Maths Quest 12. Mathematical Methods: VCE Units 3 & 4 - Solutions Manual

263 Pages·2016·7.665 MB·English

by  Margaret Swale

mathsquest12
mathematical methods
solutions manual
vce units 3 and 4

mathsquest12
mathematical methods
solutions manual
vce units 3 and 4
author
margaret swale
contributing authors
sue michell | catherine smith
First published 2016 by
John Wiley & Sons Australia, Ltd
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Table of contents
About eBookPlus vi Topic 7 — Antidifferentiation 143
Exercise 7.2 — Antidifferentiation 143
Topic 1 — Solving equations 1 Exercise 7.3 — Antiderivatives of exponential and
Exercise 1.2 — Polynomials 1 trigonometric functions 145
Exercise 1.3 — Trigonometric symmetry properties 5 Exercise 7.4 — Families of curves and applications 147
Exercise 1.4 — Trigonometric equations and general
solutions 8 Topic 8 — Integration 153
Exercise 1.5 — L iteral and simultaneous equations 14 Exercise 8.2 — The fundamental theorem of integral
calculus 153
Topic 2 — Functions and graphs 19 Exercise 8.3 — Areas under curves 159
Exercise 2.2 — Polynomial functions 19 Exercise 8.4 — Applications 162
Exercise 2.3 — Other algebraic functions 25
Exercise 2.4 — Combinations of functions 33 Topic 9 — Logarithmic functions using calculus 171
Exercise 2.5 — Non-algebraic functions 38 Exercise 9.2 — The derivative of f(x) = log(x) 171
e
Exercise 2.6 — Modelling and applications 49 1
Exercise 9.3 — The antiderivative of f(x) = 177
x
Exercise 9.4 — Applications 182
Topic 3 — Composite functions, transformations and
inverses 53
Topic 10 — Discrete random variables 189
Exercise 3.2 — C omposite functions and functional
Exercise 10.2 — Discrete random variables 189
equations 53
Exercise 10.3 — Measures of centre and spread 196
Exercise 3.3 — Transformations 56
Exercise 10.4 — Applications 204
Exercise 3.4 — Transformations using matrices 59
Exercise 3.5 — Inverse graphs and relations 62 Topic 11 — The binomial distribution 211
Exercise 3.6 — Inverse functions 64 Exercise 11.2 — Bernoulli trials 211
Exercise 11.3 — The binomial distribution 212
Topic 4 — Logarithmic functions 69
Exercise 11.4 — Applications 216
Exercise 4.2 — Logarithm laws and equations 69
Exercise 4.3 — Logarithmic scales 72 Topic 12 — Continuous probability distributions 219
Exercise 4.4 — Indicial equations 74 Exercise 12.2 — C ontinuous random variables and probability
Exercise 4.5 — Logarithmic graphs 78 functions 219
Exercise 4.6 — Applications 83 Exercise 12.3 — The continuous probability density
function 223
Topic 5 — Differentiation 87 Exercise 12.4 — Measures of centre and spread 229
Exercise 5.2 — Review of differentiation 87 Exercise 12.5 — Linear transformations 237
Exercise 5.3 — Differentiation of exponential functions 94
Exercise 5.4 — Applications of exponential functions 97 Topic 13 — The normal distribution 243
Exercise 5.5 — D ifferentiation of trigonometric Exercise 13.2 — The normal distribution 243
functions 100 Exercise 13.3 — C alculating probabilities and the standard
Exercise 5.6 — Applications of trigonometric functions 103 normal distribution 245
Exercise 13.4 — The inverse normal distribution 246
Topic 6 — Further differentiation and applications 107 Exercise 13.5 — Mixed probability application problems 247
Exercise 6.2 — The chain rule 107
Exercise 6.3 — The product rule 111 Topic 14 — Statistical inference 253
Exercise 6.4 — The quotient rule 116 Exercise 14.2 — P opulation parameters and sample
Exercise 6.5 — Curve sketching 121 statistics 253
Exercise 6.6 — Maximum and minimum problems 133 Exercise 14.3 — The distribution of p(cid:31) 253
Exercise 6.7 — Rates of change 137 Exercise 14.4 — Confidence intervals 255
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TOPIC 1 Solving equations • EXERCISE 1.2 | 1
Topic 1 — Solving equations
Exercise 1.2 — Polynomials b n2p2−4m2−4m−1
1 a 15u2−u−2=(5u−2)(3u+1) =(np)2−(4m2+4m+1)
b 6d2−28d+16=2(3d2−14d+8)=2(3d−2)(d−4) =(np)2−((2m)2+2(2m)+1)
c 3j2+12j−6 =(np)2−(2m+1)2
=3(j2+4j−2) =(np−(2m+1))(np+(2m+1))
=3(j2+4j+(2)2−(2)2−2) =(np−2m−1)(np+2m+1)
=3((j+2)2−6) 7 Let P(x)=x3−2x2−21x−18
=3((j+2)2−( 6)2) PP((−−11))==(−−11−)32−+22(1−−1)128−21(−1)−18
=3(j+2− 6)(j+2+ 6) P(−1)=0
Thus (x+1) is a factor.
