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CHAPTER 1 Graphs and polynomials • EXERCISE 1A | 1
Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
Exercise 1A — The binomial theorem
1 a ( x + 3)2 = 2
2
0 x
⎛ ⎞⎜ ⎟⎝ ⎠
+ 1 12
31
x⎛ ⎞⎜ ⎟⎝ ⎠
+ 22
32
⎛ ⎞⎜ ⎟⎝ ⎠
= x2 + 6 x + 9
b ( x + 4)5 = 55
0 x⎛ ⎞⎜ ⎟
⎝ ⎠ + 4
54
1 x⎛ ⎞⎜ ⎟
⎝ ⎠ + 3 2
54
2 x⎛ ⎞⎜ ⎟
⎝ ⎠ + 2 3
54
3 x⎛ ⎞⎜ ⎟
⎝ ⎠
+ 45
44 x
⎛ ⎞⎜ ⎟⎝ ⎠
+ 55
45
⎛ ⎞⎜ ⎟⎝ ⎠
= x5 + 5 x44 + 10 x316 + 10 x264 + 5 x256 + 1 × 1024
= x5 + 20 x
4 + 160 x
3 + 640 x
2 + 1280 x + 1024
c ( x − 1)8 = 8
8
0 x
⎛ ⎞⎜ ⎟⎝ ⎠
+ 78
( 1)1 x
⎛ ⎞−⎜ ⎟
⎝ ⎠ + 6 2
8( 1)
2 x
⎛ ⎞−⎜ ⎟
⎝ ⎠
+ 5 38
( 1)3 x
⎛ ⎞−⎜ ⎟
⎝ ⎠ + 4 4
8( 1)
4 x
⎛ ⎞−⎜ ⎟
⎝ ⎠ + 3 5
8( 1)
5 x
⎛ ⎞−⎜ ⎟
⎝ ⎠
+ 2 68
( 1)6 x
⎛ ⎞−⎜ ⎟
⎝ ⎠ + 7
8( 1)
7 x
⎛ ⎞−⎜ ⎟
⎝ ⎠ + 8
8( 1)
8
⎛ ⎞−⎜ ⎟
⎝ ⎠
= x8 − 8 x7 + 28 x6 − 56 x5 + 70 x4 − 56 x3 + 28 x2
− 8 x + 1
d (2 x + 3)4 = 44
(2 )0
x⎛ ⎞⎜ ⎟⎝ ⎠
+ 34
(2 ) 31
x⎛ ⎞⎜ ⎟⎝ ⎠
+ 2 24
(2 ) 32
x⎛ ⎞⎜ ⎟⎝ ⎠
+ 34
(2 )33
x⎛ ⎞⎜ ⎟⎝ ⎠
+ 44
34
⎛ ⎞⎜ ⎟⎝ ⎠
= 16 x4 + 96 x3 + 216 x2 + 216 x + 81
e (7 − x)4 = 44
70
⎛ ⎞⎜ ⎟⎝ ⎠
+ 3 2 24 4
7 ( ) 7 ( )1 2
x x⎛ ⎞ ⎛ ⎞
− + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
+ 3 44 4
7( ) ( )3 4
x x⎛ ⎞ ⎛ ⎞− + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= 2401 − 1372 x + 294 x2 − 28 x3 + x4
f (2 − 3 x)5 = 55
20
⎛ ⎞⎜ ⎟⎝ ⎠
+ 45
2 ( 3 )1
x⎛ ⎞
−⎜ ⎟⎝ ⎠
+ 3 25
2 ( 3 )2
x⎛ ⎞
−⎜ ⎟⎝ ⎠
+ 2 35
2 ( 3 )3
x⎛ ⎞
−⎜ ⎟⎝ ⎠
+ 45
2( 3 )4
x⎛ ⎞
−⎜ ⎟⎝ ⎠
+ 55
( 3 )5
x⎛ ⎞
−⎜ ⎟⎝ ⎠
= 32 − 240 x + 720 x2 − 1080 x
3 + 810 x
4 − 243 x
5
2 a
31
x x
⎛ ⎞+⎜ ⎟
⎝ ⎠ = x3 + 3 x2
1
x
⎛ ⎞⎜ ⎟⎝ ⎠ + 3 x
21
x
⎛ ⎞⎜ ⎟⎝ ⎠
+
31
x
⎛ ⎞⎜ ⎟⎝ ⎠
= x3 + 3 x + 3 x
+ 3
1 x
b
72
3 x x
⎛ ⎞−⎜ ⎟
⎝ ⎠ = (3 x)7 − 7(3 x)6
2
x
⎛ ⎞⎜ ⎟⎝ ⎠ + 21(3 x)
5
22
x
⎛ ⎞⎜ ⎟⎝ ⎠
− 35(3 x)4
32
x
⎛ ⎞⎜ ⎟⎝ ⎠ + 35(3 x)
3
42
x
⎛ ⎞⎜ ⎟⎝ ⎠
− 21(3 x)2
52
x
⎛ ⎞⎜ ⎟⎝ ⎠ + 7(3 x)
6 72 2
x x
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= 2187 x7 − 10 206 x5 + 20 412 x3 − 22 680 x
+ 3
15 120 6048
x x− +
5 7
1344 128
x x−
c
62 3 x
x
⎛ ⎞+⎜ ⎟
⎝ ⎠= ( x2)6 + 6( x2)5
3
x
⎛ ⎞⎜ ⎟⎝ ⎠ + 15( x
2)4
23
x
⎛ ⎞⎜ ⎟⎝ ⎠
+ 20( x2)3
33
x
⎛ ⎞⎜ ⎟⎝ ⎠ + 15( x
2)2
43
x
⎛ ⎞⎜ ⎟⎝ ⎠
+ 6( x2)
5
3 x
⎛ ⎞⎜ ⎟⎝ ⎠
+
6
3 x
⎛ ⎞⎜ ⎟⎝ ⎠
= x12 + 18 x9 + 135 x6 + 540 x3 + 1215 + 3
1458
x
+ 6
729
x
d
5
2
32 x
x
⎛ ⎞−⎜ ⎟
⎝ ⎠ =
5 4
2 2
3 35 (2 ) x
x x
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ +
32
2
310 (2 ) x
x
⎛ ⎞⎜ ⎟⎝ ⎠
−
23
2
310 (2 ) x
x
⎛ ⎞⎜ ⎟⎝ ⎠
+4 5
2
35 (2 ) (2 ) x x x
⎛ ⎞−⎜ ⎟
⎝ ⎠
= 10 7
243 810
x x− +
4
1080 720
x x− + 240 x2 − 32 x5
3 (r + 1)th
term isn
r
⎛ ⎞⎜ ⎟⎝ ⎠
(ax)n − r b
r
a ( x − 7)3
i x2is the 2nd term
⇒ = 1
Coefficient = 3
1
⎛ ⎞⎜ ⎟⎝ ⎠
x2 (−7)
1
= −21
ii x3 is first term ⇒ r = 0
term =
3
0
⎛ ⎞
⎜ ⎟⎝ ⎠ x3
70
Coefficient = 1
iii x4
Coefficient = 0
b (2 x + 1)5
i x2 is the 4
th term ⇒ r = 3
term = 5
3
⎛ ⎞⎜ ⎟⎝ ⎠
(2 x)213
Coefficient = 40
ii x3 is the third term
⇒ r = 2
term = 5
2
⎛ ⎞⎜ ⎟⎝ ⎠
(2 x)31
2
Coefficient = 80
iii x4 is the 2nd term ⇒ r = 1
term = 5
1
⎛ ⎞⎜ ⎟⎝ ⎠
(2 x)411
Coefficient = 80
c
52
3 x x
⎛ ⎞+⎜ ⎟
⎝ ⎠
i Coefficient of x2 = 0
Chapter 1 — Graphs and polynomials
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2 | CHAPTER 1 Graphs and polynomials • EXERCISE 1B
Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
ii x3 is the 5th term ⇒ r = 4
term =
15 2
4 x
⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠(3 x)4
Coefficient = 810
iii Coefficient of x4 = 0
d
62 3 x
x
⎛ ⎞−⎜ ⎟
⎝ ⎠
i Coefficient of x2 = 0
ii x3 is the 4th term ⇒ r = 3
term = 6
3
⎛ ⎞⎜ ⎟⎝ ⎠
( x2)3
33
x
⎛ ⎞−⎜ ⎟
⎝ ⎠
Coefficient = −540
iii Coefficient of x4 = 0
e
6
2
37 x
x
⎛ ⎞+⎜ ⎟
⎝ ⎠
i Coefficient of x2 = 0
ii x3 is the 2nd term ⇒ r = 1
term = 6
1
⎛ ⎞⎜ ⎟⎝ ⎠
(7 x)5
1
2
3
x
⎛ ⎞⎜ ⎟⎝ ⎠
Coefficient = 302 526iii Coefficient of x4 is 0.
4
32 5
3 x x
⎛ ⎞−⎜ ⎟
⎝ ⎠
x3 is the 2nd term ⇒ r = 1
term = 3
1
⎛ ⎞⎜ ⎟⎝ ⎠
(3 x2)2
15
x
−⎛ ⎞⎜ ⎟⎝ ⎠
= 3 × 9 x4 × 1
5
x
−
= −135 x3 The answer is A.
5 When the expression for C is expanded it does not contain an x5 term. The first three terms contain x8, x6 and x4 respectively.
All the other expressions contain an x5
term.The answer is C.
6
53
2
2 x
x
⎛ ⎞+⎜ ⎟
⎝ ⎠ = ( x
3)5 + 5( x
3)4
2
2
x
⎛ ⎞⎜ ⎟⎝ ⎠ + 10( x
3)
3
2
2
2
x
⎛ ⎞⎜ ⎟⎝ ⎠
+ 10( x3)2
3
2
2
x
⎛ ⎞⎜ ⎟⎝ ⎠ + 5( x
3)
4
2
2
x
⎛ ⎞⎜ ⎟⎝ ⎠
+
5
2
2
x
⎛ ⎞⎜ ⎟⎝ ⎠
= 1 x15 + 10 x10 + 40 x5 + 80 + 3
80
x +
2
32
x
∴ = 1 + 10 + 40 + 80 + 80 + 32
= 243
The answer is D.
7 (2 x − 3)4
= (2 x)4 − 4(2 x)33 + 6(2 x)232 − 4(2 x)33 + 34
= 16 x4 − 96 x
3 + 216 x
2 − 216 x
3 + 81
The answer is D.
