Cambridge Senior Maths Australian curriculum/VCE: Mathematical Methods 3 & 4 - Solutions Manual

844 Pages·2015·25.779 MB·English

by  Michael Evans, David Greenwood, Kay Lipson, Peter Jones

Chapter 1 – Functions and relations
Solutions to Exercise 1A
1 a 8,11 c X Y = 2,3,5,7,9,11,15,19,23
{ } ∪ { }
b 8,11 d X Y = 2,3,5,11
{ } \ { }
c 1,3,8,11,18,22,23,24,25,30 e Z Y = 2
{ } \ { }
d 3,8,11,18,22,23,24,25,30,32 f X Z = 2,7,9
{ } ∩ { }
e 3,8,11,18,22,23,24,25,30,32 g [ 2,8] X = 2,3,5,7
{ } − ∩ { }
f 1,8,11,25,30 h ( 3,8] Y = 7
{ } − ∩ { }
i (2, ) Y = 7,9,15,19,23
∞ ∩ { }
2 a 3,18,22,23,24
{ }
j (3, ) Y = (3, )
∞ ∪ ∞
b 25,30,32
{ }
c 3,18,22,23,24 5 a X Y = a,e
{ } ∩ { }
d 1,25,30 b X Y = a,b,c,d,e,i,o,u
{ } ∪ { }
c X Y = b,c,d
\ { }
3 a
−4−3−2−1 0 1 2 3 4 5 d Y X = i,o,u
\ { }
b
−4−3−2−1 0 1 2 3 4 5 6 a B C = 6
∩ { }
c
b B C = 2,4,8,10
−4−3−2−1 0 1 2 3 4 5 \ { }
c A B = 1,3,5,7,9
d
\ { }
−4−3−2−1 0 1 2 3 4 5
d A B = 1,3,5,7,9
\ { }
e A C = 2,4,5,7,8,10
\ { }
−4−3−2−1 0 1 2 3 4 5 (A B) (A C) = 1,2,
\ ∪ \ {
3,4,5,7,8,9,10
f }
−4−3−2−1 0 1 2 3 4 5
e B C = 6
∩ { }
A (B C) = 1,2,3,4,5,7,8,9,10
\ ∩ { }
4 a X Y = 7,9
∩ { }
f A B = 1,3,5,7,9
\ { }
b X Y Z = 7,9 A C = 2,4,5,7,8,10
∩ ∩ { } \ { }
1
Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
(A B) (A C) = 5,7 9 a
\ ∩ \ { }
−3−2−1 0 1 2
g B C = 1,2,3,4,6,8,9,10
∪ { }
A (B C) = 5,7 b
\ ∪ { }
−4 −3−2−1 0 1 2 3
h A B C = 6
∩ ∩ { }
c
7 a [ 3,1) 0 1 2 3
−
b ( 4,5] d
−
−4 −3−2−1 0 1
c ( √2,0)
−
e
1
d ( , √3)
−4 −3−2−1 0
− √2
e ( , 3) f
−∞ −
−2−1 0 1 2 3 4 5
f (0, )
∞
g ( ,0) 10 a
−∞
h [ 2, ) −3−2−1 0 1 2 3 4 5 6
− ∞
b
8 a ( 2,3)
−
b [ 4,1) −3−2−1 0 1 2 3 4 5 6
−
c [ 1,5] c
−
d ( 3,2] −2−1 0 1 2 3 4 5 6
−
d
−8 −6 −4 −2 0
2
Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
Solutions to Exercise 1B
1 a Domain = R c y
range = [ 2, )
− ∞
2
b Domain = ( ,2]
−∞
x
range = R
0
8
Domain = R+ 0
c Domain = ( 2,3)
∪{ }
− range = ( ,2]
range = [0,9) −∞
d y
d Domain = ( 3,1)
−
range = ( 6,2)
− (4, 4)
(1, 2)
e Domain = [ 4,0]
− x
0
range = [0,4]
Domain = [0, )
∞
range = [0, )
f Domain = R ∞
range = ( ,2) e y
−∞
2 a y 5
x
0
5
1 Domain = [0,5]
x
0
range = [0,5]
Domain = R
range = [1, ) f y
∞
b y (4, 18)
2
3 x
0
4
x Domain = [0,4]
−3 0 3
range = [2,18]
−3
Domain = [ 3,3]
−
range = [ 3,3]
−
3
Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
g y range = 4
{ }
(2, 4)
0 4 a functionDomain = R
x
range = 4
−1 −22 { }
−5
b notafunction
(−1, −5)
Domain = 2
Domain = [ 1,2]
{ }
− range = Z
range = [ 5,4]
−
c function
h y
Domain = R
range = R
4
d notafunction
−2 2
x Domain = R
0
Domain = R range = R
range = ( ,4]
−∞ e notafunction
i y Domain = [ 4,4]
−
