Hlws 3
Hlws 3
Hlws 3
b x 2 − 11x + 10 = 0 3
x=±
e
(x − 10)(x − 1) = 0
2 a 2x 2 + 5x + 2 = 0
x = 1 or 10
2
c 2x 2 + x − 3 = 0 5 r 5 4(2)(2)
x
4
(2x + 3) (x − 1) = 0
3 5 r 9
x = – 2 or 1 x
4
2 a f (x) + g (x) = 2x 3 − 3 5 r 3
4
b 2h(x) − 4g (x) + 5 f (x) = 6x 4 − 4x 2 − 10 − 8x 3
+ 4x 2 − 12x + 16 + 5x 2 x = −2 or − 1
2
− 15x + 5 2
b 3x − 10x + 3 = 0
= 6x − 8x 3 + 5x 2 − 27x + 11
4
10 r 100 36
1 x
c h( x ) 2 g ( x ) 3 4
x x2 5 4 x3 2 x2 6 x 8 6
2 5 2 2 5 5 5 5
10 r 64
3 4 4 3 3 2 6 9
= x − x − x − x− 6
2 5 5 5 10
10 r 8
6
Exercise 3A x 3 or 1
2 3
1 a 2x − 3x = 0
c 5x 2 + 3x − 2 = 0
x (2x − 3) = 0
3 3 r 9 40
x = 0 or x
2 10
b 3x 2 − 75 = 0 3 r 49
x
10
x 2 = 25 3 r 7
x
x=±5 10
c 5x 2 − 4x = 0 x 1 or 2
5
x (5x − 4) = 0 d 21x 2 + 5x − 6 = 0
4
x = 0 or 5 r 25 504
5 x
42
d 7 + 28x 2 = 0 5 r 529
x
no real roots 42
5 r 23
e 242x 2 + 2x = 0 x
42
2x (121x + 1) = 0 x or 3
2
1 3 7
x = 0 or –
121
e 9x 2 − 6x + 35 = 0
2
f 2x 8 0 6 r 36 1260
2 x
x =2 18
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 1
WORKED SOLUTIONS
3 a x 2 + 4x + 2 = 0 d x 2 − 2a2 = b2 − ax − 3ab
4 r 168 x 2 + ax + (3ab − 2a2 −b2) = 0
x
2
a r a 2 12 ab 8a 2 4 b 2
4 r 2 2 x
x 2
2
a r 9 a 2 12 ab 4 b 2
x 2 r 2 x
2
b 5x 2 − 6x − 1 = 0 a r (3a 2b )
x
2
6 r 36 20
x x = −2a + b or a − b
10
6 r 56
10 Exercise 3B
6 r 2 14 1 a x 2 − 2x − 3 = 0
10 Δ = 4 − 4(1) (−3)
3 r 14 Δ = 16 > 0
x
5
∴ 2 real roots
c 3x 2 − x − 3 = 0
b x 2 + 10x + 25 = 0
1 r 1 36
x Δ = 100 − 4(1) (25)
6
x 1 r 37 Δ= 0
6
∴ one real root
2
d 2x + 11x + 13 = 0 c 4x 2 − 3x + 2 = 0
11 r 121 104 Δ = 9 − 4(4) (2)
x
4
Δ = − 23 < 0
11 r 17
x ∴ no real roots
4
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 2
WORKED SOLUTIONS
2 a x 2 − 2x − k = 0 b 3x 2 − 5x + 1 = 0
5 1
Δ = 4 + 4k x1 x 2 x1 x 2
3 3
4 + 4k = 0 3x 12 + 3x22 = 3 [(x1 + x2)2 − 2x1x2]
k = −1
3 ª 25 2 º
b kx 2 + 3x − 2 = 0 ¬« 9 3 ¼»
19
Δ = 9 + 8k
3
9 + 8k > 0 1 3
c 5x 2 + x + 3 = 0 x1 + x2 = x1x2 =
9 5
k>− 2
5
8
1
1
2
x 2 x1
2
x 1
x
2 2 x1 x 2
2
c 3x + 5x + 2k − 1 = 0 2 2 2 2 2
x1 x2 x1 x 2
x x
1 2
Δ = 25 − 12 (2k − 1) 1 6
= 37 − 24k 25 5 = − 29
9 9
37 − 24k < 0
25
37 < 24k d x 2 − 2x + 4 = 0 x1 + x2 = 2 x1x2 = 4
37
k> 2
(x1− x2) = (x1 + x2) − 4x1x2 2
24
= 4 − 16 = −12
d x 2 − (3k + 2)x + k2 = 0 3
e 2x 2 − 4x + 3 = 0 x1 + x2 = 2 x1x2 =
Δ = (3k + 2)2 − 4k2 2
x 31 + x 32 = (x1 + x2)3 − 3x1x2 (x1 + x2)
= 5k2 + 12k + 4
5k2 + 12k + 4 = 0 = 8 − 9 (2) = − 1
2
(5k + 2)(k + 2) = 0 f 2
x + 3x + 1 = 0 x1 + x2 = −3 x1x2 = 1
2
k = − or − 2
4 4
1 1 x2 + x1
5 4
4 = 4 4
x1 x2 x1 x 2
e kx 2 + 2kx + k − 2 = 0
(x1 + x2)4 = x14 + 4x13 x2 + 6x12 x22 + 4x1x23 + x24