8 Fourth term = 6C 3 x
3 × (3 y)
3
= 20 × x3 × 27 y3
= 540 x3 y3
9 Term 3 ⇒ r = 2
=
2
79 32 4
x⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
= 278 732
16
x
= 219 683
4
x
10 x6, x3, x0
3rd term is independent of x. r = 2
=
2
4
2
6 2(3 )
2 x
x
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
= 4860
11 Powers of x are ( x2)5, ( x2)4 31
x
⎛ ⎞⎜ ⎟⎝ ⎠ , ( x
2)3
2
3
1
x
⎛ ⎞⎜ ⎟⎝ ⎠
x10, x5, x0, …
The third term is independent of x.
term = 5C 2( x
2)3
2
3
4
x
⎛ ⎞−⎜ ⎟
⎝ ⎠
= 10 × (+16)
= 160
12
4
2
2
3 x
x
⎛ ⎞+⎜ ⎟
⎝ ⎠
Powers of x are ( x2)4, ( x2)3 21
x
⎛ ⎞⎜ ⎟⎝ ⎠ , ( x
2)2
2
2
1
x
⎛ ⎞⎜ ⎟⎝ ⎠
x8, x4, x0, …The third term is independent of x.
term =
4
C 2( x
2
)
2
2
2
3
x
⎛ ⎞⎜ ⎟⎝ ⎠
= 6 × 9
= 54
13 Expand ( p + 3)5
= p5 + 5 p43 + 10 p332 + 10 p233 + 5 p34 + 35
= p5 + 15 p4 + 90 p3 + 270 p2 + 405 p + 243
∴ (2 p − 5)( p5 + 15 p4 + 90 p3 + 270 p2 + 405 p + 243)
⇒ 2 p6 + 30 p5 + 180 p4 + 540 p3 + 810 p2 + 486 p − 5 p5 − 75 p4 − 450 p3 − 1350 p2 − 2025 p − 1215
Coefficient of p4 term = 180 − 75 = 105
14 (2a − 1)n
2nd term is nC 1(2a)n − 1(−1)1
coefficient: −n × 2n − 1 = −192
n × 2n × 12
= 192
n × 2n = 384
= 3 × 27
= 3 × 2 × 26
= 6 × 26
n = 6
Exercise 1B — Polynomials
1 Polynomial expressions consist of terms which have non-negative integer powers of x only.
Not Polynomial:
ii x4 + 3 x2 − 2 x + x
iii x7
+ 3 x6
− 2 xy + 5 x
vi 2 x5 + x
4 − x
3 + x
2 + 3 x − 2
x
Polynomial:
i x3 − 2 x
iv 3 x8 − 2 x5 + x2 − 7
v 4 x6 − x3 + 2 x − 3
2 a P ( x) + Q( x) = 8 − 3 x + 2 x2 + x4 + x5 − 3 x4 − 4 x2 − 1
= x5 − 2 x
4 − 2 x
2 − 3 x + 7
b Q( x) − R( x) = x5 − 3 x4 − 4 x2 − 1 − (8 x3 + 7 x2 − 4 x)
= x5 − 3 x
4 − 4 x
2 − 1 − 8 x
3 − 7 x
2 + 4 x
= x5 − 3 x4 − 8 x3 − 11 x2 + 4 x − 1
c 3 P ( x) − 2 R( x)
3 P ( x) = 3(8 − 3 x + 2 x2 + x4)
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CHAPTER 1 Graphs and polynomials • EXERCISE 1B | 3
Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
= 24 − 9 x + 6 x2 + 3 x4
2 R( x) = 2(8 x3 + 7 x2 − 4 x)
= 16 x3 + 14 x2 − 8 x
∴ 3 P ( x) − 2 R( x) = 24 − 9 x + 6 x2 + 3 x4 − (16 x3 + 14 x2 − 8 x)
= 24 − 9 x + 6 x2 + 3 x4 − 16 x3 − 14 x2 + 8 x
= 3 x4 − 16 x3 − 8 x2 − x + 24
d 2 P ( x) − Q( x) + 3 R( x)
2 P ( x) = 2(8 − 3 x + 2 x2 + x4)
= 16 − 6 x + 4 x2 + 2 x4
3 R( x) = 3(8 x3 + 7 x2 − 4 x)= 24 x3 + 21 x2 − 12 x
2 P ( x) − Q( x) + 3 R( x)
= 16 − 6 x + 4 x2 + 2 x
4− ( x5 − 3 x
4 − 4 x
2 − 1) + 24 x
3 + 21 x
2
− 12 x
= 16 − 6 x + 4 x2 + 2 x4 − x5 + 3 x4 + 4 x2 + 1 + 24 x3 + 21 x2 − 12 x
= 17 − 18 x + 29 x2 + 24 x
3 + 5 x
4 − x
5
3 a P ( x) = x6 + 2 x5 − x3 + x2
i degree = 6
ii P (0) = 06 + 2 × 05 − 03 + 02
= 0
iii P (2) = 26 + 2 × 25 − 23 + 22
= 124
iv P (−1) = −16 + 2 × −15 − (−1)3 + (−1)2
= 1
b P ( x) = 3 x7 − 2 x6 + x5 − 8
i degree = 7
ii P (0) = 3 × 07 − 2 × 06 + 05 − 8
= −8
iii P (2) = 3 × 27 − 2 × 2
6 + 2
5 − 8
= 280
iv P (−1) = 3 × (−1)7 − 2 × (−1)6 + (−1)5 − 8
= −3 − 2 − 1 − 8
= −14
c P ( x) = 5 x6 + 3 x4 − 2 x3 − 6 x2 + 3
i degree = 6
ii P (0) = 5 × 06 + 3 × 04 − 2 × 03 − 6 × 02 + 3
= 3iii P (2) = 5 × 26 + 3 × 24 − 2 × 23 − 6 × 22 + 3
= 331
iv P (−1) = 5 × (−1)6 + 3 × (−1)4 − 2 × (−1)3 − 6 × (−1)2 + 3
= 5 + 3 + 2 − 6 + 3
= 7
d P ( x) = −7 + 2 x − 5 x2 + 2 x3 − 3 x4
i degree = 4
ii P (0) = −7 + 2 × 0 − 5 × 02 + 2 × 03 − 3 × 04
= −7
iii P (2) = −7 + 2 × 2 − 5 × 22 + 2 × 23 − 3 × 24
= −55
iv P (−1) = −7 + 2 × (−1) − 5 × (−1)2 + 2 × (−1)3 − 3(−1)4
= − 7 − 2 − 5 − 2 − 3
= −19
4 P ( x) = x8 − 3 x6 + 2 x4 − x2 + 3
P (−2) = (−2)8 − 3 × (−2)6 + 2 × (−2)4 − (−2)2 + 3
= 95
The answer is B.
5 P ( x)= 2 x7 + ax
5 + 3 x
3 + bx − 5
P (1)= 4
∴ 4 = 2 × 17 + a × 15 + 3 × 13 + b × 1 − 5
4 = 2 + a + 3 + b − 5
4 = a + b [1]
P (2) = 163
163 = 2 × 27 + a × 2
5 + 3 × 2
3 + b × 2 − 5
= 256 + 32a + 24 + 2b − 5
−112 = 32a + 2b [2]
[1] × 2 8 = 2a + 2b [3]
[2] − [3] −120 = 30a
a = −4 b = 8
6 f ( x) = ax4 + bx3 − 3 x2 − 4 x + 7
f (1) = −2
∴ −2= a × (1)4 + b × (1)3 − 3 × 12 −4 × 1 + 7
−2= a + b − 3 − 4 + 7
−2= a + b
∴ −2 − b = a
[1] f (2) = −5
∴ −5= a × 24 + b × 2
3 − 3 × 2
2 − 4 × 2 + 7
−5= 16a + 8b − 12 − 8 + 7
−5= 16a + 8b − 13
8 = 16a + 8b
8 = 8(2a + b)
1 = 2a + b [2]Substitute [1] into [2]
1 = 2(−2 − b) + b
1 = −4 − 2b + b
1 = −4 − b
b = −5
If b = −5, then [1] −2 − −5 = a.
3 = a∴ f ( x) = 3 x4
− 5 x3
− 3 x2
− 4 x + 7
7 Q( x) = x5 + 2 x
4 + ax3 − 6 x + b
Q(2) = 45
∴ 45 = 25 + 2 × 24 + 23a − 6 × 2 + b
45 = 52 + 8a + b
−7 = 8a + b
−7 − 8a = b [1]
Q(0) = −7
∴ −7 = 05 + 2 × 04
+ a × 03
− 6 × 0 + b
−7 = b [2]Substitute [2] into [1].
−7 − 8a = −7
−8a = 0
a = 0∴ Q( x) = x5
+ 2 x4
− 6 x − 7.
8 P ( x) = ax6 + bx4
+ x3
− 6
If 3 P (1) = −24
then 3 P ( x) = 3(ax6 + bx4
+ x3
− 6)
−24 = 3(a × 16 + b × 14
+ 13
− 6)
−8 = a + b + 1 − 6
−8 = a + b − 5
−3 = a + b
−3 − a = b [1]
If 3 P (−2) = 102
then 3 P ( x) = 3(ax6 + bx4
+ x3
− 6)
102 = 3[a(−2)6 + b(−2)4
+ (−2)3
− 6)
34 = 64a + 16b − 8 − 6
34 = 64a
+ 16b
− 1448 = 64a + 16b
(÷ 16) 3 = 4a + b [2]Substitute [1] into [2]
3 = 4a + (−3 − a)
3 = 4a − 3 − a
6 = 3a
2 = a
If 2 = a then b = −3 − a
b = −3 − 2
b = −5
∴ P ( x) = 2 x6 − 5 x4
+ x3
− 6
9 a P ( x) = ax4 − x3
+ 3 x2
− 5
If P (1) = −1
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4 | CHAPTER 1 Graphs and polynomials • EXERCISE 1C
Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
then −1 = a × (1)4 − (1)3
+ 3 × (1)2 − 5
−1 = a − 1 + 3 − 5
−1 = a − 3
2 = a
The answer is C.
b f ( x) = xn
− 2 x3 + x2
− 5 x
If f (2) = 10
then 10 = 2n
− 2 × 23 + 22
− 5 × 2
10 = 2n
− 16 + 4 − 10
10 = 2n − 22
32 = 2n
25 = 2n
∴ n = 5
The answer is D.
