BUKU TEKS ADDMATHS FORM 4 (ANSWERS)

HAK MILIK KPM
Complete Solution Manual

ADD MATH TEXTBOOK KSSM (DLP)
FORM 4

DOWNLOADED & COMPILED BY
CIKGU AIZUDDIN YUSOFF

CHAPTER 1 FUNCTIONS

Mind Challenge (Page 4)
The number of x-intercepts has no limits and the number of y-intercepts is at most 1.

Self Practice 1.1 (Page 5 & 6)

1. (a) Function because each object has only one image even though element 7 does
not have an object.

(b) Function because each object has only one image even though image 4 has two objects.
(c) Not a function because object r has two images, 8 and 10.

2. (a) y (b) y Vertical (c) y
0 line test
Vertical 1 Vertical
line test 02 line test

x0x x

Function because the Not a function because Function because the vertical
vertical line test cuts only the vertical line test cuts line test cuts only one point
1 point on the graph. more than 1 point on the on the graph except at x = 2
graph. where it does not cut any
point.

3. (a) h : x ˜ 1 , x ≠ 0 (b) h : x ˜  x  (c) h : x ˜ x3
x

Inquiry 1 (Page 6 & 7)

1. y
100
y y
0 10 20
90 240 Time (seconds)
72 180
54 120
36 60
18
0 1234 x
0x Number of packets of popcorn
Total price (RM)12345 x
Eaten calories
Distance from Hilal'sNumber of tickets

house (km)

Situation I Situation II Situation III

1

2. (a) The graph for Situation I is a discrete graph while the graph for Situation II and
Situation III is a continuous graph.

(b)

Situation I Situation II Situation III

Domain {1, 2, 3, 4, 5} 1<x<4 0 < x < 20

Range {18, 36, 54, 72, 90} 60 < y < 240 0 < y < 100

Self Practice 1.2 (Page 9)

1. (a) Domain = {–2, –1, 0, 2, 4}, codomain = {1, 3, 4, 5}, range = {1, 3, 4, 5}
(b) Domain = {j, k, l, m}, codomain = {2, 3, 6, 7, 10}, range = {3, 7}
(c) The domain of f is –3 < x < 5, while the codomain and range of f is 2 < f(x) < 6

2. (a)

x –2 –1 0 1 2 3 4
f(x) =  x + 1  1 0 1 2 3 4 5
(–2, 1) (–1, 0) (0, 1) (1, 2) (2, 3) (3, 4) (4, 5)
(x, y)

f(x)
f(x) = |x + 1|

5

1 4x
–2 –1 0

Thus, range of f is 0 < f(x) < 5.
(b)

x –2 –1 0 1 2 3 4
f(x) =  4 – 2x  8 6 4 2 0 2 4
(0, 4) (1, 2) (2, 0) (3, 2) (4, 4)
(x, y) (–2, 8) (–1, 6)

f(x)
8

f(x) = |4 – 2x|
4

–2 0 2 4 x
Thus, range of f is 0 < f(x) < 8.

2

(c) –2 –1 0 1 2 3 4
x 9 7 5 3 1 1 3
(–2, 9) (–1, 7) (0, 5) (1, 3) (2, 1) (3, 1) (4, 3)
f(x) =  2x – 5 
(x, y)

f(x)
9

5
f(x)=|2x – 5|

3

–2 0 52– 4 x

Thus, range of f is 0 < f(x) < 9.

Mind Challenge (Page 9) 5 2
x +
If x = 0, the function f : x = 3x + is undefined. If f(x) = x 3 , x ≠ k, then the value of k is –3.

Self Practice 1.3 (Page 10)

1. (a) g(x) = 3 + x 6 1
–

g(–5) = 3 + 6
–5 – 1
=2

g(–2) = 3 + 6
–2 – 1
=1

( )g1 =3+ 1 6
2 2
– 1

= –9

(b) g(b) = 2b

3 + b 6 1 = 2b
–

3(b – 1) + 6 = 2b(b – 1)

3b – 3 + 6 = 2b2 – 2b

2b2 – 5b – 3 = 0

(2b + 1)(b – 3) = 0

2b + 1 = 0 or b – 3 = 0
= – 21 or
b b=3

Thus, the possible values for b are – 1 and 3.
2

3

2. (a) h(x) = kx – 3
x–1

h(2) = 5

22k – 3 = 5
– 1

2k – 3 = 5

2k = 8

k=4

(b) h(x) = kx – 3
x–1

h(3) = k

33k – 3 = k
– 1

3k – 3 = 2k

k=3

(c) h(x) = kx – 3
x–1

h(k) = k

k2 – 3 =k
k – 1

k2 – 3 = k2 – k

k=3

3. (a) f(x) =  4x – 3 

f(–2) =  4(–2) – 3 

=  –8 – 3 

=  –11 

= 11

( f – 12 ) = ( 4 – 12 ) – 3 
=  –2 – 3 
=  –5 

=5

(b) f(x) = 1

 4x – 3  = 1

4x – 3 = –1   or 4x – 3 = 1
or 4x = 4
4x = 2 x=1
or
x= 2 4
4

= 1
2

(c) f(x) , 1

 4x – 3  , 1

–1 , 4x – 3 , 1

2 , 4x , 4

1 , x , 1
2

(d) f(x) . 5    4x – 3 . 5
 4x – 3  . 5 4x . 8
or x . 2
4x – 3 , –5
6 – 2x = x
4x , –2 3x = 6
x , – 21 x=2
4. g(x) =  6 – 2x 
g(x) = x
 6 – 2x  = x

6 – 2x = –x or



5. (a) f(x) = mx + c

f(2) = 7

2m + c = 7…1

f(4) = –1

4m + c = –1…2

2 – 1: 2m = –8

m = –4

Substitute m = – 4 into 1: 2(– 4) + c = 7

–8 + c = 7

c = 15

Thus, m = – 4 and c = 15.

(b) f(x) = – 4x + 15 (c) f(x) = x

f(2) = –4(2) + 15 –  4x + 15 = x

= –8 + 15 5x = 15

= 7 x = 3

Intensive Practice 1.1 (Page 11)

1. (a) This relation is a function because each object has one image.
(b) This relation is not a function because object – 4 has two images.
(c) This relation is a function because each object has 1 image.

2.

(a) y (b) y Vertical (c) y

Vertical line test

line test 0x Vertical
line test
0x
x
02

This graph is not a function This graph is not a function The graph is a function
because the vertical line test because the vertical line test because the vertical line test
cuts three points on the cuts two points on the only cuts one point on the
graph. graph. graph except on the line
x = 2.
5

3. (a) This relation is a function because each object has only one image.

(b) Domain = {–7, –6, 6, 7}
Range = {36, 49}

(c) f : x ˜ x2

4. (a) When x = 5, f(5) =  2(5) – 4 
=  10 – 4 
=  6 

=6

Thus, the value of t = 6.
(b) Range of f is 0 < f(x) < 6.

(c) f(x) < 4
 2x – 4  < 4

– 4 < 2x – 4 < 4 0 < 2x
<8
0<x<4

Thus, the range of values of x is 0 < x < 4.

5. (a) (i) H(t) = 81 – 9t2

( ) ( ) H1= 1 2
3 81 – 9 3

= 81 – 1

= 80 metres

(ii) H(1) = 81 – 9(1)2

= 81 – 9

= 72 metres

(iii) H(2) = 81 – 9(2)2

= 81 – 36

= 45 metres

(b) The stone reaches the ground when H(t) = 0

H(t) = 0

81 – 9t2 = 0

9t2 – 81 = 0

(3t – 9)(3t + 9) = 0

3t – 9 = 0   or 3t + 9 = 0

3t = 9 3t = –9

t = 3 t = –3 (Ignore)

Thus, the stone reaches the ground when t = 3 seconds.