2
Δ = 4k − 4k (k − 2) = 8k
x14 + x24 = (x1 + x2)4 − 6x12 x22 − 4x1x2 (x12 + x 22)
8k > 0
= (x1 + x2)4 – 6x12 x22 − 4x1x2 [(x1 + x2)2 − 2x1x2]
k>0
= 81 − 6 − 4 (9 − 2)
f 2kx 2 + (4k + 3)x + k − 3= 0
= 47
Δ = (4k + 3)2 − 8k (k − 3) 1 1 47
2 2
? 4
4 4
47
= 16k + 24k + 9 − 8k + 24k x1 x2 1
= 8k2 + 48k + 9 g 4x 2 − 7x + 1 = 0 x1 + x2 7
x1x2 1
4 4
8k2 + 48k + 9 < 0 x13 x 22 + x12 x 23 = x12 x 22 ( x1 + x 2 )
2
if 8k + 48k + 9 = 0, then
k
48 r 2304 288
1
16 4
7
16 7
64
48 r 12 14
4 5
16 h 7x 2 + 4x − 5 = 0 x1 + x2 x1x2
7 7
12 r 3 14
4
(x1 − x2)4 = x14 4 x13 x 2 6 x12 x 22 4 x1 x 23 x 24
−12 − 3 14 −12 + 3 14 (x1 + x2)4 = x14 + 4 x13 x 2 + 6 x12 x 22 + 4 x1 x 23 + x 24
<k<
4 4
∴ (x1 − x2)4 = (x1 + x2)4 − 8 x13 x 2 8 x1 x 23
Exercise 3C = (x1 + x2)4 − 8x1x2 ( x12 x 22 )
1 a x 2 − 3x + 2 = 0 = (x1 + x2)4 − 8x1x2 [(x1 + x2)2 − 2x1x2]
x1 + x2 = 3 4
x1 x2 = 2
§ 4 · 40
¨ 7¸
© ¹ 7 16
49
10
7
2 ( x 2 + x1 )
2
x1
+
2
x2
=
x1 x 2
=3 256
2401
40 86
7 49 24336
2401
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 3
WORKED SOLUTIONS
Exercise 3D Exercise 3F
1 a z = 3i Re (z) = 0 Im(z) = 3 2i 1
1 z1 = 1 + 4i, z2 = 2 − i, z 3 = 1 − 5 i z 4
2 2 3
b z = −7 Re (z) = −7 Im(z) = 0
−2
18 12i a
z1
= 12+−4ii × 22 ++ ii 2 i 8i 4 = +
9
i
c z Re ( z ) = 18 = 9
Im ( z ) 12 3 z2 4 1 5 5
8 8 4 8 2
z 1*
d z= 11
+i 7
Re ( z ) = 11
Im ( z ) = 7 b = 1 − 4i × 1 − 4i 1 4 i 4 i 16
= −15
−
−8
i
z2 1 + 4i 1 − 4i 1 16 17 17
4 5 4 5
(2i 1)
4i 2 −2 4 (2 i )
e z 2
Re ( z ) = 2
Im ( z ) = 2 c
z 2 .z 4 3 2 (4 i 2 2 i )
3S 3p 3p z3 1 5 3 1 5i
i
2 2
2 a 12 + 5i = 144 + 25 = 169 = 13
= 2×
3
5i
× 1 + 5i
(1 − 5i ) 1 + 5i
2 5i 25
3 1 25 50 10i
78
b −24 − 7i = 576 + 49 = 625 = 25
25 5
i
39 39
c 2 2 +i 5 = 8+5 = 13 3 z1 2 z 3 3 12i 1 5i 2 17i
d = = u1i
z2 3z4 2 i 2i 1 1i 1i
−21 + 20i 441 400
d = + =1 2 2i 17i 17
29 841 841
11
19 15
3 4 i 9 16 5 i
e 2 2
S S2 S2 S 2
z 2
(1 4 i ) 1 8i 16 15 8i
u 3 4i
1
e 2
Exercise 3E
z*
2
(2 i )
2 4 4i 1 3 4i 3 4i
45 60i 24 i 32
3 3 4i
1 z1 = 2 + 3i, z2 = − 4i, z3 = 1 − 5i, z4 9 16
2 5 13 84
i
a z1 + z3 = 3 − 2i 25 25
b z1 − 2z3 = 2 + 3i − 2 4i = −1 + 11i
3
2
2 a (2 + i) (a + ib) = 11− 2i
11 2i 2i
c z2 z4 21 16
i a + ib u
10 5 2i 2i
22 11i 4 i 2
d 5z4 − 2z2 5 2
3 4i
5
3
2
4i 12i
=4
4 1
− 3i a = 4, b = −3
e 3z1 + 4z2 − z3 − 5z4 a ib
b = −3 + 2i
= 6 + 9i + 6 − 16i − 1 + 5i − 3 − 4i 2 5i
15 7
i 23
11 i c (3i − 2) (a + ib) = 3 + 28i
2 5 5
3 + 28i
52 a + ib = × −2 − 3i 6 9i 56i 84
13 i −2 + 3i −2 − 3i 49
5 10
)( ) ( 3 + 4i
= 6 − 5i
g z 32 − 2 ( z 2 ⋅ z 4 ) = (1 − 5i )2 − 2 3
− 4i
3 3 2 5 a = 6, b = −5
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 4
