2 3
h − --------3
i 1
3
j − 2--------3
k −2
l
3
3
(4, 6)
4
2
−2 0
2 4
−2 (0, −2)
−4
x
5 a D
x=2
d Replace x with x − 2 (translation two units to right)
e
2
y = x − 2x + 4 y = x
x −2
y
3
c − -----2
4 a
6
4
1
(4, _6) y =
21
(0, − _2)
−2 0
−2
2
x−2
x2 − 2x + 4
b E
3
c C
d A
1
c − ------
6 a − 1--2
b − 3
2
7 a − -----2
b −1
c
8 a − --23
5
b − -----3
c
9 a − -----516
b
3
2
5
------2
5
c − ------------
231
------------16
231
x
4
10 a
−4
b − 5--3-
4
--3
c − 5--4-
x=2
Exercise 2B — Graphs of reciprocal
trigonometric functions
CHAPTER 2 Circular
(trigonometric) functions
1 a
y
y = 2 tan x
Exercise 2A — Reciprocal trigonometric
functions
2
0
1
sin x
cos x
tan x
cosec x
sec x
cot x
3
------2
1
------3
2
2
------3
3
a
1
--2
b
2 10
------------7
3
--7
2 10
------------3
7 10
------------20
7
--3
3 10
------------20
c
4
--5
3
--5
4
--3
5
--4
5
--3
3
--4
d
5 34
------------34
3 34
------------34
5
--3
34
---------5
34
---------3
3
--5
e
12
-----13
5
-----13
12
-----5
13
-----12
13
-----5
5
-----12
f
2
--3
5
------3
2 5
---------5
3
--2
3 5
---------5
5
------2
2 a 36.9°
b 18.2°
c 60.9°
d 30.0°
e 73.4°
f 55.0°
g 53.2°
h 9.6°
i 7.2°
b
3 a 5 3
d
7 3
---------3
b 4 3
c
8
--3
f
5 6
---------2
e
2π x
3π
—
2
y
y = tan 2x
1
0 π –π
–
8
c
4
5π
π —
3π
–π —
2 4
4
3π —
7π
—
2 4
y
y = tan –2x
1
0
15 2
------------2
π
–π
2
–π
4
–π
2
π
3π
—
2
2π x
2π
x
d
2 a
y
y
y = tan(x – –4π )
y = 3 sec x
3
1
–π
2
0
–3
e
0
3π
—
2
π
2π x
–1
b
y
–π
2
π
π
50
—
4
π 2π x
70
—
4
y
y = –21 cot x
y = cosec (x – –3π )
–1
2
1
π
0 π π
– –
4 2
f
–π
4
607
answers
Answers
y
π
30
—
2
2π x
y = 4 cosec x
0
–1
–π
3
π
c
4π
—
3
2π x
y = sec (x + –2π )
y
4
π
–π
2
0
–4
3π 2π
—
2
x
1
g
π
–π
2
0
–1
3π
—
2
2π x
y
d
y
y = cot(x – –6π )
y = cosec 2x
1
–π
2
0
–1
π
3π
—
2
2π x
0 –π
π
20
—
3
6
h
π
70
π —
6
e
y
π
50
—
3
2π
x
y = sec (x + –6π)
y
y = cot 3x
–π
3
0
π
20
—
3
π
π
40
—
3
1
π 2π x
50
—
3
0
–1
i
y
–π
3
f
π
4π
—
3
x
110
—π 2π
6
20
–π
y = cosec (x + —)
3
y
y = sec –2x
5π
—
6
1
–1
0
π
2π
x
1
0
–1
–π
3
π
4π
—
3
2π x
➔
2A
2B
answers
608
g
Answers
f
y = cot (x – –4π )
y
–π
4
0
–1
3 a C
4 a B
5 a
π
50
—
4
π
π 2π
70
—
4
b A
b C
x
c D
c E
y = tan (x + –4π ) + 1
y
y = sec (x + –3π) – 1
y
–π —
– 56π
– –3π –20 –6π –3π
g
y = 3 cosec (x – –3π) + 2
y
5
1
b
π x
d B
d D
2
π – –π π
30
–π – —
–
4 2 –4
2π
—
3
0 –π π 30π π x
4 –2 —
4
2
2π – –π π 0
–π – —
3 – –6 –1
3
–π
3
π x
π
y y = 2 tan (x – –2 ) – 1
–π – –π
2
π
–π
2
0
–1
x
h
y = 2 cot (x – –3π ) + 1
y
–3
1
c
y = 2 sec (x – –6π )
y
5π
–—
6
–π
–π
2π – –π – –π
–—
3 6
3
0 π– 2π 5π π x
—
3 —
3 6
2
π
– –2π – –3
0π
–
6
–2
d
y
–π
2
2π π x
—
3
i
y = 3 cosec (x + –3π )
y = sec (2x – –2π )
y
–π
– –2π
1
–1 0
–π
2
πx
3
–π
– –3π
0
–3
2π π x
—
3
j
e
y = cosec (3x + –2π ) – 1
y
y = cot (x – –4π ) + 2
y
1
–π
2
1
3π
–π – —
–π
4 –4
0 –π
4
0 –π –π
π – –π
6 3
—
– 20
3 –1
3
–2
π x
π
20
—
3
πx
k
y = 1–2 cot (2x – –3π ) + 1
y
π –3
—
( 70
24 , 2 )
11–2
1
2π
–π – —
3
6 3
l
– –2π
–π
–π
2
1
–1 0
–π
n
e
1
--2
πx
b −cos A
e −sin A
4 a sin A
d cos B
–π
0
– –2π
–π
2
πx
Exercise 2C — Trigonometric identities
1 a −1
b cos2x
d sin x − cos x e 1
2
2 a 0.6
d 1.25
3 a
d
− 1--22 3
---------3
4 a −2.5
d
29
− --------5
2
b 1.33
e 0.75
b
3
------2
e
3
− -----3
29
b − 5-----------29
e
c sin2x
f cosec x
c 1.67
7 a
5
------2
c – 3
2 29
------------29
29
---------2
4 15
c − -----------15
5
b − -----5
c
b
9 19
------------19
4
--3
-----e − 63
65
2 5
---------5
19
---------------c − 10
19
c cot A
f cosec B
6– 2
-------------------4
e
c
6+ 2
-------------------4
f
3–2
6– 2
-----b − 12
13
c
5
--3
-----d − 13
5
-----f − 56
65
g
33
-----56
--------h − 836
123
b −0.24
c 0.6
b
11 a –cos 2x
1
--2
sin 2x
d 0.49
c tan x
e −1
f 1 − sin 2x
12 a −0.28
b −0.96
c
13 a − 4--3-
b
d
15
b − --------15
b
6– 2
10 a 0.92
c
b C
b −cosec y
e −sec B
h cosec A
b C
8 a 2– 3
d
c tan B
f tan A
14 a
1
--4
4
--5
sin 2x
b
3
--5
24
-----7
c − --35-
4
--5
c
11 5
------------25
5–2 5
d ------------------10
2C
➔
19
8 a − --------10
5 a E
6 a cos x
d −tan A
g −tan y
7 a D
9 a
5 D
15
6 a − --------4
tan x – 1
f -------------------1 + tan x
b cos2 2x − sin2 2x
A
A
d 2 sin --- cos --2
2
3
- sin 2x
cos 2x − ( -----2 )
2 a 2 sin 3x cos 3x
2 tan 4x
c ------------------------2
1 – tan 4x
B
B
2 tan 5 A
f ---------------------------e cos2 --- − sin2 --4
4
1 – tan 2 5 A
3 a sin(2x + y)
b cos 2x
c sin 2B
d 1
tan y ( tan 2 x + 1 )
2 tan A ( 1 + tan 2 B )
e – ----------------------------------------f ---------------------------------------------1 – tan x tan y
1 – tan 2 A tan 2 B
y = –cot x
y
b x = 1.82, –1.82
d x = –1.14, 1.14
f x = –2.96, 0.18
1 a sin 2x cos y + cos 2x sin y
b cos 3x cos 2y + sin 3x sin 2y
tan x + tan 2y
c ------------------------------------- d sin y cos 4x − cos y sin 4x
1 – tan x tan 2y
πx
y = –cosec x
y
π 5π
e x = --- , -----6 6
π 7π
c x = --- , -----4 4
Exercise 2D — Compound and double
angle formulas
1
–1 0
m
3π 5π
d x = ------ , -----4
4
10 a x = 0.29, 2.85
c x = –0.93, 2.21
e x = –1.14, –2.00
y = –sec x
y
π 5π
b x = --- , -----4 4
π 5 π 7 π 11 π
f x = --- , ------ , ------ , --------6 6
6
6
2π π x
—
3
0 –π –π
– π–3
4π 5π
9 a x = ------ , -----3
3
answers
609
Answers
2D
answers
610
Answers
b 0.39
15 a 0.92
c 0.42
2+ 2
b -------------------2
2– 2
16 a -------------------2
d 2.36
y = 2 sin–1 (x) + –2π
y
b
30
–π
2
3–2 2
c
i [−1, 1]
π 3π
ii – ---, -----2 2
–π
2
Exercise 2E — Inverse circular functions
and their graphs
1 a π
--6
π
b --6
π
e – --3
f π
π
2 a --6
π
3 a --4
1
b --2
π
b --6
π
c – --4
π
d --4
π
g --3
π
c --3
π
c --3
π
h --3
π
d – --6
f
π
– --4
π
g – --3
4 a 0.5
b
3
------2
c 1
π
h – --3
d 1--2-
f π
π
g – --3
π
h – --6
π
j – --2
π
k – --4
5
a A
b D
6
a B
b D
b i [−2, 0]
ii
c i R
ii
d i [−1, 1]
ii
e i [−2, −1]
ii
f i R
ii
g i [ – 2, 2 ]
ii
h i [0, π]
ii
1
1 -----⎛ – ------, -⎞
⎝ 2 2⎠
–π
2
0
d
y
π π
– ---, --2 2
k i [0, 1]
π π
l i – ---, --6 6
i [1, 3]
π π
ii – ---, --2 2
0
1
2
x
3
– –2π
e
f(x) = 3 tan–1 (x) + –4π
y
2π
π
--6
70
π
—
4
i R
5π 7π
ii ⎛ – ------, ------⎞
⎝ 4 4⎠
π
π π
– ---, --2 2
[0, π]
π
⎛–π
---, ---⎞
⎝ 2 2⎠
π
0, --2
[0, π]
π
⎛–π
---, ---⎞
⎝ 2 2⎠
[0, π]
f
–π
2
1
--2
]
ii [0, 2π]
f(x) = 2 cos–1 (x + 1–2)
20
—π
3
0 1 1
–
–11–2 –1 – 1–
2
g
π π
– ---, --2 2
x
2
[
i − 1--2- ,
y
π
–
2
π
–
3
π
–
6
i R
π π
ii ⎛ – ---, ---⎞
⎝ 2 2⎠
ii
π
f(x) = 21–3 sin–1 (2x) + –3
0
– 1–2
h
y = tan–1(x + 1)
[
i − --32- ,
y
2π
π
ii [0, π]
y
x
–π – 50
π
—
4
π π
– ---, --4 4
ii [−1, 1]
0
x
2
y = sin–1 (2 – x)
ii
– –2π
1
–π
2
ii R
j i
–1
l
i [0, 2]
ii [0, π]
y = cos–1(x – 1)
0
ii
8 a
y
π
1x
–π
4
7 a i [0, 2]
i i
c
0
– –2π
d 0
5π
e -----6
π
e --3
3π
i -----4
–1
π
–
2
x
–2
–1
– π–2
]
π
π
---, --6 2
x
1–
2
π
f(x) = cos–1 (x + 2) – 2–
y
–3
1
--2
0
x
i [−3, −1]
ii
π π
– ---, --2 2
Chapter review
Multiple choice
1 B
2
5 B
6
9 D
10
13 D
14
17 B
18
Short answer
Analysis
f(x)
C
E
D
C
C
3
7
11
15
19
D
B
E
A
A
4
8
12
16
20
A
C
A
E
D
b
2 a
1
------3
c −2
2
4
– –π
0 –π
4
3π π x
—
4
0.75
–π
2
π
3π
2
2π
x
a = 1; b = 1
–1 h – 2
f x = cos ⎛ ------------⎞
⎝ h ⎠
π
b ⎛ ---, 2.15⎞ and 2.15 m
⎝2
⎠
d
d 3.09 m2
3
e 4.64 m or 4640 litres
π
π
3 b b = --- ; a = 2 tan -----7
14
π
π
c f(x) = 2 tan ------ cot --- ; 0.22 m
14
7
d −4.14; −0.20
π
y y = cosec (x + –3 ) – 1
2
πx
b g(x) = π cos ------ + π
12
y
dx
d ------ = − 4--3- sin --- e 6.88 m
3
dy
x–4
4 a f (x) = cos−1 ⎛ -----------⎞
⎝ 4 ⎠
2π
–—
3
–π– —
5π
6
1.5
c 2.14 m
4 –2
–2 2
b
b
c x = 1.91, 4.37
h
h
e a = --- ; b = --2
2
2 a a=b=2
y = 2 sec (x – –4π )
y
3π
–π – —
1 a a = 0.75; b = 0.75
0
1 a − --12
answers
611
Answers
– –3π 0 –π –π
–1 6 3
x
π
y
c x = 4 + 4 cos --3
2 3(
( π, –3 –—
3
–2
Technology-free questions (page 107)
1
3 a − ------
b −
π 2π
4 a x = --- , -----3 3
3π 7π
b x = ------ , -----4
4
5 a −2 −
b
3
3
6 a sin x
15
7 a − --------8
8 a
y
π
3
--2
c
π 5π
c x = --- , -----3 3
y
c −cosec x
15
c − --------7
7
--8
π
y = sin–1 (x – 1–2 ) + —
2
d
i [− 1--2- , 3--2- ] ii [0, π]
b
i R
y = tan–1 (2x) – –4π
y
(−3, −16)
3π π
ii ⎛ – ------, ---⎞
⎝ 4 4⎠
y
20
16
y = __2 − x2
x
10
–π
4
0 1
–
2
–π
2
x
−4
−2 0
−10
−20
−30
2
y = −x2
4
x
2E
➔
π
—
– 30
4
x
10
(3, −4)
y = x − 10
−10
b Asymptotes: x = 0 and y = –x2
Intercepts: x = –2, x = 2; no y-intercept
Turning points: none
0 1 1 1 x
1–2
–
2
– –4π
y = x9_ + x − 10
(1, 0) (9, 0)
4-----------------– 158
π
—
2
– 1–2
2 a Asymptotes: x = 0 and y = x – 10
Intercepts: x = 9, x = 1; no y-intercept
Turning points: (3, –4)(–3, –16)
6+ 2
b tan y
b
1 x = 0; y = 2x
7
--6
2E
answers
612
3
Answers
y
−2
11
y = f (x)
y
y = cos(x − π )
4
1
x
y = g(x) 3
y = sec(x − π )
4
3π
4
0
−1
π
7π 2π
4
x
−6
4 a
b
y
y
−1
−4
3
1
5
y=
3
4
(x − 2) − 1
x
12
0.5
(−2, −1)
−2.5
(3, −4)
2
y
y = 3 tan (x − –4π ) +1
(6, −1) x
− _π −2
4
−π
y = − 34 (x − 2) − 1
3π
4
π x
−7
3
1
13 a ------- sin A – --- cos A
2
2
5x – 3
2
3
5 a ---------------------------------- = ------------ + ----------( x + 1)( x – 3) x + 1 x – 3
5
1
x
b ------------------------- = -------------------- + -------------------2
6( x – 5) 6( x + 1)
x – 4x – 5
6 a
b
y
y=
5x − 3
(x + 1)(x − 3)
b tan B
4
14 a --5
15 a 2
y
y=
1
_
6
y= 2
x +1
2
−1
−1
3
x
y= 3
x−3
7 a V = 2x2(45 – x)
b
1
6(x + 1)
x
5
x
y=
2
x − 4x − 5
y= 5
6(x −5)
1
−_
6
12
b -----13
b cos A
c – 33
-----65
c sin x
d – 16
-----65
1+ 3
1+ 3
16 a ---------------- b ---------------1– 3
2 2
π
π
c – --17 a --b π
3
4
18 a [– 1--2- , 1--2- ] b [–1 + a, 1 + a]
π
d – --4
CHAPTER 3 Complex numbers
V
Exercise 3A — Introduction to complex
numbers
0
30
45 x
1 a 3i
e
c Vmax = 27 000 cm3 when dimensions are
30 × 30 × 30 cm.
2400
8 a P = 624 – 6w – -----------w
b
P
3 a −1 + i
e −1 + 2i
4
100
c Pmax = 384 cm2 when w = 20 cm and l = 30 cm.
9 a – 2
24
10 a – -----25
b
24
b – -----7
c – 2
11 i
2 a 9, 5
d −6, 11
g –5, 1
624
1
------3
b 5i
d 1
w
f
c 7i
7i
b 1+i
f −1 + i
4 z = −2 − 3i, w = 7 + 3i
b 15
5 a −5
f 0
e 2
6 4−i
7 a E
b C
8 1
c
10 Solution will vary.
3i
d
i
b 5, −4
e 27, 0
h 0, –17
9 a = 2, b = 5
25
– -----24
g
2
--3
h
6
--5
i
c −3, −8
f 0, 2
c 1−i
g 1 + 0i
d 0 + 0i
h 1 – 2i
c 0
g −9
d −6
h 2
c C
d E
Exercise 3B — Basic operations using
complex numbers
g
h
Imz
0
1 a
b
Im z
0
c
e
1
2
Re z
3
d
Im z
–2 –1 –10
–2
–3
–4
–5
–2 – 6i
–6
0
–1
5 – 2i
2 a 4−i
d 9 − 13i
3 a −12 + 3i
d −25 + 3i
4 a 7 − 23i
d 63 − 37i
5 a 111 + 33i
d 61
6 14 + 52i
8 a −8
d 35
1 − 14i
−12 + 4i
−19 − 8i
−50 − 48i
4 + 45i
−85 − 132i
31 − 8i
−53
7 −3
b −5
e −30
b
e
b
e
b
e
b
e
9 a x = 5, y = −2
c x = 1, y = 5
10 a E
11 a Im z
b B
c −9
f −115
16
b x = 21
------ , y = – -----41
41
d x = −2, y = −3
c C
b Im z
3 + 4i
4
c
0
–2
3
11 Re z
11 – 2i
Re z
d
Im z
Im z
0
0
–2
e
Re z
2
–3 –2 –1–10
1 2 3 Rez
–2
–3
i2z
i3z, –iz
b 5 + 9i
c 3 − 12i
e 5 − 2i
d 7 + 3i
2 Answers will vary.
3
1
--2
f
– 6 + 11 i
+ 1--2- i
4 a −i
7
- +
c − ----25
b −i
−
23
-----29
14
-----29
f
2 5 – 6 2 2 + 15
----------------------- + --------------------------- i
7
7
i
e
43
-----53
+
d
18
-----53
3–i
b ---------10
4 + 3i
c -------------25
5 – 4i
d -------------41
6 (10 + 24i)
– 3 – 2i
e -----------------13
f
23
-----10
b
17 + 9i
8 ----------------2
9
-----10
c
i
i
2+i
5 a ----------5
7 a
26
-----25
17
-----5
3 + 2i
---------------------5
-----d − 16
5
-----e − 14
5
10 −33 + 58i
9 –29
b C
11 a D
12, 13 and 14 Solution will vary.
c B
15 −16
16 a −12 + 11i
c 0
b −30 − 19i
17 Solution will vary.
18 a –4, 16, –64
19 x = −1, y = ± 2
20 a = − --12- , b =
1
--2
21 x = 2, y = 1; x = −2, y = −1
6 Re z
6 – 2i
2 – 4i
–4
f
Im z
Im z
0
0
Re z
−4 − 2i
−9 − 5i
12 + 23i
−41 − 28i
−50 − 13i
176 − 61i
22 − 48i
32 − 126i
c
f
c
f
c
f
c
f
z, i4z, –i2z
1 a 7 − 10i
2
1
0
iz, i5z 3
2
1
Exercise 3C — Conjugates and division
of complex numbers
Im z
–8
0 Re z
–88
32 – 24i
12 −3 + 3i
13
Imz
Im z
3 Re z
2
–2
0
4 – 5i
–8 + 3 i
1
–24
1 2 3 4 Re z
f
Im z
–88 + 16i
16
7 + 3i
3
2
1
0 1 2 3 4 5 6 7 Re z
Re z
32 Re z
Im z
–1
–2
–3
–4
–5
3+i
1
Imz
answers
613
Answers
Re z
10 Re z
–10
–10i
History of mathematics
1
2
3
4
5
Probability
He was a foreigner
Tutoring students and writing books
Newton
De Moivre predicted it.