Inquiry 2 (Page 12)

5. The function f [g(x)] is obtained by substituting function g into the function f.
f(x) = x + 2

f [g(x)] = f(x2)
= x2 + 2

6. The graph of f [g(x)] is a parabola that has a minimum point.

6

9. The graph of g[f(x)] is obtained by substituting function f into the function g.
g(x) = x2

g[f(x)] = g(x + 2)
= (x + 2)2
= x2 + 4x + 4

10. The graph of g[f(x)] is a parabola that has a minimum point.

Mind Challenge (Page 14)


g[f(x)] = g(x)
= x2 = x2
Thus, the function fg and gf are not necessarily always different.

Self Practice 1.4 (Page 14)

1. (a) f(x) = 3x
(b) gf(x) = 2x – 7

2. (a) f(x) = 3x, g(x) = 3 – x
fg(x) = f(3 – x)
= 3(3 – x)
= 9 – 3x
fg: x ˜ 9 – 3x
gf(x) = g(3x)
= 3 – 3x
gf: x ˜ 3 – 3x
f 2 = ff(x)
= f(3x)
= 3(3x)
= 9x
f 2: x ˜ 9x
g2 = gg(x)
= g(3 – x)
= 3 – (3 – x)
=x
g2: x ˜ x

(b) f(x) = 4 + 2x, g(x) = x2
fg(x) = f(x2)
= 4 + 2(x2)
= 4 + 2x2
fg: x ˜ 4 + 2x2
gf(x) = g(4 + 2x)
= (4 + 2x)2
= 16 + 16x + 4x2
= 4x2 + 16x + 16
gf: x ˜ 4x2 + 16x + 16

7

f 2 = ff(x)

= f(4 + 2x)

= 4 + 2(4 + 2x)

= 4 + 8 + 4x

= 4x + 12

f 2: x ˜ 4x + 12

g2(x) = gg(x)

= g(x2)

= (x2)2

= x4

g2: x ˜ x4 6
x
(c) f(x) = x + 4, g(x) = , x ≠ 0

( )fg(x) = f  6
x
6
= x + 4, x ≠ 0

fg: x ˜ 6 + 4, x ≠ 0
x

gf(x) = g(x + 4)

= x 6 4 , x ≠ – 4
+

gf: x ˜ x 6 4, x ≠ – 4
+

f 2 = ff(x)

= f(x + 4)

= x+4+4

= x+8

f 2: x ˜ x + 8

g2 = gg(x)

= g( 6 )
x

= 6
6
x
=x

g2: x ˜ x 1
–
(d) f(x) = x – 5, g(x) = x 1, x ≠ 1

( ) fg(x)= f x 1 1
–

= x 1 1 – 5
–

= 1 – 5x + 5
x–1

= 6 – 5x , x ≠ 1
x–1

fg: x ˜ 6 – 5x , x ≠ 1
x–1

8

gf(x) = g(x – 5)

= 1
x–5–1

= x 1 6, x ≠ 6
–

gf: x ˜ x 1 6, x ≠ 6
–

f 2 = ff(x)

= f(x – 5)

= x–5–5

= x – 10

f 2: x ˜ x – 10

g2 = gg(x)

= g(x 1 1 )
–

= 1 1
–
x 1 – 1

= 1
1–x+1
x–1

= x – 1 , x ≠ 2
2 – x

g 2: x ˜ x – 1 , x ≠ 2
2 – x

3. f(x) = 3x + 4, g(x) = x2 + 6

fg(x) = f(x2 + 6)

= 3(x2 + 6) + 4

= 3x2 + 18 + 4

= 3x2 + 22

gf(x) = g(3x + 4)

= (3x + 4)2 + 6

= 9x2 + 24x + 16 + 6

= 9x2 + 24x + 22

(a) f = g

3x + 4 = x2 + 6

x2 – 3x + 2 = 0

(x – 1)(x – 2) = 0

x – 1 = 0 or x – 2 = 0

x = 1 or x=2

(b) fg = gf

3x2 + 22 = 9x2 + 24x + 22

6x2 + 24x = 0

6x(x + 4) = 0

6x = 0 or x + 4 = 0

x = –4

9

4. Given f(x) = ax + b, f 2(x) = 4x – 9
f 2(x) = ff(x)

4x – 9 = f(ax + b)
4x – 9 = a(ax + b) + b
4x – 9 = a2x + ab + b
Equate the coefficients:
a2 = 4
a = ±!w4

= –2 or 2
When a = –2, –2b + b = –9

b=9
When a = 2, 2b + b = –9

3b = –9
b = –3
Thus, a = –2, b = 9 or a = 2, b = –3.

5. f(x) = 3x + k, g(x) = 2h – 3x
fg = gf

f(2h – 3x) = g(3x + k)
3(2h – 3x) + k = 2h – 3(3x + k)

6h – 9x + k = 2h – 9x – 3k
4h = –4k
h = –k

Self Practice 1.5 (Page 15)

1. (a) f(x) = 2x + 1, g(x) = x x 1, x ≠ 1
–
( )fg(x) x
= f x – 1

( )=2 x x 1 +1
–
2x + x –1
= x–1

= 3x – 1 , x ≠ 1
x–1

fg(3) = 3(3) – 1
3–1

= 8
2

=4

(b) f(x) = 5x + 6, g(x) = 2x – 1

gf(x) = g(5x + 6)

= 2(5x + 6) – 1

= 10x + 12 – 1

= 10x + 11

( ) ( ) gf – 51 = 10 – 51 + 11

= –2 + 11

=9

10

(c) f(x) = x + 1 , x ≠ 3, g(x) = x 6 2, x ≠ 2
x – 3 –

x+1
x–3
( )f 2(x) =f

= x+1 +1
x–3
x+1
x–3 –3

= x+1+x–3 ÷ x + 1 – 3x + 9
x–3 x–3

= 2x – 2 × x–3
x–3 10 – 2x

= 2x – 2
10 – 2x

= x – 1 , x ≠ 5
5 – x

f 2(4) = 4–1
5–4

=3 6
x–2
( )g2(x) = g

= 6 –2
6
x–2

= 6 ÷ 6 – 2x + 4
x–2

= 6 × x–2
10 – 2x

= 6x – 12
10 – 2x

= 3x – 6 , x ≠ 5
5–x

( ) g 21 = 3( 1 ) – 6
2 2
1
5 – ( 2 )

= –9 ÷ 9
2 2

= –1 2
–
(d) f(x) = x2 – 4, g(x) = x 2 , x ≠ 2

f 2(x) = f(x2 – 4)

= (x2 – 4)2 – 4

= x4 – 8x2 + 16 – 4

= x4 – 8x2 + 12

f 2(–1) = (–1)4 – 8(–1)2 + 12

= 1 – 8 + 12

=5

11

2
x–2
( )g2(x) = g

= 2 –2
2
x–2

= 2 ÷ 2 – 2x + 4
x–2

= 2 × x–2
6 – 2x

= x – 2 , x ≠ 3
3 – x

g2(1) = 1–2
3–1

= – 12

2. (a) f(x) = 2x – 5, g(x) = 10 , x ≠ 0
x
fg(x) = 5

f( 10 ) = 5
x

( ) 210 –5=5
x

20 – 5x = 5x

10x = 20

x = 2

(b) f(x) = x2 – 1, g(x) = 2x + 1

gf(x) = 7

g(x2 – 1) = 7

2(x2 – 1) + 1 = 7

2x2 – 2 + 1 = 7

2x2 = 8

x2 = 4
x = ±!w4

x = –2, x = 2

(c) f(x) = 3x – 2, f 2(x) = 10

f 2(x) = 10

f(3x – 2) = 10

3(3x – 2) – 2 = 10

9x – 6 – 2 = 10

9x = 18

x=2

12

(d) g(x) = x 2 2 , x ≠ 2, g2(x) = – 1
– 2
1
g2(x) = –  2

2 1
x–2 2
( )g = –

2 2 = –  1
– 2
x 2 – 2

2 ÷ 2 – 2x + 4 =– 1
x–2 2

2× x–2 = – 1
6 – 2x 2

x–2 = – 1
3–x 2

2(x – 2) = –(3 – x)

2x – 4 = –3 + x

x=1

Self Practice 1.6 (Page 16)