WORKED SOLUTIONS
38 − 2i 28 + 5
3 a
3 2i
Re ( z ) = 3
Im ( z ) = − 1 d =
4 4 2 2+i 10 + 15i
Re(z) = 0 Im ( z ) =
13
−24
13
Im ( )=0
2 − 7i
z
∴ 7a + 2b = 0
13
∴ 7Re(z) + 2 Im(z) = 0
4 z1 = 1 + 3i z2 = 3 − i
7 Let z = a + bi, z * = a − bi
a z1 · z2 + z1 · z2* = z1 (z2 + z2*) 3 − 5i 3 5i
z*
u a bi 3a 3bi 5ai 5b
= (1 + 3i )(3 − i + 3 + i ) a bi a bi a2 b2
= 6 + 18i ⎛ ⎞
Re ⎜ 3 − 5i ⎟ = 0 ∴ 3a + 2b = 0
b z1 · z2 − z1 · z2 = (z1 − z1 ) · z2
* * ⎝ z *
⎠
= (1 + 3i −(1 − 3i )) (3 − i ) ∴ 3Re(z) + 5 Im(z) = 0
= 6i (3 − i ) 8 a | z | − z = 4 + 3i. Let z = a + ib
= 6 + 18i a 2 b 2 = 4 + 3i + a + ib
c z1 · z2 + (z1 · z2)* = (1 + 3i )(3 − i ) + = (4 + a) + i (3 + b)
[(1 + 3i )(3 − i )]* equating real and imaginary parts,
= (6 + 8i ) + (6 − 8i ) a2 b 2 = 4 + a
= 12 3 + b = 0 ∴ b = −3
5 a (z + 1)i = (z + 2i )(3 + 2i ) a2 + 9 = (4 + a)2
zi + i = 3z + 2zi + 6i − 4 a2 + 9 = 16 + 8a + a2
i − 6i + 4 = 3z + zi 8a = −7
z 4 5i
u3i 12 4 i 15i 5 a 7 ∴ z = − 7 − 3i
3i 3i 9 1 8 8
7 19 i b | z | + iz = 2 − i Let z = a + ib
10 10
a b = 2 − i − i (a + ib)
2 2
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 5
WORKED SOLUTIONS
(a1a2 )2 2a1a2b1b2 (b1b2 )2 (a1b2 )2 2a1b2 a2b1 (a2b1 )2 c z1 . z2 = (a1a2 − b1b2) + i (a1b2 + a2b1)
(z1 . z2) * = (a1a2 − b1b2) − i (a1b2 + a2b1)
= a12 a22 b12b22 a12b22 a22b12
z1 * . z2 * = (a1 − ib1)(a2 − ib2)
|z1| · |z2| a12 b12 a22 b22 = (a1a2 − b1b2) − i (a1b2 + a2b1)
∴ (z1 . z2) * = z1 * . z2 *
(a12 b12 )(a22 b22 )
d z . z * = (a + ib)(a − ib) = a2 + b2 = |z|2
a12 a22 a12b22 b12 a22 b12b22 ∴ z . z * = |z|2
∴ |z1 . z2| = |z1| . |z2| e |z *| = |a − ib| a 2 ( b )2 a 2 b 2 = |z|
a1 + ib1 a2 − ib2 ( a1 a2 + b1b2 ) + i ( a2 b1 − a1b2 )
b
z1
= × = ∴ |z *| = |z|
z2 a2 + ib2 a2 − ib2 2 2
a2 + b2
z1 ( a1 a2 + b1b2 )
2
( a2 b1 − a1b2 )
2 Exercise 3G
= +
z2 2
( a2 + b2 )
2 2 2
( a2 + b2 )
2 2
1 a i5 + i8 + i14 + i19 = i + 1 − 1 − i = 0
2 2
a1 a2 + 2 a1 a2 b1b2 + b1 b2 + a2 b1 − 2 a1 a2 b1b2 + a1 b2
2 2 2 2 2 2 b i123 + i172 + i256 + i375 = −i + 1 + 1 − i = 2 − 2i
= 2
a2 + b2
2
c (2 − i53) (3 + 2i89) = (2 − 1)(3 + 2i)
2 2 2 2
a1 a2 + b1 b2 + a2 b1 + a1 b2
2 2 2 2 = 6 + 4i − 3i + 2
= 2
a2 + b2
2 =8+i
2010 2011
4i 3i 4 3i
(a 1
2
+ b1
2
)( a 2
2
+ b2
2
) =
2
a1 + b1
2
z1
d
2i
2012
5i
2013 2 5i
= 2 2 z2 4 3i
a2 + b2 u 2 i 8 i 6i 15
2 2
a2 + b2
2 5i 2 5i 4 25
z1 z1
? 7 26i
z2 z2 = +
29 29
c Let p(n) be |z n| = |z|n 2011
i (1 − i )
Step 1: When n = 1, |z n| = |z| = |z|1 so i i
2 3
i i
2011
1−i
p(1) is true. e 2 3 2011
= (1 + 2 + 3 +....+ 2011)
i . i . i i i
Step 2: Assume p(n) is true when n = k, i.e.