2 a f2−12f −28=(f −14)(f +2) x3−2x2−21x−18=(x+1)(x2−3x−18)
b g2+3g−4=(g+4)(g−1)
=(x+1)(x−6)(x+3)
c b2−1=(b−1)(b+1) 8 Let P(x)=x4−5x3−32x2+180x−144
3 a 125a3−27b3=(5a)3−(3b)3 P(1)=(1)4−5(1)3−32(1)2+180(1)−144
=(5a−3b)((5a)2+(5a)(3b)+(3b)2) P(1)=1−5−32+180−144
=(5a−3b)(25a2+15ab+9b2) P(1)=181−181
P(1)=0
b 2c3+6c2d+6cd2+d3
Thus (x−1) is a factor.
=2(c3+3c2d+3cd2+d3) x4−5x3−32x2+180x−144=(x−1)(x3−4x2−36x+144)
=2(c+d)3 Let Q(x)=x3−4x2−36x+144
c 40p3−5=5(8p3−1) Q(2)=23−4(2)2−36(2)+144≠0
=5((2p)3−13) Q(4)=43−4(4)2−36(4)+144
=64−64−144+144
=5(2p−1)((2p)2+2p+1) =0
=5(2p−1)(4p2+2p+1) Thus (x−4) is a factor.
x3−4x2−36x+144=(x−4)(x2−36)
4 a 27z3−54z2+36z−8
=(x−4)(x−6)(x+6)
=(3z)3−3(3z)2(2)+3(3z)(2)2+23 So x4−5x3−32x2+180x−144=(x−1)(x−4)(x−6)(x+6)
=(3z−2)3 9 a x4−8x3+17x2+2x−24=0
b m3n3+64 (x−4)(x−3)(x−2)(x+1)=0
=(mn)3+43 x−4=0, x−3=0 x−2=0 x+1=0
=(mn+4)((mn)2−4mn+42) x=4, x=3, x=2, x=−1
=(mn+4)(m2n2−4mn+16) b a4+2a2−8=0
Let x=a2
5 a 9x2−xy−3x+y x2+2x−8=0
=9x2−3x−xy+y (x−2)(x+4)=0
=3x(x−1)−y(x−1) x=2,−4
=(x−1)(3x−y) Substitute x=a2
b 3y3+3y2z2−2zy−2z3 a2=2, a2=−4(nosolution)
=3y2(y+z2)−2z(y+z2)
∴a=± 2
=(y+z2)(3y2−2z) 10 a 2x3−x2−10x+5=0
6 a 9a2−16b2−12a+4 x2(2x−1)−5(2x−1)=0
=9a2−12a+4−16b2 (2x−1)(x2−5)=0
=(3a)2−2(3a)(2)+22−(4b)2 (2x−1)(x− 5)(x+ 5)=0
=(3a−2)2−(4b)2 1
x= , ± 5
=(3a−2−4b)(3a−2+4b) 2
Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual
2 | TOPIC 1 Solving equations • EXERCISE 1.2
b 2a2−5a=9 b 4z2+28z+49=0
2a2−5a−9=0 (2z)2+2(2z)(7)+72=0
5± (−5)2−4×2×−9 (2z+7)2=0
a=
2×2 2z+7=0
5± 97 2z=−7
=
4 7
11 Ax3+(B−1)x2+(B+C)x+D≡3x3−x2+2x−7 z=−2
A=3, B−1=−1, B+C=2andD=−7 c 5m2+3=10m
B=0 0+C=2
5m2−10m+3=0
C=2
12 x3+9x2−2x+1≡x3+(dx+e)2+4 m=10± (−10)2−4×5×3
x3+9x2−2x+1≡x3+d2x2+2dex+e2+4 2×5
10± 40
d2=9, 2(±3)e=−2 =
10
d=±3 ±6e=−2
10±2 10
1 =
e=± 10
3
5± 10
13 a 7r3−49r2+r−7 =
5
=7r2(r−7)+(r−7) d x2−4x=−3
=(r−7)(7r2+1) x2−4x+3=0
b 36v3+6v2+30v+5 (x−3)(x−1)=0
=6v2(6v+1)+5(6v+1) x−3=0 or x−1=0
=(6v+1)(6v2+5) x=3 x=1
c 2m3+3m2−98m−147 e 48p=24p2+18
=m2(2m+3)−49(2m+3) 24p2−48p+18=0
=(2m+3)(m2−49) 4p2−8p+3=0
=(2m+3)(m−7)(m+7) (2p−1)(2p−3)=0
d 2z3−z2+2z−1 2p−1=0or2p−3=0
