Em 2019 3e4
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Quadratic Equations & Functions Exercise 1A 4
(f)
6 2 x+ =0 7 49 6 2 x2 x= 7 49 3 2 3 2 (x ) ( )2 = 7 7 49 3 2 9 2 (x ) = 7 49 49 3 1 x =± 7 7
x2
5
x = 0.81 or 0.05
(g)
x2 + 0.6x 1 = 0 x2 + 0.6x = 1 (x + 0.3)2 (0.3)2 = 1 (x + 0.3)2 = 0.09 + 1 x + 0.3 = ± 1.09 x = 0.74 or 1.34
(h)
x2 4.8x + 2 = 0 x2 4.8x = 2 (x 2.4)2 (2.4)2 = 2 (x 2.4)2 = 5.76 2 x 2.4 = ± 3.76 x = 4.34 or 0.46
(c)
(x + 2)(x 5) = 4x x2 3x 10 4x = 0 x2 7x 10 = 0 7 2 7 ) ( )2 = 10 2 2 7 2 49 (x ) = + 10 2 4 7 89 x =± 2 4
(x
(d)
x = 8.22 or 1.22
x(x 4) = 2(x + 7) x2 4x 2x 14 = 0 x2 6x 14 = 0 (x 3)2 (3)2 = 14 (x 3)2 = 9 + 14 x 3 = ± 23 x = 7.80 or 1.80
Exercise 1B 3 (e) (2x + 3)(x 1) x(x + 2) = 0 2x2 + x 3 (x2 + 2x) = 0 Math Homework Prepared by Mr Ang KS
1
x2 x 3 = 0 Using a = 1, b = 1 and c = 3, x= (f)
(b)
( 1) 2 4(1)(3) 2(1)
= 2.30 or 1.30
(4x 3)2 + (4x + 3)2 = 25 16x2 24x + 9 + (16x2 + 24x + 9) = 25 32x2 = 7 x2 =
4
(1)
7 32
x = ±0.468
3 2 1 x + 2x =0 4 2 3 1 Using a = , b = 2 and c = , 4 2
3 1 2 22 4( )( ) 4 2 x= 3 2( ) 4 = 0.230 or 2.90
(c)
5x 7 = x2 x2 5x + 7 = 0 Using a = 1, b = 5 and c = 7, x=
( 5)
( 5) 2 4(1)(7) 2(1)
There is no solution since
3
=
4
3 2
is not defined .
Exercise 1D 4
5
x( x 3) 3 = ( x 1) 2 5
(f)
5(x2 3x) = 3(x2 + 2x + 1)
2x2 21x 3 = 0
x=
( 21)
7 x 1 1 = 2 x 1 x3
( 21) 2 4( 2)(3) = 10.6 or 0.141 2( 2)
7( x 3) ( x 1)( x 1) 1 = ( x 1)( x 3) 2
2(7x + 21 x2 + 1) = x2 + 2x 3 3x2 12x 47 = 0
x=
( 12)
( 12) 2 4(3)(47) 2(3)
= 6.43 or
2.43 5
(g)
4 5 =2 ( x 2) 2 x2
Math Homework Prepared by Mr Ang KS
4 5 + =2 ( x 2) 2 x2 5( x 2) 4 =2 ( x 2) 2
5x 6 = 2(x2 4x + 4) 2
5
8
(h)
(i) (iii)
(iv)
2x2 13x + 14 = 0
x=
x 5 + =1 ( x 1) 2 x 1
(13)
( 13) 2 4( 2)(14) = 5.14 or 1.36 2( 2)
5( x 1) x =1 ( x 1) 2
6x 5 = (x2 2x + 1) x2 8x + 6 = 0
x=
60 pages/min x 60 60 + = 144 x x2 60( x 2) 60 x = 144 x ( x 2)
(ii)
120x + 120 = 144(x2 + 2x) 24: 5x + 5 = 6x2 + 12x 6x2 + 7x 5 = 0 (2x 1)(3x + 5) = 0
( 8)
( 8) 2 4(1)(6) 2(1)
= 7.16 or 0.838
60 pages/min x2
(Shown)
2 1 or x = 1 2 3 2 Reject x = 1 as x > 0 3 1 Required time taken = (2 60) 144 2
x=
(v)
11
(i)
Speed of coach = x 30, so time taken = Time taken by car =
= 6 mins
700 h x 30
700 h x
Total time taken = 20 h 700 700 + = 20 x 30 x 35 x 35( x 30) 20: =1 x ( x 30)
(ii)
70x 1050 = x2 30x x2 100x + 1050 = 0 (Shown) Using a = 1, b = 100 and c = 1050 x=
(iii)
( 100)
( 100) 2 4(1)(1050) 2(1)
= 88.079 or 11.921 = 88.08 or 11.92 Reject x = 11.92 as x = 30 becomes < 0 Time taken for return journey =
Math Homework Prepared by Mr Ang KS
700 88.079
= 7.95 h 3
Exercise 1E 1 (e) y = (3 x)(x + 2)
y = (2 x)(4 x)
(f)
y y = (3 x)(x + 2)
6
y = (2 x)(4 x)
y 2
3
0
x 8
2
y = (x + 1)2 3
(d)
y = (x 4)2 1
(f) y0 0
y
y = (x + 1)2 3
4
2
1
0
0.732
y = (x 4)2 1
x
Line of symmetry 2 is x = 1.