i (1 i ) i (1 i )
|z k| = |z|k. 2011 2023066
(2012)
(1 i ) i
Step 3: Prove p(n) is true when n = k + 1. (1 i ) i 2
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 6
WORKED SOLUTIONS
(1 + 2i − 1) = −4 55 1
i
2
1 i or 2 1
i
4 2 2
(1 − i) = (1 − i) (1 − i) = (1 − 2i − 1) 144 3 3 4 3 4
(1 − 2i − 1) = −4 e i = x + iy
∴ (1 + i) + (1 − i) = −8
4 4
i = x 2 − y2 + 2ixy
3 a 3 4i = x + iy x 2 = y2 2xy = 1
1
3 + 4i = x 2 − y2 + 2ixy y
2x
x 2 − y2 = 3 2xy = 4 1 1 1 1 1
x2 = , x4 = x r y r =±
4
y= 2
x
4x
2 4 2
2
2
2
x − 2
x2
= 3 ∴ x − 3x − 4 = 0
4 2
i= 1
+
1
i or − 1
− 1
i
2 2 2 2
2 2
(x − 4)(x + 1) = 0
∴x = ± 2 f i = x + iy
3 4i = 2 + i or −2 − i −i = x 2 − y2 + 2ixy
x 2 = y2 2xy = −1
b 12i 5 = x + iy
−1
12i − 5 = x 2 − y2 + 2ixy y=
2x
x 2 − y2 = −5 2xy = 12 As in e, x = ± 1 , i 1
1
i or 1
1
i
6 2 2 2 2 2
y= x
4 a (1 + i )2n = [(1 + i )2]n = [1 + 2i − 1]n
36
x2 − 2
= −5 ∴ x 4 + 5x 2 − 36 = 0 ∴ (1 + i )2n = (2i )n
x
(x 2 + 9)(x 2 − 4) = 0 b (1 + i )2n + 1 = (1 + i )(1 + i)2n = (1 + i ) (2i )n
x = ±2 from a ∴ (1 + i )2n + 1 = (1 + i )(2i )n
12i 5 = 2 + 3i or −2 − 3i
c
5
3i = x + iy Exercise 3H
4
1 f (x) = 2x 2 + 3x + 1 g(x) = 3x 2 − 2x − 5
5 2 2
+ 3i = x − y + 2ixy a λ ⋅ f (x) + μ ⋅ g(x) = 2λx 2 + 3λx + λ + 3μx 2
4
5
2
x −y = 2
4
, 2xy = 3 − 2μx − 5μ
3 = (2λ + 3μ)x 2 + (3λ − 2μ)x + (λ − 5μ)
∴y= 2x
2λ + 3μ = 0
9 5
x2 2 3λ − 2μ = 13
4x 4
4x 4 − 5x 2 − 9 = 0 λ − 5μ = 13
(4x 2 − 9)(x 2 + 1) = 0 λ = 3, μ = −2
3
x=± b 2λ + 3μ = 26
2
5
+ 3i = 3
+ i or − − i
3 3λ − 2μ = 26
4 2 2
λ − 5μ = 0
55
d 1i x iy λ = 10, μ = 2
144 3
55 2 a
1i x 2 y 2 2 ixy x3 −2x
144 3
2
55 1 x x 5 −2x 3
x 2 y2 2xy
144 3 2 2x 3 −4x
1
y f (x) ⋅ g (x) = x 5 − 4x
6x
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 7
WORKED SOLUTIONS
3 f (x) ⋅ g (x) = (ax 2 − 3x + 5) (7x 2 + bx − 3) f (10) = 32584 ⇒ 10000a + 1000b + 100c + 10d + 4
= 32584
= 7ax 4 + abx 3 − 3ax 2 − 21x 3 − 3bx 2 + 9x
Since a, b, c, d, ∈ + and are less than 10,
+ 35x 2 + 5bx − 15
a = 3, b = 2, c = 5, d = 8
= 7ax 4 + (ab − 21)x 3 + (−3a − 3b + 35)x 2
⇒ f (x) = 3x4 + 2x3 + 5x2 + 8x + 4
+ (9 + 5b)x − 15
7a = 14 Exercise 3I
ab − 21 = −17 x 3 3x 2 2 x 1
−3a − 3b + 35 = 23 1 a x 2 x 4 5x 3 8x 2 3x 2
9 + 5b = 19 a=2 b=2 x 4 + 2x 3
4 f (x) ⋅ g (x) = (x 3 + ax 2 − x + 2)(2x 2 + bx + c) 3x 3 + 8x 2
5 4 3 4 3
= 2x + bx + cx + 2ax + abx 3x 3 + 6x 2
2 3 2 2
+ acx − 2x − bx − cx + 4x + 2bx + 2c 2x 2 + 3x
2a + b = −5 2x 2 + 4x
c + ab − 2 = 3 −x − 2
ac − b + 4 = 5 −x − 2
−c + 2b = −8
q (x) = x 3 + 3x 2 + 2x − 1
a = −1 b = −3 c=2
x 3 3x 2 2 x 1
5 (x + px + q) = (x + px + q)(x 2 + px + q)
2 2 2
b x 2 0 x 1 x 5 3x 4 x 3 4 x 2 2 x 1
= x 4 + px 3 + qx 2 + px 3 + p2x 2 + pqx + qx 2
+ pxq + q2 x 5 + 0x 4 − x 3
= x 4 + 2px 3 + (2q + p2) x 2 + 2pqx + q2 3x 4 + 2x 3 − 4x 2
2p = 6 p=3 3x 4 + 0x 3 − 3x 2
2q + p2 = a q = ±2 2x 3 − x 2 − 2x
2pq = b If q = 2, a = 13, b = 12 2x 3 + 0x 2 − 2x
q2 = 4 If q = −2, a = 5, b = −12 −x 2 + 0x + 1
a = 13, b = 12, f (x) = (x 2 + 3x + 2)2 −x 2 + 0x + 1
or a = 5, b = −12, f (x) = (x 2 + 3x − 2)2 q (x) = x 3 + 3x 2 + 2x − 1
6 f (x) = x 3 + 12x 2 + 6x + 3 2 x 3 5x 2 4 x 1
g (x) = f (x − 2) c x 2 x 1 2 x 5 3 x 4 x 3 2 x 2 3x 1
3 2
= (x − 2) + 12(x − 2) + 6(x − 2) + 3
2x 5 + 2x 4 + 2x 3
3 2 2 3
= x + 3x (−2) + 3x(−2) + (−2) +
−5x 4 − x 3 − 2x 2
12(x 2 − 4x + 4) + 6x − 12 + 3
−5x 4 − 5x 3 − 5x 2
=x 3 − 6x 2 + 12x − 8 + 12x 2 − 48x + 48 + 6x − 9
−4x 3 + 3x 2 + 3x
g (x) = x 3 + 6x 2 − 30x + 31
4x 3 + 4x 2 + 4x
7 f (2x − 1) = 16x 4 − 32x 3 + 12x 2
y 1 −x 2 − x − 1
Let y = 2x −1, x
2 −x 2 − x − 1
4 3 2
f ( y) 16 ( y 1) 32 ( y 1) 12 ( y 1) q (x) = 2x 3 − 5x 2 + 4x − 1
16 8 4
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 8
WORKED SOLUTIONS
2 x 3 3x 2 x 3 2 f (2) = 4 × 16 – 27 × 4 + 25 × 2 – 6 = 0
2 a x 1 2 x 4 5x 3 4 x 2 4 x 3 ∴ (x – 2) is a factor.