=z2(2z−1)+(2z−1) p=1 p=3
2 2
=(2z−1)(z2+1)
f 39k=4k2+77
e 4x2−28x+49−25y2 4k2−39k+77=0
=(2x−7)2−(5y)2 (4k−11)(k−7)=0
=(2x−7−5y)(2x−7+5y) 4k−11=0ork−7=0
f 16a2−4b2−12b−9 11
=(4a)2−(4b2+12b+9) k= 4 k=7
=(4a)2−(2b+3)2 g m2+3m=4
=(4a−(2b+3))(4a+2b+3) m2+3m−4=0
=(4a−2b−3)(4a+2b+3) (m+4)(m−1)=0
g v2−4−w2+4w m+4=0orm−1=0
=v2−(w2−4w+4) m=−4 m=1
h 4n2=8−5n
=v2−(w−2)2
4n2+5n−8=0
=(v−(w−2))(v+(w−2))
=(v−w+2)(v+w−2) n=−5± (5)2−4×4×−8
h 4p2−1+4pq+q2 2×4
=4p2+4pq+q2−1 =−5± 153
8
=(2p+q)2−1
−5±3 17
=(2p+q−1)(2p+q+1) =
8
14 a 81y2=1 15 a 2x3+7x2+2x−3=0
81y2−1=0 Let P(x)=2x3+7x2+2x−3
(9y)2−12=0 P(−1)=2(−1)3+7(−1)2+2(−1)−3
(9y−1)[9y+1]=0 =−2+7−2−3
9y−1=0or9y+1=0 =0
1 1 Thus (x+1) is a factor.
y= y=−
9 9
Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual
TOPIC 1 Solving equations • EXERCISE 1.2 | 3
2x3+7x2+2x−3=0 If b3+5b2+2b−8=0
(x+1)(2x2+5x−3)=0 (b−1)(b+2)(b+4)=0
(x+1)(2x−1)(x+3)=0 b−1=0orb+2=0orb+4=0
b=1 b=−2 b=−4
x+1=0or2x−1=0orx+3=0
1 b −2m3+9m2−m−12=0
x=−1 x=2 x=−3 2m3−9m2+m+12=0
b l4−17l2+16=0 Let P(m)=2m3−9m2+m+12
(l2−1)(l2−16)=0 P(−1)=2(−1)3−9(−1)2−1+12
(l−1)(ll+1)( l−4)( +4)=0 =−2−9−1+12
l−1=0orl+1=0orl−4=0orl+4=0 =−12+12
l=1 l=−1 l=4 l=−4 =0
Thus m+1 is a factor.
c c3+3c2−4c−12=0 2m3−9m2+m+12=(m+1)(2m2−11m+12)
Let P(c)=c4+c3−10c2−4c+24
=(m+1)(2m−3)(m−4)
P(2)=(2)4+(2)3−10(2)2−4(2)+24
If 2m3−9m2+m+12=0
=16+8−40−8+24
(m+1)(2m−3)(m−4)=0
=48−48
m+1=0or2m−3=0orm−4=0
=0
3
Thus (c−2) is a factor. m=−1 m= m=4
c4+c3−10c2−4c+24=(c−2)(c3+3c2−4c−12) 2
c Let P(x)=2x3−x2−6x+3
Let Q(c)=c3+3c2−4c−12
1 13 12 1
Q(2)=23+3(2)2−4(2)−12 P =2  −  −6 +3
2 2 2 2
=8+12−8−12
1 1
=0 = − −3+3
4 4
Thus (c−2) is a factor.
=0
c3+3c2−4c−12=(c−2)(c2+5c+6) Thus (2x−1) is a factor.
=(c−2)(c+2)(c+3) 2x3−x2−6x+3=(2x−1)(x2−3)
Therefore c4+c3−10c2−4c+24=(c−2)2(c+2)(c+3) If 2x3−x2−6x+3=0
(c−2)2(c+2)(c+3)=0 (2x−1)(x2−3)=0
c−2=0or c+2=0orc+3=0
2x−1=0orx2−3=0
c=2 c=−2 c=−3
1
d p4−5p3+5p2+5p−6=0 x= x2=3
2
Let P(p)= p4−5p3+5p2+5p−6 x=± 3
P(1)=14−5(1)3+5(1)2+5(1)−6 17 a Let P(t)=3t3+22t2+37t+10
=1−5+5
P(−5)=3(−5)3+22(−5)2+37(−5)+10
=0
=−375+550−185+10
Thus (p−1) is a factor.