Line of symmetry is x = 4.
(i)
x2 8x + 5
(ii)
(iii)
Min. point = (4, 11).
(1, 3)3
7
x
4 (4, 1)
17 2.73
x
(iv)
= (x 4)2 (4)2 + 5 = (x 4)2 11
y y = (x 4)2 11
Line of symmetry is x = 4. 5
Exercise 2A 6
(g)
1 1 1 (3g + 4) (g + 1) 1 (g + 5) 5 3 3
15:
(h)
11
4
7.32
x
(4, 11)
3(3g + 4) 5(g + 1) 15 5(g + 5) 4g + 7 5g 10 9g 17 g 1
6
0 0.683
8 9
h 3 h + ) < 3( 5) 3 4 2 4h 9 h 10 4( ) < 3( ) 12 2
4(
6:
Math Homework Prepared by Mr Ang KS
2(4h + 9) < 9(h 10) 8h + 18 < 9h 90 h < 108 h > 108 4
7
1 p (2 p) 3 6 10
30:
5(2 p) 90 3p 5p 80 3p 8p 80 p 10 Largest possible value of p = 10 Exercise 2B 5 (c) 3x 3 < x 9 < 2x 3x 3 < x 9 2x < 6 x < 3 9 < x < 3 (d)
2x x + 6 < 3x + 5 2x x + 6 x6
and
x 9 < 2x x < 9 x > 9 9
and
x + 6 < 3x + 5 2x < 1 x>
8
3
1 2
1 <x6 2
0.5
6
Let x be the no. of correct answers, then (30 x) are the no. of wrong answers. 5x + (2)(30 x) 66 5x 60 + 2x 66 7x 126 x 18 Max. correct answers obtained = 18 Given:
10
1 1 x4> x 2 3
and
1 1 x+1< x+3 6 8
3x 24 > 2x 4x + 24 < 3x + 72 x > 24 x < 48 24 < x < 48 So, x = 29, 31, 37, 41, 43, 47 13
14
(d)
(d)
3d 5 < d + 1 2d + 1 3d 5 < d + 1 and 2d < 6 d<3 0d<3 2d d 1 + 5 2 5 2d d5< and 5
d + 1 2d + 1 d 0 d≥0
d5<
Math Homework Prepared by Mr Ang KS
2d d 1 + 5 2 5
5
3d <5 5 1 d<8 3
15
(c)
d 1 10 5
d ≥ 2
2 d < 8
1 3
4 7 3x 2 11 3x 5 11 5 ≥x≥ 3 3 2 2 1 x3 3 3
Required integers are 2 and 3. (d)
10 < 7 2x 1 17 < 2x 8 17 >x≥4 2 1 4x<8 2
Required integers are 4, 5, 6, 7 and 8. Exercise 3A 6 (e) (e3)5 (e2)4
7
8
= e15 e8
= e7
(f)
(4f 6)3 (2f 3)3 = 64f 18 (8f 9)
= 8f 9
(c)
(8e5f 3)2 (e3f)3 = 64e10f 6 (e9f 3)
= 64ef 3
(d)
16g8h7 (2g3h2)3
(c)
3e3 2 f
(d)
g2 3g 5 3 2 h 2h
4
27e9 11 f 6
= 16g8h7 (8g9h6)
=
3
81e12 f 11 f8 27e 9
= 2g1h
2h
= g
= 3e3f 3
8 8h 6 g 12 = 18 ( 15 ) = 27 g 3h12 27 g h
Exercise 3B 6
2
(f)
(1000) 3
7
(f)
1 3 ( f )5
8
(d)
10d = 0.01 10d = 102
Math Homework Prepared by Mr Ang KS
= ( 3 1000 )2 = (10)2 1
=
f