Exercise 1C — Division of polynomials
1 a2
3 2
3 2
2
2
2 13
2 5 24
4
2 5
2 8
13 2
13 52
50
x x
x x x x
x x
x x
x x
x
x
+ +
⎡ − + −− − ⎢
−⎢⎣
⎡ +− ⎢
−⎢⎣
−⎡− ⎢
−⎣
Q( x) = x2 + 2 x + 13
R( x) = 50
b
4 3 2
5 4 3 2
5 4
4 3
4 3
3 2
3 2
3 6 18 58
0 3 0 4 33
3
3 3
3 9
6 0
6 18
x x x x
x x x x x x
x x
x x
x x
x x
x x
− + − +
⎡ + − + + ++ − ⎢
+⎢⎣
⎡− −− ⎢− −⎢⎣
⎡ +− ⎢
+⎢⎣
2
2
18 4
18 54
58 3
58 174
171
x x
x x
x
x
⎡− +− ⎢
− −⎢⎣
+⎡− ⎢
+⎣
−
Q( x) = x4 − 3 x3 + 6 x2 − 18 x + 58
R( x) = −171
c
3 2
4 3 2
4 3
6 17 53 155
6 2 4 03
6 18
x x x
x x x x x
x x
+ + +⎡ − + − +
− − ⎢−⎢⎣
3 2
3 2
2
17 2
17 51
53 4
53 159
155 0
155 465
465
x x
x x
x x
x x
x
x
⎡ +− ⎢
−⎢⎣
⎡ −− ⎢
−⎣
+⎡− ⎢
−⎣
Q( x) = 6 x3 + 17 x2 + 53 x + 155
R( x) = 465
d
3 2
4 3 2
4 3
3 2
3 2
7 7 101
3 9 27
3 6 0 12 03 1
3
7 0 7
73
x x x
x x x x x
x x
x x
x x
− + +
⎡ − + + ++ − ⎢
+⎢⎣
⎡− +⎢−⎢− −⎢⎣
2
2
712
3
7 7
3 9
1010
9
101 101
9 27
101
27
x x
x x
x
x
⎡+⎢
⎢−⎢
+⎢⎣
⎡+⎢
− ⎢⎢ +⎢⎣
−
Q( x) = x3 − 7
3 x2 +
7
9 x +
101
27
R( x) = 20
327
−101
27
−⎛ ⎞=⎜ ⎟
⎝ ⎠
2 a i P ( x)= x3 − 2 x
2 + 5 x − 2
P (4) = 43 − 2 × 42 + 5 × 4 − 2
= 50
ii P ( x)= x5 − 3 x3 + 4 x + 3
P (−3) = (−3)5 − 3 × (−3)3 + 4 × (−3) + 3
= −171iii P ( x) = 6 x4 − x3 + 2 x2 − 4 x
P (3) = 6 × 34 − 33 + 2 × 32 − 4 × 3
= 465
iv P ( x) = 3 x4 − 6 x3 + 12 x
1
3 P
⎛ ⎞−⎜ ⎟
⎝ ⎠ = 3 ×
41
3
⎛ ⎞−⎜ ⎟
⎝ ⎠− 6 ×
31
3
⎛ ⎞−⎜ ⎟
⎝ ⎠+ 12 × −
1
3
= −320
27
b The values obtained in 2 were the same as the remainder
values obtained in 1.
3 a P (3) = 33 + 9 × 32 + 26 × 3 − 30
= 156
Since P (3) ≠ 0, x − 3 is not a factor
b P (−2) = (−2)4 − (−2)3 − 5 × (−2)2 − 2 × (−2) − 8
= 0
Since P (−2) = 0 then x + 2 is a factor.
c P (+ 4) = 4 − 9 × + 4 + 6 × 42 − 13 × (+ 4)
3
− 12 × (+ 4)4 + 3 × (+ 4)5
= 4 − 36 + 94 − 832 − 3072 + 3072
= −768
Since P (−4) ≠ 0 then 4 − x is not a factor.
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CHAPTER 1 Graphs and polynomials • EXERCISE 1C | 5
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d P 1
2
⎛ ⎞−⎜ ⎟
⎝ ⎠ = 4 ×
61
2
⎛ ⎞−⎜ ⎟
⎝ ⎠+ 2 ×
51
2
⎛ ⎞−⎜ ⎟
⎝ ⎠− 8 ×
41
2
⎛ ⎞−⎜ ⎟
⎝ ⎠
− 4 ×
31
2
⎛ ⎞−⎜ ⎟
⎝ ⎠+ 6 ×
21
2
⎛ ⎞−⎜ ⎟
⎝ ⎠− 9 × −
1
2 − 6
= 0.0625 + − 0.0625 − 0.5 + 0.5 + 1.5 + 4.5 − 6
= 0
Since P 1
2
⎛ ⎞−⎜ ⎟
⎝ ⎠ = 0 then 2 x + 1 is a factor.
4 a f ( x) = x4 − 4 x3 − x2 + 16 x − 12
A x + 1 ⇒ f (− 1)= (−1)4 − 4 × (−1)3 − (−1)2 + 16 × (−1) − 12
= 1 + 4 − 1 − 16 − 12
= −24
B x ⇒ f (0)= −12
C x + 2 ⇒ f (− 2)= (−2)4 − 4 × (−2)3 − (−2)2 + 16 × (−2) − 12
= 16 + 32 − 4 − 32 − 12
= 0
Since f (−2) = 0 then ( x + 2) is a factor.
D x + 3 ⇒ f (−3)= (−3)4 − 4 × (−3)3 − (−3)2 + 16 × (−3) − 12
= 120E x − 4 ⇒ f (4)
= 44 − 4 × 43 − 42 + 16 × 4 − 12
= 36The answer is C.
b
3 2
4 3 2
4 3
3 2
3 2
2
2
6 11 6
4 16 122
2
6
6 12
11 16 11 22
6 12
6 12
0
x x x
x x x x x
x x
x x
x x
x x x x
x
x
− + −
⎡ − − + −+ − ⎢
+⎢⎣
⎡− −− ⎢
− −⎢⎣
⎡+− ⎢+⎢⎣
− −⎡− ⎢
− −⎣
Test x = 1 into x3 − 6 x2 + 11 x − 6
= 13 − 6 × 1 + 11 − 6
= 0
∴ x − 1 is a factor.
2
3 2
3 2
2
2
5 6
6 11 61
5 11
5 5
6 6
6 6
0
x x
x x x x
x x
x x
x x
x
x
− +
⎡ − + −− − ⎢
−⎢⎣
⎡− +− ⎢
− +⎢⎣
−⎡− ⎢
−⎣
x2 − 5 x + 6 = ( x − 3)( x − 2) f ( x) factorises to
( x + 2)( x − 1)( x − 3)( x − 2)
The answer is B.
5 a P ( x) = x3 + 4 x
2 − 3 x − 18
Test x = ± 1 P ( x) ≠ 0
x = 2, P ( x) = 0
∴ ( x − 2) is a factor
2
3 2
3 2
2
2
6 9
4 3 182
2
6 3
6 12
9 18
9 18
0
x x
x x x x
x x
x x
x x
x
x
+ +
⎡ + − −− − ⎢
−⎢⎣
⎡ −− ⎢
−⎢⎣
−⎡
− ⎢ −⎣
∴ ( x − 2)( x2 + 6 x + 9)
= ( x − 2)( x + 3)2
b P ( x) = 3 x3 − 13 x2 − 32 x + 12
Test x = ± 1 P ( x) ≠ 0
x = ± 2
when x = −2, P ( x) = 0
∴ ( x + 2) is a factor.
2
3 2
3 2
2
2
3 19 6
3 13 32 122
3 6
19 32
19 38
6 12
6 12
0
x x
x x x x
x x
x x
x x
x
x
+ − +
⎡ − − ++ − ⎢
+⎢⎣⎡− −
− ⎢− −⎢⎣
+⎡− ⎢
+⎣
∴ ( x + 2)(3 x2 − 19 x + 6)
= ( x + 2)(3 x − 1)( x − 6)
c P ( x) = x4 + 2 x
3 − 7 x2 − 8 x + 12
Test x = −2, P ( x) = 0
∴ ( x + 2) is a factor.
3 2
4 3 2
4 3
2
2
0 7 6
2 7 8 122
2
0 7 8
7 14
6 12
6 12
0
x x x
x x x x x
x x
x x
x x
x
x
+ − +
⎡ + − − ++ − ⎢
+⎢⎣
⎡ − −− ⎢
− −⎢⎣
+⎡− ⎢
+⎣
∴ ( x + 2)( x3 + 0 x
2 − 7 x + 6)
Test x = 2, P ( x) = 0
∴ ( x − 2) is a factor.2
3 2
3 2
2
2
2 3
0 7 62
2
2 7
2 4
3 6
3 6
0
x x
x x x x
x x
x x
x x
x
x
+ −
⎡ + − +− − ⎢
−⎢⎣
⎡ −−⎢
−⎢⎣
− +⎡−⎢
− +⎣
∴ ( x + 2)( x − 2)( x2 + 2 x − 3)
( x + 2)( x − 2)( x + 3)( x − 1)
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6 | CHAPTER 1 Graphs and polynomials • EXERCISE 1C
Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
d P ( x) = 4 x4 + 12 x3 − 24 x2 − 32 x
Test x = −1, P ( x) = 0
∴ x + 1 is a factor3 2
4 3 2
4 3
3 2
3 2
2
2
4 8 32
4 12 24 32 01
4 4
8 24
8 8
32 32
32 32
0
x x x
x x x x x
x x
x x
x x
x x
x x
+ −
⎡ + − − ++ − ⎢
+⎢⎣
⎡ −− ⎢
+⎢⎣⎡− −
− ⎢− −⎢⎣
∴ ( x + 1)(4 x3 + 8 x2 − 32 x)
Take out factor of 4 x.
