66
Solutions
Chapter 2 Inequalities in One Unknown
Skills Assessment (P.54)
1. 21 + 3x 2 15
`
Exercise 2A (P.62)
1. (a)
3x 2 15 - 21
3x 2 -6
x 2 -2
01x18
(b) -10 G x G -4
(c)
Graphical representation:
3
-5 G x 1 4
1
1
(d) -2 3 1 x G - 2
–2
0
2. (a)
Graphical representation:
2. 2x - 14 G 16
2x G 16 + 14
2x G 30
x G 15
0
3
5
(b) Graphical representation:
Graphical representation:
0
–4
15
0
7
(c) Graphical representation:
3. (a)
1
(b) G
(c)
H
0
2
6
(d) Graphical representation:
(d) 2
0
3
(e) Graphical representation:
0
(f)
1
Graphical representation:
0
02 NSS TM 5A (E)-2P-KJ.indd 66
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67
Chapter 2: Inequalities in One Unknown
3. Solving x - 3 2 1:
x 2 4 .......................................... (i)
`
Solving x + 2 2 9:
`
x 2 7 ......................................... (ii)
According to the question, x must satisfy (i)
and (ii).
Therefore x 2 7.
According to the question, x must satisfy (i)
and (ii).
Therefore x G -3.
Graphical representation:
–3
0
Graphical representation:
0
7. Rewrite the compound inequality as
-5 1 3x - 2 and 3x - 2 1 4
we have -3 1 3x
and
3x 1 6
-1 1 x
and
x12
7
Therefore -1 1 x 1 2.
4. Solving 7x 1 16 + 3x:
`
4x 1 16
x 1 4 ............................................... (i)
Solving 9x 2 7x - 10:
2x 2 -10
`
x 2 -5 ........................................... (ii)
According to the question, x must satisfy (i)
and (ii).
Therefore -5 1 x 1 4.
Graphical representation:
Graphical representation:
–1
0
2
8. Rewrite the compound inequality as
1 - 2x 2 7 and 7 2 2x - 7
we have
-6 2 2x and 14 2 2x
-3 2 x and 7 2 x
Therefore x 1 -3.
Graphical representation:
–5
0
4
–3
5. Solving 3x H 16 - 5x:
`
8x H 16
x H 2 ............................................... (i)
Solving 13 + 4x G 17:
4x G 4
`
x G 1 ..................................... (ii)
[No values of x can satisfy both (i) and (ii).]
` This compound inequality has no solutions.
6. Solving 2x - 8 H 5x + 1:
`
-9 H 3x
x G -3 ..................................... (i)
Solving 11 - 2x G 21 - 4x:
2x G 10
`
x G 5 ..................................... (ii)
02 NSS TM 5A (E)-2P-KJ.indd 67
0
9. Rewrite the compound inequality as
15 + 2x G 8x - 3 and 8x - 3 G 2x - 9
we have 18 G 6x
and
6x G -6
3Gx
and
x G -1
Therefore the compound inequality has no
solutions.
10. Rewrite the compound inequality as
x - 2 H 16 - 5x and 16 - 5x H 11
we have 6x H 18
and
5 H 5x
xH3
and
1Hx
Therefore the compound inequality has no
solutions.
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68
Solutions
14 + 3x 2 2
3x 2 -12
x 2 -4 ......................................... (i)
11.
9 1 11 - 2x
2x 1 2
x 1 1 ................................................... (ii)
According to the question, x must satisfy (i)
and (ii).
Therefore -4 1 x 1 1.
Graphical representation:
1
2
0
and
15. Solving 3x + 2 1 6 - x:
4x 1 4
x 1 1 ........................................ (i)
`
x
Solving 2 - 5 G 2 + x:
Graphical representation:
x
-7 G 2
–4
12.
0
`
x H -14 .................................. (ii)
According to the question, x must satisfy (i)
and (ii).
Therefore -14 G x 1 1.
1
2x + 1 H 2
2x H 1
Graphical representation:
x H 21 ............................................ (i)
and 3 - 3x H 8 - 2x
-x H 5
x G -5 ........................................ (ii)
[No values of x can satisfy both (i) and (ii).]
` This compound inequality has no solutions.
13. No, they are incorrect.
3x + 4 G -5x and -5x G 5
8x G -4 and
x H -1
x G - 21
` -1 G x G - 21
2x - 1 1 3
2x 1 4
x 1 2 ............................................ (i)
14.
and
– 14
10x - 7 2 4x + 11
6x 2 18
x 2 3 ............................................ (i)
16.
and
x G 21 ......................................... (ii)
According to the question, x must satisfy (i)
and (ii).
Therefore x G 21 .
x+9
H3
4
x + 9 H 12
x H 3 ............................................. (ii)
According to the question, x must satisfy (i)
and (ii).
Therefore x 2 3.