=f
5 3
= 100
5 3
d = 2 6
10
(g)
(h)
(m 5 n 3 ) ( m 2 ) 2 ( m 1n) 2
(m 5 n 3 ) m 4 = m 2 n 2 = m5 + (4) (2)n3 2 6
(5p)3 10p 7p2 + p 3
= m3n
= 125p3 70p3 + 6p3 = 61p3 3
11
12
(f)
(g 3 h 5 ) 2 = gh 5
(e)
(4j4k) 2 2h3k 2
4
6
3
1
3
(m n
1 4
f
2
(e f )
2
=e
3 2
(e)
(f) 14
3 4 1 2
1
) 4
5
2 = e f2
=
g h
6 5
1
1
= 2j2k 2 2h3k 2
4
32m n
8
12 1
= m n 2m
j 2k h3
=
4 5
n
8 5
56
3
5 5 = m n 2
Using A = $5800, P = $5000 and n = 5 years. r 5 ) 100 r 5 5800 (1 + ) = 100 5000 1 r 58 1+ = ( )5 100 50
5800 = 5000(1 +
15
r = 3.01%
Let P be the original sum of money, then r = P + 96.60 = P(1 +
4.2 = 1.05 and n = 4. 4
1.05 4 ) 100
1.05 4 ) P = 96.60 100 1.05 4 P[(1 + ) 1] = 96.60 100
P(1 +
Math Homework Prepared by Mr Ang KS
P = $2264.09
7
Math Homework Prepared by Mr Ang KS
8
(f)
6 2 x+ =0 7 49 6 2 x2 x= 7 49 3 2 3 2 (x ) ( )2 = 7 7 49 3 2 9 2 (x ) = 7 49 49 3 1 x =± 7 7
x2
5
x = 0.81 or 0.05
(g)
x2 + 0.6x 1 = 0 x2 + 0.6x = 1 (x + 0.3)2 (0.3)2 = 1 (x + 0.3)2 = 0.09 + 1 x + 0.3 = ± 1.09 x = 0.74 or 1.34
(h)
x2 4.8x + 2 = 0 x2 4.8x = 2 (x 2.4)2 (2.4)2 = 2 (x 2.4)2 = 5.76 2 x 2.4 = ± 3.76 x = 4.34 or 0.46
(c)
(x + 2)(x 5) = 4x x2 3x 10 4x = 0 x2 7x 10 = 0 7 2 7 ) ( )2 = 10 2 2 7 2 49 (x ) = + 10 2 4 7 89 x =± 2 4
(x
(d)
x = 8.22 or 1.22
x(x 4) = 2(x + 7) x2 4x 2x 14 = 0 x2 6x 14 = 0 (x 3)2 (3)2 = 14 (x 3)2 = 9 + 14 x 3 = ± 23 x = 7.80 or 1.80
Exercise 1B 3 (e) (2x + 3)(x 1) x(x + 2) = 0 2x2 + x 3 (x2 + 2x) = 0 Math Homework Prepared by Mr Ang KS
1
x2 x 3 = 0 Using a = 1, b = 1 and c = 3, x= (f)
(b)
( 1) 2 4(1)(3) 2(1)
= 2.30 or 1.30
(4x 3)2 + (4x + 3)2 = 25 16x2 24x + 9 + (16x2 + 24x + 9) = 25 32x2 = 7 x2 =
4
(1)
7 32
x = ±0.468
3 2 1 x + 2x =0 4 2 3 1 Using a = , b = 2 and c = , 4 2
3 1 2 22 4( )( ) 4 2 x= 3 2( ) 4 = 0.230 or 2.90
(c)
5x 7 = x2 x2 5x + 7 = 0 Using a = 1, b = 5 and c = 7, x=
( 5)
( 5) 2 4(1)(7) 2(1)
There is no solution since
3
=
4
3 2
is not defined .