2x 4 + 2x 3 f (–3) = 4 × 81 – 27 × 9 – 25 × 3 – 6 = 0
3x 3 + 4x 2 ∴ (x + 3) is a factor.
3x 3 + 3x 2
x 2 + 4x
Exercise 3K
x2 + x 1 a q (x) = x 4 − x 3 + x 2 + 2x + 1 r (x) = –2
3x + 3 b q (x) = x 3 + x 2 + x − 1 r (x) = 7
3x + 3 2 f (x) = (x 2 + 2x − 1)(3x − 4) + x + 2
= 3x 3 − 4x 2 + 6x 2 − 8x − 3x + 4 + x + 2
q (x) = 2x 3 + 3x 2 + x + 3 r (x) = 0
f (x) = 3x 3 + 2x 2 − 10x + 6
2
3x 2 x 1 3 f (x) = x 5 − 4x 4 + 3x 3 + 2x 2 − 3x + a
b x 2 x 3 3x 4 4 x 3 6 x 2 2 x 6
2
f (3) = 0 243 − 324 + 81 + 18 − 9 + a = 0
4 3 2
3x + 6x + 9x a = −9
3 2
−2x − 3x − 2x 4 f (x) = x 5 − 2x 4 + 2x 3 + bx − 1
−2x 3 − 4x 2 − 6x f (1) = 0 1−2+2+b−1=0
2
x + 4x + 6 ∴b=0
2
x + 2x + 3 5 f (x) = 4x 3 + 5x 2 + ax + b
2x + 3 f (−2) = 0 f (1) = 6
q (x) = 3x 2 − 2x + 1 r (x) = 2x + 3 −32 + 20 − 2a + b = 0 4+5+a+b=6
−2a + b = 12 a + b = −3
x4 x3 x 1 a = −5, b = 2
c x 2 x 1 x 6 0x 5 0x 4 0x 3 0x 2 x 1
6 f (x) = (x 2 − 2x − 3) q (x) + ax + b
x6 + x5 + x4 f (x) = (x − 3)(x + 1) q (x) + ax + b
5
−x − x + 0x 4 3 f (3) = 2 ∴ 2 = 3a + b
−x 5 − x 4 − x 3 f (−1) = − 4 ∴ −4 = −a + b
x 3 + 0x 2 + x ∴ a = 1.5 b = −2.5
x3 + x2 + x ∴ remainder = 3 x − 5
2 2
−x 2 + 0x − 1
−x 2 − x − 1 7 f (x) = x 2011 + x 2010 + … + x + 1 = 0
x remainder = f (−1) = −1 + 1 −1 + … − 1 + 1 = 0
∴ remainder = 0
q (x) = x 4 − x 3 + x − 1 r (x) = x
8 f (x) = (x + 1)2n + (x + 2)n − 1
f (−1) = 02n + (1)n − 1 = 0 + 1 − 1 = 0
Exercise 3J ∴ f (x) is divisible by (x + 1)
1 a x3 − x2 − 4x − 5 = (x2 + 2x + 2)(x − 3) + 1, f (−2) = (−1)2n + 0n − 1 = 1 + 0 − 1 = 0
2
q (x) = x + 2x + 2 r (x) = 1 ∴ f (x) is divisible by (x + 2)
b 2x3 + 5x2 + 4x + 3 = (2x2 + 3x + 1)(x + 1) + 2, (x + 1)(x + 2) = x 2 + 3x + 2
2
q (x) = 2x + 3x + 1 r (x) = 2 ∴ f (x) is divisible by x 2 + 3x + 2
5 3 4 3 2
c x − 3x − 2x + 1 = (x – 2x + x – 2x + 2) 9 f (x) = (ax − b) q (x) + r (x)
(x + 2) – 3,
q (x) = x 4 – 2x3 + 2x2 – 2x + 2, r (x) = –3 f
b
a
ab
a + remainder
b q b
a
= (0) q + remainder
d 3x6 − 2x4 + 5x2 – 2 = (3x5 + 3x4 + x3 + x2 + b
6x + 6)(x – 1) + 4, a
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 9
WORKED SOLUTIONS
b
= (x + 2)(x + 2)(2x − 1)(x − 2)
f (x) = 12x 3 − 32x 2 + 23x − 5
c ¨
©
§ x 3 3 i ·§ x 3
2 2
¸¨
¹© 2 2
3
i ·¸¹ = x + 3x + 3
2
3 2
f (x) = 5x + 17x + 21x + 6
= (2x − 1)2 g (x)
= (x 2 + 3x + 3)(5x + 2)
= (4x 2 − 4x + 1)(3x − 5)
3
= (2x − 1)2(3x − 5) remaining zeros are − 3 − i and − 2
2 2 5
2
2 a f (x) = x (x − 1)(x − 3)(x − 5) d (x − i )(x + i ) = x + 1
2
= (x − 1)(x − 8x + 15) f (x) = x 4 − 6x 3 + 5x 2 − 4x + 4
= x 3 − 9x 2 + 23x − 15 = (x 2 + 1)(x 2 − 4x + 4)
= (x + i )(x − i )(x − 2)2
b f (x) = x (x + 2)(x + 1)(x − 1)
remaining zeros are i, 2 and 2
= x (x + 2)(x 2 − 1)
e (x −(−1 − 3i ))(x −(−1 + 3i )) = x 2 + 2x + 10
= x (x 3 + 2x 2 − x − 2)