Graphical representation:
4x + 1
G1
3
4x + 1 G 3
4x G 2
01
0
3
17. According to the question, we have
25t 2 3 000 ................................................... (i)
25t 1 5 000 .................................................. (ii)
From (i): 25t 2 3 000
t 2 120
From (ii): 25t 1 5 000
t 1 200
` The possible range of values of t is
(
120 1 t 1 200.
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2010/7/20 4:39:41 PM
Chapter 2: Inequalities in One Unknown
18. Solving
Graphical representation:
3 - 4x
G 1:
4
3 - 4x G 4
-4x G 1
0
x H - 41 ................................... (i)
`
Solving
21. (a)
2 (x + 3)
+ 2 G 3:
3
we have 3(v - 9) 1 12v
3v - 27 1 12v
-9v 1 27
v 2 -3
Therefore -3 1 v G 2.
2(x + 3) G 3
2x G -3
3
x G - 2 ....................... (ii)
[No values of x can satisfy both (i) and (ii).]
` This compound inequality has no solutions.
19. Rewrite the compound inequality as
and
and
and
and
22. (a)
2x - 4 1 6
2x 1 10
x15
x15
Graphical representation:
0
2
5
and
3
6v G 2 (v + 6)
and 12v G 3(v + 6)
and 12v G 3v + 18
and 9v G 18
and
vG2
(b) The possible values of v are -2, -1, 0, 1 and 2.
1
(6 - 3x) G 31 (2x - 4) and 31 (2x - 4) 1 2
4
3(6 - 3x) G 4(2x - 4)
18 - 9x G 8x - 16
34 G 17x
xH2
` 2Gx15
4
Rewrite the compound inequality as
3
(v - 9) 1 6v
2
2 (x + 3)
G1
3
`
69
3(0.1k - 0.5) G 1.35
0.3k - 1.5 G 1.35
0.3k G 2.85
k G 9.5 ....................... (i)
and 0.4(k + 3) G 1.3(2k - 5)
0.4k + 1.2 G 2.6k - 6.5
7.7 G 2.2k
k H 3.5 .......................... (ii)
According to the question, k must satisfy
(i) and (ii).
Therefore 3.5 G k G 9.5.
(b) From (a),
the maximum value of k = 9.5
20.
4(x - 2) H 12 - x
4x - 8 H 12 - x
5x H 20
x H 4 ........................................... (i)
and x - 10 G 6(3 - x)
x - 10 G 18 - 6x
7x G 28
x G 4 ........................................... (ii)
According to the question, x must satisfy (i)
and (ii).
Therefore x = 4.
02 NSS TM 5A (E)-2P-KJ.indd 69
the minimum value of k = 3.5
23. According to the question, we have
*
6x - 30c 2 90c ............................................. (i)
6x - 30c 1 162c .......................................... (ii)
From (i): 6x - 30c 2 90c
6x 2 120c
x 2 20c
From (ii): 6x - 30c 1 162c
6x 1 192c
x 1 32c
` The possible range of x is 20c 1 x 1 32c.
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70
Solutions
24. According to the question, we have
*
2 (2x + 10 + 25) 1 160 ................................ (i)
25 (2x + 10) H 1 200 ................................... (ii)
From (i): 2(2x + 10 + 25) 1 160
2x + 35 1 80
2x 1 45
Exercise 2B (P.70)
1. (a)
(b) x G -6 or x H 10
(c)
45
x1 2
From (ii): 25(2x + 10) H 1 200
2x + 10 H 48
2x H 38
x H 19
x 1 -4 or x 2 0
x 1 21 or x H 3
4
1
(d) x G -3 2 or x 2 - 5
2. (a)
Graphical representation:
45
` 19 G x 1 2
a x is an integer.
` The maximum value of x is 22.
0
4
6
(b) Graphical representation:
25. According to the question, we have
*
2 (x + 5 + x) H 50 ........................................ (i)
–4
2 (x + 5 + x) G 100 ...................................... (ii)
From (i): 2(x + 5 + x) H 50
2x + 5 H 25
2x H 20
x H 10
From (ii): 2(x + 5 + x) G 100
2x + 5 G 50
2x G 45
x G 22.5
` The possible range of values of x is
AB + AC 2 BC
0
(c) Graphical representation:
–5
0
(d) Graphical representation:
0
10 G x G 22.5.
26. (a)
–2
13
(e) Graphical representation:
(b) AB + BC 2 AC
(c)
From (a) and (b),
6 + x 2 17 ............................................ (i)
*
6 + 17 2 x ........................................... (ii)
From (i): 6 + x 2 17
x 2 11
From (ii): 6 + 17 2 x
x 1 23
` The solutions of the compound
–3
(f)
0
2
Graphical representation:
–4 –3 –2 –1 0 1 2 3 4
inequality are 11 1 x 1 23.
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Chapter 2: Inequalities in One Unknown
3. Solving 3x - 7 2 8:
`
71
6. Solving 5 + 3x G 9 - x:
3x 2 15
x 2 5 ........................................ (i)
`
4x G 4
x G 1 ........................................ (i)
Solving 5x + 1 1 11:
5x 1 10
`
x 1 2 ....................................... (ii)
According to the question, x must satisfy (i)
or (ii).