1. (a) Given f(x) = x – 3, fg(x) = 2x2 – 4x + 7
fg(x) = 2x2 – 4x + 7
g(x) – 3 = 2x2 – 4x + 7
g(x) = 2x2 – 4x + 10
g: x ˜ 2x2 – 4x + 10

(b) Given f(x) = x2 + 1, fg(x) = x2 + 4x + 5
fg(x) = x2 + 4x + 5

[g(x)]2 + 1 = x2 + 4x + 5
[g(x)]2 = x2 + 4x + 4
[g(x)]2 = (x + 2)2
g(x) = x + 2

g: x ˜ x + 2

2. (a) Given f(x) = x + 1, gf(x) = x2 – 2x – 3
gf(x) = x2 – 2x – 3

g(x + 1) = x2 – 2x – 3
Let y = x + 1

x=y–1
g(y) = (y – 1)2 – 2(y – 1) – 3

= y2 – 2y + 1 – 2y + 2 – 3
= y2 – 4y
Thus, g(x) = x2 – 4x or g: x ˜ x2 – 4x
(b) Given f(x) = x2 + 3, gf(x) = 2x2 + 3
gf(x) = 2x2 + 3
g(x2 + 3) = 2x2 + 3

Let y = x2 + 3
x2 = y – 3
x = !wy – 3

13

g(y) = 2(!wy – 3)2 + 3
= 2(y – 3) + 3
= 2y – 6 + 3
= 2y – 3

Thus, g(x) = 2x – 3 or g : x → 2x – 3

3. (a) Given h(x) = 8 , hg(x) = 4x
x
hg(x) = 4x

8 = 4x
g(x)
8
g(x) = 4x

= 2 , x ≠ 0
x

g: x ˜ 2 , x ≠ 0
x

(b) gh(x) = 6

g( 8 ) = 6
x

( 2 ) = 6
8
x

x =6
4

x = 24

4. (a) Given g(x) = 3x, fg(x) = 9x – 7

fg(x) = 9x – 7

f(3x) = 9x – 7

Let y = 3x
y
x= 3

( )f(y) = 9 y –7
3

= 3y – 7

Thus, f(x) = 3x – 7

(b) gf(x) = g(3x – 7)
= 3(3x – 7)
= 9x – 21

gf(2) = 9(2) – 21
= 18 – 21
= –3

14

Self Practice 1.7 (Page 18)
x
1. (a) Given f(x) = x + 1 , x ≠ –1

f 2(x) = f [ f(x)] f 3(x) = f [f 2(x)] f 4(x) = f [f 3(x)]

= f( x x 1 ) = f( 2x x 1 ) = f( 3x x 1 )
+ + +

x ( 2x x 1 ) ( 3x x )
x+1 + + 1
= x = (x = (x
x+1 1) 1)
+1 2x + 1 + 3x + 1 +

= x ÷ x+x+1 = x 1 ÷ x + 2x + 1 = x 1 ÷ x + 3x + 1
x+1 x+1 2x + 2x + 1 3x + 3x + 1

= x × x+1 = x 1 × 2x + 1 = x 1 × 3x + 1
x+1 2x + 1 2x + 3x + 1 3x + 4x + 1

= x 1, x ≠ – 1 = 3x x 1 , x ≠ – 1 = 4x x 1 , x ≠ – 1
2x + 2 + 3 + 4

(b) f 20(x) = x 1 , x ≠ – 210
20x +

f 23(x) = x 1 , x ≠ – 213
23x +

2. (a) Given f(x) = 1 , x ≠ 0
x

f 2(x) = f [f(x)] f 3(x) = f [f 2(x)] f 4(x) = f [f 3(x)]

= f( 1 ) = f(x) = f( 1 )
x x
= 1 , x ≠ 0
1 x 1
= =
( 1 ) ( 1 )
=x x =x x

(b) f 40(2) = 2

f 43(2) = 1
2

3. (a) Ar(t) = A[r(t)]

( )= A 2 t3
3

( )= 4π 2 t3 2
3

( )= 4π 4 t6
9

= 16 πt6
9

(b) When t = 2,

A(2) = 16 π(26)
9

= 113 7 π m2
9

15

4. (a) (i) v(t) = 200 + 100t

(ii) v = πr 2h

h = v
πr2

(iii) hv(t) = h(200 + 100t)

= 200 + 100t
πr2

AWpahbeinlarajedjiaursi, r = 20 cm,

h v(t) = 200 + 100t
π(20)2

= 2+t
4π

(b) When t = 20,

h(20) = 2 + 20
4π

= 11
2π

= 1.75 cm

5. (a) r(t) = 3 × t
= 3t

(b) Ar(t) is the area of the water ripple, in cm2, as a function of time, t, in seconds.
(c) A = πr2

Ar(t) = A[r(t)]

= π(3t)2
= 9 πt2

When t = 30,
Ar(30) = 9π(302)

= 8 100π cm2

Intensive Practice 1.2 (Page 19)

1. Given f(x) = 2x – 1, g(x) = x x 1 , x ≠ –1
+
x
( )(a) = +
fg(x) f x 1

( )=2 x x 1 –1
+

= 2x – x – 1
x+1

= x – 1 , x ≠ –1
x + 1

gf(x) = g(2x – 1)

= 2x – 1
2x – 1 + 1

= 2x – 1 , x ≠ 0
2x

16

(b) fg(2) = 2–1
2+1

= 1
3
( ) ( ( ) )gf 1
–  1 = 2 –  2 –1
2
2 –  1
2
=2

(c) fg = gf

x–1 = 2x – 1
x+1 2x

2x(x – 1) = (x + 1)(2x – 1)

2x2 – 2x = 2x2 – x + 2x – 1

3x = 1
1
x= 3

2. Given f(x) = x x 1 , x ≠ 1, g(x) = hx + k, g(3) = 8, gf(2) = 5
–

(a) g(3) = 8

3 h + k = 8…1

gf(2) = 5

g(2) = 5

2h + k = 5…2

1 – 2: h = 3

Substitute h = 3 into 1.

3(3) + k = 8

9+k=8

k = –1

Thus, h = 3 and k = –1.

(b) fg(a) = 3

f(3a – 1) = 3

3a – 1 = 3
3a – 1 – 1

3a – 1 = 3
3a – 2

3a – 1 = 9a – 6

6a = 5
5
a= 6

( )3. 1
Given f(x) = ax – b, g(x) = x + 4, fg(2) = 9, gf 2 =2
fg(2) = 9

f(2 + 4) = 9

f(6) = 9

6a – b = 9…1
gf ( 1 )
2 = 2

1( )g a– b =2
2

17

1 a – b + 4 = 2
2
a – 2b = –4

a = 2b – 4 …2

Substitute 2 into 1.

6(2b – 4) – b = 9

12b – 24 – b = 9

11b = 33

b=3

Substitute b = 3 into 2.

a = 2(3) – 4

=6–4

=2

Thus, a = 2 and b = 3.

4. Given f(x) = x 2 3, x ≠ 3, g(x) = hx2 + k
–

(a) Given g(2) = 5, gf(1) = –1

g(2) = 5

( 22)h + k = 5
4h + k = 5…1

gf(1) = –1

g(–1) = –1

h + k = –1…2
1 – 2: 3h = 6

h=2

Substitute h = 2 into 1.

4(2) + k = 5

k=5–8

= –3

Thus, h = 2 and k = –3.
2
x–3
( )(b) g
gf(x) =

2
x–3
( )=2 2 –3

4
x2 – 6x + 9
( )= 2 –3

= 8 – 3x2 + 18x – 27
x2 – 6x + 9

= –3x2 + 18x – 19 , x ≠ 3
(x – 3)2

5. Given f(x) = ax + b, f 3(x) = 27x + 13 18
(a) f 2(x) = f(ax + b)
= a(ax + b) + b
= a2x + ab + b
f 3(x) = f [f 2(x)]
= f(a2x + ab + b)
= a(a2x + ab + b) + b
= a3x + a2b + ab + b

By comparison,

a3x + a2b + ab + b = 27x + 13

a3 = 27 and a2b + ab + b = 13

a=3 9b + 3b + b = 13

13b = 13

b=1

Thus, a = 3 and b = 1.