=−560+560
p4−5p3+5p2+5p−6=(p−1)(p3+4p2+p+8) =0
Let Q(p)= p3−4p2+p+6 Thus t+5 is a factor.
Q(2)=23−4(2)2+2+6 3t3+22t2+37t+10=(t+5)(3t2+7t+2)
=8−16+8 =(t+5)(t+2)(3t+1)
=0 If 3t3+22t2+37t+10=0
Thus (p−2) is a factor. (t+5)(t+2)(3t+1)=0
p3−4p2+p+6=(p−2)(p2−2p−3) t+5=0ort+2=0or3t+1=0
1
=(p−2)(p+1)(p−3) t=−5 t=−2 t=−
Therefore 3
p4−5p3+5p2+5p−6=(p−1)(p−2)(p+1)(p−3) b Let P(d)=3d3−16d2+12d+16
(p−1)(p−2)(p+1)(p−3)=0 P(2)=3(2)3−16(2)2+12(2)+16
p−1=0orp−2=0orp+1=0orp−3=0 =24−64+24+16
p=1 p=2 p=−1 p=3 =64−64
16 a Let P(b)=b3+5b2+2b−8 =0
P(1)=13+5(1)2+2(1)−8=8−8=0 Thus d−2 is a factor.
Thus b−1 is a factor. 3d3−16d2+12d+16=(d−2)(3d2−10d−8)
b3+5b2+2b−8=(b−1)(b2+6b+8) =(d−2)(d−4)(3d+2)
=(b−1)(b+2)(b+4)
Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual
4 | TOPIC 1 Solving equations • EXERCISE 1.2
If 3d3−16d2+12d+16=0
(d−2)(d−4)(3d+2)=0
d−2=0ord−4=0or3d+2=0
2
d=2 d=4 d=−
3
18 a a4−10a2+9=0
(a2−1)(a2−9)=0
(a−1)(a+1)(a−3)(a+3)=0
a−1=0ora+1=0ora−3=0ora+3=0
a=1 a=−1 a=3 a=−3
b 4k4−101k2+25=0
(4k2−1)(k2−25)=0
(2k−1)(2k+1)(k−5)(k+5)=0
2k−1=0or2k−1=0ork−5=0ork+5=0
1 1
k= k=− k=5 k=−5
2 2
c 9z4−145z2+16=0
(9z2−1)(z2−16)=0
(3z−1)(3z+1)(z−4)(z+4)=0
3z−1=0or3z+1=0orz−4=0orz+4=0
1 1
z= z=− z=4 z=−4
3 3
d (x2−2x)2−47(x2−2x)−48=0
Let A=(x2−2x)
A2−47A−48=0
(A−48)(A+1)=0
(x2−2x−48)(x2−2x+1)=0
(x−8)(x+6)(x−1)2=0
x−8=0orx+6=0orx−1=0
x=8 x=−6 x=1
19 a 5z3−3z2+4z−1≡az3+bz2+cz+d
a=5, b=−3, c=4andd=−1
b x3−6x2+9x−1≡x(x+a)2−b
x3−6x2+9x−1≡x(x2+2ax+a2)−b
x3−6x2+9x−1≡x3+2ax2+a2x−b
2a=−6 b=1
a=−3
20 2x3−5x2+5x−5≡a(x−1)3+b(x−1)2+c(x−1)+d
≡a(x3−3x2+3x−1)+b(x2−2x+1)+cx−c+d
≡ax3−3ax2+3ax−a+bx2−2bx+b+cx−c+d
≡ax3+(−3a+b)x2+(3a−2b+c)x+(−a+b−c+d)
Equating coefficients
a=2 −3a+b=−5 3a−2b+c=5 −a+b−c+d=−5
−3(2)+b=−5 3(2)−2(1)+c=5 −2+1−1+d=−5
−6+b=−5 6−2+c=5 −2+d=−5
b=1 4+c=5 d=−3
c=1
Thus 2x3−5x2+5x−5≡2(x−1)3+(x−1)2+(x−1)−3
21 a k x2−3x+k=0 has no solutions if the discriminant is less than zero.
∆<0
(−3)2−4(k)(k)<0
9−4k2<0
(3−2k)(3+2k)<0
Maths Quest 12 Mathematical Methods VCE Units 3 and 4 Solutions Manual