Therefore x 2 5 or x 1 2.
Solving 5x - 1 H 3x + 13:
2x H 14
`
x H 7 ....................................... (ii)
According to the question, x must satisfy (i)
or (ii).
Therefore x G 1 or x H 7.
Graphical representation:
Graphical representation:
0
2
4. Solving 5x - 8 2 2:
`
7
7. Solving 2x + 5 H 4x - 3:
5x 2 10
x 2 2 ........................................ (i)
Solving 4x 2 -1:
`
0 1
5
x 2 - 41 ......................................... (ii)
According to the question, x must satisfy (i)
or (ii).
Therefore x 2 - 41 .
`
8 H 2x
x G 4 ........................................ (i)
Solving 6x + 7 G -3x - 2:
9x G -9
`
x G -1 .................................... (ii)
According to the question, x must satisfy (i)
or (ii).
Therefore x G 4.
Graphical representation:
Graphical representation:
0
–1
4
4
0
8. Solving 8x - (3x - 2) H 7:
5. Solving 3x - 17 1 4:
`
3x 1 21
x 1 7 ...................................... (i)
`
5x + 2 H 7
5x H 5
x H 1 ............................ (i)
Solving 14 + 4x 2 -3x:
7x 2 -14
`
x 2 -2 ................................. (ii)
Since all values of x can satisfy either (i) or (ii),
any real number is a solution of this compound
inequality.
i.e. x is any real number.
Solving 3x - 11 G 13 - 3x:
6x G 24
`
x G 4 ..................................... (ii)
Since all values of x can satisfy either (i) or (ii),
any real number is a solution of this compound
inequality.
i.e. x is any real number.
Graphical representation:
Graphical representation:
–4 –3 –2 –1 0 1 2 3 4
02 NSS TM 5A (E)-2P-KJ.indd 71
–4 –3 –2 –1 0 1 2 3 4
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72
Solutions
9. Solving 4(x - 3) 1 1 - x:
Solving 21 (6x + 4) H 11:
4x - 12 1 1 - x
5x 1 13
13
x 1 5 .................................. (i)
`
Solving 7x H 3(x - 2):
7x H 3x - 6
4x H -6
3
x H - 2 .......................................... (ii)
`
Since all values of x can satisfy either (i) or (ii),
any real number is a solution of this compound
inequality.
i.e. x is any real number.
6x + 4 H 22
6x H 18
`
x H 3 ................................ (ii)
According to the question, x must satisfy (i)
or (ii).
Therefore x 2 2.
Graphical representation:
0
12. (a)
2
No.
Graphical representation:
(b) No.
13. Solving 4x - 1 2
–4 –3 –2 –1 0 1 2 3 4
16x - 4 2 x + 1
15x 2 5
4x + 5
10. Solving - 3
H 5:
4x + 5 G -15
4x G -20
x G -5 ................................. (i)
`
Solving 7 - (3x + 8) 1 2x:
7 - 3x - 8 1 2x
-1 1 5x
x 2 - 51 ........................ (ii)
`
x+1
:
4
`
x 2 31
Solving 3(x + 1) 2 x + 5:
3x + 3 2 x + 5
2x 2 2
`
x21
Therefore x 2 31 .
Graphical representation:
According to the question, x must satisfy (i)
or (ii).
0
Therefore x G -5 or x 2 - 51 .
Graphical representation:
1
3
14. Solving 3(2x + 1) 1 2x + 1:
6x + 3 1 2x + 1
4x 1 -2
–5
– 10
5
1
11. Solving 5 (7x - 4) 2 2:
`
7x - 4 2 10
7x 2 14
x 2 2 ................................. (i)
02 NSS TM 5A (E)-2P-KJ.indd 72
`
Solving
x 1 - 21
19 - 3x
H 2x - 4:
5
19 - 3x H 10x - 20
39 H 13x
`
xG3
Therefore x G 3.
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73
Chapter 2: Inequalities in One Unknown
Graphical representation:
17. (a)
Solving
10m - 5
+ 6 1 16:
7
10m - 5
1 10
7
0
3
15. Solving 2x + 3 2 9 - x:
`
3x 2 6
x 2 2 ........................................ (i)
`
x
Solving 5 - 21 G
Solving
2x - 5 G 5 - 3x
5x G 10
`
x G 2 ..................................... (ii)
Since all values of x can satisfy either (i) or (ii),
any real number is a solution of this compound
inequality.
i.e. x is any real number.
–4 –3 –2 –1 0 1 2 3 4
x
16. Solving 2 + 2 H
`
Graphical representation:
0
3
Solving 2 (x - 3) 1 3 - x:
3(x - 3) 1 2(3 - x)
3x - 9 1 6 - 2x
5x 1 15
`
x 1 3 .................................. (ii)
Since all values of x can satisfy either (i) or (ii),
any real number is a solution of this compound
inequality.
i.e. x is any real number.