f (x) = 2x 4 + 3x 3 + 17x 2 − 12x − 10
= x 4 + 2x 3 − x 2 − 2x
= (x 2 + 2x + 10)(2x 2 − x − 1)
c f (x) = (3x + 2)(x − 1)(x − 2)(x − 3)
= (x −(−1 −3i ))(x −(−1 + 3i ))(2x + 1)(x − 1)
= (3x 2 − x − 2)(x 2 − 5x + 6)
remaining zeros are − 1 + 3i, − 1 , 1
= 3x 4 − 16x 3 + 21x 2 + 4x − 12 2
2
f (x −(−2 + i ))(x −(−2 −i)) = x + 4x + 5
3 a f (x) = (x 2 − 2)(x 2 − 3) = x 4 − 5x 2 + 6
f (x) = 2x 4 + 9x 3 + 11x 2 − 7x − 15
b f (x) = (2x + 1)(4x − 3)(x 2 − 5)
= (x 2 + 4x + 5)(2x 2 + x − 3)
= (2x + 1)(4x 3 − 3x 2 − 20x + 15)
= (x −(−2 + i))(x −(−2 − i ))
= 8x 4 − 2x 3 − 43x 2 + 10x + 15 (2x + 3)(x − 1)
c f (x) = (5x + 3)(x − (1 − 2 ))(x − (1 + 2 )) remaining zeros are − 2 − i, − 3 , 1
2
(x 3 − 3)
= (5x + 3)(x 2 − 2x − 1)(x 3 − 3) g
© 2 2
¸¨
¹© 2 2
¸
¹
§ x 1 5 i ·§ x 1 5 i · = x 2 + x +
¨ 3
2
= (5x + 3)(x 5 − 2x 4 − x 3 − 3x 2 + 6x + 3) 4 3 2
f (x) = 6x + 26x + 35x + 36x + 9
= 5x 6 − 7x 5 − 11x 4 − 18x 3 + 21x 2 + 33x + 9
= (2x 2 + 2x + 3)(3x 2 + 10x + 3)
3 2
4 a f (x) = x − 2x − 5x + 6
= (x − 1)(x 2 − x − 6) ©
2 §¨ x 1
2 2
5
i ·§¸¨¹© x 1
2
2
5
i ·¸¹ (3x 1)(x 3)
= (x − 1)(x − 3)(x + 2) remaining zeros are − 1 − 5
i, − 1 , −3
2 2 3
3 2
b f (x) = 2x − x − 7x + 6
= (x − 1)(2x 2 + x − 6)
h §x
¨
© 1
3
3
2
i ·§¸¨¹© x 1
3
3
2
i ·¸¹ x2 2 x 1
3 3
= (x − 1)(2x − 3)(x + 2) f (x) = 3x 4 − 2x 3 + 4x 2 − 2x + 1
c f (x) = 5x 4 − 12x 3 − 14x 2 + 12x + 9 = (3x 2 − 2x + 1)(x 2 + 1)
= (x − 1)(x + 1)g (x) 1 2
remaining zeros are − i , i , −i
2 2 3 3
= (x − 1)(5x − 12x − 9)
= (x − 1)(x + 1)(5x + 3)(x − 3) 2 a f (−1) = 0
∴ −1 + 13 + a = 0
Exercise 3M ∴ a = −12
1 a (x − 2i )(x + 2i ) = x 2 + 4 f (x) = x 3 − 13x − 12
f (x) = x 3 + 3x 2 + 4x + 12 = (x + 1)(x 2 − x − 12)
= (x 2 + 4)(x + 3) = (x + 1)(x − 4)(x + 3)
remaining zeros are −2i and −3 remaining zeros are −3 and 4
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 10
WORKED SOLUTIONS
b f (3) = 0 ∴ 27 − 63 + 3a − 15 = 0 3 3 3 3 3(4)
e + + + = = −2
∴ a = 17 x1 x2 x3 x4 −6
−4 + 4 − 4i − 4i + 8 + 8i + a = 0 b f (x) = 4x 6 + x 5 + 7x 4 − 3x 3 + 2x
∴ a = −8 sum = − 1 product = 0
4 3 2
4
f (x) = x + 2x − 2x − 8x − 8 3
c f (x) = 11x 10 − x 7 + 5 x 3 − πx + 22
(x −(−1 −i ))(x −(−1 + i )) = x 2 + 2x + 2 7
d
6
6
6
6 x 2 x 3 x1 x 3 x1x 2 6 5
3 = − 15 x 3, 5 , 3
x1 x2 x3 x1 x 2 x 3 2
4 2
3
e 2 2
9x1 + 9x2 + 9x3 = 9[(x1 + x2 + x3) 2 2 Exercise 3P
−2(x1x2 + x1x3 + x2x3)] 1 a 12x 3 + 17x 2 + 2x − 3 = 0
(x + 1)(12x 2 + 5x − 3) = 0
ª 2
¬
9 « 32 2
5
3
º
»¼ 9 4 10
9
3 (x − 1)(3x − 1)(4x + 3) = 0
= 34 −3 1
x = −1, ,
4 3 2 4 3
2 x − 3x + 2x − 4x − 6 = 0
b x 3 − 4x 2 − 5x + 14 = 0
a x1 + x2 + x3 + x4 = 3
(x + 2)(x 2 − 6x + 7) = 0
b x1 ⋅ x2 ⋅ x3 ⋅ x4 = −6
6 r 36 28 6r 8
c x1 ⋅ x2 + x1 ⋅ x3 + x1 ⋅ x4 + x2 ⋅ x3 + x2 ⋅ x4 + x3 ⋅ x4 = 2 x = −2 or x
2 2
=3± 2