range = [ 4,4]
−
1
5 f(x) = 2x2 +4x;
x
0
1 g(x) = 2x3 +2x 6
−
Domain = R
a f( 1) = 2( 1)2 +4( 1) = 2
Range = R − − − −
f(2) = 2(2)2 +4(2) = 16
f( 3) = 2( 3)2 +4( 3) = 6
3 a notafunction − − −
f(2a) = 2(2a)2 +4(2a) = 8a2 +8a
Domain = 1,1,2,3
{− }
range = 1,2,3,4
{ } b
g( 1) = 2( 1)3 +2( 1) 6 = 10
b function
− − − − −
Domain = 2, 1,0,1,2 g(2) = 2(2)3 +2(2) 6 = 14
{− − } −
range = 4, 1,0,3,5 g(3) = 2(3)3 +2(3) 6 = 54
{− − }
−
c notafunction g(a 1) = 2(a 1)3 +2(a 1) 6
− − − −
Domain = 2, 1,2,4 = 2(a3 3a2 +3a 1)+2a 8
{− − }
− − −
range = 2,1,2,4,6
{− } = 2a3 6a2 +8a 10
− −
d function
Domain = 1,0,1,2,3 6 g(x) = 3x2 2
{− }
−
4
Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
a g( 2) = 3( 2)2 2 = 10 c 3x 2 = 0
− − − −
g(4) = 3(4)2 2 = 46 2
− x =
3
b i g( 2) = 3( 2)2 2 = 12x2 2
− − − −
9 a f : R Rwhere f(x) = 2x+3
ii g(x 2)2 = 3(x 2)2 2 = →
− − −
3x2 12x+10
b 3y+4x = 12
−
iii g(x+2)2 = 3(x+2)2 2 = 3y = 12 4x
−
−
3x2 +12x+10 4x
y = 4
− 3
iv g(x2) = 3(x2)2 2 = 3x4 2
− − 4x
f :R Rwhere f(x) = − +4
→ 3
7 f(x) = 2x 3
− c f :[0, ) Rwhere f(x) = 2x 3
∞ → −
a f(3) = 2(3) 3 = 3
− d f :R Rwhere f(x) = x2 9
→ −
b f(x) = 11
e f :[0,2] Rwhere f(x) = 5x 3
→ −
11 = 2x 3x = 7
−
c f(x) = 4x 10 a y
4x = 2x 3 (6, 7)
−
2x = 3 (2, 3)
−
x
3 0
x = −
2
range = [3, )
d f(x) > x
∞
2x 3 > x b y
−
x > 3
8 g(x) = 6x+7 h(x) = 3x 2 x
− 0 (2, −1)
a 6x+7 = 3x 2
−
3x = 9 range = ( , 1]
− −∞ −
x = 3
−
b 6x+7 > 3x 2
−
3x > 9
−
x > 3
−
5
Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
c y h y
(4, 19)
1
1
−
2 0
x x
0 −1 0.2
(−2, −11)
(−4, −7) range = ( 11,19)
−
range = [ 7, )
− ∞
11 f(x) = 2x2 6x+1; g(x) = 3 2x
d y − −
a f(2) = 2(2)2 6(2)+1 = 3
(3, 11) − −
2
− f( 3) = 2( 3)2 6( 3)+1 = 37
3 2 − − − −
x
f( 2) = 2( 2)2 6( 2)+1 = 21
0
− − − −
b g( 2) = 3 2( 2) = 7
range = ( ,11) − − −
−∞ g(1) = 3 2(1) = 1
−
e y g( 3) = 3 2( 3) = 9
− − −
(3, 4)
c i f(a) = 2a2 6a+1
−
−1 1
x ii f(a+2) = 2(a+2)2
0
6(a+2)+1
−
range = ( ,4] = 2a2 +2a 3
−∞ −
f y iii g( a) = 3+2a
−
(6, 17)
iv g(2a) = 3 4a
1
−
3
x v f(5 a) = 2(5 a)2
−1 0 − −
6(5 a)+1
(−2, −7)
− −
range = [ 7,17] = 2a2 14a+21
−
−
g y vi f(2a) = 8a2 12a+1
−
(−5, 14)
vii g(a)+ f(a) = (2a2 6a+1)
(−1, 2) −
+(3 2a)
x −
0 = 2a2 8a+4
−
range = [2,14]
6
Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
viii g(a)− f(a) = (3−2a) d f(x) > 0
−(2a2 −6a+1) 3x2 + x−2 > x
2
from(a),the x-interceptsare −1,
= −2a2 +4a+2 3
asthe coefficientof x2 > 0
theshapeof the graphy = f(x)is
12 f(x) = 3x2 + x−2
a f(x) = 0
∴ f(x) > 0for
3x2 + x−2 = 0 2
x ∈ (−∞,−1)∪( ,∞)
usingthequadraticformula 3
(cid:112)
−(1)± (1)2 −4(3)(−2)
x =
e f(x) > x
2(3)
√
−1± 25 3x2 + x−2 > x
x =
6 3x2 −2 > 0
2
x = −1, 2