4 x( x + 1)( x2 + 2 x − 8)
∴ 4 x( x + 1)( x − 2)( x + 4)
6 a 3 x3 + 3 x
2 − 18 x = 0
3 x is a common factor
∴ 3 x( x2 + x − 6) = 0
3 x( x + 3)( x − 2) = 0
x = 0, −3, 2b 2 x
4 + 10 x3 − 4 x
2 − 48 x = 0
Take out 2 x as a common factor
∴ 2 x( x3 + 5 x2 − 2 x − 24) = 0
Factorise x3 + 5 x2 − 2 x − 24
Test x = 2, f (2) = 0
∴ ( x – 2) is a factor2
3 2
3 2
2
2
7 12
5 2 242
2
7 2
7 14
12 24
12 24
0
x x
x x x x
x x
x x
x x
x
x
+ +
⎡ + − −− − ⎢
−⎢⎣
⎡ −− ⎢
−⎢⎣
−⎡− ⎢
−⎣
∴ 2 x( x − 2)( x2 + 7 x + 12)
= 2 x( x − 2)( x + 3)( x + 4) = 0
∴ x = 2, −3, 0, and − 4
c 2 x4 + x
3 − 14 x2 − 4 x + 24 = 0
Test x = 2, f ( x) = 0
∴ ( x − 2) is a factor
3 2
4 3 2
4 3
3 2
3 2
2
2
2 5 4 12
2 14 4 242
2 45 14
5 10
4 4
4 8
12 24
12 24
0
x x x
x x x x x
x x x x
x x
x x
x x
x
x
+ − −
⎡ + − − +− − ⎢
−⎢⎣⎡ −
− ⎢−⎢⎣
⎡− −− ⎢
− +⎢⎣
− +⎡− ⎢
− +⎣
∴ ( x − 2)(2 x3 + 5 x2 − 4 x − 12)
Test x = −2, f (−2) = 0
∴ ( x + 2) is a factor
2
3 2
3 2
2 6
2 5 4 122
2 4
x x
x x x x
x x
+ −
⎡ + − −+ − ⎢
+⎢⎣
2
2
4
2
6 12
6 12
0
x x
x x
x
x
⎡ −− ⎢
−⎢⎣
− −⎡− ⎢
− −⎣
∴ ( x − 2)( x + 2)(2 x2 + x − 6)
( x − 2)( x + 2)(2 x − 3)( x + 2)
x = 2, −2, or3
2
d x4 − 2 x
2 + 1 = 0
Test x = + 1, f ( x) = 0
∴ ( x − 1) is a factor3 2
4 3 2
4 3
3 2
3 2
2
2
1
0 2 0 11
2
0
1
1
0
x x x
x x x x x
x x
x x x x
x x
x x
x
x
+ − −
⎡ + − + +− − ⎢
−⎢⎣
⎡−− ⎢−⎢⎣
⎡− +− ⎢
− +⎢⎣
− +⎡− ⎢
− +⎣
∴ ( x − 1)( x3 + x2 − x − 1)
Test x = −1, f ( x) = 0
∴ ( x + 1) is a factor2
3 2
3 2
1
11
0
1
1
0
x
x x x x x x
x
x
−
⎡ + − −+ − ⎢+⎢⎣
− −⎡− ⎢
− −⎣
∴ ( x − 1)( x + 1)( x2 − 1) = 0
( x − 1)( x + 1)( x − 1)( x + 1) = 0
x = ±1AlternativelyLet u = x2
u2 – 2u + 1 = 0
(u – 1)(u – 1) = 0u = 1 ∴ x2 = 1
x = ±1
7 If ( x − 2) is a factor then
when x = 2, f ( x) = 0
0 = x3 + ax
2 − 6 x − 4
f (2) = 0 = 23 + a2
2 − 6 × 2 − 4
0 = 8 + 4a − 12 − 4
0 = 4a − 8
8 = 4a
2 = a
8 Let
P ( x) = x3 + x2
− ax + 3
P (1) = 1 + 1 − a + 3 = 0
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CHAPTER 1 Graphs and polynomials • EXERCISE 1D | 7
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( x − 1) is a factor
a = 5
9 If ( x + 3) is a factor then
when x = −3, f ( x) = 0
f (−3) = 0 = 2(−3)4 + a(−3)3 − 3 × (−3)
+ 18
0 = 162 − 27a + 9 + 18
0 = 189 − 27a
27a = 189
a = 710 If ( x + 1) is a factor then
when x = −1, f ( x) = 0
f (−1)= 0 = −a − 4 − b − 12
0 = −a − b − 16
a = −b − 16 [1]
If ( x − 2) is a factor then
when x = 2, f ( x) = 0
f (2) = 0 = 8a − 16 + 2b − 12
0 = 8a + 2b − 28
28 = 8a + 2b
14 = 4a + b [2]Sub [1] into [2]
14 = 4(−b − 16) + b
14 = −4b − 64 + b14 = −3b − 64
78 = −3b
−26 = b
∴ a = + 26 − 16
a = 10
11 (2 x − 3) and ( x + 2) are factors of
2 x3 + ax2
+ bx + 30
P ( x) = 2 x3 + ax2
+ bx + 30
P (−2) = 2(−2)3 + a(−2)2
+ b(−2)
+ 30 = 0
−16 + 4a − 2b + 30 = 0
4a − 2b = −14 [1]
P
3
2
⎛ ⎞
⎜ ⎟⎝ ⎠ = 2
33
2
⎛ ⎞
⎜ ⎟⎝ ⎠ + a
23
2
⎛ ⎞
⎜ ⎟⎝ ⎠ + b3
2
⎛ ⎞
⎜ ⎟⎝ ⎠
+ 30 = 0
2 × 27
8 +
9
4
a +
3
2
b + 30 = 0
27 + 9a + 6b + 120 = 0
9a + 6b = −147
3a + 2b = −49 [2]
[1] + [2] 7a = −63
a = −9
Substitute into [1] 4 × −9 − 2b = −14
−2b = 22
b = −11
a = −9, b = −11.
Exercise 1D — Linear graphs
1 a 2 x + 3 y = 12
x-intercept when y = 0
2 x = 12
x = 6
y-intercept when x = 0
3 y = 12
y = 4
b 2 y − 5 x − 10 = 0
x-intercept when y = 0
−5 x = 10
x = −2
y-intercept when x = 0
2 y = 10
y = 5
c 2 x − y = 1
x-intercept when y = 0
2 x = 1
x = 1
2
y-intercept when x = 0
− y = 1
y = −1
2 a y = mx + c
y = 3 x + c
find c using the point (2, 1)
1 = 3 × 2 + c
−5 = c
∴ y = 3 x − 5
−3 x + y + 5 = 0
b y = mx + c
y = −2 x + c
find c, sub in (−4, 3)3 = −2 × −4 + c
3 = 8 + c
−5 = c
∴ y = −2 x − 5
2 x + y + 5 = 0
3 a (−3, −4), (−1, −10)
m = 10 4
1 3
− +
− +
= 6
2
−
= −3
y = −3 x + c
sub in (−3, −4)
−4 = −3 × −3 + c
−13 = c
y = −3 x − 13
3 x + y + 13 = 0
b (7, 5), (2, 0)
m = 5 0
7 2
−
−
= 5
5
= 1
y = x + csub in (2, 0)
0 = 2 + c
−2 = c
y = x − 2
− x + y + 2 = 0
4 2 y − 3 x − 6 = 0
A 2 × 6 − 3 × 2 − 6 = 0
12 − 6 − 6 = 0
B 2 × 0 − 3 × − 2 − 6 = 0
0 + 6 − 6 = 0
C 2 × 3 − 3 × 0 − 6 = 0
6 − 0 − 6 = 0
D 2 × 2 − 3 × 1 − 6 = 0
4 − 3 − 6 ≠ 0
E 2 × 9 − 3 × 4 − 6 = 0
18 − 12 − 6 = 0
The answer is D.
5 a i −2 = 5
1 2
b −
+
−2 = 5
3
b −
−6 = b − 5
−1 = b
ii y
− x
= 7
∴ y = x + 7
m = 1
1 = 5
1 2
b −
+
3 = b − 5
8 = b
b parallel to y = 3 x − 4
∴ m = 3
y = 3 x + c
sub in (4, 5)
5 = 3 × 4 + c
−7 = c
∴ y
= 3 x
− 7
0 = 3 x − y − 7
c 2 y − x + 1 = 0
2 y = x − 1
y = 1
2 x −
1
2
m = −2 gradient of perpendicular line
y − y1 = m( x − x1)
Sub in (−2, 4) y − 4 = −2( x + 2)
y − 4 = −2 x − 4
2 x + y = 0
6 i x + 2 y + 4 = 0
• x-intercept when y = 0
x = − 4• y-intercept when x = 0
2 y = − 4
y = − 2
Graph e
ii y = 3 − Graph f
iii y − 2 x − 2 = 0
• x-intercept when y = 0
−2 = 2 x
−1 = x
• y-intercept when x = 0
y = 2Graph a
iv 3 y + 2 x = 6
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Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
• x-intercept when y = 0
2 x = 6
x = 3
• y-intercept when x = 0
3 y = 6
y = 2
Graph c
v y − 2 x = 0
• x- and y-intercepts occur at the
origin.Graph b
vi x = −2 − Graph d.
7 a y ≥ −2 or [−2, ∞)
b y > −5 or (−5, ∞)
c −2 ≤ y < 3 or [−2, 3)
d −2 ≤ y ≤ 3 or [−2, 3]
e R
f −∞
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CHAPTER 1 Graphs and polynomials • EXERCISE 1E | 9
Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
c f ( x) = 10 + 3 x − x2
y-intercept x = 0
y = 10
x-intercept y = 00 = (5 − x)(2 + x)
x = 5 or −2
d f ( x) = 6 x2 − x − 12
y-intercept x = 0
y = −12
x-intercept(s) y = 0
0 = (3 x + 4) (2 x − 3)
x = −4
3 or
3
2
3 a f ( x) = x2 − 6 x + 8
= x2 − 6 x + 32 − 32 + 8
= ( x − 3)2 − 9 + 8
= ( x − 3)2 − 1TP is (3, −1)
b f ( x) = x2 − 5 x + 4
= x2 − 5 x +
2 25 5
2 2
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ + 4
=
25
2 x
⎛ ⎞−⎜ ⎟
⎝ ⎠−
25
4 + 4
=
25
2 x
⎛ ⎞−⎜ ⎟
⎝ ⎠−
25
4 +
16
4
=
25
2 x
⎛ ⎞−⎜ ⎟
⎝ ⎠−
9
4
TP is5 9
,2 4
⎛ ⎞−⎜ ⎟
⎝ ⎠
c f ( x)= 10 + 3 x − x2
= −( x2 − 3 x − 10)
= −
2 22 3 33 10
2 2 x x
⎡ ⎤⎛ ⎞ ⎛ ⎞− + − −⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
= −
23 9
102 4
x⎡ ⎤⎛ ⎞
− − −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
= −
23 9 40
2 4 4 x
⎡ ⎤⎛ ⎞− − −⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
= −
23
2 x
⎛ ⎞−⎜ ⎟
⎝ ⎠ +
49
4
TP is3 49
,2 4
⎛ ⎞+⎜ ⎟
⎝ ⎠ =
1 11 , 12
2 4
⎛ ⎞⎜ ⎟⎝ ⎠
d f ( x)= 6 x2 − x − 12
= 2
6 26
x x
⎛ ⎞− −⎜ ⎟
⎝ ⎠
=
2 22 1 16 2
6 12 12 x x⎡ ⎤⎛ ⎞ ⎛ ⎞− + − −⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
=
21 1 288
612 144 144
x⎡ ⎤⎛ ⎞
− − −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
=
21 289
612 144
x⎡ ⎤⎛ ⎞
− −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
=
21
612
x⎛ ⎞
−⎜ ⎟⎝ ⎠
−289
24
TP is1 1
, 1212 24
⎛ ⎞⎜ ⎟⎝ ⎠
4 a i y = 2 − x2
a = −1, h = 0, k = 2
TP = (0, 2)
ii Domain = R
iii Range = (−∞, 2]
b i y = ( x − 6)2
a = 1, h = 6, k = 0
TP = (6, 0)
ii Domain = R
iii Range = [0, ∞)
c i y = −( x + 2)2
a = −1, h = −2, k = 0
TP = (−2, 0)
ii Domain = R
iii Range = (−∞, 0]
d i y = 2( x + 3)2 − 6
a = 2, h = −3, k = −6
TP = (−3, −6)
ii Domain = R
iii Range = [−6, ∞)
5 Using y = A ( x + B)2 + C
a i TP = (1, − 2)
∴ B = −1 and C = −2
Assume A = 1
⇒ y = 1( x − 1)2 − 2 y = x2 − 2 x + 1 − 2
y = x2 − 2 x − 1
ii Domain = R
iii Range [−2, ∞)
b i TP = (2, −3)
∴ B = −2
C = −3
Assume A = 1
⇒ y = 1 ( x − 2)2 − 3
= x2 − 4 x + 4 − 3
= x2 − 4 x + 1
ii Domain = [−1, ∞)
iii Range = [−3, ∞)
c i TP = (1, 9)
∴ B = −1 and C = 9
Assume A = −1
⇒ y = −1( x − 1)2 + 9 y = −1( x2 − 2 x + 1) + 9
y = − x2 + 2 x − 1 + 9
y = − x2 + 2 x + 8
ii Domain = [−4, 4)
iii Range = [−16, 9]
6 a y = 2 x2 + 3
TP = (0, 3)
y-intercept when x = 0
y = 3 x-intercepts when y = 0
0 = 2 x2 + 3
There are no x-intercepts.
b y = (2 x − 5) (2 x − 3)
= 4 x2 − 16 x + 15
= 4 215
44
x x⎛ ⎞
− +⎜ ⎟⎝ ⎠
= 4 215
4 4 44
x x⎛ ⎞
− + − +
⎜ ⎟⎝ ⎠
= 4 21
( 2)4
x⎛ ⎞
− −⎜ ⎟⎝ ⎠
= 4( x − 2)2 − 1
TP = (2, −1)
y-int = (0, 15)
x-int = (1, 5, 0), (2, 5, 0)
c y = (2 x − 3)2 − 8
TP = 3
, 82
⎛ ⎞−⎜ ⎟
⎝ ⎠
y-intercept when x = 0
y = (−3)2 − 8
= 1
x-intercepts when y = 0
0 = (2 x − 3)2 − 8
= 4 x2 − 12 x + 9 − 8
= 4 x2 − 12 x + 1From the graphics calculator,
x = 2.91 and x = 0.09
7 a y = x2 − 2 x − 3
x-intercepts y = 0
0 = ( x − 3)( x + 1)
x = 3 or −1
(3, 0)(−1, 0)The answer is B.