➔
3A
3C
answers
614
Answers
Exercise 3D — Complex numbers in
polar form
1 a
Imz
z = 4 + 8i
8
b
5 a
z = 4 5
–4 –2 0
0
b 3
d 3 5
ii
–1 0
17
ii
u+z
4π
e -----3
i
37
Re z
f
9 a
Re z
6
w–u
Imz
ii
0
–2
e i
53
Im z
ii
130
9 Re z
f i
z+w–u
ii 10
Im z
z2
6
0
4 a
z1
–4 –2
π
c – --8
3π
d -----4
2π
g -----5
11 π
h --------12
π
b --6
f
6π
-----7
3π
b 5 2, -----4
e
2π
c 2, – -----3
149, – 2.182
π
2 10, 1.893 g 4, --33π
2 cis -----4
π
b 2 2 cis --6
3π
10 cis ⎛ – ------⎞
⎝ 4⎠
π
d 2 5 cis ⎛ – ---⎞
⎝ 3⎠
f
2
3π
------- cis -----4
4
3 2
b ---------- ( 1 + i )
2
c – ------5- ( 3 – i )
2
d 2–2 3 i
14
e ---------- ( 1 + i )
2
f 8i
12 B
13 D
g – 3
14 E
15 D
Exercise 3E — Basic operations on
complex numbers in polar form
π
b 20 cis --3
c
2π
6 cis -----3
π
π
d 6 5 cis ⎛ – ---⎞ e 2 7 cis ⎛ – ---⎞
⎝ 4⎠
⎝ 6⎠
8 Re z
b 42.5 square units
z2
h π
3π
1 a 6 cis -----4
Imz
6
4
2
π
g --2
2π
e cis ⎛ – ------⎞
⎝ 3⎠
11 C
–7
f 5.253
10 a – 1 + 3i
7 Re z
w+z
0
d 2.034
π
8 a 3 2, – --4
c
d i
7π
c -----4
j 0 or 2π
π
7 a – --2
ii 10
Imz
–8
3π
-----2
π
b --6
π
d 8, --6
0 1
0
u
2 4 6 8 Re z
5π
e – -----6
Re z
Im z
6
b i
65
f 5
5
Im z
z–w 4
3 a i
c i
c
e
w
6 a 0.588
4 Re z
2 a 13
b 24 square units
Imz
12
10
8
6
4
z
2
2 a –3 2+3 2 i
3 2
c – ------6- + ---------- i
2
2
z3
z4
2 4 6 8 10 Re z
e
21 – 7 i
b 10 + 10 3 i
d 3 10 – 3 10 i
5π
3 a 4 2 cis -----12
c
π
b 8 3 cis ⎛ – ---⎞
⎝ 2⎠
π
8 2 cis ⎛ – ------⎞
⎝ 12⎠
c
3π
2 cis ⎛ – ------⎞
⎝ 10 ⎠
3π
d 2 2 cis ⎛ – ------⎞
⎝ 14 ⎠
2
7π
e 3 ------- cis -----4
12
π
5 a i 3 3 cis --4
b i 16 cis π
c i 9 cis π
3π
d i 32 cis -----4
3 6 3 6
ii ---------- + ---------- i
2
2
ii −16
ii −9
ii – 16 2 + 16 2i
b − 1--8- i
3 1
d ------- – ------ i
64 64
9 aB
8 π
10 ---, – --9 6
π
2
13 ---, – --------5 120
16 a 8
c − 1--4- + 1--4- i
e 0.171 – 0.046 i f 16
7 – 64 3 – 64i
8 1
bC
cE
11 16 − 16i
12 –64 + 64i
14 and 15 Solution will vary.
b 4
c 2
d 6
Exercise 3F — Factorisation of
polynomials in C
1 a ( z + 2i ) ( z – 2i )
b (z + 7 i)(z – 7 i)
c ( z + 4 + 3i ) ( z + 4 – 3i )
7
7
3
3
d ⎛ z – --- + ------- i⎞ ⎛ z – --- – ------- i⎞
⎝ 2 2 ⎠⎝ 2 2 ⎠
e ( 2z – 1 + 4i ) ( 2z – 1 – 4i )
f – ( 3z – 4 + 4i ) ( 3z – 4 – 4i )
g (z + 2 – i)(z – 2 – i)
2+i+ 3
2+i– 3
h 2 ⎛ z – ------------------------⎞ ⎛ z – ------------------------⎞
⎝
⎠⎝
⎠
2
2
2 a ( z – 3 ) ( z + 3 ) ( z – 3i ) ( z + 3i )
b ( z + 3 )( z – 3 )( z + i )( z – i )
c ( z + 4i ) ( z – 4i ) ( z + 2i ) ( z – 2i )
d
(z +
5i )( z – 5i )( z + 2 )( z – 2 )
3 a z + 2, z – 3 ± 5 i
b z – 1, z + 1 ± 3 i
h (z + 2 – i)(z + 1)(z – 1)
b −2 − i, − --124 a 1 − i, −6
c 4 + i, 2
1
5 z + 3i, z – 1 ± ------- i
2
6 E
7 D
8 E
9 C
b −4
c −5
d 16
10 a 6
b a = −3, b = −3
11 a a = 15, b = −84
c a = −3, b = 4
b a = 12, b = 3
12 a a = −2, b = 1
c a = 1, b = −4
13 Since complex roots occur in conjugate pairs, an odd
number of roots must have at least one real root.
14 z3 − 2z2 + 16z − 32
b −1, −2, −3
15 a ±6
3 + i⎞ ⎛
3
–i
⎛
16 ( z – i ) z + --------------- z – --------------⎞
⎝
2 ⎠⎝
2 ⎠
17 a Solution will vary.
b Q(z) = z3 − 3z2i − 3z + 4 + i
c a = i, b = 4
18 ( z + 5 i ) ( z – 5 i ) ( z + 1 + 2 i ) ( z + 1 – 2 i )
19 (3z − 2 − i)(3z − 2 + i)(z + i)
20 a = 5
Exercise 3G — Solving equations in C
1 a −1 ± 2i
3
d --- ± i
2
b 4 ± 3i
c 7 ± 10i
2
e ------2- ± ------- i
2
2
11
3
2 a – 2, --- ± ---------- i
2
2
7
b 1, 1--- ± ------- i
2 2
23
c 2, 3--- ± ---------- i
4
4
11
d 4, 1--- ± ---------- i
6
6
1
e 2, 3--- ± --- i
2 2
3
3 4, 1--- ± ------- i
2 2
4 a ±4i, ±3i
c ± 1--3- , ±2i
b ±2, ±i
d ± 3---i
2
5 B
1
6 a ± --- ( 6 + 2 i )
2
b ± (6 + 5i)
1
c ± ------- ( 9 + 7i )
2
1
7 ± ------- ( 1 + i )
8 C
2
11 π
π
9 a
2 cis ⎛ – ------⎞ , 2 cis --------⎝ 12⎠
12
3D
➔
11
3
c z + 2, z – --- ± ---------- i
2
2
d z + 3, 2z – 5, z + 1
3
3
1
z ± 1, z – --- ± ------- i, z + 1--- ± ------- i
2 2
2 2
g (z + i)(z + 3)(z – 3)
11 π
b 4 cis --------12
6 a − 1--4-
e z – 3, z + 1, z + 1 ± 2i
f
π
4 a 3 cis --2
615
answers
Answers
3G
answers
616
Answers
5
6
5
--π --7π
b 2 4 cis ---, 2 4 cis ⎛ – ------⎞
⎝ 8⎠
8
c
e
f
5
4
3
2
1
8π
2π
4π
2 cis ------, 2 cis ------, 2 cis ⎛ – ------⎞
⎝ 9⎠
9
9
π
5π
π
d cis ---, cis ------, cis ⎛ – ---⎞
⎝ 2⎠
6
6
6
z2
–4 –3 –2 –1
5π
π
11 π
2 cis ⎛ – ---⎞ , 6 2 cis ------, 6 2 cis ⎛ – ---------⎞
⎝ 4⎠
⎝ 12 ⎠
12
π
π
3π
5π
cis ⎛ – ------⎞ , cis ⎛ – ------⎞ , cis ⎛ – ------⎞ , cis ⎛ ---⎞ ,
⎝ 4⎠
⎝ 12 ⎠
⎝ 12⎠
⎝ 4⎠
11 π
7π
cis ⎛ ------⎞ , cis ⎛ ---------⎞
⎝ 12 ⎠
⎝ 12 ⎠
11 a 4, −2 ± 2 3 i
z2
z3
1 2 3 Re z
5±2 2 i
1
b ± ------- ( 5 + 3i ) c – 2, 1 ± 3 i
2
Im z
2
2 3
w
z3
zw
4 Re z
−2
π
3π
5π
7π
cis ⎛ ---⎞ , cis ⎛ ------⎞ , cis ⎛ ------⎞ , cis ⎛ ------⎞
⎝ 8⎠
⎝ 8⎠
⎝ 8⎠
⎝ 8⎠
Chapter review
3 a 11 − i
12 14
------ – ------ i
17 17
π
5 7 2 cis ⎛ – 3------⎞
⎝ 4⎠
4 E
9 D
14 A
5 C
10 B
15 D
b z = 1 + 3 i, w =
2+ 2 i
2 + 6 + ( 6 – 2)i
c ------------------------------------------------------4
( 6 + 2)
( 6 – 2)
d i -------------------------ii -------------------------iii 2 – 3
4
4
e Solution will vary.
2 a (z + 8i)(z − 8i)
b (z2 + 8i)(z2 − 8i)
c Solution will vary.
d (z + 2 − 2i)(z − 2 + 2i)(z + 2 + 2i)(z − 2 − 2i)
e (z2 + 4z + 8)(z2 − 4z + 8)
3 a z = 3 ± 4i; sum = 6; product = 25
2
b a = −549, b = 296
29
π
_ apart
8
π
π
1 a cos ------ + sin ------ i
12
12
c z – ( 2 5i )z – 9
b
Re z
Analysis
7π
π
5π
3π
b cis ⎛ – ------⎞ , cis ⎛ – ------⎞ , cis ⎛ – ------⎞ , cis ⎛ – ---⎞ ,
⎝ 8⎠
⎝ 8⎠
⎝ 8⎠
⎝ 8⎠
Short answer
1 −1 − 2i
2 a −48
4
z
2
−2
2π
4π
2π
4π
13 a cis 0, cis ------ , cis ------ , cis ⎛ – ------⎞ , cis ⎛ – ------⎞
⎝ 5⎠
⎝ 5⎠
5
5
3 C
8 A
13 A
1
_z 0
w
−1
± 2, –1 ± 3 i, 1 ± 3 i
3
3 3
d ± 3, ± ------- + --- i, ± ------3- – --- i
2 2
2 2
4
−1
−√w
b ± 5, ± 5 i
Multiple choice
1 B
2 B
6 C
7 A
11 A
12 B
16 C
17 E
√w
z
–2 3
–4
12 a ±2, ±2i
c
4
1 z
z1
–4 –2
5π
d – -----12
c ( z + i ) ( z – i ) ( z – 1 + 2i )
9 a=8
11
Im z
b
z1
π
b −(31226) c −16
7 a --3
8 a (z − 8 + 5i)(z − 8 − 5i)
b 2(z + 4)(z − --12- + --12- i)(z − --12- − --12- i)
10 a
3 5
10 5i, ± 5 ------- – --- i, sum = 0
2 2
Area = 17.5
square units
Imz
c
6 5
2
d z + (3 ±
4 a i 2
ii 0
7i )z ± 3 7i
b ii 1 + i, 1 ± 2 i
c x=3
d ... –6, –2, 2, 6 ...
CHAPTER 4 Representation of
relations and regions in the
complex plane
e
0
Exercise 4A — Rays and lines
1 a
b
Im z
3π
—
4
0
c
d
Im z
e
0
7 a
– –π
6
Re z
b
Im z
c
Re z
b
Im z
Im z
0
c
Re z
2
0
d
Im z
0
3π
–—
4
–1
2
Re z
Im z
1
Re z
0
Im z Re z
0
Re z
1
Re z
0
f
0
–1
2
Im z
Re z
Im z
0 Re z
–3
0
d
Im z
e
2π
—
3
–π
2
Im z
Re z
5
–5
2 a
Re z
Re z
0
Im z
0
π
0
Im z
–3
Re z
π
– 2—
3
f
Re z
1
0
Im z
0
h
0
Im z
– –π Re z
4
0
–3
2
Re z
Im z
0 Re z
Re z
Im z
–4
g
Im z
–π
6
f
Im z
answers
617
Answers
g
h
Im z
Re z
Im z
3
π
2
–2
e
f
Im z
0
Im z
0
–1
3
Im z
–2 + i
i
0
h
Im z
c
0
–π
3
1
3 – 2i
–2
j
Im z
Im z
– –π
5
−2
0
d
Im z
0
0
e
0
–2
g
–3
Re z
f
Im z
Re z
0
Im z
0
–2
Re z
h
Im z
Re z
Im z
3
0
–1
b
5
Re z
3π
—
8
Re z
0
Im z
0
–2
Re z
d
4
Re z
1 π
–
2
Re z
Im z
0
Re z
2
b E
0
c
0
1
4 – 3i
a C
B
D
a Im z
Im z
Re z
Im z
– –π
8
–π
2
–3
3
4
5
6
4
Re z
3 Re z
0 Re z
–2
b
2π
—
3
Re z
g
0
Re z
Im z
– –π
4
5π
–—
6
0
8 a
Re z
Re z
Exercise 4B — Circles and ellipses
1 a
Im z
0
–5
2
Re z
Re z
b
Im z
1
–1 0
–1
1
Re z
Im z
4
–4
0
–4
4
Re z
➔
4A
4B
answers
618
c
Answers
d
Im z
3
0
–3
f
Im z
9
0
–9
Re z
9
2 B
4 a
10 a
c
4 Re z
3
Re z
0
–3
e
f
Im z
4 Re z
–3 + 2i
5 Re z
0 1 Re z
–4
–4 – 3i
Re z
3
–3
–3
5
g
–5
0
–8
6 E
y2
7 a x2 + ---- =1
4
x2
b ----- + y2 = 1
9
x2
y2
c ----- + ----- = 1
4
3
x2
y2
d ------ + ----- = 1
16
4
x2
e -------- + y2 = 1
f
⎛ 1---⎞ 2
⎝ 2⎠
c
Re z
–3
–3
2
1
−11 −1 − 2
2
b
−2
0
0
Re z
y
√5
0
−2
x
1
4
−√5
d
y
–1
−2 0
0
3 Re z
y
2
–2
f
2√6 x
0
y
2
−√5
4 Re z
−√3
−3
−8
2
0
−2√6
8 x
y
4
2
–4
3
−2√6
e
Re z
0
11
5
1
4
5
Re z
1 2 2 13
1–3 2–3
y
Im z
3
–2
−1 + i
Im z
d
Im z
Im z
−√3
1
–1 0 1
–2
–3 – 2i
Re z
√3
c
0 Re z
–1
–2
–3
–4
0
2√6
b
Im z
2
⎛ 1---⎞ 2
⎝ 2⎠
Im z
2 + 3i
x2
y2
-------- + -------- = 1
⎛ 1---⎞ 2
⎝ 3⎠
–9
f
h
11 a
−4
8
–6 –5 –4 –3 –2 –1
1
5 – 5i
Re z
4
4 – 3i
Re z
Im z
3
2 –43 2
Re z
Im z
3
3
Im z
3 –41
3+i
1 2 3 4 5
0
–3
0
–2
d
3
2
1
7
–7
5 C
2
–3 + 2i
Im z
0
2
1
Im z
–5 –4 –3 –2 –1
h
Im z
2
8 a
6
5
4
3
2
1
0 Re z
–4 –3 –2 –1
2
0
1
–1 0 2
–2
2
Im z
5
2–i
e
Im z
3
2
1 2+i
–9
0
–7
4
g
b
–2 + 3i
–5
3 Re z
x
y
d ----- + ------ = 1
9 25
c
–1 0
–4
2
–2
4
1
2
–1
Im z
3
7
2
Re z
–4 –2 0
–2
d
Im z
x
y
b ------ + ------ = 1
25 16
Im z
Im z
2
0 1 2
2
x
y
c ----- + ----- = 1
5 9
1
–1
2
Re z
Re z
2
3 D
b
– –1 0 –1
3
3
– –1
2
Re z
2
x
y
9 a ----- + ----- = 1
4 3
–2
Im z
0 –1
–1
0
–2
Im z
–1
2
– –1
2
10 Re z
Im z
2
–9
f
Im z
1
–10
–3
e
0
–10
Re z
3
e
Im z
10
√3 x
0
−1
−4
√5 x
x
2
2
x y
12 ----- – ----- = 1; the locus describes the right-hand
4 5
branch of the hyperbola.
c
d
Im z
Region
required
1
0
Exercise 4C — Combination graphs
and regions
1a
b
Im z
2
–3
e
Im z
Re z
0
Re z
01
Re z
–4
f
Im z
4
3
–2 0
Im z
Region
required
Region
required
answers
619
Answers
Im z
Region
required
Re z
2
0
–3
–2
Re z
3
0
–1
Re z
–3
c
d
Im z
Im z
g
2
–π
3
0
–1
e
5
–3 + 3i
Re z
h
Im z
5
4
3
2
1
Re z
–5 –3 –1 0 2 3π
(2, –1) – —
8
3 E
7 a
Re z
π
–—
2
–1
3 4 Re z
Re z
Reg.
req.
Reg.
req.
e
f
Im z
Im z
Reg. req.
0
0
Re z
0
5—
π
6
4 Re z
–3
Reg.
req.
b
Im z
2
Re z
Reg.
req.
Reg.
req.
Reg. req.
b
Reg.
req.
3–i
–π
6
0
–2
2
0
5 Re z
Im z
–1 0 1
–1
–2
Re z
Re z
Im z
3
Reg. req.
Re z
0 2
Re z
–5
–2
c
Im z
Region
required
Re z
Im z
5
–5
0
Re z
Re z
0
–7
9 a
Im z
π
–—
2
0
0
Im z
2
0
–1
Im z
–3
d
Im z
Re z
Im z
Reg. req.
–2
Re z
–6
–π
2
2—
π
3
f
2
0
–3
Reg. req.
Im z
3π
–—
4
c
Im z
Reg.
req.
Reg.
req.
e
Reg.
req.
5 D
Re z
d
Im z
Re z
5
0
–1
Re z
–7
b
Im z
Reg.
req.
Re z
Im z
Reg.
req.
Re z
4 + 3i
Re z
π
– 2—
3
π
0
0
8 a
5—
π
9
0
Im z
j
0
Re z
Im z
–π
6
0
Region 0
required
Im z
–3 –2 –1 0 1 2 3 4
b
Re z
Region
required
4 B
Im z
Reg. req.
c
i
0
–2
1
6 a
0
–5
Im z
3
2 A
2 4
– –π
4
–2 –1 0 1
Im z
9
Re z
4
f
Im z
4
–1 + 3i 3
2
1
g
0
–2
Re z
h
Im z
Region
required
–1
5
–3
–4
Reg.
req.
4C
➔
–3
0 2
d
4C
answers
620
e
Answers
Im z
Reg.
req.
2
–1 0
–2
f
Reg.
req.
b
Im z
3
1
–3 –1 0 1
Reg.
req.
2
–4 –1 0 1
–2
Im z
4
3
2
1
Re z
6
3
Re z
f
11 a
–6
–7
b
π
3—
4
0 2
–2
–2
Reg.
req.
c
f
i 3y − 4x = 12
ii
Re z
3
Im z
Reg. req.
π
2—
3
Re z
– π–4
–5
f
–3
0
Re z
0
Re z
3 a i 2y − 4x − 3 = 0
ii
0
Re z
π
–
6
– 3–4
–3 –2 –1 0 Re z
b i x+y=0
ii
0
Re z
Reg. req.
( x + 3 )2 + y2
g
( x + 4 )2 + ( y + 6 )2 h
f
ii
j
ii
– 5–8
ii
Re z
0
Re z
0
Re z
Im z
– 2–3
( 5 – x )2 + ( y + 4 )2
Im z
1
e i y = − --2
4
4
0
2
( x – 1 )2 + ( y + 5 )2
0
Im z
5–
2
d i y = 3x + 2
x2 + ( y – 5 )2
Re z
1–i
c i 2y − 8x − 5 = 0
b y+2
d y−3
e
Im z
Im z
4
Exercise 4D — Graphs of other simple
curves
1 a x−8
c x+5
Im z
3–
2
2
1 01
i x−7
2 a i x+y=4
Im z
4
−2
Im z
4
3
2
1
Reg. req.
6 Re z
Re z
0
π
–
2
e
Re z
Im z
0
Re z
d
Reg.
req.
ii
Reg.
req.
Im z
2
0
–3
Im z
– π–2
0
(1, 5)
Reg.
req.
–3 –2 0
Im z
Im z
– 1–4
4
–3 + 4i
2 – 3i
2
ii
Re z
1
x
e i y = --- − 3
2
0
Reg.
req.
d i y = 4x + 1
Re z
Im z
10
8
1
0 2
Reg.
req.
0
Im z
Im z
3
–4–3 0 3 4 Re z
–3
–4
d
Re z
–6
ii
Im z
4
Reg.
3
req.
3 Re z
Im z
e
0 3
c i 3x + 2y = 6
–3
c
Im z
Re z
Reg. req.
–6
10 a
ii
4
1
–8 –5 –2–2 0
7 Re z
3
3 – 2i
b i 2x − y = 6
Im z
–5 + i
ii
Im z
Re z
– –21
f
i x = −3
ii
c y = −2x2 − 1
Im z
Im z
Re z
0
–3
–1 0
Re z
d y = (x − 4)2
4 B
5 A
6 a E
Im z
b C
2
7 a y = --x
4 Re z
0
Re z
Im z
1 + 2i
0
e y = −(x + 2)2
Im z
Re z
–1 – 2i
2
b y = – --x
0
answers
621
Answers
–2
Im z
–1 + 2i
f
Re z
y = 2(x + 1)2 + 5
Im z
0
–1 + 5i
1 – 2i
c
3
y = ---------------( x – 1)
0
Im z
Re z
2 + 3i
g y = −1 − (x − 2)2
0 1
Im z
Re z
–3
2 – i Re z
0
1
d y = ----x2
Im z
–1 + i
1+i
Re z
0
9 a
b
Im z
3
1 + i Re z
e
4
y = – ----x2
–1 + 2i
0
–1 – i
Im z
Im z
Re z
0
Re z
0
c
–1 – 4i
f
2
y = ------------------( x + 1 )2
1 – 4i
2
Re z
0
e
Im z
3
4
1
--2
b i y = 1--4- x2 − 1
Re z
2
3 Re z
Re z
Im z
4
3
1 + 2i
2
1
1+ 5
1– 5 –1 0 1 2 3 Re z
10 a i y = − 1--2- x2+
Im z
0
–3
f
2+i
Re z
0
Im z
3
–3
1
–1 0
Im z
b y = x2 + 2
Re z
–2
–2 + 2i
8 a y = x2
0
–2
Im z
d
Im z
2
c i y = − --16- x2 +
ii Maximum at (0, 1--2- )
ii Minimum at (0, –1)
3
--2
ii Maximum at (0, --32- )
0
➔
4D
4D
answers
622
Answers
11 a
b
Im z
Analysis
1 a
Im z
–1 + 4i
Re z
Re z
0
0
Region
required
c
Region
required
d
Im z
–1 + i
e
f
Im z
Re z
Region
required
Im z
–1 + i
Re z
2
2
–2
–1 – i
d
Im z
Re z
2
Re z
–1
–1
1
–1 + i
–2
e
Re z
–2
Im z
–2 –1
0
Re z
Im z
2
1 + i Re z
0
Region 0 2
required
1+i
1
Region
required
0 1
2
Re z
Im z
4
The circle is rotated 90n° clockwise about 0 + 0i as
the coefficient of z is multiplied by i n.
1 + 3i
Re z
0
–1
3
Region
required
2 a i (x + 1)2 + y2 = 4
ii (x − 2)2 + (y + 2)2 = 72
Chapter review
b Answers will vary.
Multiple choice
1 E
2 B
3 C
4 A
5 D
6 A
7 B
8 E
9 C
10 C
11 D
12 A
Short answer
1
Im z
5 + 3i
0
2
2π
–—
5
0
Re z
Re z
0
–1 1 2 3 4
–2
–3
3 – 3i
–4
Re z
01 2 3 4
2 – 2i
Region
required
Im z
Region
required
4+i
Re z –2 1 1 2 3 4
0
–1
2 – 3i
−3 −2 −1−10
z2
−2
−3
Im z
–1 + 2i
4
3
2
1
Re z
– 3 – 1 –1 0 3 – 1
{
–1
–2
–3
–4
–5
Im z
3
z
2 1
1
Re z
π
5 {z: ⏐z − 3i⏐ = 2} ∪ z: Arg (z + 3 − i) = − --2
Im z
1
3 a and b
c y = –x
d
4
Im z
1
c 68π square units.