Graphical representation:
8
(b) The maximum value of m is 8.
18. (a)
Solving
30 (2k - 1) - 1
G -2:
8
30(2k - 1) - 1 G -16
60k - 31 G -16
60k G 15
x+5
:
3
3x + 12 H 2(x + 5)
3x + 12 H 2x + 10
x H -2 ..................................... (i)
2m + 5
G 3:
7
2m + 5 G 21
2m G 16
`
mG8
Therefore m G 8.
5 - 3x
:
10
Graphical representation:
10m - 5 1 70
10m 1 75
m 1 7.5
k G 41 ................. (i)
`
k-4
Solving 21 a 8 + 8 k 2 4:
k-4
+424
16
k-4
20
16
k-420
k 2 4 .................. (ii)
`
According to the question, k must satisfy
(i) or (ii).
Therefore k G 41 or k 2 4.
Graphical representation:
–4 –3 –2 –1 0 1 2 3 4
0 1
4
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4
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74
Solutions
(b) The numbers that can satisfy the compound
inequality in (a) are:
-2, 0, 81 , 41 , 8.
From the sketch,
y G 0 when -4 G x G 3.
` The required solutions are -4 G x G 3.
Graphical representation:
19. According to the question, we have
10 + x 1 12 or 10 + x 2 65
` x 1 2 or
x 2 55
20. (a)
–4
According to the question, we have
4 000 + 0.1x 1 6 000
0.1x 1 2 000
x 1 20 000
or 4 000 + 0.1x 2 18 000
0.1x 2 14 000
x 2 140 000
` The range of sales revenue of those to
be interviewed is less than $20 000 or
0
3
2. The corresponding quadratic function is
y = (x - 2)(3x + 1).
When y = 0, we have
(x - 2)(3x + 1) = 0
` The x-intercepts of the graph of the
quadratic function are 2 and - 31 .
Since the coefficient of x2 is 3 (2 0), the
parabola opens upward.
Sketch:
greater than $140 000.
y
(b) Graphical representation of the range of
values of x in (a):
0 20 000
y = (x – 2)(3x + 1)
140 000
–1
0
3
x
2
Exercise 2C (P.79)
From the sketch,
1. The corresponding quadratic function is
y = (x - 3)(x + 4).
When y = 0, we have
(x - 3)(x + 4) = 0
` The x-intercepts of the graph of the
quadratic function are 3 and -4.
Since the coefficient of x 2 is 1 (2 0), the
parabola opens upward.
Sketch:
y
02 NSS TM 5A (E)-2P-KJ.indd 74
` The required solutions are x 1 - 31 or
x 2 2.
Graphical representation:
–1 0
3
y = (x – 3)(x + 4)
–4
y 2 0 when x 1 - 31 or x 2 2.
0
x
3
2
3. The corresponding quadratic function is
y = 2x(x - 5).
When y = 0, we have
2x(x - 5) = 0
` The x-intercepts are 0 and 5.
2010/7/20 4:40:02 PM
Chapter 2: Inequalities in One Unknown
Since the coefficient of x 2 is 2 (2 0), the
parabola opens upward.
Sketch:
Graphical representation:
–4
y
y = 2x(x – 5)
0
5
From the sketch,
y H 0 when x G 0 or x H 5.
` The required solutions are x G 0 or x H 5.
y = x2 - 2x - 8.
When y = 0, we have
x2 - 2x - 8 = 0
i.e. (x - 4)(x + 2) = 0
` The x-intercepts are 4 and -2.
Since the coefficient of x 2 is 1 (2 0), the
parabola opens upward.
Sketch:
y
y = x2 – 2x – 8
Graphical representation:
0
4
5. The corresponding quadratic function is
x
0
75
5
x
0
–2
4
2
4. The inequality can be written as x - 16 1 0.
The corresponding quadratic function is
y = x2 - 16.
When y = 0, we have
x2 - 16 = 0
i.e. (x + 4)(x - 4) = 0
` The x-intercepts are -4 and 4.
Since the coefficient of x 2 is 1 (2 0), the
parabola opens upward.
Sketch:
y
0
Graphical representation:
0
4
6. The corresponding quadratic function is
x
4
From the sketch,
y 1 0 when -4 1 x 1 4.
` The required solutions are -4 1 x 1 4.
02 NSS TM 5A (E)-2P-KJ.indd 75
x 2 4.
–2
y = x2 – 16
–4
From the sketch,
y 2 0 when x 1 -2 or x 2 4.
` The required solutions are x 1 -2 or
y = 3x2 - 17x + 10.
When y = 0, we have
3x2 - 17x + 10 = 0
i.e. (x - 5)(3x - 2) = 0
` The x-intercepts are 5 and 23 .
Since the coefficient of x2 is 3 (2 0), the
parabola opens upward.