d x1 ⋅ x2 ⋅ x3 + x1 ⋅ x2 ⋅ x4 + x1 ⋅ x3 ⋅ x4 + x2 ⋅ x3 ⋅ x4 = 4 x = −2, 3 ± 2
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 11
WORKED SOLUTIONS
(x + 3)(x 2 − 1) = 0 x ]−∞
∞, −4[ ]−4, 1[ ]1, ∞[
(x − 1)2 + + +
(x + 3)(x − 1)(x + 1) = 0
(x + 4) − + +
other roots are 1, −1 f (x) − + +
3 ax 3 − 7x 2 + bx + 4 = 0 ∴ x ∈ ]−∞, −4] or x = 1
a (x − 2)(x − 2)(ax + 1) = 0 c x 3 + 3x 2 − 4x − 12 < 0
2
(x − 4x + 4)(ax + 1) = 0 (x − 2)(x 2 + 5x + 6) < 0
ax 3 + x 2 − 4ax 2 − 4x + 4ax + 4 = 0 (x − 2)(x + 2)(x + 3) < 0
ax 3 + (1 − 4a) x 2 + (4a − 4)x + 4 = 0 x ]−∞
∞, −3[ ] −3, −2[ ] −2, 2[ ]3, ∞[
1 − 4a = −7 ∴a=2 x−2 − − − +
4a − 4 = b b=4 x+2 − − + +
x+3 − + + +
b remaining root = − 1 = − 1
a 2 f (x) − + − +
4 Let x = a be an integer zero ∴ x ∈ ]−∞, −3 [ ∪ ] −2, 2[
∴ a + 5a + p = 0
3
d 2x 3 − 5x 2 − 18x + 45 > 0
2
p = −a (a + 5) (x − 3)(2x 2 + x − 15) > 0
∴ a and a 2 + 5 are factors of p (x − 3)(2x − 5)(x + 3) > 0
∴ p is not prime x ]−∞
∞, −3[ ]−3, 2.5[ ]2.5, 3[ ]3, ∞[
∴ If p is prime there are no integer zeros x−3 − − − +
5 f (x) = x 3 + ax 2 + bx + c 2x − 5 − − + +
a Let the 2 zeros be p, −p x+3 − + + +
f (x) − + − +
§ ·
f (x) = (x − p)(x + p) ¨ x c2 ¸
© p ¹ ∴ x ∈ ]−3, 2.5 [ ∪ ] 3, ∞[
§ · e 12x 3 + 17x 2 + 2x − 3 ≤ 0
= ( x p ) ¨ x c2 ¸
2 2
(x + 1)(12x 2 + 5x − 3) ≤ 0
© p ¹
c
(x + 1)(4x + 3)(3x − 1) ≤ 0
x3 x 2 p2x c
p2 x ]−∞
∞, −1[ ]−1, −3 [ ] −3 , 13 [ ] 13 , ∞[
c 4 4
∴a b = −p 2
p2 x+1 − + + +
⎛ ⎞ 4x + 3 − − + +
∴ ab = ⎜ − c2 ⎟ (− p 2 ) = c ∴ ab = c
3x − 1 − − − +
⎝ p ⎠
f (x) − + − +
c c
b third zero = = –a
p2 b ∴ x ∈ ]−∞, −1] ∪ [ 43 , 31 ]
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 12
WORKED SOLUTIONS
f x 3 − 4x 2 − 5x + 14 > 0 3 x 7 − 2x 2 − 1 ≥ 0
a
6 r 36 28
3r 2 x ∈ [−1, −0.921] ∪ [1.26, ∞[
(x + 2)(x 2 − 6x + 7) > 0
2 b x 9 − 2x 8 + 2x 5 + x ≤ 0
]−2, x ∈ ]−∞, 0]
]3− 2 ,
x ]−∞
∞, −2[ ]3+ 2 , ∞[ 4 a x3 + x − 2 > 0
3− 2 [ 3+ 2 [
x+2 − + + + x3 > 2 − x
− − − + x ∈ ]1, ∞[
x − (3+ 2 )
y
x − (3− 2 ) − − + + y = x3
y=2–x
f (x) − + − + 2
(1, 1)
x @ 2, 3 2 > @ 3 2, f >
O x
2
g 3x 3 − 13x 2 + 11x + 14 < 0
(3x + 2)(x 2 − 5x + 7) < 0
2 b −2x 3 + 3x + 1 ≥ 0
x
3 3x + 1 ≥ 2x 3
x ∈ ]−∞, −1] ∪ [−0.366, 1.37]
x @ f, 32 > @ 32 , f > y
y = 2x3
3x + 2 − +
x 2 − 5x + 7 + +
f (x) − +
x ∈ ] − ∞, −2
[ –0.366
3
–1 O 1.37 x
h x 4 − x 3 − 11x 2 + 9x + 18 ≥ 0
y = 3x + 1
(x + 1)(x − 2) g (x) ≥ 0
(x 2 − x − 2)(x 2 − 9) ≥ 0 c x 4 + 2x + 1 ≤ 0
(x + 1)(x − 2)(x − 3) (x + 3) ≥ 0 x 4 ≤ −2x − 1
x ∈ [−1, −0.544]
x ]−∞
∞, −3[ ]−3, −1[ ]−1, 2[ ]2, 3[ ]3, ∞[ y
x+1 − − + + +
x−2 − − − + +
y = x4
x−3 − − − − +
x+3 − + + + +
f (x) + − + − +
x
∴ x ∈ ]−∞, −3] ∪ [−1, 2] ∪ [3, ∞[ –1 –0.544
y = –2x – 1
2 f (x) > g (x)
4x 3 − 17x 2 + 30x + 5 > −2x 3 Exercise 3R
+ 8x 2 + 9x − 5 1 a 2ix + (2 + 3i)y = i a = 2i b = 2+ 3i e = i
6x 3 − 25x 2 + 21x + 10 > 0 (1 + i)x + 2y = 3 c=1+i d=2 f=3