Im z
–4
8
0 1
c
0
5
––
2
4
6
1
1–i
0
Region
required
3
0
–1
1 – 4i
Im z
2
Im z
1+i
1
Re z
0
g
b
Im z
2
1
7 y = 2x − --2
}
1 2 3 Re z
Im z
3
z
2 1
1
−3 −2 −1−10
z2
−2
−3
1 2 3 Re z
f For any two complex numbers z1 and z2, the locus
of the points described by the complex relation
z – z 1 = z – z 2 is a perpendicular bisector of the
line segment joining z1 and z2.
2
2
x
25
y
1
4 a ------ – --------- = -----b x < – -----25 231 16
64
c Hyperbola with centre (0, 0) and asymptotes
231
y = ± ------------- x
5
d
c
Im z
5
−_
4
2
5
a = 54, b = 3
6
x = − 1--2- , y = − 5--2-
8
0
2 4 6 8 10 12 14 Re z
1 2 3 4 5 Re z
2π 5π
b ------ , -----3
3
7a 6
2
Im z
(4, 33)
0
Re z
5 a x + y = 16
b A circle with centre at 0 + 0i and radius 4
π
c 6cis ⎛ – ---⎞
⎝ 3⎠
Im z
5
Im z
4
p
−_
4
0
−4
Re z
4 Re z
−5
−4
c
d
Im z
40
30
20
10
623
answers
Answers
2
2
9a
x +y =a
d A circle with centre at 0 + 0i and radius a ; ‘If
z = x + yi, the graph of zz = a , where a ∈ R+ is a
circle with centre at 0 + 0i and radius
2
b
Im z
Im z
5p
6
0
Re z
a .’
2 Re z
2
e ( x – 2) + ( y – 3) = 4
c
f A circle with centre at 2 + 3i and radius 2
d
Im z
Im z
Im z
p
_
3
5
4
3
2
1
0
2
g ( x – u) + ( y – v) = a
P
(−1, 3)
( z – w ) ( z – w ) = a , where a ∈ R is a circle
with centre at u + vi and radius
−1 0
a .’
Technology-free questions (page 195)
10
20
30
40
1 2 3 4 5 Re z
(4, −33)
1 Re z
b –8
b a = –3, b = 0, c = –50
1
−1
2
3
4
R (5, 0)
5 Re z
−2
Q
(−1, −3) −4
e { z: z + 1 = 3 }
{ z: Re ( z ) = – 1 }
–5 + 12i, –5 – 12i b –10
169
d z2 – 6z + 11
2i
( z – i ) ( 3z + 2i ) ( 2i – 3z ), i, ± ----3
π
14 a 2 cis ⎛ ---⎞ , 3 2 cis ⎛ – π
---⎞
⎝ 6⎠
⎝ 4⎠
d
12 a
c
13
Im z
(−8, −96)
2
−3
2
iv –1 – 2i
3
iv 9 – 3i
3
iv –12 – 3i
c 1 + 38i d 4
c 4 + 33i d 13
−8 −7 −6 −5 −4 −3 −2 −1
3
1
+
i –1 + 2i
ii –1
iii
i 9 + 3i
ii 9
iii
i –12 + 3i ii –12 iii
–1 – 2i
b 11 – 12i
4 – 33i
b –8 – 96i
b
Im z
−2
4
h A circle with centre at u + vi and radius a ;
‘If z = x + yi and w = u + vi, the graph of
1a
b
c
2a
3a
4a
−5
(3, −2)
10 a 8 cis(–π)
11 a –1 + 3i
c
Im z
1 2 3 4 Re z
2
Re z
3
−2
Re z
0
20
40
60
80
100
b 144 cis(0), 144
c 2 3 , –2 3, 2 3 i, – 2 3 i
➔
4D
4D
answers
624
15 a
Answers
1
b y = – --x
Im z
Re z
1
c Horizontal translation of 2 in the positive
direction. Vertical translation of 2 in the positive
direction.
d { z : Im ( z – 2 ) Re ( z – 2 ) = – 1 }
1
e y = – ----------+2
x–2
f
Im z
y=−
3i
1
−2
x+2
2i
i
−3 −2 −1
−i
−2i
Region
required
1
2
3 Re z
Region
required 2
Re z
CHAPTER 5 Differential calculus
Exercise 5A — The derivative of tan kx
5 sec2 5x
8 sec2 2x
−12 sec2 (−12x)
−6 sec2 3x
x
j --15- sec2 --5
b
d
f
h
2
c 21 sec 7x
e −3 sec2 (−3x)
g −12 sec2 (−4x)
i 10 sec2 (−2x)
x
sec2 --2
2x
m --27- sec2 -----7
2 4x
8
o --3- sec -----9
2 8x
q 8 sec -----5
3x
sec2 -----5
3x
n 6 sec2 -----4
2 5x
5
p − --2- sec -----6
2 7x
r −14 sec -----2
1
--2
l
2
24
-----7
6x
sec2 -----7
i 2 tan x sec2 x
j 8 sec2 2x tan3 2x
2
2
5 a 2x tan 3x + 3x sec 3x
4x
4x
b −2 sin 2x tan ------ + 4--5- cos 2x sec2 -----5
5
c (15x2 − 6) tan 4x + 4(5x3 − 6x) sec2 4x
d 4e4x tan (−3x) − 3e4x sec2 (−3x)
2
x sec x – 2 tan x
--------------------------------------3
2x
2
b Parabola
1 a 4 sec2 4x
2
h --- +
x
g 4 cos 4x − 3 sec2 3x
2
Im z
2
d 5 sec2 5x cos (tan 5x)
5
2
e −2 sec2 2x sin (tan 2x) f --- sec 5x
2
f
c (0, –1), Minimum
k
c 3e3x sec2 (e3x)
2
16 a y = 1--4- x2 − 1
−1
b 3 sec2 3x etan 3x
e 8xe4x tan (8x) + 8e4x sec2 8x
−3i
d
f
g −4 sec2 (7 − 4x)
3 a C
12
4 a ----------------sin 12x
i
−1
−i
2
--- sec2 8x
x
h −10x sec2 (1 − 5x2)
b A
2
e 3 sec ( 3x + 2 )
3
--5
2 a (2x + 3) sec (x + 3x) b sec2 (x + 2)
c 5 sec2 (5x − 4)
d (4x − 3) sec2 (2x2 − 3x)
3
2
12x tan 2x – 8x sec 2x
g ---------------------------------------------------------------2
tan 2x
h −4 sin 4x
6 a (0, 0) and ( π , 0)
π 1
5 π –1
b --- , ------- and ------ , ------6
6
3
3
π
2π
c --- , 3 and ------ , – 3
3
3
d approx. (1.249, 3) and (1.893, −3)
–π
π
7 a ------ , −4 and --- , 4
b (0, 0)
2
2
(
(
) (
) (
)
(
) ( )
)
c No points, as gradient cannot equal zero.
8 y = 24x − 4π + 3 3
π
π
– 2x
x π2
9 i y = (2 + π ) --- – ----ii y = ------------ + --------------- + --2 + π 4 + 2π 4
2 8
11 sec2 x > 0 for all x.
10 sec2 x ≠ 0
12 a (0, 0) a stationary point of inflection
b f ′(x) = sec2 x − 1 ≥ 0 for all x as sec2 x ≥ 1.
Exercise 5B — Second derivatives
b 30x − 2
1 a 8
2
d 12x + 12x
g 24x −
1
--2
x
i 2 + 60x−5
2 a B
3
– --2
c 0
4
2
e 30x + 24x
h
63
-----2
j
3
--2
5
---
x2 −
x
5
– --2
20
-----3
x
f
35
-----4
3
---
x2 +
3
--4
x
– 1--2
1
– --3
+ 6x−3 k 12x−4 + 2x−3
b E
–1
ii -----x2
1
3 a i --x
1
ii 2 + ----x2
ii −4 sin 2x − 4 cos x
1
b i 2x – --x
c i 2 cos 2x − 4 sin x
1
--4 3
--- x
j (0, 0) a stationary point of inflection and
27
- ) a local min.
(− 3--4- , − -------256
2 a
x
ii − 1--2- sin --- + 9
2
ii 50e5x − 3ex + 1
ii 36e−3x + 18x
ii 8 sin 2x sec3 2x
ii −96 sin (−4x) sec3 (−4x)
0
5
---
2
x
0
–2
f
–3
2 sin x
+ 4 cos x e
ii −25e5x sin (e5x) − 25e10x cos (e5x)
2
cos 2x
– 2sin 2x – cos 2 2x
m i -----------------ii ----------------------------------------------sin 2x
sin 2x sin 2x
–1
0 1
g
h
g(x) =
3
5
---
x
0
(0, 3)
3
–
i
j
y
y
h(x) = x3 − 3x
(–1, 2)
y = x 4 + x3
x
0
–1
3
– 3
1 a (2, −4) a local min.
b (0, 12) a local max.
3 a D
4 A
i
i
i
i
i
i
i
x
b B
5 C
None
None
None
None
None
(0, 0)
None
ii
ii
ii
ii
ii
ii
(0, 0)
(1, 0)
(3, 2--9- )
16
-)
(− --23- , 1 ----27
(−2, −2e−2)
1
-)
( --14- , − -------128
5A
➔
6 a
b
c
d
e
f
g
0
27
–– )
(– 3–4 , 256
(1, –2)
d (0.55, −0.63) a local min.,
(−1.22, 2.11) a local max.
e Approx. (1.43, −16.9) a local min.,
(−2.097, 5.049) a local max.
f Approx. (0.43, −16.9) a local min.,
(−3.097, 5.049) a local max.
g (0, 0) a stationary point of inflection.
h (0, 3) a stationary point of inflection.
i (−1, 2) a local max., (1, −2) a local min.
x
0
3
g A
Exercise 5C — Analysing the behaviour
of functions using the second
derivative
) a local max.
f(x) = x 3+ 3
y
x3
(0, 0)
32
-----27
x
(0.43, –16.9)
y
3
c (0, 0) a local min., (− 4--3- ,
2
–16
3
---
1
---
0
–4 –2
x
3
(1.43, –16.9)
n i – 10 sin 4x ( cos 4x ) 2
ii 60 sin 2 4x ( cos 4x ) 2 – 40 ( cos 4x ) 2
o i −tan x
ii −sec2 x
4 a C b F c I d B e D f G
9 k = 5 or −5
10 k = −1 or 4
11 a 3t2 − 2t − 1
b 6t − 2
c t = 1 s and x = 6 cm d − 4--- cm/s
x
f(x) = x3 + 4x2 – 4x – 16
y
–9
5x)
1
(0.55, –0.63)
(–3.097, 5.049)
(–2.097, 5.049)
l i −5e sin (e
5x
y
(–1.22, 2.11) y = x(x – 1)(x + 2)
h(x) = x(x – 3)(x + 3)(x + 1)
y
5
---
x
y = x2(x + 2)
(0, 0) 0
k i 2 cos x e2 sin x
ii −2 sin x e
d
y
e
3
---
2 3
g(x)
c
—2
2
2
2
2
2
15 2 2
------ x sin 2 x + 4 x sin2 x – 10 x sin 2 x cos 2 x + 8 x cos 2 x
4
-----------------------------------------------------------------------------------------------------------------------------------------------------3
sin 2 x
0
–2 3
(2, –4)
5
--2x 2
2 sin x
f(x) = 12 – x2
(0, 12)
x
4
——
(− 4—
, 32
)
3 27
sin 2x –
cos 2x
2
j i ------------------------------------------------------sin 2 2x
ii
y
f(x) = x 2 – 4x
– 2x sin 2x – cos 2x
i i ----------------------------------------------x2
– 4x 2 cos 2x + 2 cos 2x + 4x sin 2x
ii ----------------------------------------------------------------------------------x3
1
---
b
y
2
– --4
--- x 3
x
d i cos --- + 3
2
e i 10e5x − 3xx + 1
f i −12e−3x + 9x2
g i 2 sec2 2x
h i 12 sec2 (−4x)
3
5 --2--- x
answers
625
Answers
5C
answers
626
Answers
ii [−2 −
[−2 +
h i None
2 , (6 + 4 2 ) e−2 −
2
12
] and
y = x 4 – x 2 – 12
−2 + 2
2 , (6 − 4 2 ) e
]
ii None
–2
8 (0, 1) a local min.
1
2
9 a i (− ------- , ---------- ) a local max.,
3 3 3
1
2
( ------- , − ---------- a local min.
3 3 3
ii (0, 0)
iii
y
(– 1–3,
2
–––
3 3)
(0, –12)
( 1–6 , –12 365–)
(– 1–2 , –12 1–4 )
( 1–2 , –12 1–4 )
y
y = x 3 + 8x
9
(0, 0)
0
x
1
c i Approx. (−1.26, −1.89) a local max.
ii none
iii
y
y = x – 1–2
x
1
b --------------------25 – x 2
–1
e --------------------16 – x 2
4
h ----------------16 + x 2
1
k -----------------5 – x2
–1
n -----------------7 – x2
1
c --------------------64 – x 2
–1
f --------------------36 – x 2
7
i ----------------49 + x 2
1
l -------------------------0.04 – x 2
0.8
p ---------------------0.64 + x 2
1
q -----------------6 – x2
10
r ----------------10 + x 2
5
b -----------------------1 – 25x 2
3
c --------------------1 – 9x 2
–4
e -----------------------1 – 16x 2
f
–6
-----------------------1 – 36x 2
– 10
h --------------------------1 – 100x 2
i
3
----------------1 + 9x 2
–7
g -----------------------1 – 49x 2
x
y = – x1–2
9
-------------------1 + 81x 2
b
3 a -----------------------1 – b2 x2
4 a D
j
(–1.26, –1.89)
d i (−1, 3) a local min.
1
---
ii ( 2 3 , 0)
iii
y
y = x2
0
3
( 2, 0)
11
3
--2
5 a A
6 A
y = x2
(–1, 3)
– x2–
x
y = – x2–
t t2
b 40 + --- – ---2 8
d No, it gets within 33 1--3- m of it.
1
1 a ----------------4 – x2
–1
d -----------------9 – x2
2
g -------------4 + x2
3
j -------------9 + x2
–1
m -------------------------6.25 – x 2
2
2 a --------------------1 – 4x 2
8
d -----------------------1 – 64x 2
y=x
0
x
Exercise 5D — Derivatives of inverse
circular functions
x
iii
2
1
(– –6, –12 365– )
c 40 1--2- m/s
y = x3 – x
2
( 1–3 , – 3–––
3)
ii (0, 0)
0
1
13 a 383 --3- m
0
b i None
y
–3
7 a -----------------------16 – 9x 2
–5
d --------------------------64 – 25x 2
14
g -------------------4 + 49x 2
j
5
-----------------------4 – 25x 2
4
k -------------------1 + 16x 2
–b
b -----------------------1 – b2 x2
3
o -------------3 + x2
5
-------------------1 + 25x 2
b
c -------------------1 + b2 x2
l
b C
b E
–7
b --------------------------16 – 49x 2
–9
c --------------------------25 – 81x 2
20
e -----------------------25 + 16x 2
f
24
-------------------64 + 9x 2
45
h -----------------------25 + 81x 2
i
2
--------------------9 – 4x 2
6
k --------------------------49 – 36x 2
l
8
--------------------------25 – 64x 2
b
8 a -------------------------a2 – b2 x2
–b
b -------------------------a2 – b2 x2
ab
c ---------------------a2 + b2 x2
2
9 a ----------------------------------1 – ( 2x + 3 ) 2
3
b ----------------------------------1 – ( 3x – 5 ) 2
–4
c ----------------------------------1 – ( 4x – 3 ) 2
–5
d -----------------------------------1 – ( 5x + 8 ) 2
3
e -------------------------------1 + ( 3x + 2 ) 2
f
1
g -------------------------------4 – ( x + 3 )2
–2
h -----------------------------------9 – ( 2x + 1 ) 2
6
------------------------------1 + ( 6x – 7 ) 2
20
---------------------------------25 + ( 4x – 3 ) 2
j
–3
----------------------------------1 – ( 4 – 3x ) 2
2
k ----------------------------------1 – ( 7 – 2x ) 2
l
–5
------------------------------1 + ( 8 – 5x ) 2
–4
m -------------------------------------25 – ( 3 – 4x ) 2
3
n -------------------------------------49 – ( 6 – 3x ) 2
i
– 12
o ---------------------------------16 + ( 2 – 3x ) 2
10 a
3x 2
2
– --------------------1 – 4x 2
f
x
sec sin –1 ---⎞
⎝
3⎠
k ------------------------------9 – x2
2⎛
11 a
1
--4
12 2y = x + π
14 a (0, 0)
–1
h 3
j
l
7
9 --2
+ --- x 2 + --------------------2
9 – 4x 2
1
-----------------------------------------2
2 ( 1 – x ) sin –1 x
5
e 7e 7 x + 3 + -------------------1 + 25x 2
i 0
cos –1
6x
2x – --------------------1 – 4x 2
–x
-----------------1 – x2
x
–4 sin ⎛ tan –1 ---⎞
⎝
4⎠
-----------------------------------16 + x 2
b 4y = x
3x π
3
13 y = ---------- + --- – ------6 3
3
b 1--a
Exercise 5E — Antidifferentiation
involving inverse circular functions
x
1 a tan−1 --- + c
6
x
2 a sin−1 --- + c
2
x
g 4 sin−1 --- + c
5
x
h 3 cos−1 --- + c
3
x
i 2 tan−1 --- + c
3
x
j 5 tan−1 --- + c
4
3 a
1
--2
sin−1 2x + c
b
1
--4
tan−1 4x + c
4 a
1
--3
sin−1 3x + c
b
1
--4
sin−1 4x + c
c
1
--2
cos−1 2x + c
d
1
--5
cos−1 5x + c
e tan−1 3x + c
g
1
--5
sin−1 5x + c
i tan−1 8x + c
k
5
--2
cos−1 4x + c
f tan−1 5x + c
h
1
--6
cos−1 6x + c
j
6
--7
sin−1 7x + c
l 12 tan−1 2x + c
b C
x
b cos−1 --- + c
3
x
c sin−1 --- + c
5
7 a
1
--3
3x
sin−1 ------ + c
4
2x
cos−1 ------ + c
3
3x
−1
e --16- tan ------ + c
2
c
1
--2
b
d
f
1
--5
5x
sin−1 ------ + c
3
5x
cos−1 ------ + c
2
4x
−1
1
------ + c
------ tan
20
5
1
--5
5x
g sin−1 ---------- + c
5
1
i ------- tan−1 7x + c
7
6x
h cos−1 ---------- + c
6
2 7x
sin−1 ------------- + c
7
m sin−1 (x + 3) + c
3 10x
cos−1 ---------------- + c
10
n cos−1 (x − 2) + c
x–1
o sin−1 ----------- + c
2
x
s 2 sin−1 --- + c
4
x+4
p cos−1 ------------ + c
3
−1 x + 5
1
r --4- tan ------------ + c
4
−1 x
t
2 cos --- + c
4
x
u tan−1 --- + c
6
x
v 6 tan−1 --- + c
5
k
1
--2
q tan−1 (x − 3) + c
8 a x + tan−1 x + c
j
l
4
15x
---------- tan−1 ------------- + c
5
15
1
--3
3
x
b x + --- tan−1 --- + c
2
2
c
1
--2
x2 + tan−1 x + c
d
1
--2
x2 + 5 tan−1 x + c
b cos−1 x + c
e
1
--2
x
x2 + tan−1 --- + c
3
f
3 2
x
--- x − 4 tan−1 --- + c
2
2
x
d cos−1 --- + c
4
1
x
g x2 − --- tan−1 --- + c
4
4
5D
➔
x
c sin−1 --- + c
3
x
f tan−1 --- + c
5
6 D
1
b 8x – -----------------9 – x2
d x
g
x
e tan−1 --- + c
2
5 a B
4
c 4 cos 4x + ----------------16 + x 2
x 3
8 (tan –1 --- )
2
-------------------------4 + x2
627
answers
Answers
5E
answers
628
Answers
9 sin−1 x2 + c 10 2 cos−1
x + c 11 2 tan−1
13 y = x3 +
12 y = 2x + sin−1 x
1
--2
x+c
tan−1 2x −
1
--8
History of mathematics
1 Women were not allowed to attend lectures or
matriculate.