3 x2 >
i(cid:40)nsetn(cid:41)otation  √ 3  √ 
−1, 2 x ∈ −∞, −√ 2∪ √2,∞
3 3 3
b f(x) = x f f(x) ≤ −2
3x2 + x−2 ≤ −2
3x2 + x−2 = x 1
from(c),the x-interceptsare − ,0
3x2 = 2 3
asasthe coefficientof x2 > 0
x2 = 2 theshapeof the graphy = f(x)is
3
(cid:112)
x = ± 2/3
insetnotation
 (cid:114) (cid:114)  (cid:34) (cid:35)
− 2, 2 ∴ f(x) ≤ −2for x ∈ −1,0
 3 3 3
c f(x) = −2 13 f(x) = x2 + x
3x2 + x−2 = −2
a f(−2) = (−2)2 +(−2) = 2
3x2 + x = 0
x(3x+1) = 0 b f(2) = (2)2 +(2) = 6
∴ either x = 0or 3x+1 = 0 c f(−a) = (−a)2 +(−a) = a2 −a
−1
x = 0,
3 d f(a)+ f(−a) = (a2 +a)+(a2 −a)
insetnotation
(cid:40) −1(cid:41) = 2a2
0,
3
7
Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
e f(a)− f(−a) = (a2 +a)−(a2 −a) f 1 = 6
g(x)
= 2a
1 = 6g(x)
f f(a2) = (a2)2 +(a2) = a4 +a2 1 = 6(3x−2)
1 = 18x−12
14 g(x) = 3x−2 18x = 13
a g(x) = 4 x = 13
18
3x−2 = 4
x = 2
15 a f(x) = kx−1
b g(x) > 4 3 = 3k−1
3x−2 > 4 k = 4
3
x > 2
in setnotation b f(x) = x2 −k
{x:x > 2}
3 = 9−k
c g(x) = a k = 6
3x−2 = a
c f(x) = x2 +kx+1
a+2
x = 3 = 9+3k+1
3
−7
d g(−x) = 6 k =
3
−3x−2 = 6
k
d f(x) =
x = −8 x
3 k
3 =
3
e g(2x) = 4
k = 9
6x−2 = 4
x = 1 e f(x) = kx2
3 = 9k
1
k =
3
8
Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
f f(x) = 1−kx2 c 1 = 9
x2
3 = 1−9k
1
x = ±
9k = −2 3
−2
k = d x = 1 = 2
9 x
x2 −2x+1 = 0
16 a 5x−4 = 2 (x−1)2 = 0
6
x = x = 1
5
e (x+1)(x−2) = 2
1
b = 5 ∴ either x+1 = 0or x−2 = 0
x
x = −1 x = 2
1
x = ∴ x = −1,2
5
9
Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
Solutions to Exercise 1C
1 a Thefunctionswhichare one-to- 6 a Domain: R range: R
oneare bandc
b Domain: R+ 0 range: R+ 0
∪{ } ∪{ }
2 a Thefunctionswhichare one-to- c Domain: R range: [ 2, )
− ∞
oneare b,dandf
d Domain: [ 4,4] range: [0,4]
−
3 a Thegraphsoffunctionsarei,iii,iv, e Domain: R 0 range: R 0
\{ } \{ }
vi,vii,andviii.
f Domain: R range: ( ,4]
−∞
b Thegraphsofone-to- onefunctions
g Domain[3, ) range: [0, )
areiii,andvii.
∞ ∞
4 y2 = x+2,x 2 7 a Domain: R range: R
≥ −
y = √x+2 b Domain: R range: [ 2, )
− ∞
±
c Domain[-3,3] range: [0,3]
twopossiblefunctions f andgare
f :[ 2, ) R f(x) = √x+2 d Domain: R 1 range: R 0
\{ } \{ }
− ∞ →
rangeof f :[0, ) = R+ 0
∞ ∪{ } 8 a R 3
\{ }
g:[ 2, ) Rg(x) = √x 2
− ∞ → − − b ( , √3] [√3, )
−∞ − ∪ ∞
rangeofg:( ,0] = R 0
−
−∞ ∪{ } c R
d [4,11]
5 a y
e R 1
\{− }
f h(x) = √(x+1)(x 2)
−
Domain: ( , 1] [2, )
2
−∞ − ∪ ∞
x
0
g R 1,2
\{− }
b twopossiblefunctionsarethe right
h Domain: ( , 2) [1, )
half −∞ − ∪ ∞
g :[0, ) R g (x) = x2 +2 1
1 1 i f(x) = √x(1 3x)Domain: 0,
∞ →
andthelefthalf − 3
(cid:31) (cid:30)
g :( ,0) R g (x) = x2 +2
2 2 j [ 5,5]
−∞ →
−
10
Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.