Exercise 1D 4
5
x( x 3) 3 = ( x 1) 2 5
(f)
5(x2 3x) = 3(x2 + 2x + 1)
2x2 21x 3 = 0
x=
( 21)
7 x 1 1 = 2 x 1 x3
( 21) 2 4( 2)(3) = 10.6 or 0.141 2( 2)
7( x 3) ( x 1)( x 1) 1 = ( x 1)( x 3) 2
2(7x + 21 x2 + 1) = x2 + 2x 3 3x2 12x 47 = 0
x=
( 12)
( 12) 2 4(3)(47) 2(3)
= 6.43 or
2.43 5
(g)
4 5 =2 ( x 2) 2 x2
Math Homework Prepared by Mr Ang KS
4 5 + =2 ( x 2) 2 x2 5( x 2) 4 =2 ( x 2) 2
5x 6 = 2(x2 4x + 4) 2
5
8
(h)
(i) (iii)
(iv)
2x2 13x + 14 = 0
x=
x 5 + =1 ( x 1) 2 x 1
(13)
( 13) 2 4( 2)(14) = 5.14 or 1.36 2( 2)
5( x 1) x =1 ( x 1) 2
6x 5 = (x2 2x + 1) x2 8x + 6 = 0
x=
60 pages/min x 60 60 + = 144 x x2 60( x 2) 60 x = 144 x ( x 2)
(ii)
120x + 120 = 144(x2 + 2x) 24: 5x + 5 = 6x2 + 12x 6x2 + 7x 5 = 0 (2x 1)(3x + 5) = 0
( 8)
( 8) 2 4(1)(6) 2(1)
= 7.16 or 0.838
60 pages/min x2
(Shown)
2 1 or x = 1 2 3 2 Reject x = 1 as x > 0 3 1 Required time taken = (2 60) 144 2
x=
(v)
11
(i)
Speed of coach = x 30, so time taken = Time taken by car =
= 6 mins
700 h x 30
700 h x
Total time taken = 20 h 700 700 + = 20 x 30 x 35 x 35( x 30) 20: =1 x ( x 30)
(ii)
70x 1050 = x2 30x x2 100x + 1050 = 0 (Shown) Using a = 1, b = 100 and c = 1050 x=
(iii)
( 100)
( 100) 2 4(1)(1050) 2(1)
= 88.079 or 11.921 = 88.08 or 11.92 Reject x = 11.92 as x = 30 becomes < 0 Time taken for return journey =
Math Homework Prepared by Mr Ang KS
700 88.079
= 7.95 h 3
Exercise 1E 1 (e) y = (3 x)(x + 2)
y = (2 x)(4 x)
(f)
y y = (3 x)(x + 2)
6
y = (2 x)(4 x)
y 2
3
0
x 8
2
y = (x + 1)2 3
(d)
y = (x 4)2 1
(f) y0 0
y
y = (x + 1)2 3
4
2
1
0
0.732
y = (x 4)2 1
x
Line of symmetry 2 is x = 1.
Line of symmetry is x = 4.
(i)
x2 8x + 5
(ii)
(iii)
Min. point = (4, 11).