2 Teaching mathematics at primary school
3 She was a woman.
2 9 + 6 loge x
3 a 3x2 − 8x + 1
b 6x − 8
4 a (1, 4) a local max and (4, −23) a local min.
b (2 1--2- , − 9 1--2- )
5
y
y = x 4 – 3x 3 + 2x 2
(0.61, 0.20)
Exercise 5F — Implicit differentiation
(0, 0) 0
1 D
dy
x
c ----- = – -dx
y
dy
4x
f ----- = -----dx
y
dy
24x
+ 5i ------ = ----------------dx
2y
2
dy
2x
dy = – ------3x - b ----3 a ----- = – -------2
2y
dx
dx
3y
2
dy
x
d ------ = ----dx
y
4 E
dy
g ------ =
dx
dy
i ------ =
dx
x
2
(1.64, –0.62)
dy
1
b ------ = --dx
3
dy
x
e ----- = -dx
y
dy
6
h ------ = --dx
y
dy
1
2 a ------ = --dx
2
dy
x
d ----- = – -----dx
5y
dy
6
g ----- = --dx
y
dy
5 a ----- =
dx
dy
c ----- =
dx
dy
e ------ =
dx
1
2
x
dy
c ------ = – ----2
dx
y
2
dy
dy
15x + 4
x+3
e ------ = --------------------- f ------ = ----------2y
dx
dx
2y
1 – 2y
b
--------------2x
4 – 3y
d
--------------3x + 7
3
(1 + y )
f
– --------------------2
1 + 3xy
3
1 – 2xy
h
-------------------2 2
3x y
3
2(1 – y )
---------------------2
6xy + 5
3x + y
b
– --------------x + 5y
2y ( 3x + 1 ) – -----------------------------------
dy
1 – 2y
------ = --------------dx
2x – 1
2
dy
1 – 2y
------ = -----------------4xy – 1
dx
2
dy
2x – 3y
------ = -------------------6xy – 1
dx
5
5
13 ⎛ 2 5, – -------⎞ , ⎛ – 2 5, -------⎞
⎝
2⎠
2⎠ ⎝
Analysis
1 a $144
2 2
dy
1 – 9x y
------ = ----------------------3
dx
6x y + 1
Short answer
1 a 2e2x sec2 (e2x)
E
B
E
E
C
4
9
14
19
24
B
D
B
B
E
2( x2
c
d
e
f
g
1
t
0
Chapter review
3
8
13
18
23
b $494
144
x
c A = --------- + 2 + -----d $13.44
x
10
e $9.59 with 37.95 litres
2 a
b 70.5%
p
dy
dy
3x – y
6 a ------ =
------ = --------------dx
dx
x – 5y
dy
c ------ =
2
dx
3x + 2x + 10y
dy
– ( 1 + 6y – 4x )
d ------ = ----------------------------------dx
3 ( 2x – 3y )
a
dy
1 – 4xy
−
7
8 ------ = -----------------+ --------------------2
dx
2
2x + y
16 – a
Multiple choice
1 A
2 C
6 A
7 D
11 E
12 C
16 A
17 C
21 B
22 D
3
– --1
6 a -----------------------b x ( 144 – x 2 ) 2
2
144 – x
4
– 128x
b ---------------------------7 a --------------------2
1 + 16x
( 1 + 16x 2 ) 2
3
1
b -------------------------------8 a -----------------------------------2
1 – ( 3x + 4 )
1 – ( 4 – x )2
2
c -------------------------------1 + ( 2x + 5 ) 2
π
x
2x
10 f(x) = tan−1 --- – 5 – -----9 2 cos−1 ------ + c
16
4
9
5x
dy
12y – 9x
11 f(x) = 2--5- sin−1 ------ + 0.93 12 ----- = -----------------------3
dx
16y – 12x
5
10
15
20
2x
+ 1)
b -------------- – ---------------------tan 2x
sin 2 2x
A
C
D
C
70.5%
2 years
–0.064 per year
At the start, i.e. t = 0
After 6 years
3 a f′(x) = 6x2 – 30x + 24
= 6(x – 1)(x – 4)
f′(x) = 0 when x = 1, 4
f″(x) = 12x – 30
f″(1) < 0, thus local maximum at x = 1
f″(4) > 0, thus local minimum at x = 4
b Coordinates of local maximum (1, 15)
Coordinates of local minimum (4, –12)
c The point C has coordinates (5.50, 15.00)
d y
15
A
4
0 0
1
C
2
3
4 5 x
B
Area of region ABC
=
∫
5.5
3
2
( 15 – 2x + 15x – 24x – 4 ) dx
1
( x3 – 5 )3
g --------------------- + c
3
h 2 x 3 + 2x 2 + c
i ex + c
j −cos(x2 + 3x − 2) + c
k sin(x3 + 5x) + c
l
– sin 5 x
m --------------- + c
5
( log e x ) 2
n -------------------+c
2
4
5.5
4
3
2
x
= – ----- + 5x – 12x + 11x
2
1
≈ 68.34
4 a f ′(x) = 2axe–bx – abx2e–bx
f ′(x) = 0
2
x = --b
There is only one stationary point where x ∈ R+
f ″(x) = 2ae–bx – 4abxe–bx + ab2e–bx
2
f ″( --- ) = 2ae–2 – 8ae–2 + 4ae–2
b
a
= – 2 ----- < 0
2
e
2
The stationary point at x = --- is a maximum.
b
2 4a
b Coordinates of stationary point: ⎛ ---, ----------⎞
⎝ b 2 2⎠
e b
–1 x
5 a f ( x ) = tan --a
–
4 a --------------------- + c
3
( x3 – 1 )8
c --------------------- + c
24
f ″(x) = 0 when x = 0
1
There is a maximum gradient at x = 0, f′ ( 0 ) = --- .
a
b When x = 0, f″(x) = 0, also x < 0, f″(x) < 0 and
x > 0, f″(x) > 0. Hence there is a point of
inflection in the function f(x) at x = 0.
dy
6 a ------ = – --y
dx
x
dy
b ------ = – --y
dx
x
dy
c ------ = – --y
dx
x
d The answers to a, b and c are the same.
CHAPTER 6 Integral calculus
Exercise 6A — Substitution where the
derivative is present in the integrand
x 2 + 3 ) 5- + c
1 a (--------------------5
( x3 – 2 )6
c --------------------- + c
6
3
---
1
d – ---------------------------- + c
2 ( x 2 + 4x ) 2
f
1
– ---------------------------- + c
3 ( x 2 – 3x ) 3
( 4 – x2 )4
b – --------------------- + c
8
( x 2 + 6x – 2 ) 5
d ---------------------------------- + c
10
4 x 2 + 3x + c
3
--( x 2 – 5x + 2 ) –5
( 4 – 3x + x 3 ) 2
g – ------------------------------------ + c h 2------------------------------------+c
5
9
3
ex + 2
j ------------- + c
3
sin ( 6x – x 3 )
cos ( x 2 + 2x – 3 )
k – ---------------------------------------- + c l – ----------------------------- + c
2
3
− x + 3)
+c
cos 5 2x
m – ----------------- + c
10
3 3x
n sin
---------------- + c
9
( log e 3x ) 2
o ----------------------+c
4
p [loge(x2 − x)]2 + c
7
---
5 a
c
e
g
i
( x2 + 1 )2
---------------------- + c
7
( 3 + 2e x ) 5
------------------------- + c
10
– cos x 3
------------------ + c
3
[ log e ( sin x ) ] 2
---------------------------------- + c
2
2
⎛ cos –1 --x-⎞ + c
⎝
3⎠
3
---
( 1 – x2 )2
b – --------------------- + c
3
1
d ----------------- + c
2 cos 2 x
f −ecosx + c
( 1 – e3 x )3
h – ----------------------- + c
9
3
---
j
2( x + x2 – 3 )2
---------------------------------- + c
3
sin ( x 2 + 4x )
k ------------------------------ + c
2
l 2e
( sin –1 4x ) 2
m ------------------------- + c
8
( tan –1 x ) 2
n ---------------------- + c
2
x+1
+c
– log e ( 1 – 4x 2 )
o ----------------------------------+c
8
2 1 + 3sin x
6 a -------------------------------- + c
3
3
---
2 ( 2 + tan x ) 2
b -------------------------------- + c
3
5F
➔
2 ( x 2 + 5x ) 2
e ---------------------------- + c
3
1
b ------------------------- + c
2( 6 – x2 )2
c E
( x 2 + 2x + 3 ) –3
e – ------------------------------------ + c f
6
2
2ax
f ″ ( x ) = – -----------------------2
2 2
(x + a )
b B
2 )6
i 3e(x
a
f ′ ( x ) = ----------------2
2
x +a
( sin –1 x ) 3
p ---------------------+c
3
tan 4 x
o ------------- + c
4
2 A
3 a D
( x3
sin 4 x
------------ + c
4
answers
629
Answers
6A
answers
630
1
d – ------------------------ + c
2( e2 x – 3 )
sec2x + c
1
--2
c
Answers
1
e ------------------------------- + c
2 ( 5 – tan x ) 2
( log e x ) 4
-------------------+c
4
g
i
2
k
1
--5
ex
e–x
+
f 4loge(logex) + c
h etanx + c
+c
cos x − cos3x + c
5
1
--3
j −loge(sin x + cos x) + c
1
--5
l
x2 + 5 – 2
9 g(x) = (logex2)2 − 2
1 a i
b i
c i
∫
∫
8 f ( x ) = 2e
u
∫
i ( u – u ) du
∫
i ( 4u + 12u ) du
∫
u –u
i ---------------- du
∫ 2
u –u
i ---------------- du
∫ 3
e
f
g
h
i i
j i
k i
l i
m i
n i
o i
4
3
5
5
4
6
5
⎛ --72
--- ⎜ u 4
3
∫⎝
∫
∫
∫
∫
3
---⎞
+ 2u 4⎟ du
⎠
4
---
1
---
u 3 – 7u 3
-------------------- du
4
3
1
⎛ u --2- – 3u --2-⎞ du
⎝
⎠
∫
∫
4
3---
3---⎞
⎛ 5--du
⎝ u 2 + 6u 2⎠
5---⎞
⎛ 7--⎝ u 2 – 7u 2⎠
-------------------------- du
4
⎛ 1--⎝ 2u 2
+
1
– ---⎞
12u 2⎠ du
3
---
3
---
5
---
7
---
13
------
10
------
7
---
7
---
11
------
15
------
f
84– 4--7- ( 1 – x ) 4 + ----( 1 – x ) 4 – ----(1 – x) 4 + c
11
15
ii 4loge(x − 3) + c
g
2
--- ( x
7
------ ( x – 2 ) 2 + 6 ( x – 2 ) 2 + c
– 2 ) 2 + 12
5
2 log e ( 3x + 5 )
ii ---------------------------------+c
3
h
2
--- ( x
7
------ ( x + 1 ) 2 + 32
------ ( x + 1 ) 2 + c
+ 1 ) 2 – 16
3
5
10 g(x) = 1 − 4cos4x
j
3
---
( 10x – 1 ) ( 2x + 1 ) 5
ii ---------------------------------------------- + c
60
( 18x + 1 ) ( 1 – 3x ) 6
ii – ---------------------------------------------- + c
126
ii
8
--------- ( 3x
231
7
---
– 2 ) 4 ( 21x + 8 ) + c
--4-
3 ( 8x – 21 ) ( 2x + 7 ) 3
ii ------------------------------------------------- + c
112
3
---
2( x – 2)( x + 3)2
ii ---------------------------------------- + c
5
ii
2-------( 3x
135
--3-
– 4 ) 2 ( 9x + 8 ) + c
5
--4)2
2 ( 5x + 22 ) ( x –
ii ---------------------------------------------- + c
35
7
--1)2
( x – 4 ) ( 2x +
ii ---------------------------------------- + c
9
1
--6)2
------ ( x + 2 ) 3 + c
+ 2 ) 3 – 6--5- ( x + 2 ) 3 + 12
7
7
---
5
---
7
---
5
---
3
---
3
---
i loge(ex + 1) + c
( 3 – 2x ) 2
ii – ---------------------- + c
3
( 4x – 1 ) ( x + 1 ) 4
ii ---------------------------------------- + c
20
2 ( 5x + 3 ) ( x – 3 ) 5
ii ------------------------------------------- + c
15
1---
u 2 + 4u 2
--------------------- du
9
5
---
3----(x
13
+1
( 4x +
ii ----------------------- + c
6
u
– ------- du
2
d i
7
---
– 1 ) 2 + 4--5- ( x – 1 ) 2 + 2--3- ( x – 1 ) 2 + c
e
x
3--1)2
∫ ------4- du
2
--- ( x
7
12d – 6 ( 3 – x ) 2 + ----( 3 – x ) 2 – --27- ( 3 – x ) 2 + c
5
Exercise 6B — Linear substitution
4
--- du
u
2
------ du
3u
∫
1
------ ( 5 – x ) 4 – --- ( 5 – x ) 6 + c
b 2 ( 5 – x ) 5 – 25
4
6
c
sin5x − 1--7- sin7x + c
[ log e ( tan x ) ] 2
-+c
m --------------------------------2
7 f ( x) =
1
--⎛ --1– --1-⎞
ii –2 ( x + 16 ) ( 8 – x ) 2 + c
⎝ 3u 2 – 24u 2⎠ du
b E
2 a C
a
B
b
D
3
4
16
1
7
- ( x – 4 )5 + c
4 a --7- ( x – 4 ) + --3- ( x – 4 ) 6 + ----5
p i
4 ( x + 12 ) ( x –
ii ------------------------------------------- + c
3
2
--- ( x
5
5
---
3
---
1
---
+ 1 ) 2 – 4--3- ( x + 1 ) 2 + 2 ( x + 1 ) 2 + c
--1-
--3-
--5-
k – 36 ( 3 – x ) 2 + 8 ( 3 – x ) 2 – 4--5- ( 3 – x ) 2 + c
l
2
--- ( x
9
9
---
7
---
5
---
3
---
– 1 ) 2 + 6--7- ( x – 1 ) 2 + 6--5- ( x – 1 ) 2 + 2--3- ( x – 1 ) 2 + c
7
---
5
---
3
---
1---
------ ( x + 4 ) 2 + 32 ( x + 4 ) 2 – 128 ( x + 4 ) 2 + c
m 2--7- ( x + 4 ) 2 – 24
5
1
---
3
---
5
---
7
---
12
4
- ( 1 – x ) 2 + --- ( 1 – x ) 2 + c
n – 4 ( 1 – x ) 2 + 4 ( 1 – x ) 2 – ----5
7
5
2
3
o log e ( x – 2 ) – ----------+ c p ---------------------- – ------------ + c
x–2
2( x + 1 )2 x + 1
8
q 4 log e ( x + 2 ) + -----------+c
x+2
1
r ( x – 1 ) + 2 log e ( x – 1 ) – ----------+c
x–1
s
2
--- ( x
5
5
---
3
---
--5-
t – 2--5- ( 2 – x ) 2 + c
v
1
--2
1
---
+ 2 ) 2 + 4--3- ( x + 2 ) 2 + 2 ( x + 2 ) 2 + c
u ex + 2 − 2loge(ex + 2) + c
(ex −1)2 + 2(ex − 1) + loge(ex − 1) + c
3
---
1
---
5 a f ( x ) = 2--3- ( 5 – x ) 2 – 20 ( 5 – x ) 2 + 32 2--35
---
3
---
b x≤5
1
---
6 a f ( x) = ( x + 1)2 – 2( x + 1)2 + ( x + 1)2 + 1
b x ≥ −1
3
7 a g ( x ) = 2 log e ( x – 1 ) – ----------- + 3
x–1
b x>1
8 a g ( x ) = e x + 1 – log e ( e x + 1 )
b R
5 a cos x − 2--3- cos3x + c
Exercise 6C — Antiderivatives
involving trigonometric identities
1 a
1
--2
x + --14- sin 2x + c
b
1
--2
x − 1--8- sin 4x + c
c x+
c
x+
1
-----20
sin 10x + c
x−
1
-----24
sin 12x + c
e
f
1
--2
g
1
--2
(x + sin x) + c
l
1
--2
1
--2
sin 2x − 1--3- sin32x + c
cos 3x − 2--9- cos33x + c d
1
--4
sin 4x − 1--6- sin34x + c
x
x
f 3sin --- − 2sin3 --- + c
3
3
b − 1--8- cos42x + c
x
c − 1--3- cos6 --- + c
2
2x
x − 3--4- sin ------ + c
3
x
i 3--2- x + 9--2- sin --- + c
3
x
j x − 2sin --- + c
2
4x
k 1--2- x + 3--8- sin ------ + c
3
h
1
--2
6 a − 1--5- cos5x + c
d 2x − --13- sin 6x + c
1
--2
b
x
x
e 2cos --- − --43- cos3 --- + c
2
2
sin 8x + c
1
--8
1
--3
631
d
1
-----15
sin53x + c
2x
sin8 ------ + c
3
e
5
--7
x
sin7 --- + c
5
f
3
-----16
7 a
1
--5
cos5x − 1--3- cos3x + c
b
1
--3
c
1
-----10
d
1
--9
sin3x − 1--5- sin5x + c
cos52x − 1--6- cos32x + c
sin33x −
1
-----15
sin53x + c
x
x
cos5 --- − 2--3- cos3 --- + c
2
2
3x
5 3x
2
- sin ------ + c
f 2--9- sin3 ------ − ----15
2
2
5x
3x
12
g -----5- cos --- − 4cos --- + c
3
3
3 5x
5 5x
8
24
- sin ------ + c
h − --5- sin ------ + ----25
4
4
x − 1--6- sin 3x + c
2 a sin2x + c (or − --12- cos 2x + c)
b − 1--2- cos 4x + c
e
2
--5
i
7
5
1
1
--- cos x − --- cos x + c
7
5
5
7
1
1
------ sin 2x − ------ sin 2x + c
10
14
8
6
1
1
--- cos 2x − --- cos 2x + c
8
6
7
9
2
2
- sin 3x + ------ sin 3x + c
– ----21
27
c
1
- cos 6x + c
− ----12
j
d
1
--8
cos 8x + c
k
e
1
--8
f
1
--8
g
x−
1
-----32
sin 4x + c
l
x−
1
-----64
sin 8x + c
m
1
--4
x−
1
-----64
sin 16x + c
h
1
--4
x−
1
-----48
sin 12x + c
i
3
--4
x − sin 2x + c
8 a
1
--2
tan 2x
j
1
--2
4x
x − --38- sin ------ + c
3
c
1
--3
tan3x
x
b 3tan --3
d 1--4- tan4x
k
1
--8
x−
e
1
--3
tan62x
f
l
− 1--4- x +
g
1
--3
tan3x + 1--5- tan5x
h tan32x + 3--5- tan52x
i
4
--3
x
x
tan3 --- + 4--5- tan5 --2
2
j
x
x
cos9 --- − --87- cos7 --- + c
2
2
8 3x
10 3x
1
1
- sin ------ − ------ sin ------ + c
n ----12
15
2
2
3
--8
3 a C
1
-----80
sin 10x + c
3
-----64
16x
sin --------- + c
3
b A
b
sin 2x − sin 2x
c – --32- cos 4x + --12- cos 4x
d
4
--3
sin 3x − --49- sin33x
x
x
k tan5 --- + --57- tan7 --5
5
cos37x − 1--7- cos 7x
f
1
--6
sin 6x −
l
4 a
cos x − cos x
c E
8
--9
1
--2
1
--3
3
3
e
1
-----21
x
x
g 2 cos3 --- − 6 cos --2
2
answers
Answers
1
--6
1
-----18
3
sin36x
x
x
h 6 sin --- − 2 sin3 --3
3
2
--9
3x
3x
cos3 ------ − 2--3- cos -----2
2
j
2
--5
5x
sin ------ −
2
k
4
--9
3x
3x
cos3 ------ − 4--3- cos -----4
4
l
3
--4
4x
4x
sin ------ − 1--4- sin3 -----3
3
2
-----15
5x
sin3 -----2
1
--4
x
tan5 --2
tan43x + 1--6- tan63x
tan66x + 1--2- tan86x + 1--5- tan106x
– cos n + 1 x
sin n + 1 x
b ------------------- + c
9 a ----------------------- + c
n+1
n+1
cos n + 3 x cos n + 1 x
tan n + 1 x
d -------------------- – -------------------- + c
c ------------------- + c
n+3
n+1
n+1
n
+
1
n
+
3
sin
x sin
x
e ------------------- – ------------------- + c
n+1
n+3
6B
➔
i
1
--3
16
-----5
6C
answers
632
Answers
10 f (x) = −2 cos3x +
x
11 f (x) = --- −
2
12 g(x) =
2
--7
1
-----16
f 3loge(x + 1) − 2loge(x + 6) + c
g 2loge(x + 3) + 5loge(x − 3) + c
h 3loge(x + 1) + 4loge(x − 1) + c
1
--4
7π
sin 8x + -----8
7x
5x
cos --- − cos --2
2
2
--5
Exercise 6D — Antidifferentiation using
partial fractions
1 a
c
e
g
a = 1, b = 2
a = 3, b = 10
a = −2, b = 3
a = −6, b = −1, c = 1
1
1
2 a ------------ – -----------x+1 x+2
9
3
c ------------------- + -------------------2( x + 3) 2( x – 1)
b
d
f
h
a = −1, b = 2
a = 2, b = −1
a = –5, b = 7
a = −1, b = 2, c = 2
3
3
b ----------– -----------x–2 x+2
2
1
d ----------+ -----------x–2 x+1
1
e -----------x+2
f
4 – -----------------3 g ----------x + 2 ( x + 2 )2
5
1 h ----------– -----------------x – 5 ( x – 5 )2
3
2
----------- – -----------x–4 x+4
i
1
--2
loge(2x + 1) + 2loge(x − 2) + c
j
4
--3
loge(3x − 2) − 2loge(x + 3) + c
k 2loge(x − 2) −
3
--2
loge(2x − 1) + c
2+x
l loge ------------ + c, –2 < x < 2
2–x
m −loge(4 − x) − 2loge(4 + x) + c
n 2loge(1 + x) − 3loge(5 − x) + c
7 a x − 6loge(x + 5) + c
b x + 5loge(x − 2) + c
c x−
1
--3
loge x −
8
--3
loge(x + 3) + c
d
e
f
g
x + 7loge(x − 4) − loge x + c
x + loge(x + 1) − 6loge(x + 3) + c
x + 4loge(x − 3) − loge(x + 1) + c
x + loge(x + 2) + 2loge(x − 2) + c
h
1
--2
x2 + x − 10loge(x + 2) + 4loge(x + 1) + c
j
2
5
----------- + -----------x–2 x+3
i
1
--2
x2 + 4x + 22loge(x − 5) + 3loge(x + 1) + c
4
2
k -------------- + ----------2x + 1 x – 3
l
–3
4
--------------- + -----------3x – 2 x + 1
j x2 + x + 3loge(x + 1) − loge(x – 1) + c
1
4
m ----------+ -----------2–x x+3
5
3
n ----------– ----------1–x 3–x
i
–2
3
------ + ----------x x–2
x+1
3 a log e ⎛⎝ ------------⎞⎠ + c
x+2
c
d
e
f
x–2
b 3log e ⎛ ------------⎞ + c
⎝ x + 2⎠
1
--- loge[(x + 3)9(x − 1)3] + c, x > 1
2
2loge(x − 2) + loge(x + 1) + c
loge(x + 2) + c
3loge(x − 4) − 2loge(x + 4) + c
3
g 4loge(x + 2) + ------------ + c
x+2
1
h 5loge(x − 5) + ----------- + c
x–5
i 3loge(x − 2) − 2loge x + c
j 2loge(x − 2) + 5loge(x + 3) + c
k 2loge(2x + 1) + 2loge(x − 3) + c
l 4loge(x + 1) − loge(3x − 2) + c
m 4loge(x + 3) − loge(2 − x) + c
n 3loge(3 − x) − 5loge(1 − x) + c
b B
4 a D
5 C
6 a 5logex − 2loge(x + 2) + c
b 2loge(x − 2) + 3loge(x + 1) + c
c 2loge(x + 1) − loge(x + 2) + c
d 5loge(x − 6) + loge(x + 1) + c
e 4loge(x − 3) + loge(x − 1) + c
4
k x − 3loge(x + 1) − ------------ + c
x+1
2
l 2x + 3loge(x − 3) + ----------- + c
x–3
8 a
b
c
d
e
2loge x − 3loge(x + 2) + c
5loge(x − 3) + 4loge(x + 4) + c
3loge(x − 5) + 2loge(x + 5) + c
x + 2loge(x − 3) − 2loge(x + 3) + c
x + 4loge(x − 4) + loge(x + 2) + c
f x−
g
1
--2
11
-----4
loge(x + 7) + 3--4- loge(x − 1) + c
4
x2 + 5x – ----------- + 12 loge(x − 2) + c
x–2
h 2x + 4--7- loge(2x + 3) +
i
1
--2
24
-----7
loge(x − 2) + c
x
loge(x2 + 4) + --12- tan−1 --- + c
2
x
j 2loge(x2 + 9) − 2--3- tan−1 --- + c
3
k −4loge(x − 1) +
33
-----15
loge(x + 2) +
34
-----5
loge(x − 3) + c
l 2loge(x + 1) + 3loge(x − 2) − 4loge(x + 3) + c
1
m 2loge(x − 1) − loge(x + 2) − ------------ + c
x+2
n loge(x2 + 1) − tan−1 x − 3loge(2 − x) + c
6( x – 1)
9 a f ( x ) = 3log e -------------------x+1
b x>1
5
( log e 3 ) 2
g ------------------2
( x – 3)
10 a g(x) = x + --12- loge -------------------5( x + 1)
b x>3
j
Exercise 6E — Definite integrals
1 a
b
c
d
e
f
g
h
i
j
i (−3, 3)
i (−2, 2)
i (−4, 4)
iR
i R\{0}
i R\{–1, 0}
i R\{–2, –3}
i R\{1}
i R\{2, 6}
iR
k i (− --13- , --13- )
l i
ii The integral does not exist.
ii The integral does exist.