(b) f 4(x) = f [f 3(x)]

= f(27x + 13)

= 3(27x + 13) + 1

= 81x + 39 + 1

= 81x + 40

6. (a) A(x) = x2
V(A) = 10A

(b) VA(x) = V[A(x)]
= 10x2

= 10A

7. Given f(x) = x + 6

(a) fg(x) = 2x2 – 3x – 7

g(x) + 6 = 2x2 – 3x – 7

g(x) = 2x2 – 3x – 13

(b) gf(x) = x2 + 4

g(x + 6) = x2 + 4

Let y = x + 6

x=y–6

g(y) = (y – 6)2 + 4

= y2 – 12y + 36 + 4

= y2 – 12y + 40

Thus, g(x) = x2 – 12x + 40

(c) gf(x) = 8 – x

g(x + 6) = 8 – x

Let y = x + 6

x=y–6

g(y) = 8 – (y – 6)

= 14 – y

Thus, g(x) = 14 – x

8. (a) g : x ˜ x –1
3
x–1
3
( )(b) f = x2 – 3x + 6

Let y = x –1
3
x = 3y + 1

f(y) = (3y + 1)2 – 3(3y + 1) + 6

= 9y2 + 6y + 1 – 9y – 3 + 6

= 9y2 – 3y + 4

19

Substitute y with x,
f(x) = 9x2 – 3x + 4
Thus, f : x ˜ 9x2 – 3x + 4

9. Given f(x) = px + q, f 3(x) = 8x – 7

(a) f(x) = px + q

f 2(x) = f [f(x)]

= f(px + q)

= p(px + q) + q

= p2x + pq + q

f 3(x) = f [f 2(x)]

= f(p2x + pq + q)

= p(p2x + pq + q) + q

= p3x + p2q + pq + q

By comparison,

p3x + p2q + pq + q = 8x – 7

p2q + pq + q = –7

p = 2 4q + 2q + q = –7

7q = –7

q = –1

(b) f 4(x) = f [f 3(x)]

= f(8x – 7)

= 2(8x – 7) – 1

= 16x – 14 – 1

= 16x – 15

(c) f(x) = 2x – 1

f 2(x) = 4x – 3

f 3(x) = 8x – 7

f 4(x) = 16x – 15

Thus, f n(x) = 2nx + 1 – 2n

10. CN(t) = C(100t – 5t2)
= 15 000 + 8 000(100t – 5t2)

= 15 000 + 800 000t – 40 000t2

Inquiry 3 (Page 20)

3. Yes, the graphs of each function and its inverse are symmetrical about the line h(x) = x, which
is the line y = x.

Self Practice 1.8 (Page 21)

1. (a) f(4) = –5 (b) f –1(–1) = 6 (c) f –1(2) = –2 (d) f –1(–5) = 4

2. Given g(x) = 2 5 x, x ≠ 2 and h(x) = 3x + 6
–

(a) g(12) = 5
2 – 12

= –  1
2

20

(b) Let a = g–1(4)

g(a) = 4

2 5 a = 4
–

5 = 8 – 4a

4a = 3
3
a= 4

Thus, g–1(4) = 3
4

(c) h(–1) = 3(–1) + 6

=3

(d) Let a = h–1(9)

h(a) = 9

3 a + 6 = 9

3a = 3

a=1

Thus, h–1(9) = 1

Inquiry 4 (Page 22) g–1
1. (a) x
g
x2

–1 1 l –1
–2 –2
4
l 4 l
2 B 2
h
A BA
x+2 h–1
(b) x

–2 0 0 –2
02 20
24 42
46 64

CD DC

2. g–1 is not a function because there are objects that have two images. h–1 is a function
because each object has only one image.

3. The type of functions that can have an inverse is a one-to-one function.

Inquiry 5 (Page 23)

2. f(1) = 1 g(1) = 1 gf(1) = g(1) = 1 gf(3) = g(5) = 3
f(2) = 3 g(3) = 2 fg(1) = f(1) = 1 fg(5) = f(3) = 5
f(3) = 5 g(5) = 3
f(4) = 7 g(7) = 4 gf(2) = g(3) = 2 gf(4) = g(7) = 4
fg(3) = f(2) = 3 fg(7) = f(4) = 7

21

3. fg(x) = x where x in the domain of g and gf(x) = x where x in the domain of f.

Inquiry 6 (Page 23)

4. Graph Domain Range

Graph of function f 0 < x < 8 0 < f(x) < 4

Graph of function g 0 < x < 4 0 < f(x) < 8

5. If two functions f and g are inverses of each other, then the domain of f = the range of g and
the domain of g = the range of f.

6. The graph of g is the reflection of the graph of f on the line y = x.

Inquiry 7 (Page 24)

4. For any real numbers, a and b, if the point (a, b) is on the graph of f, then the point (b, a) is
on the graph of g, that is the inverse of the graph of f. The point (b, a) on the graph of g is

the reflection of the point (a, b) on the graph of f on the line y = x.

Self Practice 1.9 (Page 26)

1. (a) This function maps one element in a domain to only one element in a codomain. This
inverse function only maps each element in the codomain to only 1 element in the
domain. Therefore, this function has an inverse.

(b) This function maps one element in the domain to only one element in the codomain but
there are elements in the codomain that is mapped to two elements in the domain.
Therefore, this function does not have an inverse function.

(c) When the horizontal line test is done, the horizontal line cuts two points on the graph.
This shows that this function is not a one-to-one function. Therefore, this function does
not have an inverse function.

(d) This function is a one-to-one function. Therefore, this function has an inverse.
(e) This function is a many-to-one function and its inverse would be a one-to-many

function. Therefore, this function does not have an inverse function.
(f) The graph for function f(x) = 4 – x2 is sketched as follows:

f(x)
4

Horizontal x
line test

–2 0 2

When the horizontal line test is done, the horizontal line cuts two points on the graph.
This shows that this function is not a one-to-one function and does not have an inverse
function.

22

(g) The graph of function f(x) = (x 1 , x . 2 is sketched as follows:
– 2)2
f(x)

Horizontal
line test

02 x

When the horizontal line test is done, the horizontal line cuts only one point on the
graph. This shows that this function is a one-to-one function and has an inverse.

(a) Given f(x) = 3x – 2, g(x) = x+2
( ) 2. 3
fg(x) = f x + 2 gf(x) = g(3x – 2)
3
( )= x + 2 3x –2 + 2
3 3 –2 = 3

=x =x

Thus, the functions of f and g are inverses of each other.

(b) Given f(x) = x 2x , g(x) = 3x
–3 x–2

3x 2x
x–2 x–3
( )fg(x) = f ( )gf(x) =g

= 2( x 3x ) = 3( x 2x )
–2 –3
3x 2x
( x –2 ) – 3 ( x –3 ) – 2

= 6x ÷ 3x – 3x + 6 = 6x ÷ 2x – 2x + 6
x–2 x–2 x–3 x–3

= 6x × x–2 = 6x × x–3
x–2 6 x–3 6

=x =x

Thus, the functions of f and g are inverses of each other.

(c) Given f(x) = x 2 3 , g(x) = 3x – 2
– x

3x – 2 2
x x–3
( )fg(x) = f ( )gf(x) =g

( )= 2 3( 2 ) – 2
3x – 2 x – 3
x –3 = 2
( – )
=2÷ 3x – 2 – 3x x 3
x
x = 6 – 2x + 6 ÷ 2
= 2× –2 x–3 x–3

= –x = 12 – 2x × x–3
x–3 2

=6–x

Thus, the functions of f and g are not inverses of each other.
23

(d) Given f(x) = 2 + 5x, g(x) = x–5
2

x–5 gf(x) = g(2 + 5x)
2
( )fg(x) =f 2 + 5x – 5
2
x–5 =
2
( )=
2+5 5x – 3
2
= 4 + 5x – 25 =
2

= 5x – 21
2

Thus, the functions of f and g are not inverses of each other.