(x − 2)(6x 2 − 13x − 5) > 0 2i 2 3i
(x − 2)(3x + 1)(2x − 5) > 0 D= = 4i − (1 + i)(2 + 3i)
1 i 2
= 4i − 2 − 5i + 3 = 1 − i
x @ f, 13 > @ 13 , 2 > @ 2, 52 > @ 52 , f >
i 2 3i
x−1 − − + + Dx = = 2i − 3 (2 + 3i) = −6 −7i
3 2
3x + 1 − + + +
2i i
2x − 5 − − − + Dy = = 6i − i (1 + i) = 5i + 1
f (x) − g (x) − + − + 1 i 3
6 7i 1 i 6 6i 7i 7
1 5
x= 1i
u
1i
= 1 − 13 i
x @ , 2> @ , f> 2 2 2
3 2 1 5i 1 i 1 i 5i 5
y u = −2 + 3i
1i 1i 2
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 13
WORKED SOLUTIONS
17 9i
(4) + (5) are inconsistent ∴ no solution
y u 4 13i 68 221i 36i 117
=1−i
4 13i 4 13i 185 3 a x + 2y + z = 0 (1) (1) − (3) z − kz = −2
2 a x + y = −1 (1) x + y = −1 (1) 2x + y + 2z = 1 (2) (1 − k) z = −2
x+z=4 (2) (2) − (3) x − y = 3 (4) x + 2y + kz = 2 (3) no solution if k = 1
2
y + z =1 (3) If k ≠ 1, z
1 k
(1) + (4) 2x = 2 ∴x=1 2
sub in (1) x 2 y (4)
1 k
sub in (1) 1 + y = −1 y = −2
4 5k
sub in (3) −2 + z = 1 z=3 (1, −2, 3) sub in (2) 2 x y 1 (5)
1 k 1 k
b x − 5y + 3z = −1 (1) (1) + 3(3) 7x − 2y 5k 4
(5) − 2(4) 3 y
1 k 1 k
= 5 (4)
1 k
3x − y + 2z = 4 (2) (2) + 2(3) 7x + y 3 y ? y 1
1 k 3
= 8 (5)
2 6 2 (1 k )
2x + y − z = 2 (3) sub in (4) x 2 8 2k
1 k 3 3(1 k ) 3(1 k )
(5) − (4) 3y = 3, y = 1 ∴ no unique solution if k = 1
sub in (5) 7x + 1 = 8 x = 1 b x + y + z = 1 (1) (2) − 2(1) (k − 2) y + z
sub in (3) 2+1−z=2 z=1 (1, 1, 1) = −4 (4)
c 2x + y + 2z = 0 (1) 4(1) + (2): 2x + ky + 3z = −2 (2) (3) − 3(1) 2y + (k − 3) z
14x + 3z = −2 (4) = −4 (5)
6x − 4y − 5z = −2 (2) (3) − (1): 3x + 5y + kz = −1 (3)
2x − 5z = 2 (5) For no unique solution
4x + y − 3z = 2 (3) k2 1
2 k 3
8
(4) − 7(5) 38z = − 16 z
19 (k − 2)(k − 3) = 2
40 1
sub in (5) 2x + = 2, x k 2 − 5k + 4 = 0
19 19
(k − 4)(k − 1)= 0
sub in (1)
2
19
y 16
19
0 y 18
19 1 18 8
, ,
19 19 19 k = 1 or 4
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 14
WORKED SOLUTIONS
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 15
WORKED SOLUTIONS
1 1 1 6
∴ 2
+ 2
+ 2
= 2
=6= 2
α 2γ + βγ 2 + β 2γ )
D E J 3 9 3
(α + β + γ)(γ β + βγ + αγ ) = 3αβγ + αβ 2 + αγ 2 +
10 a Let a = 3 7 50 , b = 3 7 50 , x = a + b α2β + α 2γ + βγ 2 + β 2γ
x 3 + 3x − 14 = (a + b)3 + 3(a + b) − 14 ∴αβ 2 + αγ 2 + α 2β + α 2γ + βγ 2 + β 2γ
= a 3 + 3a 2b + 3ab 2 + b 3 + 3 (a + b) − 14
= a 3 + b 3 + 3ab(a + b) + 3(a + b) − 14 3
5 2
3 3
4
3
26
9
a 3 + b 3 + 3(a + b)(ab + 1) − 14
∴ α3 + β 3 + γ 3 125
27
6 3
4
3
26
9
ab = 3
7
50 7 50 = 3 49 50 = 3 1 = −1 143
=
∴ x + 3x − 14 = a + b 3 + 3(a + b)(−1 + 1) − 14
3 3 27
= a 3 + b 3 − 14 = 0 5 f (x) = x 7 + 35x 6 − 97x 5 + 33x 2 + 4
7 50 7 50 14 = 0 smallest zero = 0.833
© Oxford University Press 2012: this may be reproduced for class use solely for the purchaser’s institute Worked solutions: Chapter 3 16