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76
Solutions
Sketch:
From the sketch,
7
y 1 0 when 0 1 x 1 8 .
y
y = 3x2 – 17x + 10
7
` The required solutions are 0 1 x 1 8 .
Graphical representation:
0
x
5
2
3
0
7
8
2
8. The inequality can be written as 4x + 25x H 0.
From the sketch,
y G 0 when 23 G x G 5.
` The required solutions are 23 G x G 5.
Graphical representation:
The corresponding quadratic function is
y = 4x2 + 25x.
When y = 0, we have
4x2 + 25x = 0
i.e. x(4x + 25) = 0
25
` The x-intercepts are 0 and - 4 .
0 2
Since the coefficient of x 2 is 4 (2 0), the
parabola opens upward.
Sketch:
5
3
2
7. The inequality can be written as 8x - 7x 1 0.
The corresponding quadratic function is
y = 8x2 - 7x.
When y = 0, we have
8x2 - 7x = 0
i.e. x(8x - 7) = 0
7
` The x-intercepts are 0 and 8 .
Since the coefficient of x 2 is 8 (2 0), the
parabola opens upward.
Sketch:
y
2
y = 4x + 25x
0
– 25
4
x
From the sketch,
25
y H 0 when x G - 4 or x H 0.
y
y = 8x2 – 7x
25
` The required solutions are x G - 4 or
x H 0.
0
x
7
8
Graphical representation:
– 25
4
02 NSS TM 5A (E)-2P-KJ.indd 76
0
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77
Chapter 2: Inequalities in One Unknown
9. The corresponding quadratic function is
Sketch:
2
y = 7x - x .
When y = 0, we have
7x - x2 = 0
i.e. -x(x - 7) = 0
` The x-intercepts are 0 and 7.
Since the coefficient of x2 is -1 (1 0), the
parabola opens downward.
Sketch:
y
x
0
–3
2
y = 6 – x – x2
y
x
0
7
From the sketch,
y 2 0 when -3 1 x 1 2.
` The required solutions are -3 1 x 1 2.
Graphical representation:
y = 7 x – x2
–3
From the sketch,
y G 0 when x G 0 or x H 7.
` The required solutions are x G 0 or x H 7.
Graphical representation:
0
7
10. The corresponding quadratic function is
y = 6 - x - x2.
When y = 0, we have
6 - x - x2 = 0
i.e. -(x + 3)(x - 2) = 0
` The x-intercepts are -3 and 2.
Since the coefficient of x2 is -1 (1 0), the
parabola opens downward.
0
2
11. The corresponding quadratic function is
y = 4 + 3x - x2.
When y = 0, we have
4 + 3x - x2 = 0
i.e. -(x + 1)(x - 4) = 0
` The x-intercepts are -1 and 4.
Since the coefficient of x2 is -1 (1 0), the
parabola opens downward.
Sketch:
y
y = 4 + 3x – x2
–1
0
x
4
From the sketch,
y H 0 when -1 G x G 4.
` The required solutions are -1 G x G 4.
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78
Solutions
Since the coefficient of x 2 is 1 (2 0), the
parabola opens upward.
Sketch:
Graphical representation:
–1
0
4
y
2
12. The corresponding quadratic function is
y = x + 7x + 15
y = 2 - 3x - 2x2.
When y = 0, we have
2 - 3x - 2x2 = 0
i.e. -(x + 2)(2x - 1) = 0
0
` The x-intercepts are -2 and 21 .
Since the coefficient of x2 is -2 (1 0), the
parabola opens downward.
Sketch:
a y 2 0 for all values of x.
` x can be any real number.
14. The corresponding quadratic function is
y
y = 2 – 3x – 2x2
–2
x
x
0 1
2
From the sketch,
y = 3x2 - 8x + 6.
When y = 0, we have
3x2 - 8x + 6 = 0
Since the discriminant of 3x2 - 8x + 6 = 0 is
(-8)2 - 4(3)(6) = -8 1 0,
y = 3x2 - 8x + 6 has no x-intercepts.
Since the coefficient of x2 is 3 (2 0), the
parabola opens upward.
Sketch:
y
y 1 0 when x 1 -2 or x 2 21 .
y = 3x2 – 8x + 6
` The required solutions are x 1 -2 or
x 2 21 .
Graphical representation:
0
–2
0
1
2
x
a y G 0 is not true for all values of x.
` The inequality has no solutions.
13. The corresponding quadratic function is
y = x2 + 7x + 15.
When y = 0, we have
x2 + 7x + 15 = 0
Since the discriminant of x2 + 7x + 15 = 0 is
72 - 4(1)(15) = -11 1 0,
y = x2 + 7x + 15 has no x-intercepts.
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Chapter 2: Inequalities in One Unknown
15.
5x 1 2(2x - 1) + x2
5x 1 4x - 2 + x2
0 1 x2 - x - 2
2
x -x-220
The corresponding quadratic function is
y = x2 - x - 2.
When y = 0, we have
x2 - x - 2 = 0
i.e. (x + 1)(x - 2) = 0
` The x-intercepts are -1 and 2.