π
b – --6
loge --52-
1
--6
e 2loge4
π
j -----36
3 a C
4 a E
5 a
∫
n 6+ e − e
∫ ⎝u
⎛
3
--2
0
log e 3
d
1
---⎞
+ u 2⎠ du
f
0
0
∫ (u
11
– u 10 ) du
j
1
k
m
∫
1
u 3 du
l
0
0
∫ ( 1 – u ) du
2
n
1
6 a 832
16 2
d ------------3
∫
–1
−6
2u –2 du
–2
∫
1
--8 u2
m – 2--3-
4
⎛
∫ ⎝u
∫ –e
3
--2
1
1
--– ---⎞
– 2u 2 + u 2⎠ du
0
∫
--12
1 1--u2
u
du
∫
∫
du
0
2
π
----4 sin
0
u
----------- du
2
e + 1 – 1--u 2
du
2
b 1
2 2–1
c ------------------3
1
e 1 ----15
1
f 5 ----15
1
--------132
l Approx. 0.89
n 2 e + 1 – 2 2 or approx. 1.028
7 a 2 − 2e4
c
b 18
4
--3
8 2 – 10
d ---------------------3
e
π
g --3
h 1.249 (or tan−13 or −tan−1 (−3))
31
--------160
π
j
– --2
l loge6 (or 0.981)
n
4
--3
-----f loge 27
5
2
k −loge ------2
m 4−π
π
p ------ + 3 – 1
12
4
-----15
i
o e2 − 1
------ (or 10.626)
r 8 + 3loge 12
5
π
8 --4
π 1
10 --- + --8 4
2π
3
9 ------ + ------3
2
11 a = 1
12 a = −1
13 a = 2
14 a =
2
Exercise 6F — Applications of
integration
1 a i x=0
b i x=1
ii 18 sq. units
ii 1 sq. unit
c i x = 0, 1
1
- sq. units
ii 17 ----15
d i x=
ii loge2.88 sq. units
2
--3
π
ii --- sq. units
6
e i None
f
----- du
0 2
1
h
u du
1
--2
c B
b
u2
----- du
1 2
6
1
--2
1
---
2
e
i
l
9 3
---------5
--- du
3
2
∫
∫
g 2loge6
b A
b D
c
g
f
loge 4--3-
10 u
1
d Approx. 1.893
i
1
--4
k
-----q 23 11
15
ii The integral does exist.
m i R\{2}
n iR
2 a
2
--3
ii The integral does not exist.
, ∞)
[ 1--2-
ii The integral does exist.
ii The integral does exist.
ii The integral does not exist.
ii The integral does exist.
ii The integral does exist.
ii The integral does exist.
ii The integral does exist.
ii The integral does not exist.
ii The integral does not exist.
ii The integral does exist.
1
---
h e2 – 1
answers
633
Answers
π
i x = (2n + 1) --- , n is an integer
2
π
ii --- sq. units
4
( 2n + 1 ) π
g i x = 0, ± ------------------------ , n ∈ J + ∪ {0}
2
ii Approx. 0.8897 sq. units
h i None
e+2
ii loge ⎛ ------------⎞ or approx. 0.453 sq. units
⎝ 3 ⎠
ii 2 2--3- sq. units
2 a i x = y2
b i x=1±
y
ii 7 2--3- sq. units
c i x = sin y
ii
d i x = ey
e i x = cos y
ii (e2 − 1) sq. units
ii 1 sq. unit
1
---
f i x = y3
g i x = tan y
1
--2
sq. units
ii 12 sq. units
ii loge 2 or 0.347 sq. units
➔
6D
6F
answers
634
Answers
8 2
b ---------- sq. units
3
3 a 5 1--3- sq. units
c Approx. 0.464 sq. units
d (e2 + 1) sq. units
e 3π sq. units
f 1 1--3- sq. units
4 a D
5 a E
6
6 3--4-
b A
b C
sq. units
0
10 a
–π
2
0
–3
21 1--3-
9
11 20 --56- sq. units 12
1
--3
0
sq. units
b i
0
c i
x
0
d i
13 2 sq. units
1
– 1–2 –1 0
b 24π cubic units
c Answers will vary.
x
256 π
2 a ------------ cubic units
3
3 a
y
y = x–1
2
0
1
5 x
x
3
–7
y
2
e i
x2 + y2 = 4
22 π
ii --------- cubic units
3
x
0 1 2
8π
ii ------ cubic units
3
y
2
y = 2–x
2–
3
3 x
01
Exercise 6G — Volumes of solids of
revolution
2
ii 12π cubic units
y2 = 2x + 1
x
6
x
2
–2
19 a Answers will vary.
π–1
b ------------ or approx. 0.535 sq. units
4
0
y
b −1 + 2loge2 or approx.
0.386 sq. units
y = 3x
32 π
ii --------- cubic units
5
7
f i
y
x
4
0 x
1
–1 0 1
4
y = x2
–2 –1
y
1–x
y = x––––
+1
1 a
1
y
b x=0
c Approx. 2.929 sq. units
sq. units
15 π
ii --------- cubic units
2
2
1
(–2, 1)
18 a
y
y= x
b Answers will vary.
c ( 2 – 1 ) sq. units
π
14 a --- – 1 + e –1 or approx. 0.939 sq. units
2
b 1.276 sq. units.
15 abπ sq. units
16 3π sq. units
b y=x+3
17 a
y
y = e x +22
c (e2 − 5) sq. units
e
–2
x
2
sq. units
y = cos x
y=x+3
3
10 2--3-
x
π
y
y= 9–x
y
3
1
b
y = sin x
1
5 a i
π
b --- cubic units
2
26 π
y= x + 1 ii --------- cubic units
3
c C
7
8 a Answers will vary.
9 a
y
16 π
4 a --------- cubic units
15
g i
π2
ii ----- cubic units
2
y
1
– –π
0
2
y = cos x
h i
206 π
b ------------ cubic units
15
ii Approx. 4.787
cubic units
y = ex + 1 y
2
–2 –1
b Answers will vary.
x
–π
2
28 π
6 a --------- cubic units
3
c 8π cubic units
0
x
124 π
b ------------ cubic units
5
( 224 7 + 4 ) π
d ------------------------------------ or 124.96 cubic units
15
( 32 – 12 3 ) π
e ---------------------------------- or 11.744 cubic units
3
f 8π cubic units
b D
c A
7 a B
π
2π
b --- cubic units
8 a ------ cubic units
6
15
9 π cubic units
10 π (loge4 − 3--4- ) or approx. 1.999 cubic units
5 f (x) = − 2--3- cos3x +
c
(4 – 2 3)π
1
--3
2
6 f (x) = sin–1x + x 1 – x – 3
π
7 a --8
8 a
π2
11 ⎛ ----- – π⎞ or approx. 1.793 cubic units
⎝2
⎠
227 π
12 ------------ cubic units
30
1
13 a ⎛⎝ 1, -------⎞⎠
3
635
b – --23b (e2 − 1) sq. units
y
answers
Answers
y = loge x
2
0
π
b --- loge 3 cubic units
4
1
--6
9
cubic units
14 a 24π cubic units
48 π
15 a --------- cubic units
5
2π
16 a ------ cubic units
35
625 π
17 ------------ cm3
3
b 16π cubic units
π
b ------ cubic units
10
140 35 π
18 ----------------------- cm3
3
b 320π 2 cm3
Exercise 6H — Approximate evaluation
of definite integrals and areas
a 1.090
b 2.0523 c 2.605
d 3.9292
a 1.117
b 1.896
c 2.564
d 3.805
a 6
b 11.4375 c 4.34
d 44
a 6
b 13.125 c 4.5625 d 40
a D
b B
a C
b D
a 139.8
b 144.2
a 9.374
b 9.314
a 2.575 sq. units
b 2.553 sq. units
2.041 sq. units
9+6 2
11 a 4.5 sq. units
b ------------------- or 4.37 sq. units
4
1
2
3
4
5
6
7
8
9
10
π
12 a --- sq. units
2
x
π
10 (1 − e−2) --- cubic units
2
sq. unit
11 63 π cm3
b 4.8π cubic units
19 130π cm3
2
2
20 a x + (y − 10) = 16
e2
1
12 a i 42
ii 44
b The midpoint answer is closer in this question as
the exact answer is 42 2--3- .
Analysis
1 a 0.4388 sq. units b 1.7932 cubic units
c 0.7641 cubic units
2 a 7.2 cm b π (e2 − 1) or 20.07 cm3 c 318.27 cm3
3 a 3.75 m b 5.96 m2 c 13.39 m3 d 119.2 m3
4 a a = 1--9- ; b = 9
b
P (3, 3)
y
0
3
432
d --------- π cm
35
b 2.06 cm2
c 4.5 cm2
5 a
y
1
a
−b
b 2.106 sq. units
x
b
x
Chapter review
Multiple choice
1 B
2 D
6 A
7 E
11 A
12 D
16 C
17 A
21 E
22 A
3
8
13
18
Short answer
1 a esin x + c
2
2
--3
A
E
E
C
4
9
14
19
C
C
E
D
5
10
15
20
E
D
B
B
c V = 2π
∫
5
9
---------------------dx
2 2
0 (9 + x )
Technology-free questions (page 313)
1 a cos x – xsin x
b
1
--3
(loge x)3 + c
(x − 2) x + 1 + c
–x
c -----------------2
1–x
2cosx
2 a -------------3
sin x
c –sec2x
–3
b ---------------------------------------2
– 3 + 12x – 9x
1
b – ----- [ sin ( log e x ) + cos ( log e x ) ]
2
x
6G
➔
x
x – sin x
b -------------------- + c
3 a --- + 1--8- sin 4x + x
2
8
4 x + 4loge(x − 8) + loge(x + 1)
d ≈1.54 cm3
6H
answers
636
Answers
3 Vertical asymptote at x = 0,
minimum turning point at
(2, 1 + loge2) and no x-intercept.
4
17 a
y
4
3
2
1
2
y = x + loge x
0
1 2 3 4 x
y
(2, 1 + loge2)
−1
0
4
--3
b
y = x 2 − 2x
square units
x
1 2
(1, −1)
y
y = f −1(x)
3
c
y=x
(2, 3)
16 π
--------- cubic units
15
18 a
y=
0
0
5
c
7 a
1
y1 = e − 1
(3, 1 )
2
2
------ + π
--⎝ 3⎠ 6
– ( x + 2y )
----------------------2x + y
2–x
-----------y+1
sin
6 a
1
f(x)
y 2
b ⎛ ------⎞
⎝ 2x⎠
3
cos 2x – 1--2- cos 2x b
2
–1 x
log e ( x + 4 ) + tan --2
3
3
--2
---
2
x–2
2
- ( x – 1 ) ( 3x + 2 )
c 3log e ⎛ -----------⎞ , x > 2 d ----15
⎝ x – 1⎠
1
3 –4 x –1
e --x- – ------ sin8x
f --x + --- e
8 64
e 2
π- + ------38 a 4log e 3 – 4
b ----12 16
c 4--9d log e 1--2-
9 a
b tan ( x + 1 ) – x + c
1
- cos6x + c
– ----12
3
c
1 x +5
--- e
3
1
--3
d
+c
3
tan x + c
1
– 1--3- cos 2x – ----24
11
x + 2x loge x,
1 2
--- ( e
2
–1
–1
2
x
cos x – ------------------ , x cos x – 1 – x + c
2
1–x
3
8
- (or 9.3)
13 a ----b 9 ----10
15
c
14
log e 2.4
5
-----12
b
y
8
1
- square units
3 ----12
2
−1
y=
0
1
1 + x2
x
1
dy
2x
b ----- = – ---------------------; equation of tangent is
2 2
dx
(1 + x )
y = − 1--2- x + 1.
1
--4
( π – 3 ) square units
CHAPTER 7 Differential
equations
Exercise 7A — Differential equations:
related rates
8πr
2 a --------3
3 a 6r
x
e
y = x3
16 a (0, 0), (4, 2)
(0, 1)
1
2
2
d – -------5x 2
y = x 2 + 2x
−2−1
−1
y1 = 1 + x 2
dA
1 a ------dL
square units
15 a
y = −1
+ 1)
12
1
--2
x
1
π 2
c --- ( e – 3 ) cubic units
2
19 a (–1, –1), (1, 1)
4π
b i ------ ( 2 – 1 ) cubic units
3
4π
ii ------ cubic units
3
20 a
y
c
3
10
0
−1
x
3
– 1 ⎛ 2x⎞
1
--6
(1, e − 1)
e−1
y = f(x)
1
b 1 square unit
y2 = e x − 1
y
(3, 2)
2
b
8π
------ cubic units
3
h
4 a 24
dV
b ------dr
dP
c ------dh
da
d -----dt
c 0.04te 0.01t
b 0.9 A
2
e 4π r + 20π
b 4.5L2
c 3.2π
d 12 + 8r
f 0.2
2
g -----πr
1
h --------16L
b 14.4
c
1
--8
e3
8
d – --3
5
6
9
12
13
a B
b E
C
7 100π m2/h
8 80π cm2/min
0.1 cm/s
10 0.9 cm/s
11 7.5 beats/min/s
a 2.4π cm2/min
b 27π cm3/min
Approx. −0.000 44°C/s
14 −0.125 amp/s
16
15 Approx. 0.0025 cm/s
17 a D =
9 + x2
2
-----15
m/s
b 16 km/h
12h – 3h 2
18 a V = ------------------------ b i −10 mm/min ii −7 mm/min
2
History of mathematics
1 Calculus
2 Very few women received more than a basic
education.
3 The University of Bologna
4 The witch of Agnesi
Exercise 7D — Differential equations of
d2 y
the form --------2- = f ( x )
dx
1 a y = 1--2- x2 − x3 + cx + d
Exercise 7B — Verifying solutions
c y=
1
3
5
7
d y = 1--2- x−1 + cx + d
Answers will vary.
C
a = − --32- or 4
a = −3, b = 1
2 B
4 k = −1 or −2
6 a = −3 or 3
8, 9 and 10: Answers will vary.
Exercise 7C — Differential equations of
dy
the form ------ = f ( x )
dx
1 a y = x2 + x + c
c y = 1--3- x3 + 1--2- x2 − 6x + c
b y = 1--4- x4 − 2x2 + c
d y = 2--3- e3x + c
e y = 1--3- x3 + 3 log e x + c
f y=
1
--3
3
---
( x2 – 2 )2 + c
g y = 2 tan 2x + c
3
---
k y=
1
--2
1
--4
c y = 1 + 2 sin 3x
1
--3
x
tan−1 --- + c
2
1
--2
3
---
( 2x – 1 ) 2 – 5
3)2
(t +
f h = log e ---------------- , t > −1
t+1
t
e h = sin−1 --- + 3
4
g P = n2 – 7 + 2
h L = 2σ − sin 2σ − π
i M = σ + tan σ − 3
π
x
j u = x − 1 + --- − 2 tan−1 --2
2
k V = 2et
2+
2
+ e2 l y = 2 log e (x2 + 1) + tan−1 x + 7
3
--2
m L = 2 ( 1 + sin σ ) – 2 n y = −log e (2 − ex) + 3
6 y = 2x + 1
7 y = 3 cos x – π
8 h = 7.5t + 6 − 3 sin 0.5t
9 a V = 10 000(e0.1t − 1) b 6487.2 kL
c 2321.1 kL/h
2
(t2
3
--2
+ 4) – 8
b 3.46 s
1
--4
cos 2x + cx + d
b A
2 a E
3 C
4 a y = x2 − 3x + 1
c y = (x − 3)4 + 2
b y = 2x3 + 4x − 3
x
d 3 sin–1 --- − x + 2 3 – π
2
1 4x
π
e y = x − --- + sin 2x f f (x) = --2- e
6
g f (x) = 2 − cos (3x − 1) h g(x) = x − log e x
7
--2
b 125 m
(5 – x)
x
1
7 a y = – ------------------- − ------ + -----30 000
60 48
b Approx. 0.046 m (or 46 mm)
L4
8 ----------------------- , 0.444 m or 44.4 cm
5 760 000
Exercise 7E — Differential equations of
dy
the form ------ = g ( y )
dx
1 a y = ± x–c
1
--- ( x
b y = e3
c
y =
– c)
ex – c
e
Ae x
– 3 or
3x − 3c
d y= e
1
--3
x
---
or Ae 3 where A = e
+
5
--3
y = ± x+k −
1
--- ( c
g y = 40 – e 5
– x)
i y = tan−1 (x − c)
2 B
3 a D
b C
1
---
4 a i y = –3 ± 2 x 4
b i y =
1
--2
π
tan ⎛ 2x + ---⎞
⎝
2⎠
3
– 3 where A = e –c
or Ae3x +
7
--2
– --c-
where A =
1
--3
e−3c
7
y = --- ± k – x
4
f
h
j
5
--3
2
y = – ----------x–c
y = cos−1 (ec − x)
ii (0, ∞)
⎧ π
⎫
3π
ii R\ ⎨ ± ---, ± π, ± ------, … ⎬
2
2
⎩
⎭
7A
➔
10 a V =
–1
h y = 1--2- x2 +
4
4 A
b y = − --12- e−2x +
d y=
f f (x) = −2 tan−1 x + cx + d
g y = --13- x3 + cos x + cx + d
6 a h = −5t + 50t
sin4 σ + c
2 D
3 C
5 a y = x3 + x + 3
7
---
x 2 + cx + d
e f (x) = ex − e−x + cx + d
2
l y = – 1--4- cos 4x − 2 sin x + c m h = − 1--5- (5 − 2t)5 + c
n x=
24
-----35
x4 + --16- x3 − 3x2 + cx + d
5 y = --16- x4 − --13- x3 − 8x +
1
2x 2
h f (x) = – -------- + -------- + c
2
3
2x
i f (x) = x2 − 7x + 20 log e (x + 3) + c
x
j y = cos–1 --- + c
3
1
-----12
b y=
answers
637
Answers
7E
answers
638
Answers
c i y = 3 + 2x
ii (0, ∞)
d i y =
ii R\{−5}
3
3x + 15 – 1
Input
concentration
(kg/L)
Input
rate
(L/min)
Output
rate
(L/min)
Volume
at any
time
(Litres)
Output
concentration
(kg/L)
π
2 cos ⎛ x + ---⎞
⎝
4⎠
6t
3e
f i h = ---------------e 6t + 2
π 3π
ii ⎛ – ---, ------⎞
⎝ 4 4⎠
v 0.4
5
6
100 − t
Q
---------------100 – t
ii R
vi 0.5
12
10
100 + 2t
Q
--------------------100 + 2t
g i v=
ii ( --12- , ∞)
e i y =
1
--2
log e (2x − 1)
h i y = log e (2e − 1)
2
ii R\{loge 2}
i i y = -------------2 – ex
6e 4 x – 2
ii R
j i u = -------------------3e 4 x + 1
3
3x – 10
5 y = ------------------ – 1 or y = -----------------13 – 3x
13 – 3x
b Approx. 46 weeks
6 a P = 10e0.1t
b V = 50 cm3
7 a V = 10 3t + 1
8 46.44 weeks
Exercise 7F — Setting up and solving
differential equations
dP
dy
dy
b ------ = k 3 y c ------- = – k P
1 a ----- = x2
dt
dx
dx
dv
k
dv
dh
d ------ = -f ------ = k t 3
e ------ = kh
dt
x
dt
dt
2 65 726
3 0.693 years
4 3 hours 19 minutes
dP
b 1949
5 a ------- = 0.02P + 50
dt
6 17.39 weeks
7 46°C
b 86.3°C
8 a 14 minutes 22 seconds
b −2°C
9 a 4 minutes 22 seconds
10 5 minutes 20 seconds
Exercise 7G — Input/output of mixing
problems
b V = 100 + t c V = 100 − 2t
1 a V = 100
d V = 100 + 5t e V = 100 + t f V = 100 − 3t
b 12 g/min
2 a V(t) = 60 + t
5x
dx
5x
c -------------- g/min
d ------ = 12 – -------------60 + t
dt
60 + t
3 a
Input
concentration
(kg/L)
dQ
Q
i ------- = 5 – -----dt
10
b
ii (loge 1--2- , ∞)
x
Input
rate
(L/min)
Output
rate
(L/min)
Volume
at any
time
(Litres)
Output
concentration
(kg/L)
i 0.5
10
10
100
Q
--------100
ii 0.4
8
8
100
Q
--------100
iii 0.2
5
4
100 + t
Q
----------------100 + t
iv 0.3
4
2
100 + 2t
Q
-------------------100 + 2t
dQ
ii ------- = 3.2 – 2Q
------dt
25
dQ
4Q
iii ------- = 1 – ---------------dt
100 + t
dQ
Q
iv ------- = 1.2 – ------------dt
50 + t
dQ
6Q
v ------- = 2 – ---------------dt
100 – t
dQ
5Q
vi ------- = 6 – ------------dt
50 + t
dx
x
4 a ------ = – --------b Approx. 33.16 kg
dt
160
c Approx. 110 minutes 54 seconds
dQ
240 – 3Q
5 a ------- = ---------------------b Approx. 28.14 kg
200
dt
c Approx. 83 minutes 31 seconds
dx
30 – x
6 a ------ = -------------- b Approx. 27 minutes 28 seconds
dt
25
7 31.15 g/L
8 2 hours
dQ
3Q
dx
4x
10 ------- = 0.8 – ---------------9 ------ = 1 – ----------------dt
250 – t
dt
300 + t
Exercise 7H — Numerical solutions by
Euler’s method
1 a y = x2, y(2) = 4
b y = loge x, y(2) = 0.6931
2
c y = x + 8 , y(2) = 3.4641
d y = ex − 1, y(2) = 6.3891
2 a i 3.5
ii 3.8
b i 0.8333
ii 0.7457
The approximate solutions are more accurate with
h = 0.2.