3. Given f(x) = x3 for the domain –1 < x < 2.

f(x)

8

f

2 f −1 8x
1
−1 0 –11 2

The domain of the function f –1 is –1 < x < 8 and the range is –1 < f –1(x) < 2.

4. Given h(x) = x2 – 2 for the domain 0 < x < 3.
(a) y

7h

3 h−1

−2 0 3 x
−2 7

(b) The domain of function h–1 is –2 < x < 7.

(c) h(x) = h–1(x)
x2 – 2 = !wx + 2

(x2 – 2)2 = x + 2
x4 – 4x2 + 4 = x + 2
x4 – 4x2 – x + 2 = 0
When x = 1, LHS = 14 – 4(1)2 – 1 + 1

= –3
≠ RHS
When x = 2, LHS = 24 – 4(2)2 – 2 + 2

=0
= RHS

Thus, the value of x is 2.

24

( )5. 1
(a) P 2 , –2

(b) Q(–3, 1)

(c) R(5, 4)

(d) S(–8, –6)

6. (a) y

y = f −1(x) Bʹ(10, 3)

0 Aʹ(2, −1) x

(b) a = 1, b = 4

Self Practice 1.10 (Page 28)

1. (a) Given f(x) = 2x – 5

Let y = 2x – 5

2x = y + 5
y+5
x= 2

Since x = f –1(y),

f –1(y) = x
y+5
= 2

Substitute the variable y with x,
x + 5
f –1(x) = 2

Thus, f –1 : x → x + 5
2

(b) Given f(x) = 3 , x ≠ 0
x

Let y = 3
x

x= 3
y

Since x = f –1(y),

f –1(y) = x

= 3
y

Substitute the variable y with x,

f –1(x) = 3
x

Thus, f –1 : x ˜ 3 , x ≠ 0.
x

25

(c) Given f(x) = x 4 1 , x ≠ 1
–

Let y= 4
x–1

x–1= 4
y

x= 4+y
y

Since x = f –1(y),

f –1(y) = x
4+y
= y

Substitute the variable y with x,

f –1(x) = 4 + x
x

Thus, f –1 : x ˜ 4 + x , x ≠ 0.
x
5x
(d) Given f(x) = x– 6 , x ≠ 6

Let y= 5x
x–6

y(x – 6) = 5x

xy – 6y = 5x

xy – 5x = 6y

x(y – 5) = 6y
6y
x= y–5

Since x = f –1(y),

f –1(y) = x

= 6y
y–5

Substitute the variable y with x,

f –1(x) = 6x
x–5
6x
ThereTfhoures, f –1 : x → x –5 , x ≠ 5.

(e) Given f(x) = x + 9 , x ≠ 8
x – 8
x + 9
Let y = x – 8

y(x – 8) = x + 9

xy – 8y = x + 9

xy – x = 8y + 9

x(y – 1) = 8y + 9

x = 8y + 9
y–1

26

Since x = f –1(y),

f –1(y) = x

= 8y + 9
y–1

Substitute the variable y with x,
8x + 9
f –1(x) = x–1

MThauksa, f –1 : x ˜ 8x + 9 , x ≠ 1.
x–1

(f) Given f(x) = 2x – 3 , x ≠ 1
2x – 1 2
2x – 3
Let y= 2x – 1

y(2x – 1) = 2x – 3

2xy – y = 2x – 3

2xy – 2x = y – 3

x(2y – 2) = y – 3
y – 3
x = 2y – 2

Since x = f –1(y),

f –1(y) = x
y–3
= 2y – 2

Substitute the variable y with x,

f –1(x) = x–3
2x – 2
x–3
MTahkuas, f –1 : x ˜ 2x – 2 , x ≠ 1.

2. (a) Given f(x) = 3– x, x ≠ 0
2x

Let y = 3–x
2x

2xy = 3 – x

2xy + x = 3

x(2y + 1) = 3
3
x = 2y + 1

Since x = f –1(y),

f –1(y) = x
3
= 2y + 1

Substitute the variable y with x,

f –1(x) = 3 1, x ≠ – 1
2x + 2

Thus, f –1(4) = 3 1
2(4) +

= 3
= 9
1
3

27

(b) f(x) = f –1(x)

3–x = 3 1
2x 2x +

(3 – x)(2x + 1) = 3(2x)
6x + 3 – 2x2 – x = 6x

2x2 + x – 3 = 0

(2x + 3)(x – 1) = 0

2x + 3 = 0   or x – 1 = 0
3
x = –  2 or x=1

The possible values of x are – 3 and 1.
2
5
3. Given h(x) = 4x + a and h–1(x) = 2bx + 8
h(x) = 4x + a

Let y = 4x + a

4x = y – a
y–a
x= 4

Since x = h –1(y),

h–1(y) = x
y–a
= 4

Substitute the variable y with x,
x – a
h–1(x) = 4

Compare with the function h–1,

x–a = 2bx + 5
4 8

x – a = 2bx + 5
4 4 8

1 = 2b and – a = 5
4 4 8

8b = 1 –8a = 20

b= 1 a = – 280
8

= –  5
2
4. (a) f –1(x) = 6x + 7
Let x = 6y + 7

6y = x – 7
x–7
y= 6

Thus, f : x ˜ x – 7
6
2–x
(b) f –1(x) = 5

Let x = 2–y
5

5x = 2 – y

y = 2 – 5x

Thus, f : x ˜ 2 – 5x

28

(c) f –1(x) = 3x , x ≠ 3
x–3

Let x= 3y
y–3

xy – 3x = 3y

y(x – 3) = 3x

y= 3x
x–3

MTahkuas, f : x ˜ 3x , x ≠ 3
x–3

5. Given g–1(x) = 2 4 x , x ≠ k
–
(a) 2 – x = 0

x=2

Thus, k = 2.
4
(b) g–1(x) = 2–x

Let x= 4
2–y

2x – xy = 4

xy = 2x – 4

y= 2x – 4
x

g(x) = 2x – 4 , x ≠ 0
x

( )MThakuas, g1 = 2( 1 ) – 4
2 2
1
( 2 )

= –6

Intensive Practice 1.3 (Page 29)

1. (a) f (2) = 5 f(x) = 2x –31 (b) g(5) = 8 (c) gf(2) = 8
(d) f –1(5) = 2 (e) g–1(8) = 5 (f) f –1g–1(8) = 2

2. (a) (b) (c)
y y
y f(x) = –12 x3 – 5x + 1
f(x) = –1x
Horizontal Horizontal
line test line test

0x 0x 0x

Horizontal
line test

This function has an inverse This function has an inverse This function does not have
function because the function because the an inverse function because
horizontal line cuts only one horizontal line cuts only one the horizontal line cuts more
point on the graph. point on the graph. than one point on the graph.

29

3. (a) (b) (c)
y y y
0
(2, 4) (4, 8)

f −1 (4, 2) 4 f (8, 4)
f x f f −1 x

2

f −1

0 0 x
24

Domain of f –1 is 0 < x < 2 Domain of f –1 is 0 < x < 4 Domain of f –1 is 0 < x < 8

4. (a) Given f(x) = 2x + h , x ≠ 3 dan f(4) = 13
x–3
f(4) = 13

2(4) + h = 13
4–3

8 + h = 13

h=5
2x + 5
(b) Let y= x– 3

y(x – 3) = 2x + 5

xy – 2x = 3y + 5

x(y – 2) = 3y + 5

x = 3y +5
y –2

Since x = f –1(y),

f –1(y) = x

= 3y + 5
y–2

Substitute the variable y with x,

f –1(x) = 3x + 5 , x ≠ 2
x–2

MThauksa, f –1(3) = 3(3) + 5
3–2

= 14
(c) f –1(m) = 2

3m + 5 = 2
m–2

3m + 5 = 2m – 4

m = –9

30

5. Given h–1(x) = 3 2 x , x ≠ 3
–

(a) Let x= 2
3–y

3x – xy = 2

xy = 3x – 2

y= 3x – 2
x

Thus, h(x) = 3x_–__2_ , x ≠ 0
x

(b) h(x) = 2

3x – 2 =2
x

3x – 2 = 2x

x=2

6. Given f(x) = 4x – 17, g(x) = 5 7 , x ≠ 3 1
2x – 2

f(x) = 4x – 17 g(x) = 5
2x – 7
Let y = 4x – 17
5
4x = y + 17 Let =y 2x – 7

x = y + 17 2xy – 7y = 5
4
2xy = 5 + 7y
Since x = f –1(y),
5 + 7y
f –1(y) = x x= 2y
y + 17
= 4 Since x = g–1(y),