Since the coefficient of x 2 is 1 (2 0), the
parabola opens upward.
Sketch:
Sketch:
y
y = x2 – 4x + 4
2
From the sketch,
y = 0 when x = 2.
` The required solution is x = 2.
Graphical representation:
y = x2 – x – 2
–1
x
0
y
0
x
0
79
2
2
17.
From the sketch,
y 2 0 when x 1 -1 or x 2 2.
` The required solutions are x 1 -1 or
x 2 2.
Graphical representation:
x(x - 1) 2 (2x - 5)(x + 1)
x2 - x 2 2x2 - 3x - 5
0 2 x2 - 2x - 5
2
x - 2x - 5 1 0
The corresponding quadratic function is
y = x2 - 2x - 5.
When y = 0, we have
x2 - 2x - 5 = 0
2
x=
–1
16.
0
2
2x(x + 6) H (3x + 2)(x + 2)
2x2 + 12x H 3x2 + 8x + 4
0 H x2 - 4x + 4
2
x - 4x + 4 G 0
The corresponding quadratic function is
y = x2 - 4x + 4.
When y = 0, we have
x2 - 4x + 4 = 0
i.e. (x - 2)2 = 0
`
x = 2 (repeated)
Since the coefficient of x 2 is 1 (2 0), the
parabola opens upward.
02 NSS TM 5A (E)-2P-KJ.indd 79
=
-(-2) ! (-2) - 4 (1)(-5)
2 (1)
2 ! 24
(or 1 !
2
6)
Since the coefficient of x 2 is 1 (2 0), the
parabola opens upward.
Sketch:
y
y = x2– 2 x – 5
2 – 24
2
0
x
2 + 24
2
2010/7/20 4:40:11 PM
80
Solutions
19. (a)
From the sketch,
y 1 0 when
2 + 24
2 - 24
1x1
.
2
2
` The required solutions are
The corresponding quadratic function is
y = 2x2 - 9x - 21.
When y = 0, we have
2x2 - 9x - 21 = 0
2
2 - 24
2 + 24
1x1
2
2
(or 1 -
6 1x11+
x=
=
6 ).
Graphical representation:
0
2 – 24
2
- (-9) ! (-9) - 4 (2)(-21)
2 (2)
9 ! 249
4
Since the coefficient of x2 is 2 (2 0), the
parabola opens upward.
Sketch:
y
2 + 24
2
y = 2 x2– 9 x – 21
18.
4x - 5 G (2x + 1)(2x - 1)
4x - 5 G 4x2 - 1
0 G 4x2 - 4x + 4
2
x -x+1H0
The corresponding quadratic function is
y = x2 - x + 1.
When y = 0, we have
x2 - x + 1 = 0
Since the discriminant of x2 - x + 1 = 0 is
(-1)2 - 4(1)(1) = -3 1 0,
y = x2 - x + 1 has no x-intercepts.
Since the coefficient of x 2 is 1 (2 0), the
parabola opens upward.
Sketch:
9 – 249
4
0
x
9 + 249
4
From the sketch,
y 2 0 when x 1
x2
9 - 249
or
4
9 + 249
.
4
` The required solutions are
x1
9 - 249
9 + 249
or x 2
.
4
4
Graphical representation:
y
y = x2 – x + 1
9 – 249 0
4
9 + 249
4
(b) From the sketch in (a), we have
0
a y 2 0 for all values of x.
` x can be any real number.
Graphical representation:
x
y H 0 when x G
xH
9 - 249
or
4
9 + 249
.
4
` The required solutions are
xG
9 - 249
9 + 249
or x H
.
4
4
–4 –3 –2 –1 0 1 2 3 4
02 NSS TM 5A (E)-2P-KJ.indd 80
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81
Chapter 2: Inequalities in One Unknown
20. (a)
-4x2 G 49 - 28x
4x2 - 28x + 49 H 0
The corresponding quadratic function is
y = 4x2 - 28x + 49.
When y = 0, we have
4x2 - 28x + 49 = 0
i.e.
(2x - 7)2 = 0
Since the coefficient of x2 is -1 (1 0),
the parabola opens downward.
Sketch:
y
y = 1 + 3 x – x2
7
`
x = 2 (repeated)
Since the coefficient of x2 is 4 (2 0), the
parabola opens upward.
Sketch:
3 – 13
2
y
y = 4x2 – 28x + 49
x
0
3 + 13
2
From the sketch,
y H 0 when
3 - 13
3 + 13
GxG
.
2
2
` The required solutions are
3 - 13
3 + 13
GxG
.
2
2
x
0
7
2
Graphical representation:
a y H 0 for all values of x.
` x can be any real number.
3 – 13 0
2
Graphical representation:
3 + 13
2
(b) From the sketch in (a), we have
–4 –3 –2 –1 0 1 2 3 4
(b)
21. (a)
y 1 0 when x 1
2
-4x 2 49 - 28x
4x2 - 28x + 49 1 0
The corresponding quadratic function is
y = 4x2 - 28x + 49.
a y 1 0 is not true for all values of x.