3 a 5
d 4
b 118.9375
e 5.6054
b −1.5159
d 2.8
4 a 0.4501
c 1.3692
5 C
6 D
9 a
7 B
0
2
4
yn
15
10
5
0
8 0.8505
c The graphs are
very similar for
small values of
h, say h = 0.1.
yn
10
5
10
c 5.0625
f 7
1
2
3
xn
xn
Exercise 7I — Direction field of a
differential equation
c
1
—
y = _6 x3 + 17
6
4
P (1, 3)
3
1a
y
x
–3
–2
–1
0
1
2
3
dy
-----dx
–6
–4
–2
0
2
4
6
2
x
–4
–2
0
2
4
−4 −3 −2 −1
0
1
x
2 3
y
y= 4
y= 3
y= 2
y= 1
x y = −1
y = −2
y = −3
y = −4
y
–3
9
--2
–2
2
–1
1
--2
0
0
1
1
--2
2
2
3
9
--2
2a
−4 −3 −2 −1
2
x
0
1
2
3 x
1
5
9
13
17
21
E
B
B
A
E
E
2
6
10
14
18
B
D
C
B
A
3
7
11
15
19
A
A
D
D
B
Short answer
1 5.4 cm3/s 2 Answers will vary.
4 f (x) = --13- x3 + --12- x2 − 6x +
4
8
12
16
20
C
E
E
C
E
3 y = 2x2 − 1
1
--6
1
1
5
5 y = −2 log e x + --------- sin 4π x + (4 − --- )x − 4 + --π
π
4π2
b (−∞, 2]
6 a y = sin−1(e x − 2)
7 32.3°C
8 71 days
9 Approx. 50 minutes 7 seconds
10 4.2
11
dy
y
x
-----dx
–2
6
–1
2
0
0
−2 −1 0
1
2 x
1
0
2
2
Analysis
0.4
1 a N = 121(T − 15)
c 23.1°C
y
6
y = x2 + 2
5
4
3
P (1, 3)
2
1
0
−3 −2 −1 0
1
Multiple choice
c
dy
-----dx
17
—)
6
Chapter review
dy
-----dx
4
2
0
2
–4
x
(0,
1
0
−3 −2 −1 0
b
b
y
answers
639
Answers
1
2
y
12
10
8 (0, 3e)
y = 3e1 − x
6
4
P(1, 3)
2
0
1 2 3 4 5 6 x
x
b 357
d 40.2°C
2 a 10 minutes 8 seconds b x = 10 t + 1 – 10
c 9 minutes 20 seconds d Yes, by 48 seconds
3 a 8 dQ
b
------- + 10Q = 10
dt
c t = 0.80 s
d
t
– -------- ⎞
⎛
e Q ( t ) = VC ⎜ 1 – e RC ⎟
⎝
⎠
−0.1733t
b
4 a D = 50 e
c 9 hours 17 minutes
d
2
Q(t ) = 1 – e
5t– ---4
1
17.68 mL
10 hours 20 minutes
3
7F
➔
3k x k x
5 a y ( x ) = ----------- – -------- since a = 1.5 and y(0) = 0 and
4
6
dy
------ = 0 when x = 0.
dx
b k ≈ –0.027 56
–1
dy
c ----- = – 0.023 25 ⇒ θ = tan – 0.023 25 ⇒ θ = 1.33°
dx
7I
answers
640
Answers
t
⎛
– --- ⎞
6 a v ( t ) = 49 ⎜ 1 – e 5 ⎟
⎝
⎠
b t = 2.62 s when v = 20 m/s
c Terminal velocity = lim v(t) = 49 m/s
7 a 5m
d
b 2 s, 9 m
x (m)
c 0 m/s
e
(2, 9)
t=5
O
5
t=4
t=0
5
t→∞
2 m/s
1
20k
d t = – --- log e ⎛ 1 – ---------⎞ when v = 20 m/s
⎝
k
9.8 ⎠
e Terminal velocity =
– kt
9.8
9.8
lim v ( t ) = lim ------- ( 1 – e ) = ------- m/s
k
t→∞
t→∞ k
CHAPTER 8 Kinematics
Exercise 8A — Differentiation and
displacement, velocity and
acceleration
b i
1 a iii
ii
2 a i 3
b i 2
ii
c i −1
ii
d i 0
ii
3 a i 1.5 units/s ii
b i 2.5 units/s ii
c i 3 units/s
ii
d i 4 units/s
ii
4 a E
ii
5 a i −6
b i 10
ii
c i 4
ii
d i 4
ii
i −12 cm
6 a
iv 21 cm/s
vii 18 cm/s
i 2 cm
b
iv 25 cm/s
vi −12 cm/s
i 0 cm
c
iv
vi
viii
i
d
iv
vi
viii
i
e
iv
vii
f
i
iv
vi
c iv
t=2
x
9
d ii
6
10
12
16
2
iii
4
iii
−2
iii
2
iii
0.5 units/s
1 unit/s
−0.5 units/s
0.5 units/s
b A
18
5
2 2--34
ii 15 cm
iii 36 cm
v v(t) = 6t
vi 0 cm/s
viii 0 s, −12 cm
ii −7 cm
iii 18 cm
v v(t) = 3t2 − 12
vii 15 cm/s
viii 2 s, −14 cm
ii 3 cm
iii 3.2 cm
4
0.2 cm/s
v v(t) = -----------------( t + 1 )2
4 cm/s
vii 0.25 cm/s
Not stationary
0 cm
ii 3 cm
iii 3.58 cm
t+2
v v(t) = ------------------3
0.58 cm/s
(t + 1)2
2 cm/s
vii 0.63 cm/s
Not stationary
−3 cm
ii 1.09 cm
iii 30.60 cm
29.51 cm/s v v(t) = et − 5 vi −4 cm/s
15.09 cm/s viii 1.61 s, −7.05 cm
0 cm
ii −0.54 cm iii −0.29 cm
0.25 cm/s
v v(t) = 4 cos 4t
4 cm/s
vii 3.38 cm/s
π
viii t = (2n + 1) --- , n ∈ J+ ∪ {0}, 1 cm when
8
n = 0, 2, 4, … and −1 cm when n = 1, 3, 5, …
0
t(s)
5
8 a i 10
ii t = 0
iii 18 cm
b i 1
ii t = 1, t = 3
iii 8 cm
c i 1 − 4 loge 4 or −4.55
ii t = 3
iii 4 loge 4 − 3 or 2.55 cm
d i 1
ii t = 0, 2, 4, 6, 8, …
iii 3 cm
e
π
i 4 + --- or 4.79
4
π
iii --- or 0.79 cm
4
ii N/A
9 a i 3t 2 + 4t − 5
ii 6t + 4
c i e t − 2e−t
1
ii ---------------------3
4(t + 2)2
ii e t + 2e −t
1
d i 2t – ----------t+3
1
ii 2 + -----------------2
(t + 3)
1
e i ----------------2
9–t
t
ii --------------------3
(9 – t2)2
f i 10 + 6t − 3t2
ii 6 − 6t
1
b i 4 – -----------------2 t+2
10 a 20 m/s
b 3s
c 95 m
d −10 m/s2
11 a x(0) = 0, v(0) = 8 m/s, a(0) = −6 m/s2
b t = 2 s, x = 6 --23- m and t = 4 s, x = 5 --13- m
c t = 3 s, x = 6 m
d ii v (m/s)
8
3
3
–1 0
ii
2
4
5 t(s)
a (m/s2)
4
0
3
5
t(s)
_6
12 a t = 1, 2 and 3 s
b v = 3t 2 − 12t + 11
6– 3
6+ 3
c t = ---------------- s and ---------------- s d −1 cm/s
3
3
e v (cm/s)
f 6 cm
11
0
–1
2
4 t(s)
13 a Min. x = 1, max. x = 5
π+1
b t = 1--3d a = −9x + 27
c t = -----------3
14 a v = −10t + 2 − 3t2, a = −10 − 6t
2
π
πt
πt
b v = 2π cos ----- , a = − ----- sin ----2
4
4
d v = 3e t − 2e −t, a = 3e t + 2e−t
e v = 20t3 − 6t2 + 4, a = 60t2 − 12t
f v = 2 cos 3t − 6t sin 3t, a = −12 sin 3t − 18t cos 3t
g v = 8 − 5e , a =
5
--2
e
–( t + 2 )
2(t + 3)
h v = ------------------- , a = ------------------t3
t4
15 a v = 6 cos 3t − 9 sin 3t, a = −18 sin 3t − 27 cos 3t
b x(0) = 3, v(0) = 6, a(0) = −27
c a = −9x
d Approx. 3.61
t
– --4
16 a v = 50 e − 40, a = −12.5 e
b h(0) = 0 m, v(0) = 10 m/s, a(0) = −12.5 m/s2
v
c Approx. 4.297 m
d a = – --- – 10
4
Exercise 8B — Using antidifferentiation
1 a x = 2t 3 − t 2 − 8t + 2
c t=
4
--3
d −2 m/s2
10 a v = −2 cos 2t
b x = −sin 2t
d a=2 4–v
11 a i v = 3t 2 − 4t + 1
3
b i v = --- t 2 + 4t + 6
2
t
--2
c i v = 4e − 5
ii 8 e − 5t − 8
1
1
d i v = ----------------- + --2
2
4–t
t t
ii x = sin−1 --- + --- − 2
2 2
3
e i v = ----------- – 3
t+1
ii a = 3 loge(t + 1) − 3t + 5
et + 1
f i v = loge ⎛ -------------⎞
⎝ 2 ⎠
12 a 20 --23- m
c Down
b 5 1--3- m
13 a 30 s
c v = 16 m/s, t = 15 s
b 305.6 m
3
3
14 a x = ---------------------- + 3t – --- b 4.8 cm
2 ( 2t + 1 )
2
15 a v = 3e t − 3
b x = 3e t − 3t − 3
t
t π
16 a x = 2 cos ⎛⎝ --- + ---⎞⎠ or −2 sin --3
3 2
b x = 12t − 2
d 7m
2
ii x = t 3 − 2t 2 + t
1 3
ii x = --- t + 2t 2 + 6t + 0.8
2
t
--2
t
– --2
t
– --4
b v = 4 m/s, t = 0
c a = −4x
1
–1
c v = ----------- , a = -----------------2
t+1
(t + 1)
t
– --2
9 a x = 4 tan−1 t + 1
– 8t
c a = -------------------( 1 + t 2 )2
answers
641
Answers
b 2m
Exercise 8C — Motion under constant
acceleration
e 14 m/s2
2 a x = --23- t3 − --92- t 2 + 4t − 1 b ẍ = 4t − 9
d 7 cm
c t = 1--- , t = 4
1
u
v
a
s
t
a
0
10
2
25
5
b
30
45
2
281.25
7.5
c
−6
8
0.70
20
20
d
18.65
18
−1
12
0.65
e
4
10
1.5
28
4
f
50
48.54
−3
24
0.49
2
e 3.5 cm/s
2
--3
f
cm/s
3 a v = 6t − 6t 2
b x = 3t 2 − 2t 3 + 4
c t = 0, x = 4 and t = 1, x = 5
d 29 m
4 a v = 2e2t − t 2 + 3t
b x = e2t −
1
--3
t3 +
3
--2
t2 − 1
c 406.93 cm
5 a E
6 C
7 a 0 m/s
b B
b x=
1 – t2
c a = -------------------( t 2 + 1 )2
1
--2
loge(t 2 + 1)
d 0.5 m/s
e − 1--8- m/s2
8 a x = 10e−t − 10
b
v (m/s)
v = 10e–t_10
0
_10
b B
2
c D
d A
3 a 4 m/s
b 32 m
4 a 6.5 m/s2
b Approx. 1.54 s
5 a 80 m/s
b Approx. 8.94 s
6 a 18 m/s
b −0.6 m/s2
7 a 58.8 m/s
b 176.4 m
8 a 19.6 m
b 4s
9 a 2.83 s
10 a 78.4 m
11 a 90 m
b 27.72 m/s
b 39.2 m/s
b 7.14 s
c 42 m/s
8A
➔
c 11 s
t(s)
2 a E
8C
answers
642
Answers
13
14
15
16
17
18
19
20
6
20 10
---------------3
12 21.08 m/s or
a 2 m/s2
b 144 m
c 7.5 s
a 12.5 m/s
b 4.8 s
a −5 m/s2 b 22 m
c 3.2 s
d 25.6 m
12 m/s
7.62 m/s, average speed = 7 m/s
a 24.5 m
b 17.5 s
a 8.04 m/s
b 1.1 m
c 1.77 s
a 16.1 m/s
b 27.23 m
Exercise 8D — Velocity–time graphs
1 a
b
c
d
e
f
2 a
ii
ii
ii
ii
ii
ii
b
v (m/s)
300 m
575 m
420 m
500 m
150 m
160 m
d
v (m/s)
t(s)
0
30 t(s)
20
f v (m/s)
v (m/s)
15
0 15
0 11
35 43 t(s)
21 28 t(s)
3 a
b
i v = 0.8t
ii a = 0.8
iii x = 0.4t 2
i v = −1.4t + 14
ii a = −1.4
iii x = −0.7t 2 + 14t
c
3t 2
t3
i v = ------- – 2
ii a = 3t
iii x = ---- – 2t
2
2
2
19
------ t + 10 ii a = − --- t + -----i v = − 1--3- t 2 + 19
3
3
3
e
f
19 2
------ t
6
+ 10t
i v = 1 + 3 sin 2t
iii x = t +
i v = 8e
3
--2
40
9 90 s
−
3
--2
ii a = 6 cos 2t
i 0.375 m/s2
ii 15 m/s
60 t(s)
10 6 m/s
11 30.25 s
1
--2
t+1
2t
1
--2
ii a = 4 e
b x=7
c 3 cm
b v=e
c 2 cm/s2
t+1
e t + 1 ( t + 1 – 1)
c a = --------------------------------------------3
4(t + 1)2
4 2 m/s2
b C
5 a D
7 a a = 8x + 2x3
2t
e t+1
b v = -----------------2 t+1
c E
π
b x = sin(4t + --- )
2
π
b x = 2 tan(2t + --- )
4
c 16 cm/s
8 a x = −1
b Whenever x > −1, the acceleration is positive,
which means the velocity is increasing.
c v=x+1
d 1 m/s
9 3 cm/s
10 a v = 2et − 2
b a = 2e t
c x = 2e t − 2t − 2
11 a v = 4 − e2 − t b e−1
30 ( e t – 1 )
12 a v = -----------------------et + 1
60e t
b a = -------------------( et + 1 )2
et + 1
c x = 60 loge ⎛ -------------⎞ – 30t
⎝ 2 ⎠
13 a The velocity is always increasing since a > 0.
Since initially the velocity is zero and the velocity
is always increasing, the velocity is always
positive.
cos 2t
t
--2
b
13 Cart B by 6 2--3- m
6 a a = −16x
19
iii x = − 1--9- t 3 +
8 50 s
3 a x= e
8
d
0
2 a x= e −
18
22
11 t(s)
Total area
= 450
v
t(s)
15
v (m/s)
15
e
8
v (m/s)
1 a x=3+
0
t(s)
0
10
7 a
2
Exercise 8E — Applying differential
equations to rectilinear motion
12
0
0
14 65 s or 1 min 5 s
15 12 m
16 a 2100 m
b 60 s, 63 km/h
c 150 m
17 a 50 m/s or 180 km/h
b 156.25 m
c 27.7 m/s or 100 km/h
18 8 s
v (m/s)
14
c
8
12 3 m/s
1.2 m/s2
0.5 m/s2
0.533 m/s2
−0.4 m/s2
−0.667 m/s2
0.8 m/s2
i
i
i
i
i
i
68 m
v (m/s)
t
--2
t
--2
iii x = 16 e − 16
4 a B
b C
c A
d C
5 a B
b A
c D
d C
b v=
x
----------2–x
c x = 1.8
14 20 m/s
5 a
t
– --5
b v = 5g(1 − e )
d 90 m
15 a Answers will vary.
c 5g m/s
( g + ku 2 )e –2kx – g
-------------------------------------------k
b v=
16 a Answers will vary.
bt=
2
1
-----2k
c
g + ku
loge ⎛ -----------------⎞
⎝ g ⎠
d Approx. 42.2 m
Chapter review
Multiple choice
1 D
2 A
3 B
4 E
5 C
6 B
7 D
8 A
9 A
10 E
11 C
12 B
13 E
14 C
15 D
16 A
17 D
18 D
19 C
20 E
Short answer
1 a t = 4, x = −15
c
t= 5
v (m/s)
30
25
20
15
10
5
0
8 1--3-
(12.5, 25)
643
(50, 25)
(12.5, 17.5)
0
10
20
30
40
50
60 t(s)
s and 50 s
c t = 19.38, distance = 328.3 m
t = 80.62, distance = 1859.3 m
2
x
uΔ t
1
6 a -- = --b x = --------- + --18- a Δ t
s
2
4
7 a 25 m
b 22.36 m/s
8 a k = 4.96 × 1012
answers
Answers
c When a = 0
12
6
4.96 × 10
b v ( r ) = 2 ⎛ --------------------------- – 2.69 × 10 ⎞
⎝
⎠
r
c 1.84 × 106 m
Technology-free questions (page 421)
b 0<t<4
4
1 a y = x + --- + c
x
8x
1
b y = Ae , where A = --k
t=4
t=3
–15
2
4
5
6
7
–9 –7
t=2
O 5
21 33 x
1
−75
3 1 --- cm
8
a 30.625 m
b t ≈ 1 s and 4 s
a 0.8 m/s2 b 30 m
c 17.5 s
a 2 min 14 s
b 16.2 s
a v (m/s)
b 36.25 s
c 145 m
B
5
d 16 m
A
4
0
6
8.5
t(s)
8 a The acceleration is negative between x = 0 and
x = 4, which means the velocity is decreasing
between these positions.
b v=4
9 a = −1
5π
20
10 a – -----b 10 + ------ or approx. 16.37
4
π
Analysis
1 a Car A
b 8.4 m
c i −2 m/s
d i 23.62 m/s
2
2 a
c
3 a
d
e
ii 3.58 m/s2
ii 28.33 m/s
30
3
29.4 m/s
b v(t) = ----------------- – --12t + 1 5
7.08 s
d 7.33 m
1.02 s
b 14.6 m
c 20 m
First ball 17.15 m/s down; second ball 2.85 m/s up
v = 18.5 m/s; t = 1.89 s
2
1
- cos 4x + cx + d
c y = – 1--4- sin 2x + ----16
t=1 t = 0
–2 x
+ cx + d
–1 – x
x+1
2 a y = cos ( e ) b y = log e ⎛ ------------⎞
⎝ 2 ⎠
c
–1
y = 2 sin x + x
k = – 1, 5
3
4 Check with your teacher
1
1
5 y 2 = 1 + ---------- + --- for x 2 = 0.5 + 0.5 = 1
2 2 3
6 160 days
dy
7 a ------ =
dx
8
x
b y = 2--3- x x – 4--3-
2
------ m/h
3π
– 0.1t
dx
x
9 a ------ = – -----b x = 20 e
dt
10
10 The weld will be effective as the metal rod’s
temperature is below 30° ( 27°C ).
11 a t = 3 seconds
b a = 2 m/s2 at all times
c 10 metres
d Velocity is –2 m/s when t = 2 and the particle is at 0
12 The velocity is 3 m/s and the position is 32 m right
of 0.
13 a 72 km/h
b The brakes are applied after 26 --23- seconds
c 0.75 m/s2
14 a 5 metres
b –5 m/s2
c 10 metres
d 9 metres
15 a 9.8 m/s downwards
b 9g metres
c –29.4 m/s
8D
➔
( t + 1) + 1
4 b x ( t ) = --------------------------- ; t ≥ 0
2
x
d y = e – --14- e
8E
answers
644
16 a
Answers
23 Displacement, velocity, force
24 Speed, time, length
25 1 magnitude and 2 angles (N–S) and (E–W)
v(m /s)
16
Exercise 9B — Position vectors in two
and three dimensions
100
0
T1−100
T1
T t(s)
1 a 3, 4, −2
b 2 minutes and 5 seconds
17 v =
2
--3
m/s, a = – 2--9- m/s
2
72
2 a i
18 60 metres
19 25 metres
20 30 m/s
b
c
d
3 a
CHAPTER 9 Vectors
4 a 5 2
Exercise 9A — Vectors and scalars
1 a i
~s
~r
iii
~s
ii
r +~s
~
~s
~r
~s
5
7
~s
~r − ~s
9
–r
~
~s – ~r
b
i Same as 1 a i except scaled by a factor of 2.
ii Same as 1 a ii except scaled by a factor of 2.
iii
–4r
3s~
~s
10
15
~
3s~ – 4r
~
~r
b s + t + u + v c –s – t
2 a s+t
˜ ˜ ˜ ˜
˜ ˜
˜ ˜
d –v – u – t
e –u – t – s
˜ ˜ ˜
˜ ˜ ˜
3 C
4 a A to C b D to B c B to D d A to C
5 D
b s+t
6 a r+s
˜ ˜
˜ ˜
d r+s
e t–s
˜ ˜
˜ ˜
g r+s+t
h –s–t
˜ ˜ ˜
˜ ˜
7 a, b
c 500 km
d 53.1° clockwise from N
16
17
c r–s
˜ ˜
f s+t–r
˜ ˜ ˜
400 km E
~
300 km
N
~
R
~
Flight
path
8 512.1 km; find bearing using trigonometry
9 721.1 km, 326.3° (clockwise from N)
10 Each part of answer has
15
coordinate labelled a, b, . . . j.
a
i a
The original vectors a and b
5 ~
˜
˜
b
c
are also drawn.
~
–15 g –5 d 5
–5
11 Magnitude = 10.77, direction
j e
68.2° True.
–15
19 One can deduce that x and y
components can be added/
subtracted/multiplied separately.