Thus, f –1(x) = x + 17 g–1(y) = x
4 5 + 7y
= 2y

Thus, g–1(x) = 5 + 7x
2x
f –1(x) = g–1(x)

x + 17 = 5 + 7x
4 2x

2x(x + 17) = 4(5 + 7x)

2x2 + 34x = 20 + 28x

2x2 + 6x – 20 = 0

x2 + 3x – 10 = 0

(x + 5)(x – 2) = 0

x + 5 = 0   or x – 2 = 0

x=2

7. (a) Given f(x) = 17 (220 – x)
20

Let y = 17 (220 – x)
20
20
17 y = 220 – x

x = 220 – 20 y
17

31

Since x = f –1(y),

f –1(y) = x

= 220 – 20 y
17

MTahkuas, f –1(x) = 220 – 20 x
17

(b) f(16) = 17 (220 – 16)
20

= 173.4

8. Given V = 4 πr3
(a) 3

r V = –34 πr3
f

f –1

(b) 4 πr3 = 1
3 2

( )r3 = 1 3
2 4π

= 3
8π

!wr = 3 3
8π

= 0.49 cm

Mastery Practice

1. (a) (i) 1
(ii) 6, 8, 9

(b) Yes, because each object has only one image.
(c) Domain = {2, 6, 7, 8, 9}

Codomain = {1, 4, 5}
Range = {1, 4}

2. (a) (–2)2 – 1 = 3
22 – 1 = 3
42 – 1 = 15
62 – 1 = 35

Thus, m = 35.
(b) h : x ˜ x2 – 1

32

3.

y Vertical y Horizontal

line test line test

0x 0x

This graph is a function because the vertical line test cuts only one point on the graph but
not a one-to-one function because the horizontal line test cuts more than one point on the
graph.

4. Given f(x) =  x – 3  with domain –1 < x < 7,
(a) y

4 f (x) = Խx − 3Խ

−10 3 7x

The range for the function f is 0 < f(x) < 4.

(b) f(x) < 2

 x – 3  < 2

–2 < x – 3 < 2

1 < x < 5

(c) y y = 2x−3
f (x)
= Խx − 3Խ

−10 3 7x
−3

 x – 3  = 2x – 3

x – 3 = –(2x – 3) or x – 3 = 2x – 3
x=0
3x = 6

x=2

Thus, x = 2.

5. DGibverni f(x) = hx + k , x ≠ 0
x
(a) f(2) = 17

2h + k = 17
2

4h + k = 34

k = 34 – 4h…1

f(3) = 23

33

3h + k = 23
9h + 3k = 69…2

Substitute 1 into 2.

9h + 34 – 4h = 69

5h = 35

h=7

Substitute h = 7 into 1.

k = 34 – 4(7)

= 34 – 28

=6

Thus, h = 7 and k = 6.
6
(b) f(6) = 7(6) + 6

= 43

6. Given f(x) = x + 2 , x ≠ 2, g(x) = mx + c, g–1(2) = f(3), gf –1(2) = 5
x – 2
x+2
f(x) = x–2

KatakLaent y = x+2
x–2

xy – 2y = x + 2

xy – x = 2 + 2y

x(y – 1) = 2 + 2y

x= 2 + 2y
y–1

MThauksa, f –1(x) = 2+ 2x , x ≠ 1
x– 1

g(x) = mx + c

Katakan y = mx + c

mx = y – c

x= y–c
m

MThauksa, g–1(x) = x–c
m

g–1(2) = f(3)

2–c = 3+2
m 3–2

2 – c = 5m

5m + c = 2…1

gf –1(2) = 5

g(6) = 5

6m + c = 5…2

2 – 1: m = 3

Substitute m = 3 into 1.

5(3) + c = 2

c = –13

Thus, m = 3 and c = –13.

34

7. (a) (i) f(x) = 3x – 2

Let y = 3x – 2

3x = y + 2

x= y+2
3
ThereTfohrues, the function that x+ 2.
maps set B to set A is f –1(x) = 3

(ii) gf(x) = 6x + 1

g(3x – 2) = 6x + 1

Let y = 3x – 2

3x = y + 2
y+2
x= 3

y+2
3
( )g(y) = 6 +1

= 2y + 5

Thus, g(x) = 2x + 5

(b) fg(x) = 4x – 3

f(2x + 5) = 4x – 3

3(2x + 5) – 2 = 4x – 3

6x + 15 – 2 = 4x – 3

2x = –16

x = –8
m
8. Given f(x) = x–1 + n, x ≠ k, f(2) = 3, f(3) = 2

(a) k = 1

(b) f(2) = 3
m
2–1 + n = 3

m + n = 3…1

f(3) = 2

m + n = 2
3–1
m + 2n = 4…2

2 – 1: n = 1

SGubasnttiitkuaten n = 1 iknetoda1la.m 1.
m+1=3

m=2

Thus, m = 2 and n = 1.

(c) f 2(x) = f [f(x)]

2
x–1
( )= f +1

= 2 2 1 – 1 + 1
–1 +
x

( )= 2 +1
2
x–1

= x–1+1

=x

35

(d) f(x) = 2 +1
x–1

Ley y = x 2 1 + 1
–

y–1= 2
x–1

x–1= 2
y–1

x= 2 +1
y–1

f –1(x) = 2 + 1, x ≠ 1
x–1

Thus, f –1(2) = 2 2 1 + 1
–

=3

9. (a) (i) Continuous function
(ii) –4 < f(x) < 4

(b)

y

4

Horizontal

1 line test

–3 0 1 x

f(x) = –x3 – 3x2 – x + 1

–4

Graph of f does not have an inverse function because the horizontal line test cuts more
than one point on the graph.
10. (a) The functions f(x) =  x  and f(x) = x4 are one-to-one functions with the condition that the
domain of f is x > 0.

1

(b) f –1(x) = x, f –1(x) = x 4

11. The graph does not necessarily need to intersect at the line y = x if the graph for a function
and its inverse intersect. Both of these graphs might intersect on other lines.

12. Given f(x) = ax + b
cd + d
ax + b
Let y = cd + d

cxy + dy = ax + b

x(cy – a) = b – dy
b – dy
x= cy – a

f –1(x) = b – dx , x ≠ a 36
cx – a c

(a) (i) f(x) = x + 8 , where a = 1, b = 8, c = 1 and d = –5
x – 5

f –1(x) = 8+ 5x , x ≠ 1
x– 1

(ii) f(x) = 2x – 3 , where a = 2, b = –3, c = 1 and d = 4
x+4

f –1(x) = –3 – 24x, x ≠ 2
x –

(b) f = f –1 if a = –d.

13. (a) (i) (ii)

3 y=x
f –1 f

2 ( 0, 2 )

1

( 2, 0 )

–1 0 1 3

–1

The range of f is –1 < f(x) < 3 and the domain of f –1 is –1 < x < 3.
(b) (i) Yes

(ii) Yes. Any point (b, a) on the graph of f –1 is the reflection of point (a, b) on the graph
on the line y = x.

14. (a) p= 100 – 1 x
4
1
4 x = 100 – p

x = 400 – 4p

danand C = ! x + 600
25

ApWabhielna x = 400 – 4p,

C= ! 400 – 4p + 600
25

= ! 4(100 – p) + 600
= 25

2! 100 – p + 600
25

Thus, C = 2! 100 – p + 600
25

37

(b) When p = 36,

C= 2! 100 – 36 + 600
25

= 0.64 + 600

= 600.64

15.

14 Θ Ιh(x) = 2π –41–0x– 0.5

12 Θ Ιg(x) = 2π –x–1+–0–4– 0.5
Θ Ιf(x) = 2π –1x–0– 0.5
10

8

6

4

2

–5 0 5 10 15

The period of the pendulum, T depends on the length of the pendulum, l. If the length
increases, the period also increases.