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10 | CHAPTER 1 Graphs and polynomials • EXERCISE 1E
Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
b y = x2 − 2 x − 3
y = x2 − 2 x + 12 − 12 − 3
y = ( x − 1)2 − 4
TP = (1, −4)The answer is C.
8 f ( x) = −( x + 3)2 + 4
TP = (−3, 4)
∴ range ( ,4]−∞
The answer is D.
9 y = ( x − 4)2
x ∈ [0, 6]TP = (4, 0)
When x = 0 y = (−4)2 = 16
∴ range [0, 16]
When x = 6 y = (6 − 4)2 = 22 = 4
But x = 4 y = 0The answer is A.
10 a f ( x) = ( x − 2)2
− 4TP = (2, −4)
y-int x = 0
y = (0 − 2)2 − 4
y = (−2)2 − 4
y = 0
x-int y = 0
0 = ( x − 2 − 2)( x − 2 + 2)
0 = ( x − 4)( x)
∴ x = 4 or 0
b f ( x) = −( x + 4)2 + 9
TP = (−4, 9)
y-int x = 0
y = −(0 + 4)2 + 9
y = −16 + 9
y = −7
x-int y = 0
0 = 9 − ( x + 4)2
0 = (3 − ( x + 4))(3 + ( x + 4))
0 = (3 − x − 4) (3 + x + 4)
0 = (− x − 1)(7 + x)
x = −1 or −7
c y = x2 + 4 x + 3
y = x2 + 4 x + 4 − 4 + 3
y = ( x + 2)2 − 1
TP = (−2, −1)
y-int x = 0
y = 3
x-int y = 0
0 = ( x + 2 − 1)( x + 2 + 1)
0 = ( x + 1)( x + 3)
∴ x = −1 or −3
d y = 2 x2 − 4 x − 6
y = 2[ x2 − 2 x − 3]
= 2[ x2 − 2 x + 1 − 1 − 3]
= 2[( x − 1)2 − 4]
y = 2( x − 1)2 − 8
TP = (1, −8)
y-int x = 0
y = −6
x-int y = 0
0 = 2[( x − 1 − 2)( x − 1 + 2)]
0 = 2( x − 3)( x + 1)
∴ x = 3 or −1
11 a y = x2 − 2 x + 2 x ∈ [−2, 2]
y = x2 − 2 x + 1 − 1 + 2
y = ( x − 1)2 + 1
∴ TP = (1, 1)
i Domain = [−2, 2]ii Range:
When x = −2
y = 10When x = 2 y = 2
but TP = (1, 1)
∴ [1, 10]
b y = − x2 + x − 1 x ∈ R+
y = −( x2 − x + 1)
y = −
2 2
2 1 1 12 2
x x⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ − + − + ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
y = −
21 3
2 4 x
⎡ ⎤⎛ ⎞− +⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
y = −
21
2
x⎛ ⎞
−⎜ ⎟
⎝ ⎠
−3
4
∴ TP = 1 3
,2 4
⎛ ⎞−⎜ ⎟
⎝ ⎠
i Domain = R+
ii Range = (−∞, −3
4]
c f ( x)= x2 − 3 x − 2 x ∈ [−10, 6]
f ( x)= x − 3 x +
23
2
⎛ ⎞⎜ ⎟⎝ ⎠
−
23
2
⎛ ⎞⎜ ⎟⎝ ⎠
− 2
=
23
2 x
⎛ ⎞−⎜ ⎟
⎝ ⎠ −
9
4 −
8
4
=
23
2 x
⎛ ⎞−⎜ ⎟
⎝ ⎠−
17
4
∴ TP = 3 17
,
2 4
⎛ ⎞−⎜ ⎟
⎝ ⎠
i Domain = [−10, 6]ii Range:
When x = −10 y = 128
∴ 17 ,1284
⎡ ⎤−⎢ ⎥⎣ ⎦
d f ( x) = −3 x2 + 6 x + 5 x ∈ [−5, 3)
=
2 5
3 2 3 x x
⎡ ⎤
− − −⎢ ⎥⎣ ⎦
= 25
3 2 1 13
x x⎡ ⎤
− − + − −⎢ ⎥⎣ ⎦
= 28
3 ( 1)3
x⎡ ⎤
− − −⎢ ⎥⎣ ⎦
= −3( x − 1)2 + 8
TP = (1, 8)
i Domain = [– 5, 3)ii Range:
When x = −5, y = −100
∴ [– 100, 8]
12 V (t ) = 2t 2 − 16t + 40 t ∈ [0, 10]
V (t ) = 2(t 2 − 8t + 20)
= 2[t 2 − 8t + 16 − 16 + 20]
= 2[(t − 4)2 + 4]
= 2(t − 4)2 + 8
∴ TP = (4, 8)
When t = 0 V (t ) = 40
When t = 10
V (t ) = 2 × 62 + 8
= 80
a minimum V = 8 m3
b maximum V = 80 m3
13 h(t ) = −3t 2 + 12t + 36
h(t ) = −3[t 2 − 4t − 12]
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CHAPTER 1 Graphs and polynomials • EXERCISE 1F | 11
Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
= −3[t 2 − 4t + 4 − 4 − 12]
= −3[(t − 2)2 − 16]
= −3(t − 2)2 + 48
∴ TP = (2, 48)
a maximum height = 48 m
b When h(t ) = 0
0 = −3[(t − 2 − 4)(t − 2 + 4)]
0 = −3(t − 6)(t + 2)
∴ t = 6 or −2
Since time 0≥ ⇒ 6 secondsc Domain [0, 6]Range [0, 48]
14 a h(t ) = t 2 − 12t + 48, t ∈ [0, 11]The lowest point is the
y-coordinate of the turning pointh(t ) = t 2 − 12t + 36 − 36 + 48
= (t − 6)2 + 12
TP = (6, 12)Lowest point is 12 m above theground.
b Time taken is the x-coordinate ofthe turning point.
t = 6 secondsc Check the end points of the domain.
h(0) = 48
h(11) = 112 − 12 × 11 + 48
= 37The highest point above the ground
is 48 m.d Domain = [0, 11]
Range = [12, 48]e
Exercise 1F — Cubic graphs1 a Positive cubic so a = 1. Goes
through origin so x is a factor. y = x( x − a)( x − b)
= x( x + 6)( x − 5)b Positive cubic in form
y = a( x − m)( x − n)2
∴ a = 1, m = 1, n = −2
∴ y = 1( x − 1)( x + 2)2
2 a Positive cubic in form
y = ( x − l )( x − m)( x − n)
l = −3, m = 1, n = 4
∴ y = ( x + 3)( x − 1)( x − 4)
∴ (v)
b Negative cubic in form
y = a( x − m)( x − n)2
∴ a = −1, m = 5, n = − 2
∴ y = −1( x − 5)( x + 2)2
y = (5 − x)( x + 2)2 (iv)
c Negative cubic in form
y = a( x − l )( x − m)( x − n)
a = −1, l = −3, m = 1, n = 4
∴ y = −1( x + 3)( x − 1)( x − 4)
y = ( x + 3)(1 − x)( x − 4)
(ii)d Positive cubic in form
y = a( x − t )3
a = 1, t = 3
∴ y = ( x − 3)3 (i)
e Positive cubic in form
y = a( x − l )( x − m)( x − n)
a = 1, l = −4, m = −2, n = 1
∴ y = ( x + 4)( x + 2)( x − 1)(vi)
f Positive cubic in form
y = a( x − m)( x − n)2
a = 1, m = 5, n = −2
y = ( x − 5)( x + 2)2 (viii)
g Negative cubic in form
y = a( x − t )3
a = −1, t = 3
∴ y = −1( x − 3)3
y = (3 − x)3 (vii)
h Negative cubic in form
y = a( x − l )( x − m)( x − n)
a = −1, l = − 4, m = −2, n = 1
∴ y = −1( x + 4)( x + 2)( x − 1)
y = ( x + 4)( x + 2)(1 − x)(iii)
3 a y = x3 + x2 − 4 x − 4
y-intercept
x = 0 y = −4Factorise to find x-intercepts
Test x = −1, y = 0
∴ x + 1 is a factor2
3 2
3 2
4
4 41
4 4
4 4
0
x
x x x x
x x
x
x
−
⎡ + − −+ − ⎢
+⎢⎣
− −⎡− ⎢
− −⎣
∴ y = ( x + 1)( x2 − 4)
y = ( x + 1)( x − 2)( x + 2)
If y = 0, x = −1, 2, or −2
b y = 2 x3 − 8 x2 + 2 x + 12
y-int x = 0
y = 12Factorise to find x-intercepts
Test x = −1 so y = 0
∴ ( x + 1) is a factor
2
3 2
3 2
2
2
2 10 12
2 8 2 121
2 2
10 2
10 10
12 12
12 12
0
x x
x x x x
x x
x x
x x
x
x
− +
⎡ − + ++ − ⎢
+⎢⎣
⎡− +− ⎢
− −⎢⎣
+⎡− ⎢
+⎣
∴ y = ( x + 1)(2 x2 − 10 x + 12)
y = 2( x + 1)( x − 2)( x − 3)
If y = 0, then x = −1, 2 or 3
c y = −2 x3 + 26 x + 24
y-int x = 0
y = 24
Factorise to find x-interceptsTest x = −1 so y = 0
∴ ( x + 1) is a factor.2
3 2
3 2
2
2
2 2 24
2 0 26 241
2 2
2 26
2 2
24 24
24 24
0
x x
x x x x
x x
x x
x x
x
x
− + +
⎡− + + ++ − ⎢
− −⎢⎣
⎡ +− ⎢
+⎢⎣
+⎡− ⎢
+⎣
∴ y = ( x + 1)(−2 x2 + 2 x + 24)
y = 2( x + 1)(− x − 3)( x − 4)
If y = 0, then x = −1, −3 or 4.
d y = − x3 + 8 x2 − 21 x + 18
y-int x = 0
y = 18
x-intercept
Factorise:Test x = 3 so y = 0
∴ ( x − 3) is a factor.2
3 2
3 2
2
2
5 6
8 21 183
3
5 21
5 15
6 18
6 18
0
x x
x x x x
x x
x x
x x
x
x
− + −
⎡− + − +− − ⎢
− +⎢⎣
⎡ −− ⎢
−⎢⎣
− +⎡− ⎢
− +⎣
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12 | CHAPTER 1 Graphs and polynomials • EXERCISE 1F
Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
∴ y = ( x − 3)(− x2 + 5 x − 6)
= −( x − 3)( x2 − 5 x + 6)
y = −1( x − 3)( x − 3)( x − 2)
∴ x = 3 or 2
4 a x3 + 6 x2 + 12 x + 8 = y
Test x = −2 so y = 0
∴ ( x + 2) is a factor.2
3 2
3 2
2
4 4
6 12 82
2
4 12
4 8
4 8
4 8
0
x x
x x x x
x x
x x
x x
x
x
+ +
⎡ + + ++ − ⎢
+⎢⎣
⎡ +− ⎢
+⎣
+⎡− ⎢
+⎣
∴ y = ( x + 2)( x2 + 4 x + 4)
= ( x + 2)( x + 2)2
y = ( x + 2)3
The answer is B.
b In form y = a( x − t )3
a = 1, t is intercept
The answer is E.