(1, 3)3
7
x
4 (4, 1)
17 2.73
x
(iv)
= (x 4)2 (4)2 + 5 = (x 4)2 11
y y = (x 4)2 11
Line of symmetry is x = 4. 5
Exercise 2A 6
(g)
1 1 1 (3g + 4) (g + 1) 1 (g + 5) 5 3 3
15:
(h)
11
4
7.32
x
(4, 11)
3(3g + 4) 5(g + 1) 15 5(g + 5) 4g + 7 5g 10 9g 17 g 1
6
0 0.683
8 9
h 3 h + ) < 3( 5) 3 4 2 4h 9 h 10 4( ) < 3( ) 12 2
4(
6:
Math Homework Prepared by Mr Ang KS
2(4h + 9) < 9(h 10) 8h + 18 < 9h 90 h < 108 h > 108 4
7
1 p (2 p) 3 6 10
30:
5(2 p) 90 3p 5p 80 3p 8p 80 p 10 Largest possible value of p = 10 Exercise 2B 5 (c) 3x 3 < x 9 < 2x 3x 3 < x 9 2x < 6 x < 3 9 < x < 3 (d)
2x x + 6 < 3x + 5 2x x + 6 x6
and
x 9 < 2x x < 9 x > 9 9
and
x + 6 < 3x + 5 2x < 1 x>
8
3
1 2
1 <x6 2
0.5
6
Let x be the no. of correct answers, then (30 x) are the no. of wrong answers. 5x + (2)(30 x) 66 5x 60 + 2x 66 7x 126 x 18 Max. correct answers obtained = 18 Given:
10
1 1 x4> x 2 3
and
1 1 x+1< x+3 6 8
3x 24 > 2x 4x + 24 < 3x + 72 x > 24 x < 48 24 < x < 48 So, x = 29, 31, 37, 41, 43, 47 13
14
(d)
(d)
3d 5 < d + 1 2d + 1 3d 5 < d + 1 and 2d < 6 d<3 0d<3 2d d 1 + 5 2 5 2d d5< and 5
d + 1 2d + 1 d 0 d≥0
d5<
Math Homework Prepared by Mr Ang KS
2d d 1 + 5 2 5
5
3d <5 5 1 d<8 3
15
(c)
d 1 10 5
d ≥ 2
2 d < 8
1 3
4 7 3x 2 11 3x 5 11 5 ≥x≥ 3 3 2 2 1 x3 3 3
Required integers are 2 and 3. (d)
10 < 7 2x 1 17 < 2x 8 17 >x≥4 2 1 4x<8 2
Required integers are 4, 5, 6, 7 and 8. Exercise 3A 6 (e) (e3)5 (e2)4
7
8
= e15 e8
= e7
(f)
(4f 6)3 (2f 3)3 = 64f 18 (8f 9)
= 8f 9
(c)
(8e5f 3)2 (e3f)3 = 64e10f 6 (e9f 3)
= 64ef 3
(d)
16g8h7 (2g3h2)3
(c)
3e3 2 f
(d)
g2 3g 5 3 2 h 2h
4
27e9 11 f 6
= 16g8h7 (8g9h6)
=
3
81e12 f 11 f8 27e 9
= 2g1h
2h
= g
= 3e3f 3
8 8h 6 g 12 = 18 ( 15 ) = 27 g 3h12 27 g h
Exercise 3B 6
2
(f)
(1000) 3
7
(f)
1 3 ( f )5
8
(d)
10d = 0.01 10d = 102
Math Homework Prepared by Mr Ang KS
= ( 3 1000 )2 = (10)2 1
=
f
=f
5 3
= 100
5 3
d = 2 6
10
(g)
(h)
(m 5 n 3 ) ( m 2 ) 2 ( m 1n) 2
(m 5 n 3 ) m 4 = m 2 n 2 = m5 + (4) (2)n3 2 6
(5p)3 10p 7p2 + p 3
= m3n
= 125p3 70p3 + 6p3 = 61p3 3
11
12
(f)
(g 3 h 5 ) 2 = gh 5
(e)
(4j4k) 2 2h3k 2
4
6
3
1
3
(m n
1 4
f
2
(e f )
2
=e
3 2
(e)
(f) 14
3 4 1 2
1
) 4
5
2 = e f2
=
g h
6 5
1
1
= 2j2k 2 2h3k 2
4
32m n
8
12 1
= m n 2m
j 2k h3
=
4 5
n
8 5
56
3
5 5 = m n 2
Using A = $5800, P = $5000 and n = 5 years. r 5 ) 100 r 5 5800 (1 + ) = 100 5000 1 r 58 1+ = ( )5 100 50
5800 = 5000(1 +
15
r = 3.01%
Let P be the original sum of money, then r = P + 96.60 = P(1 +
4.2 = 1.05 and n = 4. 4
1.05 4 ) 100
1.05 4 ) P = 96.60 100 1.05 4 P[(1 + ) 1] = 96.60 100
P(1 +
Math Homework Prepared by Mr Ang KS
P = $2264.09
7
Math Homework Prepared by Mr Ang KS
8
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