38

CHAPTER 2 QUADRATIC FUNCTIONS

Inquiry 1   (Page 36)

4. The x-coordinate for point A and point B is the solution for the equation y = 3x2 + 11x − 4
when y = 0. This x-coordinate is called the roots for the quadratic equation 3x2 + 11x − 4 = 0.

Mind Challenge   (Page 38)

ax2 + bx + c = 0 Divide both sides of the
equation by a
x2 + b x = – c
a a
( ) ( )x2 +bb c b2
a x + 2a 2= – a + 2a Add both sides of the equation

coefficient of x 2
2
( )x+ b 2 = – 4ac + b2 ( )with
2a 4a2
± b2 – 4ac
x + b = 2a
2a

x = –b + b2 – 4ac or x = –b – b2 – 4ac
2a 2a

In general, the formula for solving a quadratic equation is

x = –b ± b2 – 4ac
2a

Self Practice 2.1 (Page 38)

1. (a) x2 + 4x – 9 = 0

x2 + 4x = 9

( ) ( )x2 + 4x +4 42
2 2=9+ 2

x2 + 4x + 22 = 9 + 22

(x + 2)2 = 13

x + 2 = ± 13

x = – 13 – 2 or x = 13 – 2
= 1.606
= −5.606
1
(b) x2 − 3x – 5 = 0

x2 − 3x = 5

( ) ( )x2 − 3x +–3 2 = 5 + –3 2
2 2
( )x − 3 29
2 2= 4

x− 3 =± 29
2 4

x=– 29 + 3 or x= 29 + 3
4 2 4 2

= −1.193 = 4.193

(c) –x2 – 6x + 9 = 0

x2 + 6x = 9

( ) ( )x2 + 6x +6 62
2 2=9+ 2

x2 + 6x + 32 = 9 + 32

(x + 3)2 = 18

x + 3 = ± 18

x = – 18 – 3 or x = 18 – 3

= −7.243 = 1.243

(d) 2x2 − 6x + 3 = 0

2x2 − 6x = −3

2(x2 – 3x) = −3
3
x2 – 3x = – 2

( ) ( )x2 − 3x +–3 2=– 3 + –3 2
2 2 2
( )x − 3 3
2 2= 4

x− 3 =± 3
2 4

x=– 3 + 3 or x= 3 + 3
4 2 4 2
= 0.634 = 2.366

(e) 4x2 − 8x + 1 = 0

2x2 − 8x = −1

4(x2 – 2x) = −1
1
x2 – 2x = – 4

( ) ( )x2 − 2x +–2 2 =– 1 + –2 2
2 4 2
3
(x − 1)2 = 4

x−1=± 3
4

x=– 3 +1 or x= 3 +1
4 4
= 0.134 = 1.866

(f) −2x2 + 7x + 6 = 0

2x2 − 7x = 6

( )2 x2 − 7 x =6
2
7
x2 − 2 x = 3

−( ) ( )x27x+ –7 2 = 3+ –7 2
2 4 4
2

( )x − 7 2= 97
4 16

x− 7 =± 97
4 16

x=– 97 + 7 or x= 97 + 7
16 4 16 4

= −0.712 = 4.212

2. (a) x2 – 4x – 7 = 0 with a = 1, b = −4 and c = −7.

x = –b ± b2 – 4ac
2a

x = –(–4) ± (– 4)2 – 4(1)(–7)
2(1)

= 4 ± 44
2

x= 4 – 44 or x= 4 + 44
2 2
= −1.317 = 5.317

(b) 2x2 + 2x – 1 = 0 with a = 2, b = 2 and c = −1.

x = –b ± b2 – 4ac
2a

x = –(2) ± (2)2 – 4(2)(–1)
2(2)

= –2 ± 12
4

x= –2 – 12 or x= –2 + 12
4 4
= −1.366 = 0.366

(c) 3x2 – 8x + 1 = 0 with a = 3, b = −8 and c = 1.

x = –b ± b2 – 4ac
2a

x = –(–8) ± (–8)2 – 4(3)(1)
2(3)

= 8 ± 52
6

x= 8 – 52 or x= 8 + 52
6 6
= 0.131 = 2.535

(d) 4x2 – 3x – 2 = 0 with a = 4, b = –3 and c = –2.

x = –b ± b2 – 4ac
2a

x = –(–3) ± (–3)2 – 4(4)(–2)
2(4)

= 3 ± 41 3
8

x= 3 – 41 or x= 3 + 41
8 8
= –0.425 = 1.175

(e) (x – 1)(x – 3) = 5

x2 – 4x + 3 = 5

x2 – 4x – 2 = 0 with a = 1, b = –4 and c = –2.

x = –b ± b2 – 4ac
2a

x = –(–4) ± (– 4)2 – 4(1)(–2)
2(1)

= 4 ± 24
2

x= 4 – 24 or x= 4 + 24
2 2
= –0.449 = 4.449

(f) (2x – 3)2 = 6

4x2 – 12x + 9 = 6

4x2 – 12x + 3 = 0 with a = 4, b = –12 and c = 3.

x = –b ± b2 – 4ac
2a

x = –(–12) ± (–12)2 – 4(4)(3)
2(4)

= 12 ± 96
8

x= 12 – 96 or x= 12 + 96
8 8
= 0.275 = 2.725

3. (a) Let x be the length and (x – 2) be the width. x cm
x2 + (x – 2)2 = 102
10 cm (x – 2) cm
x2 + x2 – 4x + 4 = 100
2x2 – 4x – 96 = 0
x2 – 2x – 48 = 0

(x – 8)(x + 6) = 0

x = 8 or x = –6 (Ignore)

Thus, the rectangle has a length of 8 cm and width of 6 cm.

(b) 2x + 2y = 26…
xy = 40…

From , y = 40 …
x

Substitute ᕣ into ᕡ.

( )2x + 240 = 26
x
2x2 + 80 = 26x

2x2 – 26x + 80 = 0

x2 – 13x + 40 = 0 4

(x – 5)(x – 8) = 0
x = 5 or x = 8

Substitute x = 5 into ᕣ.

y = 40
5
=8

Substitute x = 8 into ᕣ.

y = 40
8
=5

Thus, the measurement of the rectangle is 5 cm × 8 cm.

4. 1 × [(x + 3) + (3x + 2)] × (x – 1) = 17
2 (4x + 5) × (x – 1) = 34

4x2 + x – 5 = 34

4x2 + x – 39 = 0

(4x + 13)(x – 3) = 0 13
4
x = – (Ignore) or x=3

Thus, the value of x is 3.

Self Practice 2.2 (Page 41) (b) x2 – (α + β )x + αβ = 0
x2 – (–1 + 4)x + (–1)(4) = 0
1. (a) x2 – (α + β )x + αβ = 0 x2 – 3x – 4 = 0
x2 – (2 + 6)x + (2)(6) = 0
x2 – 8x + 12 = 0 (d) x2 – (α + β )x + αβ = 0
( ) ( )x2 –1 1
(c) x2 – (α + β )x + αβ = 0 5 –5 x+ 5 (–5) = 0
x2 – (–4 – 7)x + (–4)(–7) = 0
x2 + 11x + 28 = 0 x2 + 24 x – 1 = 0
5
2. α + β = –(p – 5), αβ = 2q 5x2 + 24x – 5 = 0
–(p – 5) = –3 + 6
–p + 5 = 3
p=2
2q = (–3)(6)
2q = –18
q = –9

3. 5x2 – 10x – 9 = 0
9
x2 – 2x – 5 =0

Thus, α + β = 2, αβ = – 9
5

5

(a) (α + 2) + (β + 2) = α + β + 4 , (α + 2)(β + 2) = αβ + 2α + 2β + 4
= αβ + 2(α + β) + 4
=2+4