` The inequality has no solutions.
The corresponding quadratic function is
y = 1 + 3x - x2.
When y = 0, we have
1 + 3x - x2 = 0
x2 - 3x - 1 = 0
2
x=
=
02 NSS TM 5A (E)-2P-KJ.indd 81
-(-3) ! (-3) - 4 (1)(-1)
2 (1)
3 ! 13
2
3 - 13
3 + 13
or x 2
.
2
2
` The required solutions are
x1
22. (a)
3 - 13
3 + 13
or x 2
.
2
2
x + 7 G (x - 2)(3x + 4)
x + 7 G 3x2 - 2x - 8
0 G 3x2 - 3x - 15
2
x -x-5H0
The corresponding quadratic function is
y = x2 - x - 5.
When y = 0, we have
x2 - x - 5 = 0
2
x=
=
-(-1) ! (-1) - 4 (1)(-5)
2 (1)
1 ! 21
2
2010/7/20 4:40:20 PM
82
Solutions
Since the coefficient of x2 is 1 (2 0), the
parabola opens upward.
Sketch:
y
y = x2– x – 5
x
0
1 – 21
2
1 + 21
2
2
2. x + 13x + 36 G 0
(x + 4)(x + 9) G 0
` x + 4 H 0 and x + 9 G 0, i.e. x H -4
and x G -9, this compound inequality has
no solutions.
or x + 4 G 0 and x + 9 H 0, i.e. x G -4
and x H -9, the solutions of this
compound inequality are -9 G x G -4.
` The solutions of the quadratic inequality
x2 + 13x + 36 G 0 are -9 G x G -4.
Graphical representation:
From the sketch,
y H 0 when x G
–9
1 - 21
1 + 21
or x H
.
2
2
` The required solutions are
1 – 21
2
0
1 + 21
2
(b) From the sketch in (a), we have
1 - 21
1 + 21
y 1 0 when
1x1
.
2
2
` The required solutions are
1 - 21
1 + 21
1x1
.
2
2
0
3. A table is created below.
1 - 21
1 + 21
xG
or x H
.
2
2
Graphical representation:
–4
3x + 4
x-7
(3x + 4)(x - 7)
x 1 - 43
-
-
+
x = - 43
0
-
0
- 43 1 x 1 7
+
-
-
x=7
+
0
0
x27
+
+
+
From the above table, x G - 43 or x H 7 when
(3x + 4)(x - 7) H 0.
` The solutions of (3x + 4)(x - 7) H 0 are
Exercise 2D (P.90)
x G - 43 or x H 7.
1. (x - 2)(x - 6) 2 0
` x - 2 2 0 and x - 6 2 0, i.e. x 2 2 and
x 2 6, the solutions of this compound
inequality are x 2 6.
or x - 2 1 0 and x - 6 1 0, i.e. x 1 2 and
x 1 6, the solutions of this compound
inequality are x 1 2.
` The solutions of the quadratic inequality
(x - 2)(x - 6) 2 0 are x 1 2 or x 2 6.
Graphical representation:
–4 0
3
7
Graphical representation:
0
02 NSS TM 5A (E)-2P-KJ.indd 82
2
6
2010/7/20 4:40:25 PM
83
Chapter 2: Inequalities in One Unknown
4.
5x2 + 8x 1 4
5x2 + 8x - 4 1 0
i.e. The inequality can be written as
(x + 2)(5x - 2) 1 0.
A table is created below.
x+2
x 1 -2
-
6.
` The critical points are x = - 17 and x = 3.
5x - 2 (x + 2)(5x - 2)
-
20x + 3 G 7x2
0 G 7x2 - 20x - 3
2
7x - 20x - 3 H 0
Factorizing 7x2 - 20x - 3 H 0, we have
(7x + 1)(x - 3) H 0
i.e.
+
interval 1 interval 2 interval 3
–1
3
7
x = -2
0
-
0
-2 1 x 1 25
+
-
-
x = 25
+
0
0
x = - 17 and x = 3.
x 2 25
+
+
+
In interval 2: a- 17 1 x 1 3 k
7x2 - 20x - 3 = 0 when x = - 17 or x = 3.
i.e. The (partial) solutions of the inequality are
Take x = 0, then [7(0) + 1](0 - 3) = -3 1 0.
From the above table, -2 1 x 1 25
Therefore the required solutions are x G - 17
when (x + 2)(5x - 2) 1 0.
or x H 3.
∴ The solutions of 5x2 + 8x 1 4 are
-2 1 x 1 25 .
7.
Graphical representation:
6x2 - 5x - 4 1 0
(2x + 1)(3x - 4) 1 0
` - 21 1 x 1 43
–2
0
2
5
Graphical representation:
2
5. Factorizing 2x - 8x 1 0, we have
2x(x - 4) 1 0
` The critical points are x = 0 and x = 4.
i.e. interval 1 interval 2 interval 3
0
4
2x2 - 8x = 0 when x = 0 or x = 4.