5 Function graph is a negative cubic so
a = −1
Point of inflection (2, 2)
The answer is C.
y = (2 − x)3 + 2
6 f ( x) = 5( x + 1)3
− 3Point of inflection (−1, −3)
Graph is a positive cubic.
Τhe answer is A.
7 Positive cubic
Turning point at (1, 0) because ofrepeated factor
x-intercept at (−3, 0)
y-intercept at (0, 6)
The answer is B.
8 Negative cubic B or D
Point of inflection is (a, b)
a < 0 so,
y = −( x + a)3 + b
The answer is D.9 y = −h( x − a)2 ( x − c)
x = 0, y = −ha2(−c)
b = ha2c
h = 2
b
a c
y = 2
b
a c− ( x − a)2( x − c)
The answer is E.
10 a f ( x) = x3 + x
2 − 10 x + 8 x ∈ [2, ∞)
a > 1 positive
b > 1 so 2 turning points.
y-intercept x = 0
y = 8
When x = 2
y = 23 + 22 − 20 + 8
= 0
Closed end point = (2, 0)
i Domain [2, ∞)
ii range [0, ∞)
b f ( x) = 3 x3 − 5 x2 − 4 x + 4 for
x ∈ [−2, −1]
a > 1 so positive
b ≠ 0 ∴ 2 turning points.
y-intercept x = 0
y = 4
When x = −2
y = 3 × (−2)3 − 5 × (−2)2 − 4 × −2
+ 4
= −32
When x = −1
y = 3 × (−1)3 − 5 × (−1)2 − 4 × −1
+ 4
= 0
Closed end point (−2, −32)
Closed end point (−1, 0)
i Domain [−2, −1]
ii Range [–32, 0]
c f ( x) = −3 x3 + 4 x2 + 27 x − 36
x ∈ (0, 1]
a
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CHAPTER 1 Graphs and polynomials • EXERCISE 1G | 13
Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
0 = 4a + 2a − 36
0 = 6a − 36
36 = 6a
6 = a
∴ b = 12 − 36
= −24
12 0 = −1 − 2 − a + 10
0 = 7 − a [1]
∴ a = 7
∴ y = 6 + (7 + b) x − 4 x2 − x3 0 = 6 + (7 + b) × (−1) − 4 × 1
− (−1)
0 = 6 − 7 − b − 4 + 1
0 = −4 − b
b = −4
13 a f ( x) = a( x + b)3 + c
point of inflection (3, 3)
⇒ b = −3 and c = 3
f ( x) = a( x − 3)3 + 3
When x = 2, f ( x) = 0
0 = a(2 − 3)3 + 3
0 = a × (−1)3 + 3
0 = −a + 3∴ a = 3
∴ f ( x) = 3( x − 3)3 + 3
b Point of inflection due to
reflection = (−3, 3)
∴ g ( x) = −3( x + 3)3 + 3
domain [−4, −2]
range [0, 6]
c When f ( x) = 3.375
3.375 = 3( x − 3)3 + 3
x = 3.5
∴ width = 3.5 × 2
= 7 cm
14 d (t ) = at 2(b − t )
a (2, 3) and (5, 0)
∴ 3 = 4a(b − 2) [1]
0 = 25a(b − 5) [2]
3 = (4ab − 8a) × 25
0 = (25ab − 125a) × 4
75 = 100ab − 200a
0 = 100ab − 500a
∴ 75 = 300a
1
4 = a
Sub1
4 = a into [1]
3 = 4 × 1
4 (b − 2)
3 = 1(b − 2)
5 = b
b ∴ Rule: d (t ) = 2
4
t (5 − t ) for
domain = [0, 5]
c
d d (t ) =2 3
5
4 4
t t −
d ′(t ) = 2
10 3
4 4
t t −
Let d ′(t ) = 0 = 2
10 3
4
t t −
0 = 10t − 3t 2
0 = t (10 − 3t )
∴ t = 0 or 10 − 3t = 010 = 3t
∴ time is1
33
hours
When time is1
33
,
d (t ) = 2(3.3)
4
× 1.6
Maximum distance = 4.6 km
Exercise 1G — Quartic graphs
1 a y = ( x − 2)( x + 3)( x − 4)( x + 1)
y-intercept when x = 0 y = −2 × + 3 × −4 × + 1
y = 24
x-intercepts when y = 0
x = −3, −1, 2 and 4
b y = 2 x4 + 6 x3 − 16 x2 − 24 x + 32
y-intercept when x = 0
y = 32
x-intercept when y = 0Test x = 1, y = 0
∴ ( x − 1) is a factor
Test x = + 2 so y = 0
∴ ( x − 2) is a factor
( x − 1)( x − 2) = x2 − 3 x + 2
2
4 3 22
4 3 2
3 2
3 2
2
2
2 12 16
2 6 16 24 323 2
2 6 4
12 20 24
12 36 24
16 48 32
16 48 32
0
x x
x x x x x x
x x x
x x x
x x x
x x
x x
+ +
⎡ + − − +− + −⎢
− +⎢⎣
⎡ − −− ⎢
− +⎢⎣
⎡ − +− ⎢ − +⎢⎣
∴ ( x − 1)( x − 2)(2 x2 + 12 x + 16) = y
( x − 1)( x − 2)2( x2 + 6 x + 8) = y
2( x − 1)( x − 2)( x + 4)( x + 2) = y
When using N.F.L, x = 1, 2, −4, −2
then y = 0.
c y = x4 − 4 x2 + 4
y-intercept when x = 0
y = 4
x-intercept when y = 0
Let a = x2
y = a2 − 4a + 4
y = (a − 2)2
Sub a = x2 back in
y = ( x2 − 2)2
∴ x = ± 2
d y = −2 x4 + 15 x3 − 37 x2 + 30 x
y-intercept, when x = 0
y = 0
x-intercepts when y = 0
Test x = 2 ∴ y = 0
Test x = 3 ∴ y = 0
( x − 2)( x − 3) are factors
( x − 2)( x − 3) = x2
− 5 x + 62
4 3 22
4 3 2
3 2
3 2
2 5
2 15 37 305 6
2 10 12
5 25 30
5 25 30
0
x x
x x x x x x
x x x
x x x
x x x
− +
⎡− + − +− + − ⎢
− + −⎢⎣
⎡ − +− ⎢
− +⎢⎣
∴ ( x − 2)( x − 3)(−2 x2 + 5 x)= y
x( x − 2)( x − 3)(−2 x + 5) = y
Using N.F.L, x = 0, 2, 3 or5
2
e y = 6 x4 + 11 x3 − 37 x2 − 36 x + 36
y-intercept when x = 0
y = 36
x-intercepts when y = 0
Test x = −3 ∴ y = 0
Test x = 2
( x + 3)( x − 2) are factors
( x + 3)( x − 2) = x2
+ x − 62
4 3 22
4 3 2
3 2
3 2
2
2
6 5 6
6 11 37 36 366
6 6 36
5 36
5 5 30
6 6 36
6 6 36
0
x x
x x x x x x
x x x
x x x
x x x
x x
x x
+ −
⎡ + − − ++ − −⎢
+ −⎢⎣
⎡ − −− ⎢
+ −⎢⎣
⎡− − +− ⎢
− − +⎢⎣
∴ ( x + 3)( x − 2)(3 x − 2)(2 x + 3) = y
Using N.F.L
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14 | CHAPTER 1 Graphs and polynomials • EXERCISE 1G
Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
x = −3, 2,2
3, −
3
2
2 a y = x2( x − 2)( x − 3)
y = 0, x2( x − 2)( x − 3) = 0
Turning point (0, 0)
Intercepts at x = 2 and x = 3
y-intercept when x = 0
y = 0(−2)(−3)
= 0Positive quarticMaximum turning point at (1.16, 2.08)
Minimum turning points at (0, 0)
and (2.59, −1.62)
b y = −( x + 1)2( x − 1)
2
x = 0, y = 1 y-intercept is 1
y = 0, ( x + 1)2( x − 1)2 = 0
x = −1, x = 1
Minimum turning points at (−1, 0)and (1, 0).
Maximum turning point (0, 1)
c y = ( x − 1)2( x + 1)( x + 3)
x = 0, y = 3 y-intercept is 3
y = 0, ( x − 1)2( x + 1)( x + 3) = 0
x = 1, −1, −3
Positive quarticMinimum turning points
(−2.28, −9.91) and (1, 0)Maximum turning point
(−0.22, 3.23)
d y = ( x + 2)3(1 − x)
x = 0, y = 8 y-intercept is 8
y = 0, ( x + 2)3(1 − x) = 0
x = −2, 1
Point of inflection (2, 0)
x-intercept is 1
Negative quartic (1 − x) = −( x − 1)
Maximum turning point (0.25, 8.54)
3 a f ( x) = x4 − 8 x2 + 16
Let x2 = a
f ( x) = a2 − 8a + 16
= (a − 4)2
Substitute x2 = a back in:
f ( x)=
( x
2
−
4)
2
∴ ( x − 2)2( x + 2)
2
The answer is E.
b f ( x) = x4 − 8 x2 + 16
y-int when x = 0
y = 16 (0, 16)
x-int when y = 0
2, −2
∴ The answer is B.
c range = [0, 16]
The answer is A.
d 25 = x4 − 8 x
2 + 16
0 = x4 − 8 x2 − 9
Let x2 = a
0 = a2
− 8a − 90 = (a − 9)(a + 1)
Substitute x2 = a back in
0 = ( x2 − 9)( x2 + 1)
∴ x = ± 3
∴ (−3, +3) is the restricted domain.The answer is D.
e f (−1) = 1 − 8 + 16 = 9
f (0) = 16Range: (9, 16)
Answer is D.