=6 = – 9 + 2(2) + 4
5
31
= 5

Quadratic equation: x2 – 6x + 31 = 0
5
5x2 – 30x + 31 = 0

(b) 5α + 5β = 5(α + β) , (5α)(5β) = 25αβ

= 5(2) = 25(– 9 )
5
= 10 = –45

Quadratic equation: x2 – 10x – 45 = 0

(c) (α – 1) + (β – 1) = α + β – 2 , (α – 1)(β – 1) = αβ – α – β + 1
= αβ – (α + β) + 1
= 2–2

=0 = – 9 –2+1
5
14
= – 5

Quadratic equation: x2 – 0x – 14 =0
5
5x2 – 14 = 0

( )( )(d)α+ β = α +β , α β = αβ
3 3 3 3 3 9
9
= 2 = – 5
3
9
1
= – 5

Quadratic equation: x2 – 2 x – 1 = 0
3 5
15x2 – 10x – 3 = 0

4. 2x2 + 5x = 1

2x2 + 5x – 1 = 0

x2 + 5 x – 1 =0
2 2
5 1
Thus, α + β = – 2 , αβ = – 2

1 1 α+β 1 1 1
α β αβ α β αβ
( )( )(a)+ = , =

= (– 5 ) = 1
(– 2
1 ) (– 1 )
=5 2 = –2 2

Quadratic equation: x2 – 5x – 2 = 0

6

( ) ( )(b)α+1 + β + 1 = α + β + α+β , ( )( )α+ 1 β + 1 = αβ +1+1+ 1
β α αβ β α αβ

=– 5 +5 = – 1 +1+1–2
2 2
5 1
= 2 = – 2

Quadratic equation: x2 – 5 x – 1 =0
2 2
2x2 – 5x – 1 = 0

(c) α2 + β2 = (α + β)2 – 2αβ , (α2)(β2) = (αβ )2

= (– 5 )2 – 2(– 1 ) ( )= – 1 2
2 2 2
25 1
= 4 + 1 = 4

= 29
4
29 1
Quadratic equation: x2 – 4 x + 4 =0

4x2 – 29x + 1 = 0

α β α2 + β 2 α β
β α αβ β α
( ) ( ) ( )( )(d)+ = , =1

= (α + β)2 – 2αβ
αβ

= ( 29 )
4
(– 1 )
2

= – 29
2
29
Quadratic equation: x2 + 2 x + 1 = 0

2x2 + 29x + 2 = 0

5. 2x2 = 6x + 3

2x2 – 6x – 3 = 0

x2 – 3x – 3 =0
2
3
Thus, p + q = 3 , pq = – 2

p2q + pq2 = pq(p + q) , (p2q)(pq2) = p3q3

= – 3 (3) = (pq)3
2
9 ( )= 3
= – 2 – 2 3

= – 27
8
9 27
Quadratic equation x2 + 2 x – 8 = 0

8x2 + 36x – 27 = 0

7

Inquiry 2   (Page 41)

2. Graph sketching method:
•  The roots are x = –1 and x = 3
y

y>0 y>0 x
x>3
x < –1
–1 O 3
y<0

–1 < x < 3

Number line method:

•  The roots are x = –1 and x = 3

Test point –2: Test point 0: Test point 4:
(4 + 1)(4 – 3) > 0
(–2 + 1)(–2 – 3) > 0 (0 + 1)(0 – 3) < 0

+– +

x < –1 –1 –1 < x < 3 3 x > 3

Tabular method: x –1 Range of values of x –1 x 3
– x3 +
(x + 1) – + –
(x – 3) + + –
(x + 1) (x – 3) +

3. When (x + 1)(x – 3) . 0, the range of values of x is x , –1 or x . 3 and when
(x + 1)(x – 3) , 0, the range of x is 1 , x , 3.

Self Practice 2.3   (Page 44)

1. (a) x2 , 4
x2 – 4 , 0
(x – 2)(x + 2) , 0
y
x = –2 and x = 2 –2 0 2

Graph sketching method: –4
Since x2 – 4 , 0, the range of x is determined on the
x

graph below the x-axis.

Thus, the solution for this quadratic inequality is
–2 , x , 2.

8

Number line method:

Test point –3: Test point 1: Test point 3:

(–3)2 – 4 0 (1)2 – 4 0 (3)2 – 4 0

+–+
x

–2 2
Since x2 – 4 , 0, the range of x is determined on the negative part of the number line.

Thus, the solution for this quadratic inequality is –2 , x , 2.

Tabular method: x , –2 –2 , x , 2 x.2
– – +
(x – 2) – + +
(x + 2) + – +
(x – 2)(x + 2)

Since x2 – 4 , 0, the range of x is determined on the negative part of the table.
Thus, the solution for this quadratic inequality is –2 , x , 2.

(b) (2 – x)(8 – x) , 0 y
x – 2   and  x = 8 02 8x

Graph sketching method:
Since (2 – x)(8 – x) , 0, the range of x is determined

on the curve of the graph below the x-axis.
Thus, the solution for this quadratic inequality is

2 , x , 8.

Number line method:

Test point 1: Test point 3: Test point 9:

(2 – 1)(8 – 1) 0 (2 – 3)(8 – 3) 0 (2 – 9)(8 – 9) 0

+–+
x

28
Since (2 – x)(8 – x) , 0, the range of x is determined on the negative part of the

number line.
Thus, the solution for this quadratic inequality is 2 , x , 8.

9

Tabular method: x,2 2,x,2 x.8
+ – –
(2 – x) + + –
(8 – x) + – +
(2 – x)(8 – x)

Since (2 – x)(8 – x) , 0, the range of x is determined on the negative part of the table.
Thus, the solution for this quadratic inequality is 2 , x , 8.

(c) x2 < 4x + 12
x2 – 4x – 12 < 0
(x + 2)(x – 6) < 0
y

x = –2   and  x = 6

Graph sketching method: –2 0 6 x
Since x2 – 4x – 12 < 0, the range of x is determined
–12
on the curve of the graph below the x-axis.
Thus, the solution for this quadratic inequality is

–2 < x < 6.

Number line method:

Test point –3: Test point 1: Test point 7:

(–3)2 – 4(–3) – 12 0 (1)2 – 4(1) – 12 0 (7)2 – 4(7) – 12 0

+–+
x

–2 6
Since x2 – 4x – 12 < 0, the range of x is determined on the negative part of the number

line.
Thus, the solution for this quadratic inequality is –2 < x < 6.

Tabular method: x < –2 –2 < x < 6 x>6
– + +
(x + 2) + – +
(x – 6) + – +
(x + 2)(x – 6)

Since (x + 2)(x – 6) < 0, the range of x is determined on the negative part of the table.
Thus, the solution for this quadratic inequality is –2 < x < 6.

10

(d) x(x – 2) > 3 y
  x2 – 2x – 3 > 0
(x + 1)(x – 3) > 0
x = –1   and  x = 3

Graph sketching method: –1 0 3x
Since x2 – 2x – 3 > 0, the range of x is determined –3

on the curve of the graph above the x-axis.
Thus, the solution for this quadratic inequality is x < –1

and x > 3.

Number line method: Test point 2: Test point 4:
Test point –2:

(–2)2 – 2(–2) – 3 0 (2)2 – 2(2) – 3 0 (4)2 – 2(4) – 3 0

+–+

–1 3
Since x2 – 2x – 3 > 0, the range of x is determined on the positive part of the number line.
Therefore, the solution for this quadratic inequality is x < –1 and x > 3.

Tabular method: x < –1 –1 < x < 3 x>3
– + +
(x + 1) – – +
(x – 3) + – +
(x + 1)(x – 3)

Since (x + 1)(x – 3) > 0, the range of x is determined on the positive part of the table.
Thus, the solution for this quadratic inequality is x < –1 and x > 3.

(e) (x + 2)2 , 2x + 7 y
x2 + 4x + 4 , 2x + 7
x2 + 2x – 3 , 0 –3 0 1 x
(x + 3)(x – 1) , 0 –3
x = –3   and  x = 1

Graph sketching method:
Since x2 + 2x – 3 , 0, the range of x is determined

on the curve of the graph below the x-axis.
Thus, the solution for this quadratic inequality is

–3 , x , 1.

11