20 B
21 D
22 0
˜
h
f
b
15
b 6, 0, −3
c 3.4,
2,
1
--2
ii 45°
i 65
i 4.88
i 320.16
45°
b 330.3°
ii 119.7°
ii 225.8°
ii 358.2°
c 224.2° d 91.8°
c
b 5 2
53
---------2
(3.64)
d 11
e 7 14
f
3
6 C
–50i – 50 3 j
˜ – 383.3
˜ j
8 –60.6i + 109.3 j
248.9i
˜
˜
˜ 3 4
˜
3
4
a --5- i + --5- j
b --5- i – --5- j
c 4--5- i + 3--5- j
˜ ˜
˜ ˜
˜ ˜
d – 4--5- i + 3--5- j e 1--- i + ------2- j – ------6- k f – 0.49 i + 0.81 j – 0.32k
3˜ 3
˜
˜
˜
˜ ˜
˜ 3˜
3
-j
B
12 –0.98i – 0.20 j 13 – 1--2- i – -----˜
˜ 2˜
˜
a i 4i – 7 j
ii
65
˜
˜
b i 3i + j
ii 10
˜ ˜
c i –4i + 7 j
ii 65
˜
˜
d i –3i – j + 2k
ii 14
˜ ˜
˜
e i 2i – 6k
ii 2 10
˜
˜
f i – 4i – 2k
ii 2 5
˜
˜
a –4i + 7 j
b –3i – j
c 4i – 7 j
˜
˜ ˜
˜
˜
˜
d 3i + j – 2k e – 2i + 6k
f 4i + 2k
˜ ˜
˜
˜
˜
˜
˜
4
7
3
1
- i – ---------- j
- i + ---------- j
a --------b --------65 ˜
65
10 ˜
10
˜
˜
4
7
1
2
3
- j + ---------- k
c – ---------- i + ---------- j
d – ---------- i – --------65 ˜
65
14
14 ˜
14 ˜
˜
˜
1
3
2
1
f – ------- i – ------e ---------- i – ---------- k
10 ˜
10 ˜
5˜
5
18 a
b
29
13
3
2
- i + ---------- j
d – --------13
13 ˜
˜
e 3i + j
f
10
˜ ˜
g Reject, because magnitudes different.
b 30
19 a 26
c
5
---------- i
29 ˜
–
2
---------- j
29
˜
1
- ( 5i – j + 2k )
d --------3i + 4 j + k )
30 ˜ ˜
˜
˜
˜ ˜
e 2i + 3 j + 3k
f
22
˜
˜
˜
g Reject, because magnitudes different.
c
20 a
1
---------- ( –
26
2
21 a 3i
˜
d 31.0°
b
74
b 5j
˜
34
e
c 3i + 5 j
˜
˜
22 329.0°
b –3i + 118 j c 358.5°T
˜
˜
b 486i + 14 j c 88.35°T
˜
˜
c 2 6
d 62
13
23 a 3.61 km/h
24 a 7.62 km/h
25 a
35
b
26 CD = i – 7 j – 2k , EF = 2i – 14 j – 4k
˜
˜
˜
˜
˜
˜
c 53.1°
27 a 20i , – 15 j , 20i + 15 j b 25
˜
˜
˜
˜
a
–4i
+
12
j
b 341.6°
29
˜
c 0.0417 h ˜or 2.5 minutes
10
---------------c i – 13
10
17
---------------ii – 26
17
13
d i – 6-----------13
ii – --65-
-----e i – 23
5
26
---------------ii – 23
13
2 a i
b i
c i
d i
30 a 14.34i + 20.48 j + 8k
˜
˜
˜
c 21.3°
b 26.25 m
31 a − 3i – 9 j – 8k
˜
˜
˜
1
b ------------- ( – 3i – 9 j – 8k )
˜
˜
˜
154
iii
e i
c 5 2 km
d 744.6 km/h
Exercise 9C — Multiplying 2 vectors —
the dot product
1 23.99
2 Dot product = 24; more accurate, since no angle
needed
b 12
c −36
d −26
3 a 45
e 1
f −20
g 0
h 0
4 E
5 C
6 −36
8 12
11 D
12 B
13 E
14 a −12
b 2
c 7
15 a 107°
16 D
b 87°
17 E
c 81°
19 a = − --35-
20 4i + 8 j
˜
˜
d −25
d 109°
18 a = 1
21
64
------ i
5˜
f i
iii
3 a 3.6 km
b 0.2 km or 200 metres
4 316 metres
Exercise 9F — Time-varying vectors
x
1 a y = – --2
b y = –3x – 3
x3
d y = ----8
c y = 4( x – 3 )2
2 B
3 a y = 4x2 + 2x
b
y
48
-j
– ----5
˜
Exercise 9D — Using vectors in
geometry
4 a –v
b
˜
2
d b .b = u +
˜ ˜
13 p = 7
14 p = 10
15 a No
b
iii
3
1
51
- i – ------ j iii v
------ i + ------ j
ii v || = ----= 17
10 ˜
10
10 ˜
10
˜
˜⊥
˜
˜
iii
v
=
v
=
8i
+
10
j
ii
0
||
2 41
⊥
˜
˜
˜
˜
˜
iii v ⊥ = – 3 i + 4 j
ii v || = 0
0
˜
˜
˜
˜
˜
2
------ii v || = 2--3- i + 2--3- j + 2--3- k
3
˜ ˜
˜
˜
v ⊥ = 4--3- i + 1--3- j – 5--3- k
˜
˜
˜ ˜
21
21
– --------- ( 2i + 3 j + 4k )
ii v || = – ----29
29
˜
˜
˜
˜
4
- ( 25i – 6 j – 8k )
v ⊥ = ----29
˜
˜
˜
˜
5
5
5
--------------- i + ------ j – ------ k
ii v || = 15
11 ˜
11
11 ˜
11
˜
˜
17
28
------ i + ------ j – ------ k
v ⊥ = – 15
11 ˜
11
11 ˜
˜
˜
1
---------10
u +v
c u –v
˜
˜
˜
˜
2
2
2
v , c .c = u + v
˜ ˜
4 a y = x − 4x + 3
b
y
Note: x ≥ 1
0
3 x
–1 1
(2, –1)
5 a x2 + y2 = 1
History of mathematics
1 Mathematics lecturer
2 Alice’s Adventures in Wonderland and Through the
Looking-Glass
3 The daughter of the Dean of his college.
4 13 cats
Exercise 9E — Resolving vectors —
scalar and vector resolutes
ii
23 41
---------------41
b i
17 29
---------------29
ii
17 10
---------------10
b
c Period = π
y
1
–1
0
1 x
–1
6 a x2 + y2 = 9
y
3
b
–3
0
c Period = π
3 x
–3
9A
➔
23 13
---------------13
x
0
2
Yes
1 a i
answers
645
Answers
9F
answers
646
Answers
7 a (x − 1)2 + (y + 2)2 = 1
b
Analysis
1 a 6i – 12 j + 12k
b --13- ( i – 2 j + 2k )
˜
˜
˜
˜
˜
˜
m
c ---- ( i – 2 j + 2k ) , 0 < m < 18
3 ˜
˜
˜
1
d --3- ( 10i + 16 j + 11k )
˜
˜
˜
e 7.28 km
f 1080 km/h
g 720 km/h
2 a (3, 5.5, 0)
b – 3i + 4k
˜
˜
d 66 cm3 = 0.066 litres
e 3i + 5.5 j + 4k
f 7.43 cm
g 84.5°
˜
˜
˜
3 a y
y
0
–1
2 x
1
–2
–3
c Period = 2π
y
x2
8 ----- + y 2 = 1
9
1
–3
3 x
0
–1
y
x2 y2
9 ----- + ------ = 1
4 16
X (2, 7)
4
–2
0
2i~ + 7j
~
2 x
10 a y = 1 – x ; u ( 0 ) = i ; t = 2, u =
˜
˜
˜
b y = (x – 1)2, x ≥ 2
+
1
--- i
2˜
1
--- j
2
b 2i + 7 j, – 5i
c 7i + 7 j, –3i + 7 j
˜
˜
˜
˜
˜
˜
˜
2 29
2i
d -----------e
74.1°
f
29
˜
g (−2, 7)
h 35 square units
4 a
N
˜
( x – 3 )2 ( y + 1 )2
c ------------------- + ------------------- = 1
4
9
19
-----7
,y=
5
2
--7
3
12 y = x2; No, since v is always ‘ahead’ of u .
˜
˜
b 5.83 m/s
c 59°T
d 360 m
e 600 m
f Same result, except bearing = 180° − 59° = 121°
5 b 109.5°
1
1
c i p = ------- ; r = – ------2
2
Chapter review
Multiple choice
1 A
5 E
9 B
2 B
6 C
10 B
3 A
7 D
11 D
4 D
8 E
5 2
ii ---------3
Short answer
1 200 3i + 200 j + 40k
˜
˜
˜
2 a ( 5 + 5 2 )i + ( 5 + 5 2 ) j
˜
˜
3 1.3909
4 a – i – 4j
˜
˜
c −17
6 a
y
X (3, 5)
Y (9, 5)
b 17.07 km
Z (6, 0)
b 7i – 6 j
˜
˜
3
5
--------- i – ---------- j
d
34 ˜
34
˜
x
O
b YX = – 6i ; ZY = 3i + 5 j
˜
˜
˜
e 135°
c OY = 9i + 5 j, ZX = –3i + 5 j
˜
˜
˜
˜
d 91.9°
e 29.1°
f 5j
˜
g ( 12 3--8-, 10 5--8- )
5 56.1°, 111.8°, 42.0°
9 ± 69
6 ------------------2
8
9 p = --3
h 31 7--8- square units
3
1
- j – ---------- k
+ --------14
14 ˜
˜
1
b v || = --2- ( 2i + 3 j – k ) , v ⊥ = 1--2- ( – j – 3k )
˜
˜
˜
˜ ˜ ˜
˜
y
2
11 y = ----- – 2 , hyperbolic
x2
10 a
5i~ Z (5, 0) x
O
–4
11 x =
Y (7, 7)
2
---------- i
14 ˜
1
0
–2
CHAPTER 10 Vector calculus
Exercise 10A — Position, velocity and
acceleration
1 a 5
b
d
2
x
2426
5 + ( log e 11) ≈ 3.28
c 1
b i – 16 j
c 0
2 a 2 2j
˜
˜
d 2i + ˜( log e 11 ) j – 2k
˜
˜
˜
b 2t i – 2 j + 3t 2 k
3 a –2 j
˜
˜
˜
˜
c 3 cos3t i + 8 sin ( –2t ) j
d – 8e –2t i + 2t –2 j
˜
˜
˜
˜
e ( t – 1 ) –1 i + ( 9t 2 + 2 ) j + 2tk
˜
˜
˜
f ( t 2 cos t + 2t sin t )i + ( t + 1 )e t j
˜
˜
g –6 sin 2t i + 6 cos 2t j
˜
˜
4 a 2 t2 + 1
i 2 5
ii 2i – 2 j
˜
˜
4
2
b 9t – 2t + 5
i 141 ii 3i – 2 j + 2k
˜
˜
˜
3 ⎞2
⎛
i 4.02
ii 0.973i – 4 j
16 + -------------c
⎝ 3t + 1⎠
˜
˜
b y = 4 log e x
c
O
4 a D
5 a E
(e 2, 8)
y
1
647
Dom = [1, e2];
Ran = [0, 8]
x
b B
c C
b A
3t 2
6 a i r ( t ) = ⎛ ------- + 2⎞ i + ( 5t – 3 ) j + ( 2t + 1 )k
⎝
⎠˜
2
˜
˜
˜
ii a ( t ) = 3i
˜
˜
b i r ( t ) = – ( 3 cos t + 2 )i + ( 1--2- sin 2t + 3 ) j
˜
˜
˜
ii a ( t ) = 3 cos t i – 2 sin 2t j
˜
˜
˜
5t 2
c i r ( t ) = ( t 3 + 2 )i + ⎛ ------- + 1⎞ j – --12- log e ( 2t + 1 )k
⎠˜
˜
˜ ⎝ 2
˜
ii a ( t ) = 6t i + 5 j + 2 ( 2t + 1 ) –2 k
˜
˜
˜
˜
d 4
i 4
ii –0.573i – 0.495 j
2 + e 3t
d i r ( t ) = ---------------- i + 5 ( 1 – e –t ) j
˜
˜
3 ˜
˜
b 2i
˜
5 a 0
˜
ii a ( t ) = 3e 3t i – 5e –t j
c –9 sin 3t i – 16 cos ( – 2 t ) j – 6tk
˜
˜
˜
˜
˜
3t 2 + 7
˜
---------------e
i
r
(
t
)
=
i
+
(
5t
– 4) j
d 16e –2t i – 4t –3 j
e – ( t – 1 ) –2 i + 18t j – 2t –3 k
2 ˜
˜
˜
˜
˜
˜
˜
˜
ii a ( t ) = 3i
t
t
˜
˜
f – sin t i + ( te + 2e ) j
t
5t 2- – 13⎞
˜
˜
f i r ( t ) = ⎛ 1--2- log e --- + 2⎞ i + ⎛ -----j + ( 4t – 5 )k
⎝
⎠˜
⎝
⎠
b B
c C
6 a C
2
2
˜
˜
˜
1- i + 5 j
b A
7 a E
ii a ( t ) = – -----˜
2t 2 ˜
˜
8 a 24i + 2--3- j ; 24.01
b ( 3t 2 – 1 )i – 2t –2 j
1
– cos 3t
˜ ˜
˜
˜
g i r ( t ) = ----------------------- i + 2 ( 1 – sin ( – 2 t ) ) j
3
1
˜
˜
˜
d t = ------c 11.01
ii a ( t ) = 3 cos 3t i + 8 sin ( – 2t ) j
3
˜
˜
˜
b 3 π cos 3 π t i + j
9 a 3j; 3
t2
˜
--v
(
t
)
=
–
4.9t
i
+
j
7
a
i
˜
˜
˜ 2˜
˜
c 9 π 2 + 1 = 9.48
d 9 π 2 + 1 = 9.48
3
2 )i + ⎛ 1 + t----⎞ j
(
–
=
r
(
t
)
1
2.45t
ii
–
1
⎝
6⎠˜
˜
˜
b ------------------ i + j
10 a 1--3- i + 2 j ; 2.03
2˜
–
2t
(
t
+
1
)
˜
˜
˜
b i v ( t ) = ----------- i + 3t 2 j
c 1.01
d Various answers
t + 1˜
˜
˜
ii r ( t ) = [ 1 – 2t + 2 log e ( t + 1 ) ]i + ( 1 + t 3 ) j
b 6i + 6t j
11 a 6t i + 3t 2 j – 6k
˜
˜
˜
˜
˜
˜
˜
c 45°
d t=0 ˜
1 – e –4t
c i v ( t ) = 3i + ------------------ j
4 ˜
˜
˜
π
c --d 90°
12 a Various answers b 12
e –4t + 4t – 1
4
ii r ( t ) = 3t i + ----------------------------- j
16
˜
˜
˜
Exercise 10B — Cartesian equations
d i v ( t ) = – 5 sin 5t i + 5 cos 5t j
˜
˜
˜
and antidifferentiation of vectors
ii r ( t ) = ( 4 + cos 5t )i + sin 5t j
˜
˜
˜
2
1 a 2y = x
b 90y = x − 60x
8 a r˙˙( 0 ) = 0.1i – 9.8 j
˜
˜
˜
t
2
x
– -----c 2y = x + 1
d y = 2e − 1
b r˙( 0 ) = ( 0.1t + 3 )i + ( 98e 10 – 96 ) j
e 2y = −loge x
˜
˜
f x2 + y2 = 1
˜
– -----tc r ( t ) = ( 0.05t 2 + 3t )i + ( – 980e 10 – 96t + 1000 ) j
2 a r ( 0 ) = j ; v ( 0 ) = 2i – 4 j ; a ( 0 ) = 0
˜
˜
˜
˜
˜ ˜ c y ˜
˜ ˜
b y = 1 − 2x
1–
Dom = [0, 10],
9 a r˙( t ) = 12i + ( 26 – 10t ) j
1 2
˜
˜
˜
Ran = [−19, 1]
x
0
b r ( t ) = ( 12t + 20 )i + ( 26t – 5t 2 + 5 ) j
˜
˜
˜
2
10 a r = ( t + 4t )i + 5t j
(10, –19)
˜
˜
˜
b i (12, 10)
ii (32, 20)
3 a r ( 0 ) = i ; v ( 0 ) = 1--2- i + 2 j ; a ( 0 ) = 1--4- i
c Distance 3840 m; height 300 m
˜
˜
˜
˜ ˜
˜ ˜
answers
Answers
➔
10A
10B
answers
648
d
Answers
e
y
7 a v ( 2 ) = 2i – 12 j ; v ( 3 ) = 2i – 18 j
˜
˜
˜
˜ ˜
˜
b 2i – 15 j
˜
˜
c a ( 2 ) = –6 j , a ( 3 ) = –6 j
d –6 j
e 6.3°
˜
˜ ˜
˜
8 a log e 11i + 50 j
˜
˜
b 0.2398i + 5 j
c t = 0.3803 s
˜
˜
–
t
9 a r ( t ) = ( 4 – 3e )i + ( 3 – 2e –2t ) j ; 3.9797,
˜
˜
˜
4.9999, 5.000
51.3°
y2
4y
x=
+
5
25
(32, 20)
(3840, 300)
(12, 10)
0
11 a
b
c
x
2
r = ( 8t – 2t )i + 4t j
˜
˜
˜
i (0, 16)
ii (−120,
40)
Distance 12.16 km; height 320 m.
At t = 2, distance is 8 m and height is 8 m.
d
x = 2y −
(−12 160, 320)
(−120, 40)
y2
8
10
10 a The displacement is r ( 10 ) – r ( 0 ) = – ------ i + 10 j .
11 ˜
˜
˜
˜
(0, 16)
(8, 8)
0
e
b 4i + 3 j
c 17.17°
˜
˜
2
d 9y = −2x + 16x − 5
y
b – 0.0909i + j
c Various answers
˜ ˜
11 a 3t i + ( 2t – 4 ) j , ( t 2 – 4 )i + t j
˜ ˜
˜
˜
b Various answers
x
i 179.27° from positive i direction
˜
ii 178.49° from positive i direction
˜
Exercise 10C — Applications of vector
calculus
1 a v ( t ) = 10i – 10t j
˜
˜
˜
x2 y
b y = – -----20 O
12 a 3.59 m
b 0°
c t=0
13 a Various answers
b The collision occurs at noon.
14 a The boats will meet after 22 minutes.
x
b 0.29i – 5.82 j
˜
˜
(20, –20)
2
3
4
5
c v ( 0 ) = 10i, 10, 0° ; v ( 1 ) = 10i – 10 j, 10 2
˜
˜
˜
˜
˜
– 45° ; v ( 2 ) = 10i – 20 j, 10 5, – 63°
˜
˜
˜
d a ( t ) = – 10 j , constant acceleration
˜
˜
a 1, 0.71, 1.10
1
1
b v = ----------- i – 2e –2t j , a = – ------------------ i + 4e –2t j
t + 1˜
˜
( t + 1 )2˜
˜ ˜
˜
c 167°, 143°, 153°
a v ( t ) = ( 1 + sin t )i + cos t j, a ( t ) = cos t i – sin t j
˜
˜
˜
˜ ˜
˜
3π
3π
3 π⎞
⎛
b t = -----c r ------ = ------ i – j
2
2˜ ˜
˜⎝ 2 ⎠
3 π⎞
⎛
d a ------ = j
˜⎝ 2 ⎠
˜
e The acceleration is the derivative of velocity with
respect to time. Velocity is a vector quantity so
that when its magnitude (the speed) is at a
minimum, the acceleration may still be non-zero.
a, b Various answers
c Even though the speed is constant, the
acceleration is non-zero because the direction of
the velocity changes.
a,b,c and d Various answers
e The constant speed is 6 m/s.
6 a v ( 1 ) = –1i + 4 j ; v ( 2 ) = – --14- i + 4 j
˜
˜
˜
˜
˜ ˜
b – 1--2- i + 4 j c a ( 1 ) = 2i ; a ( 1 ) = 1--4- i
˜
˜
˜ ˜
˜
˜
c They collide when t = 4 at 12i + 4 j .
d 2i
˜
˜
˜
2
51
- = 3.57
=
x
+
4
e t = --------f
3y
=
2x
−
12,
y
2
2
2
x
y
15 a ------ + ---- = 1;
16 9
y
x2
y2
+
=1
16
9
Start r (0) = (4, 0)
~
3
0
−4
b r˙ = – 4 sint i + 3cost j
˜
˜
˜˜
c 127.6° from positive i direction
˜
π
d Maximum speed is 4 when t = --- ; minimum
2
speed is 3 when t = 0
e r˙˙ = – 4cost i – 3sint j
˜
˜
˜
2
2
x
y
16 a ---- + ------ = 1,
9 16
y
x2
y2
+
=1
9
16
4
Start r (0) = (3, 0)
~
0
−3
3
−4
d
3
--- i
4˜
x
4
−3
b r˙ = – 6 sin ( 2t )i + 8 cos ( 2t ) j
˜
˜
˜
x
c –90° from positive i direction; speed = 8
˜
π
d Minimum speed = 6 at t = --- ;
4
maximum speed = 8 at t = 0
e r˙˙ = – 12 cos ( 2t )i – 16 sin ( 2t ) j
˜
˜
˜
2
649
d i 3
4 10
ii ------------3
20 a y = 4x – 4
y
2
( x + 1) ( y – 3)
17 a ------------------- + ------------------- = 1,
1
4
y
(−1, 5)
(2, 4)
(x + 1)2 (y − 3)2
+
=1
1
4
(−1, 3)
(−2, 3)
answers
Answers
(1, 0)
0
(0, 3)
1
2
3
x
b Domain [1, ∞) and range [0, ∞)
Start r (0) = (−1, 1)
~
(−1, 1)
0
Exercise 10D — Projectile motion
x
b r˙ = 3 cos ( 3t )i + 6 sin ( 3t ) j
˜
˜
˜
117
c 73.9° from positive i direction; speed = ------------2
˜
d Minimum speed = 3 when t = 0;
π
maximum speed = 6 when t = --6
r˙˙ = – 9 sin ( 3t )i + 18 cos ( 3t ) j
˜
˜
˜
π π
f t = 0, --- , --- , …
6 3
e
2
y = 4x 2 + 4
y
18 a y = 4x + 4,
4
Start r (0) = (0, 4)
~
x
0
2
2
b r˙ = sec t i + 8 tant sec t j
˜
˜
˜
c 82.9° from positive i direction; speed = 2 65
˜
d Minimum speed = 1 when t = 0
2
2
2
e r˙˙ = 2 tant sec t i + 8sec t ( 1 + 3tan t ) j
˜
˜
˜
f i 8 at 90° from positive i direction
˜
ii 4 257 at 86.4° from positive i direction
˜
2
2
y
x y
19 a ----- – ----- = 1
4 9
y= 3x
2
(2, 0)
0
2
b Domain [2, ∞) and range [0, ∞)
2
Chapter review
Multiple choice
1 D
2 C
5 E
6 A
9 C
10 B
3 A
7 E
4 E
8 B
10C
➔
c v = 2 tan t sect i + 3 sec t j
˜
˜
˜
x
1 a 78.78 m/s, 13.89 m/s
b 78.89 m/s
c r ( t ) = 78.78t i + ( 13.89t – 4.9t 2 ) j , 79.29 m
˜
˜
˜
2 a v ( t ) = 75i + ( 129.9 – 9.8t ) j
b 133.38 m/s
˜
˜
˜
2
c r ( t ) = 75t i + ( 129.9t – 4.9t ) j
˜
˜
˜
d 283.19 m
e 13.26 s after firing
3 a 44.02i + 9.36 j
b 44.02i + ( 9.36 – 9.8t ) j
˜
˜
˜
˜
c 4.47 m, r ( 0.955 ) = 42.04i + 4.47 j
˜
˜
˜
d 1.91 s
e 84.08 m
4 a 1.2 m
b 9.8 m/s
c 8.5 m
d 9.8 m/s
e −30°
5 a 17.61°
b 29.15 m/s
6 a 90i + 105.9 j
˜
˜
b v ( t ) = 30i + ( 50 – 9.8t ) j ; 36.39 m/s
˜
˜
˜
c a ( t ) = – 9.8 j
˜
˜
d t = 5.1s. This occurs when the object is at the
apex of its motion.