f f (0) = 16
f (2) = 0
Range is y ≥ 0 or [0, ∞ )
Answer is C.4 a x = −2, −1, 1, 3
y = a( x + 2)( x + 1)( x − 1)( x − 3) y-int (0, 6)
6 = a(0 + 2)(0 + 1)(0 − 1)(0 − 3)
6 = a × 6
a = 1
y = ( x + 2)( x + 1)( x − 1)( x − 3)
b x = 4, 2, −1
Repeated factor at x = −2.
y = a( x − 4)( x − 2)2( x + 1)
y-int (0, 8)
8 = a(0 − 4)(0 − 2)2(0 + 1)
8 = −16a
a = −1
2
y = 1
2
−( x − 4)( x − 2)2( x + 1)
5 a y = (2 − x)( x2 − 4)( x + 3) x ∈ [2, 3]
y-int when x = 0
y = 2 × −4 × 3
= −24
x-int when y = 0
x = 2, −2, −3.
when x = 2, y = 0 closed end point
x = 3, y = −30 closed end point
i Domain [2, 3]
ii Range [−30, 0]
b y = 9 x4 − 30 x
3 + 13 x
2 + 20 x + 4 x ∈
(−2, −1]
y-int when x = 0
y = 4 x-int when y = 0
Test x = 2 y = 0
∴ ( x − 2) factor
3 2
4 3 2
4 3
3 2
3 2
2
2
9 12
9 30 13 20 42
9 18
12 13
12 24
11 20
11 22
2 4
2 4
0
x x
x x x x x
x x
x x
x x
x x
x x
x
x
− −
⎡ − + + +− − ⎢
−⎢⎣
⎡− +− ⎢
− +⎢⎣
⎡− +− ⎢
− +⎢⎣
− +⎡− ⎢
− +⎣
11 2 x −
y = ( x − 2)(9 x3 − 12 x2 − 11 x − 2)
Try for 2nd factor of x − 2
Test x = 2, 9 x3 − 12 x2 − 11 x − 2 = 0
x − 2 is a factor. So turning pointat (2, 0)
2
3 2
3 2
2
2
9 6 1
2 9 12 11 2
9 18
6 11
6 12
2
x x
x x x x
x x
x x
x x
x
+ +
− − − −
−
−
−
−
y = ( x − 2)2(9 x2 + 6 x + 1)
y = ( x − 2)2(9 x2 + 6 x + 1)
= ( x − 2)2(3 x + 1)2
Turning points at (2, 0) and
1, 0
3
⎛ ⎞−⎜ ⎟
⎝ ⎠
when x = −2, y = 400 open end point
when x = −1, y = 36 closed end point
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i Domain (−2, −1]ii Range [36, 400)
c y = −( x − 2)2( x + 1)2 x ∈ (−∞, −2]
y-int when x = 0 y = −1 × 4 × 1
y = −4
x-int when y = 0
0 = −1( x − 2)2( x + 1)
2
∴ x = −1 or 2
When x = −2, y = −16.
y = −(−2 − 2)2(−2 + 1)2
y = −(−4)2(−1)
2
y = −16 × 1
y = −16
i Domain (−∞,−2]
ii Range (−∞,−16]
d y = − x4 + 4 x2
x ∈ [−3, −2]
y-int when x = 0
y = 0
x-int when y = 0
0 = x2(− x2 + 4)
0 = x2(4 − x2)
0 = x2(2 − x)(2 + x)
∴ x = 0 ± 2When x = −3, y = −45
x = −2, y = 0
Both of these are closed end points.
i Domain [−3, −2]
ii Range [−45, 0]
6 f ( x) = x4 + ax3 − 4 x2 + bx + 6
(2, 0):
0 = 24 + 23a − 4 × 22 + 2b + 6
16 + 8a − 16 + 2b + 6 = 0
6 + 8a + 2b = 0
8a + 2b = −6
Divide both sides by 2:
4a + b = −3
−4a − 3 = b [1]
4a + 3 = − b
(−3, 0):
0 = (−3)4 + (−3)3a − 4 × (−3)2 − 3b + 6
0 = 81 − 27a − 36 − 3b + 6
0 = 51 − 27a − 3b
0 = 17 − 9a − b [2]
Sub [1] into [2]
0 = 17 − 9a + 4a + 3,
0 = 20 − 5a
5a = 20
a = 4
If a = 4 then
b = −4 × 4 − 3
b = −16 − 3
b = −19
7 f ( x) = x4 + ax3 + bx2 − x + 6( x − 1) is a factor
P (1) = 1 + a + b − 1 + 6
a + b = −6 [1]
( x + 3) is a factor
P (−3) = 81 − 27a + 9b + 3 + 6 = 0
27a − 9b = 90
3a − b = 10 [2]
[1] + [2] 4a = 4
a = 1
b = −7
8 y = (a − 2b) x4 − 3 x − 2
Sub in (1, 0):
0 = (a − 2b)1
4
− 3 − 20 = a − 2b − 5
5 = a − 2b
5 + 2b = a [1]
Sub (1, 0) into equation.
y = x4 − x3 + (a + 5b) x2 − 5 x + 7
0 = 1 − 1 + (a + 5b)1 − 5 + 7
0 = a + 5b + 2
−2 = a + 5b [1]
Sub [1] into [2]
−2 = 5 + 2b + 5b
−7 = 7b
−1 = b
If −1 = b then 5 − 2 = a
∴ 3 = a
Exercise 1H — Solving systemsof equations
1 9 0ax y+ = [1]
3 ( 6) 0 x a y+ − = [2]
From [1] 9 y ax= −
9
ax y = − [3]
From [2] ( 6) 3a y x− = −
3
6
x y
a= −
− [4]
Lines which are parallel have no
solutions while lines which are co-incident have infinitely many solutions.In both cases the gradients are thesame. Equate the gradients for [3] and
[4].
2
3
9 6
( 6) 27
6 27 0
( 9)( 3) 0
9 or 3
a
a
a a
a a
a a
a a
− =−
− =
− − =
− + =
= = −
Check: When 3a = − then there are
co-incident lines andsubsequently infinitely manysolutions because equations
become
3 9 0
3 9 0
x y
x y
− + =
− =
3 ( 3 6) 0
3 9 0
x y
x y
+ − − =
− =
When 9a = then there are again
co-incident lines because equations become
9 9 0
0
x y
x y
+ =
+ =
3 (9 6) 0
3 3 0
0
x y
x y
x y
+ − =
+ =
+ =
For a unique solution then
\{ 3,9}.a R∈ −
2 a 5 10mx y− = [1]
3 ( 2) 6 x m y− − = [2]
From [1] 5 10 y mx− = −
25
mx y = − [3]
From [2] 3 6 ( 2) x m y− = −
3 6
2 2
x y
m m= −
− − [4]
Equate gradients
2
3
5 2
( 2) 15
2 15 0( 5)( 3) 0
5 or 3
m
m
m m
m mm m
m m
=−
− =
− − =− + =
= = −
Check: When 3m = − there are
parallel lines and
subsequently no solutions asequations become
3 5 10
3 5 10
x y
x y
− − =
+ = −
3 ( 3 2) 6
3 5 6
x y
x y
− − − =
+ =
When 5m = there are co-incident lines
and subsequently infinitely manysolutions as equations become
5 5 10
2
x y
x y
− =
− =
3 (5 2) 6
3 3 6
0
x y
x y
x y
− − =
− =
− =
b
5 10 [1] 3
3 ( 2) 6 [2]
3 15 30 [3]
3 ( 2) 6 [4]
mx y
x m y m
mx y
mx m m y m
− = ×
− − = ×
− =
− − =
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Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
2
3 15 30 [3]
3 ( 2) 6 [4]
15 ( 2) 30 6
( 15 ( 2)) 30 6
( 2 15) 30 6
6( 5)
( 5)( 3)
6 , 53
mx y
mx m m y m
y m m y m
y m m m
y m m m
m y
m m
y mm
− = −
− − =
− + − = −
− + − = −
− − = −
− −=
− +
= − ≠+
Substitute6
3
ym
= −+
into [1]
65 10
3
3010
3
mxm
mxm
⎛ ⎞− − =⎜ ⎟
+⎝ ⎠
+ =+
3010
3
10( 3) 30
3
103
10
( 3)
10, 0, 3
3
mxm
mmx
m
mmxm
m x
m m
x m mm
= −+
+ −=
+
=+
=+
= ≠ ≠ −+
For a unique solution10
,3
xm
=+
6
3 y
m= −
+ providing
\ [ 3, 0, 5}m R∈ −
3 a 9 x y z + + = (1)
2 3 15 x y z − + − = − (2)5 3 29 x y z + + = (3)
(1) + (2) 3 2 6 y z − = − (4)
(2) + (3) 7 14 y =
2 y =
Substitute 2 y = into (4)
3(2) 2 6 z − = −
6 2 6 z − = −
2 12 z − = −
6 z =
Substitute 2, 6 y z = = into (1)
2 6 9 x + + =
8 9 x + =
1 x =
b
1 1 1 9
2 2 3 15
1 5 3 29
x
y
z
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
1
2
6
x
y
z
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥
=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
4 a 5 x y z − − = − (1)
6 2 5 2 x y z + − = − (2)
3 4 13 x y z − + + = (3)
(1) 2× 2 2 2 10 x y z − − = − (4)
(2) (4)+ 8 7 12 x z − = − (5)
(3) 2× 6 2 8 26 x y z − + + = (7)
(2) (7)− 12 13 28 x z − = − (8)
(5) 3× 24 21 36 x z − = − (9)
(8) 2× 24 26 56 x z − = − (10)
(10) (9)− 5 20 z − = − 4 z =
Substitute 4 z = into (5)
8 7(4) 12 x − = −
8 28 12 x − = −
8 12 28 x = − +
8 16 x =
2 x =
Substitute 2, 4 x z = = into (1)
2 4 5 y− − = −
2 5 y− − = −
3 y− = −
3 y =
b
1 1 1 5
6 2 5 2
3 1 4 13
x
y
z
− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥
− = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
2
3
4
x
y
z
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥
=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
5 3 2 1 x y z + − = [1]
2 x y z + + = [2]
1kx y z + − = [3]
3 2 1
1 1 1
1 1
ma
k
−⎡ ⎤⎢ ⎥ →⎢ ⎥⎢ ⎥−⎣ ⎦
det (ma) = 3k − 5
(using a calculator)
⇒ when5
3k ≠
there is a unique solution.
Therefore, when5
3k = , there is no
solution.
6 a Let a be the number of adult
tickets bought.
Let c be the number of children’s
tickets bought.
Let s be the number of seniors’
tickets bought.
200a c s+ + = (1)
9.5 4.5 3.5 1375a c s+ + = (2)
3c s= (3)
Substitute (3) into (1) and (2)
3 200a s s+ + =
4 200a s+ = (4)
9.5 4.5(3 ) 3.5 1375a s s+ + =
9.5 13.5 3.5 1375a s s+ + =
9.5 17 1375a s+ =
95 170 13 750a s+ =
19 34 2750a s+ = (5)
(4) 19× 19 76 3800a s+ = (6)
(6) (5)− 42 1050 s =
105025
42 s = =
Substitute 25 s = into (3)
3(25) 75c = =
Substitute 25, 75 s c= = into (1)
75 25 200a + + =
100 200a + =
100a =
b
1 1 1 200
9.5 4.5 3.5 1375
0 1 3 0
a
c
s
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥
=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
100
75
25
a
c
s
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥
=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
There were 100 adult tickets,75 children’s tickets and 25 seniors’tickets sold on the opening night.
7 a 255 F C P + + = (1)
6 4.5 1067.5 F C P + + = (2)
50C F = + (3)
Substitute (3) into (1)
50 255 F F P + + + =
2 205 F P + = (4)
Substitute (3) into (2)
6 4.5( 50) 1067.5 F F P + + + =
6 4.5 225 1067.5 F F P + + + =
10.5 1067.5 225 F P + = −
10.5 842.5 F P + = 105 10 8425 F P + = (5)
(4) 10× 20 10 2050 F P + = (6)
(5) (6)− 85 6375 F =
637575
85 F = =
Substitute 75 F = into (3)
75 50 125C = + = Substitute 75, 125 F C = = into (1)
75 125 255 P + + =
200 255 P + =
55 P =
b 255 F C P + + = (1)6 4.5 1067.5 F C P + + = (2)
50 F C − + = (3)
1 1 1 255
6 4.5 1 1067.5
1 1 0 50
F
C
P
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥
=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
75
125
55
F
C
P
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥
=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
75 portions of flake, 125 portions of
chips and 55 potato cakes were soldduring the ‘lunch special’ period.
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8 a 3 2 y ax bx cx d = + + +
When 0, 4; x y= =
3 24 (0) (0) (0)a b c d = + + +
4 d =
3 2 4 y ax bx cx= + + +
When 2, 62; x y= − = −
3 262 ( 2) ( 2) ( 2) 4a b c− = − + − + − +
66 8 4 2a b c− = − + −
33 4 2a b c− = − + − (1)
When 2, 26; x y= = −
3 226 (2) (2) (2) 4a b c− = + + +
30 8 4 2a b c− = + +
15 4 2a b c− = + + (2)
When 5, 64; x y= =
3 264 (5) (5) (5) 4a b c= + + +
60 125 25 5a b c= + +
12 25 5a b c= + + (3)
(1) (2)+ 48 4b− =
12 b− =
(1) (3)+ 21 21 7a b− = +
3 3a b− = + (4)
Substitute 12b = − into (4)
3 3 12a− = −
9 3a=
3 a=
Substitute 12, 3b a= − = into (3)
12 25(3) 5( 12) c= + − +
12 75 60 c= − +
12 15 c= +
3 c− =
b 4 2 33a b c− + − = − (1)
4 2 15a b c+ + = − (2)
25 5 12a b c+ + = (3)
4 2 1 33
4 2 1 15
25 5 1 12
a
b
c
− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥
= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
3
12
3
a
b
c
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥
= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
The equation of the cubic function is
3 23 12 3 4. y x x x= − − +
9 a When 0, 2; x y= =
2 20 2 (0) 2 0a b c+ + + + =
4 2 0b c+ + =
2 4b c+ = − [1]
When 6, 2; x y= =
2 26 2 6 2 0a b c+ + + + =
36 4 6 2 0a b c+ + + + =
6 2 40a b c+ + = − [2]
When 3, 1; x y= = −
2 2
3 ( 1) 3 0a b c+ − + − + =
3 10a b c− + = − [3]
From (1) 2 4c b= − − [4]
Substitute (4) into (2)
6 2 2 4 40a b b+ − − = −
6 36a = −
6a = −
Substitute (4) into (3)
3 2 4 10a b b− − − = −
3 3 6a b− = −
2a b− = −
Substitute 6a = − into (5) 6 2b− − = −
4b− =
4b = −
Substitute 4b = − into (1)
2( 4) 4c− + = −
8 4c− + = −
4c =
b 2 4b c+ = − (1)
6 2 40a b c+ + = − (2)
3 10a b c− + = − (3)
0 2 1 4
6 2 1 403 1 1 10
a
bc
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥
= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
6
4
4
a
b
c
−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥
= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
10 2 24a b c d e+ − + − =
2 3 2 3 34a b c d e+ − − − =
2 3 2 31a b c d e− + + − + = −
3 5 2 2 3 18a b c d e+ − − + =
4 2 3 5a b c d e− − − + =
1 2 1 1 1 24
2 3 2 1 3 342 1 3 2 1 31
3 5 2 2 3 18
4 2 1 3 1 5
a
bc
d
e
− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥
− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥
− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
7
3
2
5
4
a
b
c
d
e
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
Chapter review
Short answer
1 a (2 y − 3 x)5
= (2 y)5 + 5(2 y)4(−3 x) + 10(2 y)3(−3 x)2
+ 10(2 y)2(−3 x)
3 + 5(2 y)(−3 x)
4 + (−3 x)
5
= 32 y5 + 5 × 16 × (−3) y4 x + 10 × 8
× 9 y3 x2 + 10 × 4 × (−27) y2 x3 + 5
× 2 × 81 yx4 − 243 x5
= 32 y5 − 240 y4 x + 720 y3 x2
− 1080 y2 x3 + 810 yx4 − 243 x5
b
82
2
x
x
⎛ ⎞−⎜ ⎟
⎝ ⎠ =
8
2
x⎛ ⎞⎜ ⎟⎝ ⎠
+ 8
72
2
x
x
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
+ 28
6 22
2
x
x
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠+ 56
5 32
2
x
x
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
+ 70
4 42
2
x
x
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠+ 56
3 52
2
x
x
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
+ 28
2 62
2
x
x
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠+ 8
72
2
x
x
⎛ ⎞⎛ ⎞−⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
+
82
x
⎛ ⎞−⎜ ⎟
⎝ ⎠
= 8 6
256 8
x x− +
47
4
x − 14 x2 + 70
− 2
224
x +
4
448
x −
6
512
x +
8
256
x
2 ( x2 − 1) = ( x − 1)( x + 1)
∴ solutions are x = −1 or 1
∴ If x = −1 then 0 = −7 − a + 5 − 15 + b
0 = −17 − a + b
17 + a = b [1]
If x = 1, then 0 = −7 + a + 5 + 15 + b
0 = 13 + a + b [2]
Sub [1] into [2]
0 = 13 + a + 17 + a
0 = 30 + 2a
−30 = 2a
−15 = a
∴ b = 17 − 15
b = 2
3 a x3 − 12 x2 + 17 x + 90 = y
Test x = −2. y = 0
∴ ( x + 2) is a factor2
3 2
3 2
2
2
14 45
12 17 902
2
14 17
14 28
45 90
45 90
0
x x
x x x x
x x
x x
x x
x
x
− +
⎡ − + ++ − ⎢
+⎢⎣
⎡− +− ⎢
− −⎢⎣+⎡
− ⎢+⎣
∴ ( x + 2)( x2 − 14 x + 45)
( x + 2)( x − 9)( x − 5)
b 2 x4 + 7 x3 − 31 x2 + 0 x + 36 = y
Test x = −1 ∴ y = 0
Test x = 2 ∴ y = 0
∴ ( x + 1)( x − 2) are factors
∴ ( x2 − x − 2) is a factor
2
4 3 22
4 3 2
3 2
3 2
2 9 18
2 7 31 0 362
2 2 4
9 27 0
9 9 18
x x
x x x x x x
x x x
x x x
x x x
+ −
⎡ + − + +− − − ⎢
− −⎢⎣
⎡ − +− ⎢
− −⎢⎣
2
2
18 18 36
18 18 36
0
x x
x x
⎡− + +−⎢
− + +⎢⎣
∴ ( x + 1)( x − 2) (2 x2 + 9 x − 18)
( x + 1)( x − 2) (2 x − 3)( x + 6)
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Maths Quest 12 Mathematical Methods CAS Second Edition Solutions Manual
4 a (−5, 6), (1, −1)
m = 6 1
5 1
+
− −
= 7
6−
∴ y = 7
6
− x + c
Sub in (1, −1) to find c:
−1 = 7
6− × 1 + c
−1 +7
6 = c
1
6 = c
∴ y = 7
6
− x +
1
6
6 y = −7 x + 1
7 x + 6 y − 1 = 0
b 2 x − y + 10 = 0
2 x + 10 = y
Perpendicular ⇒ m = 1
2
−
∴ y = 1
2− x + c
Sub in point (3, 3)
3 = 3
2
− + c
3 +3
2 = c so c =
9
2.
∴ y = 1
2
− x +
9
2
2 y = − x + 9
x + 2 y − 9 = 0
5 y = − x2 − 2 x + 8
y-int when x = 0∴ y = 8
x-int when y = 0
0 = −1( x2 + 2 x − 8)
0 = −1( x + 4)( x − 2)
∴ x = −4 or 2
TP ⇒ y = −1( x2 + 2 x − 8)= −1( x
2 + 2 x + 1 − 1 − 8)
= −1[( x + 1)2 − 9]
= −1( x + 1)2 + 9
TP = (−1, 9)
Domain = R
Range = (−∞, 9]
6 y = 3 x2 + 8 x − 3
x ∈ [−3, 0)
y-int when x = 0
y = −3
x-intercepts when y = 0
0 = 3 x2 + 8 x − 3
0 = (3 x − 1)( x + 3)
∴ x = 1
3 or −3
TP ⇒ y = 28
3 13
x x⎛ ⎞
+ −⎜ ⎟⎝ ⎠
Now x2 + 8
3 x − 1
=
22 8 8 64 36
3 6 36 36 x x
⎛ ⎞⎛ ⎞+ + − −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
⎝ ⎠
∴ y =
28 100
36 36
x⎡ ⎤⎛ ⎞
+ −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
=
28 100
36 12
x⎛ ⎞
+ −⎜ ⎟⎝ ⎠
TP = 8 100
,6 12
− −⎛ ⎞⎜ ⎟⎝ ⎠
= 4 25
,3 3
− −⎛ ⎞⎜ ⎟⎝ ⎠
Domain [−3, 0)
Range:
When x = 3, y = 0 closed end point
When x = 0, y = −3 open end point
25, 0
3
−⎡ ⎤⎢ ⎥⎣ ⎦
7 a f ( x) = − x3 + bx2 + ax − 18
0 = −(−3)3 + b(−3)2 + a × (−3) − 18
0 = 27 + 9b − 3a − 180 = 9 + 9b − 3a
0 = 3 + 3b − a
a = 3 + 3b [1]
g ( x) = ax2 + bx − 75
0 = a(−3)2 + b × (−3) − 75
0 = 9a − 3b − 75
25 = 3a − b [2]
Sub [1] into [2]
25 = 3(3 + 3b) − b
25 = 9 + 9b − b
16 = 8b
2 = b
∴ a = 3 + 6= 9
b f ( x) = − x3 + 2 x2 + 9 x − 18
y-int when x = 0
∴ y = −18
x-int when y = 0
( x + 3) is a factor2
3 2
3 2
2
2
5 6
2 9 183
3
5 9
5 15