7 a 1m
b v ( t ) = 4.9i + ( 22 – 9.8t ) j
˜
˜
˜
c 22.54 m/s, 77.4°
d 22.97 m/s, −77.7° (from i direction)
˜
e 22.24 m
8 a 25i + 40 j
b y = 1.6x − 0.007 84x2
˜
˜
c 204.06 m
d 81.63 m
9 a 44.1 m
b 30 m
c 31.05 m/s, −71.2°
d r ( t ) = 10t i – 4.9t 2 j
e y = −0.049x2
˜
˜
˜
10 a 0.17 m
b 0.329°
11 a 5.0i + 4.0 j
b r ( t ) = 5.0t i + ( 4.0t – 4.9t 2 ) j
˜
˜
˜
˜
˜
c 0.82 s
d 4.1 m
2
13 a y = 50 + 0.417x – 0.00034x
b 177.6 m c 1336.8 m d 11.1 s e −26.2°
14 y = 0.41x − 0.0021x2; 4.0 s
10D
answers
650
Answers
Short answer
1 18 m/s2
2 x ( t ) = 20t i + ( – 4.9t 2 + 49t + 20 ) j
˜
˜
˜
3 a r ( t ) = 5--3- sin 3t i – 4 cos 3t j
˜
˜
˜
b a ( t ) = –15 sin 3t i + 36 cos 3t j
˜
˜
˜
nπ
c t = ------ for n = 0, 1, 2, . . .
6
e 12 m/s
d 144x2 + 25y2 = 400
–3 3
c ------------- i + j , 158.9° from positive i direction
2 ˜ ˜
˜
d Minimum speed = 2 when t = 0;
π
maximum speed = 3 when t = --2
e r˙˙ = – 3 cost i – 2sint j
˜
˜
˜
10 a r = 10 2t i + ( 10 2t – 4.9t 2 + h ) j
˜
˜
˜
s = ( 40 – 15 3t )i + ( 15t – 4.9t 2 ) j
˜
˜
˜
b t = 1 and h = 0.86 m
2
4 a r = ( 3t + 6t )i + 4t j
˜
˜
˜
b i (72, 16)
ii (360, 40)
c Distance 30.6 km; height 400 m
d y
x=
Analysis
1 a r .⋅ s = 2 cos t + 2 sin t
˜ ˜
3π
t = -----4
3y2 3y
+
16
2
(30 600, 400)
(360, 40)
(72, 16)
x
0
e 0.75° from positive i direction
˜
5 12 5 units
6 a Speed = 8 m/s
b Since v ( t ) .⋅ a ( t ) = 0 for all times and neither
˜
˜
v ( t ) nor a ( t ) = 0i + 0 j then the two vectors are
˜
˜
˜
˜
perpendicular.
7 a r ( 0 ) = 4i + j ; v ( t ) = – 6e –2t i – 8e –4t j and thus
˜
˜ ˜ ˜
˜
˜
v ( 0 ) = –6i – 8 j ; a ( t ) = 12e –2t i + 32e –4t j and
˜
˜
˜
˜
˜
˜
thus a ( 0 ) = 12i + 32 j
˜
˜
˜
b 163.7°
c i– j
y
˜ ˜
Initial
d 9y = 2x2 − 4x − 7
8 a 4α m/s
1
position
0
–1
1
4 x
Final position
c
–4
0
b x + 2y = 0; x2 + y2 = 16
B
4
–4
2
x
A
8
4
d ⎛ -------, – -------⎞
⎝ 5
5⎠
e 2.27
2
x
y
9 a ----- + ----- = 1
9 4
y
x2
y2
+
=1
9
4
2
−3
R
1
S
–1 0
–1
1 2
3
4
x
3
4
–2
e
. π–
~s( 2 )
y
2
. π– 1
~r( 2 )
–1 0
–1
.
~r(0)
.
~s(0)
1 2
x
–2
2 a 15 2i + 30 2 j
b 180 m, 8.5 s
˜
˜
c 57.5° or 71.2°
d For 57.5°, the second object must be launched
1.74 s after the first object; for 71.2°, it must be
launched 0.88 s before the first object is launched.
3 a i −cosec2 x
ii –cot x
π
π
b r = ⎛ 3 + tan ⎛ t + ---⎞ ⎞ i + cot ⎛ t + ---⎞ j
⎝ 4⎠ ⎠ ˜
⎝ 4⎠ ˜
⎝
˜
y
1
x=3
c y = ----------x–3
Start r (0) = (3, 0)
~
0
−2
2
This shows that the directions of motion are
π
perpendicular at t = 0 and t = --- .
2
y
4
y2
b r˙ .⋅ s˙ = 0 for all t
c i x 2 + ----- = 1
4
˜ ˜
( 2 – x )2
2
ii ------------------- + ( y – 1 ) = 1 or
4
( x – 2 )2
------------------- + ( y – 1 ) 2 = 1
4
d
y
3
y=
x
1
x−3
Start r (0) = (4, 1)
~
1
b r˙ = – 3 sin t i + 2 cos t j
˜
˜
˜
0
3 4
x
4 10
d ------------- , –18.4° from i direction
3
˜
b i
97
c i –9 i + 4 j
˜
˜
5 a i i + 2j
˜
˜
b i 5
2
π
π
e r˙˙ = 2 tan ⎛ t + ---⎞ sec ⎛ t + ---⎞ i
⎝ 4⎠
⎝ 4⎠ ˜
˜
2
π
π
+ 2 cot ⎛ t + ---⎞ cosec ⎛ t + ---⎞ j
⎝ 4⎠
⎝ 4⎠ ˜
iii
ii 6i + 3 j
˜
˜
iii
ii
(
ii
c i 63.4°
8 246
f ---------------- , 6.3° from i direction
9
˜
g 24.8°
ii 3 5
Exercise 11A — Force diagrams and
the triangle of forces
b
D
~
Book
Ball
W
~
W
~
c
d
~N
~F
Car
~A
~N
F
~
A
Boat ~
W
~
e
ii 0°
ii – 50 i + 5 j
˜ ˜
ii 4i – 12 j
˜
˜
ii 2.9 N at S26.9°E
b i 50i – 5 j
˜
˜
c i – 4 i + 12 j
˜
˜
d i 2.9N at N26.9°W
7 a 10 N
b 120°
8 79.2°
Z = 2 sin 60° × X
˜
˜
= 3X
˜
b 37.6 N c
9 a 77°
f
h
F
~
j
~D
~D
W
~
3 a C
b E
(2
23.5 N
C
d A
0N
39 392.3 N
8g N
~
~Tright
~Tleft
b T1 = 4g N;
T2 = 4 3 g N
b 41.4°
c 245 N
d –278 i + 245 j N;
278 i ˜+ 245 j ˜ N
˜ N f ˜8.0 m
e 370.6
W
~
~N
~A = 120 N at
40° to
vertical
~F
W
~ = 40g
~
2 – 3 5) i – j
˜ ˜
b Av = 91.9 N down
AH = 77.1 N left
c R = ( F friction – 77.1 )i + ( N – 483.9 ) j = 0i + 0 j
˜
˜
˜
˜
˜
with all forces in N.
11A
➔
ii –6i – 3 j iii
˜
˜
30°
~T1
60°
Speaker
6 a
W
~
b E
c
c
b
c
T
~
~i
5 a
Ball moving
down
2 a C
T
~
Ship
~T2
W
~
Ball moving
up
Tugboat
due to
water
4 a
~F
Sliding
body
W
~
4 a i 9i + 4 j
˜
˜
1 a 23.5 N b 8.6 N
b B
2 a D
3 a
Drag force
j
~
~N
Body at rest
i
Exercise 11B — Newton’s First Law
of Motion
W
~
~N
~F
Bubonic plague broke out and the university closed.
Leibniz
The laws of mechanics and gravitation
He reformed the coinage and introduced measures to
prevent counterfeiting.
Tugboat
~A
Accelerating car
W
~
g
1
2
3
4
N
~
~F
Sliding object
~F
History of mathematics
W
~
N
~
5 – 2 2) i + j
˜ ˜
2 – 1) i
˜
2–1
CHAPTER 11 Mechanics
~N
(3
d i – i – 2j
ii ( 1 – 2 ) i
˜
˜
˜
ii 61.4 N at S9.4°E
6 a i 61.4 N at N9.4°W
4 a CS = 6i + 2 j , 6.3 km
˜
˜
ii P (8.4, –2.8)
b i 56m – 40m2
iii 1.3 km
c i r c = ( 15t + 2 )i + ( – 5t – 6 ) j
˜
˜
˜
ii r s = ( 12t + 8 )i + ( –3 cos t – 8t – 1 ) j
˜
˜
˜
d r c ( 2 ) – r s ( 2 ) = – 0.25 j
˜
˜
˜
1 a
54 – 12 10
answers
651
Answers
11B
Answers
d 77.1 N
7 a
~N
W
~
N
~
Mass
H
~
W
~
9 a
Ball moving
up
~A
2
0
0.2 0.4 0.6 0.8 1.0 1.2 1.4
a (m/s2)
b – 7.4i – 12.7 j Ν
˜
˜
c 7.4i – 4.3 j N
˜
˜
d 17 j N
˜
e Various answers
6 a B
b E
j
~
~i
b 2.96 N down
c 1.96 N down
N
~
~D
8 a −9.3 m/s
b 25 m
c 11.25 m/s
1 a −2.5 m/s2
2 a The resultant force acts in a direction opposite to
the car’s motion during that time. Without further
information, nothing can be said about the
magnitude of the force.
b The resultant force acts down the slope, in the
direction of the skier’s motion. Further
information is needed to calculate the magnitude
of the resultant force acting on the skier.
2
5
5
0.50
10
2
2
1
4 a i 1.6 m/s2
ii 3.2 m/s
iii 3.2 m
b i 3.2 m/s2
ii 6.4 m/s
iii 6.4 m
2
c i 4.8 m/s
ii 9.6 m/s
iii 9.6 m
d i 6.4 m/s2
ii 12.8 m/s
iii 12.8 m
b Newton’s second law states
that the sum of the forces
acting on a particle is equal to
T
Block ~
the product of its mass and its
W
acceleration. The frictional
~
force acts in the opposite
direction to the applied force
MQ Spec fig 9.059 and the acceleration is in the
same direction as the applied
force. Therefore, T – F = ma.
c 5.0 × 10−5 s
iii 4.53 m/s2 at N20.4°W
c i –4.3i – 22.1 j N
ii 22.5 N
˜
˜
2
iii –1.8i – 9.2 j m/s
˜
˜
12 a j
b R = 388.9i
~
˜
˜
c 588.9 N
N
i
~
~
~D
~A
Acceleration
(m/s2)
10
2
11 a i 6i + 3 j N
ii 3 5 N
˜
˜
2
iii 3.0i + 1.5 j m/s
˜
˜
b i 339.4 N at N20.4°W ii 339.4 N
W
~
13 a
~N
b R = ( A – D )i + ( N – W ) j
˜
˜
˜
= max i + may j
˜
˜
c a = 7.33i + 0 j
˜
˜
˜
d N = W = 588 N
˜ N˜
e 440
9 a −2.4 × 10 m/s b 7.2 × 105 N
10 10.64 kg
Mass (kg)
d C
b 11 112 N
7
Exercise 11C — Newton’s Second Law
of Motion
F
~
~A
2
W
~
5 a
c D
7 a
e 0.96 N down
Force (N)
The intercept
1.2 N
The gradient
1.8 kg
1
Ball moving
down
3
3
W
~
W
~
A
~
d
di
ii
e i
ii
4
b 0i + 0 j ; R = 0 N
˜
˜
c – 7.4 i – 12.7 j N
˜
˜
d 7.4 N
e 12.7 N
~T
Mass
8 a
c
e 483.9 N
T (N)
answers
652
Drag
Buoyancy
b i 0 N, 0.078 N
ii 4.5 × 10−4 N, 0.077 N
Ball
Weight
14 a
b 29.4 N upward
c 19.6 m/s2
~N
Mass
W
~
15 a
D
~
Anita
a
b
c
d
e
408 N
588 N
180 N
588 N upward
W
~
16 a
N
~
Inclined plane
Cart
W
~
b 3.0 N acting down
the plane
c 11.4 N acting into
the plane
d 3.0 N acting down
the plane
e The component of the weight force perpendicular
to the plane is balanced by the normal force. The
resultant force is therefore the component of the
weight force acting parallel to the plane.
f −2.5 m/s2
g 1.8 m
17 a
b −3.03 m/s2
~N
Cart
Inclined
plane
W
~
~F
18 a
f
g
h
i
j
k
N
~
~F
b 10 N
A = 12 N
~
N
~ ~
30°
0.75 kg
mass
c 0.51
10 a
18°
2.27 N
2.27 N
6.4 N
2.1 N
c
e
g
i
6.99 N
0.32
3.7 N
2.1 m/s2
11 a
5 a A
b D
6 a B
b E
b 85 N
~N
Crate
~F
8°
W
~
12 82 N
13 6.2 m
14 a
N
~
1.0 kg mass
1.0 kg mass
~N
~F
~F
W
~
Moving up the plane
W
~
Moving down the plane
b 0.45 m
c 1.63 s
d 0.27
Exercise 11E — Applications of
Newton’s first and second laws to
connected masses
1 a
N2
~
N1
~
~T2
~T1
12 N
Mass 2
W
~2
Mass 1
W
~1
~N2
~N1
~T2
~F2
~T1
15 N
F
Mass 2 ~ 1
3 a
~N2
~T2
Mass 1
W
~1
W
~2
18 N
~N1
~T1
30°
Mass 2
Mass 1
W
~1
W
~2
4 a
~F2
~N2
~T2
~T1
W
~2
5 a B
b E
1.7 m/s2
6.9 N
1.7 m/s2
5.1 N
b
c
d
e
f
0.18 m/s2
10.72 N
1.24 m/s2
0.18 m/s2
4.3 N
b
c
d
e
2.6 m/s2
13 N
2.6 m/s2
2.6 N
~N1
~F N
~F1
Mass 2
b
c
d
e
b F = 9.8 N
˜
c 3.9 N
Mass 1
W
~1
c C
d D
11C
➔
i 32 N acting upwards, perpendicular to the
plane
ii 3.2 N acting up the plane
iii 2.4 m/s2 down the plane
b i 26 N acting upwards, perpendicular to the
plane
ii 2.6 N acting up the plane
iii 5.6 m/s2 down the plane
c i 32 N acting upwards, perpendicular to the
plane
ii 6.4 N acting up the plane
iii 1.5 m/s2 down the plane
8°
~P
2 a
b 9.95 m
Crate
~F
W
~
b 13.6 N
c 10.4 N
d 0.76
4 a 0.082
b Assuming that
Sally applies her
force parallel to
the plane, she
must apply a force
of 70 N.
~N
~P
W
~
7 a
b
d
f
h
9 6.9 N
Exercise 11D — Applications of
Newton’s First and Second Laws
of Motion
~F
8 a
W
~
b N = m(g + a) where
m = 80 kg and N is the
normal contact force.
Person
c Various answers
d 160 N
W
~
e 784 N; 944 N
The magnitude of the normal contact force is
larger than the magnitude of the weight force
because the lift is accelerating upwards.
160 N
784 N; 624 N
The magnitude of the normal contact force is
smaller than the magnitude of the weight force
because the lift is accelerating downwards.
The person will have a normal contact force equal
to twice their weight when a = 9.8 j m/s2.
˜
˜
The person will have a zero normal contact force
when a = – 9.8 j m/s2.
˜
˜
2 7.84 N
3 a
i 64 N acting upwards, perpendicular to the
plane
ii 6.4 N acting up the plane
iii 2.4 m/s2 down the plane
c 1.48 m
~N
1 a 19.6 N
d
answers
653
Answers
11E
answers
654
Answers
6 a
b 3.8 m/s2
c 11.4 N
~T2
~N1
~T1
W1
~
Mass 1
7 a
b 6.0 m/s2
c 11.4 N
~T1
~T2
W
~2
Mass 2
d The tension T in the string (answer c) is the same
in both cases, because the expression for it is
gm 1 m 2
- . The
symmetrical in m1 and m2: T = -----------------m1 + m2
acceleration is greater in question 7 because the
greater weight is hanging vertically.
b 5.5 m/s2
c 6.5 N
d 1.5
~T2
~N1
T1
~
~F
W
~1
Mass 1
m2 – m1
-j
b a = g -----------------m1 + m2˜
˜
c m1 = m2
d Tension = m 1 g
or m 2 g
~T2
~T1
W
~1
W
~2
Mass 2
Mass 1
c 204 m
b 96 kg m/s
c 0.04 kg m/s
1 a 30 kg m/s
d 30 000 kg m/s e 20 000 kg m/s
c 9500 kg m/s
3 a 260 kg m/s; 390 kg m/s
b 130 kg m/s
4 D
5 a 9.0 s
b 67.5 m
c 4000 N
d 4000 N
6 1.5 kg m/s south; 1.35 kg m/s north
7 a C
b A
8 a
~N
b 1.1 j kg m/s
˜
b 1.9 m/s2
c m1 × (7.86 m/s2)
d 1.66
~T2
~T1
~F
W
~1
Mass 1
W
~2
Mass 2
Exercise 11F — Variable forces
t4
b x = 2--3- t3 − ---4
1 a v = 2t2 − t3
t2 2
t
2 a v = ---- − --- sin --6
3
2
------ m/s
3 a 2 11
12
4
t
t3
4
b x = ------ + --- cos --- − --18 3
2 3
------ m
b 4 11
16
π+2
π2 + 2π
4 a ------------ m/s
b ------------------ m
12
48
5 a t=5s
b (20 log e 2 − 10) or 3.86 m
6 a 2000 − 40t
W
~
9 a
Force exerted by
table on glass
Glass of
orange juice
Force exerted by
Earth on glass
10 a
b i 9.5 m/s ii 24 1--6- m
b i Gravitational force exerted by Earth on glass
and gravitational force exerted by glass on Earth
and ii Normal contact force of table on glass
and normal contact force of glass on table.
c The gravitational force exerted by the Earth is
balanced by the normal contact force of the table.
They are equal in magnitude and opposite in
direction because the glass is in equilibrium.
d The table interacts with both the glass and the
Earth. It is in equilibrium because the
gravitational forces exerted by the Earth and the
glass are balanced by the normal contact force
from the floor and glass respectively.
10 a The weight of the apple is the gravitational force
exerted on it by the Earth.
b 1.2 N
c 2.0 × 10−25 m/s2 towards the apple
Chapter review
7 ±3
Multiple choice
4 3
8 ± ---------3
9 a 5 m/s2
1
5
9
13
b 20 m/s
−0.2t
10 800(3 + 5e
b 60 m/s
W
~2
Mass 2
9 a
~N
12 a 60 ( 1 – e )
b 44 s
2 a −40 kg m/s b 48 kg m/s
d −3556 kg m/s
W
~1
Mass 1
8 a
– --t3
Exercise 11G — Momentum and
Newton’s Third Law of Motion
W
~2
Mass 2
~N2
11 a 12 500(9e−0.05t + 1)
)
B
D
D
C
2 E
6 D
10 D
3 C
7 C
11 B
4 A
8 A
12 C
Short answer
1 138.6°
2 a T2
7 12
~
b 26.3 N; 52.6 N
c 26.6°; 63.4°
d Various answers
~T1
8 a
1
--- ( 15
2
c –4 3i + 4 j
˜
˜
3 a 1.5 m/s2
b 12 000 N
4 a 0.364
b 1.82 m/s2
5 a Various answers
AB = 1--2- i + 1--2- j
˜ ˜
c 108 m
b β=0
6 a 301 kg
b 656 N
c 0.865 m/s2
7 a μ = 0.27
b 2.4 m/s2
c 9.6 N
8 ± 5
DC = 1--2- i + 1--2- j
˜ ˜
14
10 a – ---------29
2
~N
A
0.5 N
30°
6g
~
b
c
d
e
1.19 m/s2
51.7 N
1.54 m/s
1.13 m
5g
~
2 a i 7482 N
ii 602 N
b 0.63 s
c 12 people of mass 80 kg.
2
a = 2i + 3t j
˜
˜
˜
3
b i v = 2t i + t j
˜
˜
˜
3 a
2
4
( --14- t
ii r = ( t + 2 )i +
˜
˜
iii v ( 2 ) = 4i + 8 j ;
˜
˜
˜
c
d
4 a
b
c
d
+ 1) j
˜
r ( 2 ) = 6i + 5 j
˜
˜
˜
13 a a ( t ) = 2i + 6t j
˜
˜
˜
2
3
b r ( t ) = ( t + 5 )i + ( t – 6 ) j
˜
˜
˜
14 a v ( t ) = 4 sin4t i – 4 cos4t j + 8 j
˜
˜
˜
˜
b r ( t ) = ( 5 – cos4t )i + ( 8t – sin4t ) j
˜
˜
˜
15 a R = 8i + 4 j
b 4 5
˜
˜
˜
c F 3 = – 8i – 4 j
˜
˜
˜
16 3g N
20
17 a – ------ m/s2
3
20
b ------ m
3
2
r = ( t + 2 )i + ( 8t – 11 ) j
˜
˜
˜
They collide when t = 7 at 51i + 45 j
˜
˜
Coefficient of friction is 0.038
193.5 m
12.33 m/s
48.2 m
Technology-free questions (page 597)
1 600 m
1
2 a – --- ( 3i – 4 j )
5 ˜
˜
1
c – ------------- ( 6i + 4 j )
˜
2 13 ˜
3 a
2
b v ( t ) = ( 3t + 2 )i + 2t j
˜
˜
˜
29 units/s
c
~T
14
b – ------ ( 2i + 5 j )
29 ˜
˜
11 y = x + 11x + 28
12 a 12i + 4 j
˜
˜
9 9375 kg m/s
~T
B
b – 6 2i – 6 2 j
˜
˜
d 5i + 5 3 j
˜
˜
9 OQ = i + j
˜ ˜
W
~
Analysis
1 a
3i + 15 j )
˜
˜
answers
655
Answers
13
3i – j
˜ ˜
4 N30° W or 330° true
c
1
--6
4
3
t 2t
19 a ---- – ------4
3
5
4
t
t
b ------ – ---20 6
5
20 – --- i m/s
3˜
1
b – ------- ( – i + 5 j )
˜
6 ˜
1
d − ---------- ( –2i – 3 j )
˜
˜
13
b
17
d
10
11F
➔
5 11 units
6 a –46
b –40
18
11G