In interval 2: (0 1 x 1 4)
Take x = 1, then 2(1)(1 - 4) = -6 1 0.
Therefore the required solutions are 0 1 x 1 4.
–
1
2
4
3
2
8. -2x + 5x + 3 G 0
2x2 - 5x - 3 H 0
(2x + 1)(x - 3) H 0
` x G - 21 or x H 3
Graphical representation:
–
02 NSS TM 5A (E)-2P-KJ.indd 83
0
1 0
2
3
2010/7/20 4:40:29 PM
84
9.
Solutions
x(x - 2) 2 4(x - 2)
x(x - 2) - 4(x - 2) 2 0
(x - 2)(x - 4) 2 0
` x 1 2 or x 2 4
13. (a)
Graphical representation:
0
10.
2
2
(b) For the quadratic equation x + kx - 3k =
4
0 to have no real roots, the discriminant of
the equation = k2 - 4(1)(-3k) 1 0.
k2 + 12k 1 0
k(k + 12) 1 0
` -12 1 k 1 0
2
3x G -2(7x + 4)
2
3x + 14x + 8 G 0
(x + 4)(3x + 2) G 0
` -4 G x G - 23
14. Substituting x = 3 into (x - k)(x - 2k) 1 2,
Graphical representation:
–4
11.
For the quadratic equation x2 + kx - 3k =
0 to have two distinct real roots, the
discriminant of the equation =
k2 - 4(1)(-3k) 2 0.
k2 + 12k 2 0
k(k + 12) 2 0
` k 1 -12 or k 2 0
we have
(3 - k)(3 - 2k) 1 2
9 - 3k - 6k + 2k2 - 2 1 0
2k2 - 9k + 7 1 0
(k - 1)(2k - 7) 1 0
–2 0
3
x2 + 14 H 6x
x2 - 6x + 14 H 0
x2 - 6x + 14 = x2 - 6x + 9 + 5
= (x - 3)2 + 5
a For all real values of x, (x - 3)2 + 5 H 0
holds.
` The solutions of x2 + 14 H 6x can be
any real number.
Graphical representation:
7
` 11k1 2
15. According to the question, we have
x(x - 7) 2 98
x2 - 7x - 98 2 0
(x + 7)(x - 14) 2 0
` x 1 -7 (rejected) or x 2 14
` The range of values of the larger number x
is x 2 14.
16. (x - 2)(2 - x) H 0
(x - 2)(x - 2) G 0
(x - 2)2 G 0
` x=2
–4 –3 –2 –1 0 1 2 3 4
12.
4x2 1 4x - 3
4x2 - 4x + 3 1 0
4x2 - 4x + 3 = 4(x2 - x) + 3
Graphical representation:
2
= 4 a x - x + 41 k + 2
0
2
= 4 a x - 21 k + 2
2
a For all real values of x, 4 a x - 21 k + 2 1 0
does not hold.
` 4x2 1 4x - 3 has no solutions.
2
02 NSS TM 5A (E)-2P-KJ.indd 84
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85
Chapter 2: Inequalities in One Unknown
20. Let the smaller number be x, then the larger
3x (x - 3)
1 1 - 2x
2
17.
number is x + 1.
According to the question, we have
x2 + (x + 1)2 G 25
x2 + x2 + 2x + 1 - 25 G 0
2x2 + 2x - 24 G 0
x2 + x - 12 G 0
(x + 4)(x - 3) G 0
`
-4 G x G 3
a x is a positive integer.
` The possible values of the smaller number
3x(x - 3) 1 2 - 4x
2
3x - 9x + 4x - 2 1 0
2
3x - 5x - 2 1 0
(3x + 1)(x - 2) 1 0
` - 31 1 x 1 2
Graphical representation:
–1 0
are 1, 2 and 3.
2
3
21. Let the length of the rectangular field along the
18. (3x + 2)(3x - 4) - (3x - 4)(2x + 3) H 0
(3x - 4)(3x + 2 - 2x - 3) H 0
(3x - 4)(x - 1) H 0
river be x m, then the width of the field is
a 100 - x k m, i.e. a 50 - x k m.
2
2
The area of the field must be at least 800 m2.
` x G 1 or x H 43
i.e.
Graphical representation:
x a 50 - 2 k H 800
x
50x - 21 x2 - 800 H 0
0
19.
1
4
3
(x + 3)(2x - 1) 1 6x2 + 5x + 1
2x2 + 6x - x - 3 1 6x2 + 5x + 1
0 1 4x2 + 4
4x2 + 4 2 0
2
a For all real values of x, 4x + 4 2 0 holds.
` The solutions of (x + 3)(2x - 1) 1
6x2 + 5x + 1 can be any real number.
Graphical representation:
–4 –3 –2 –1 0 1 2 3 4
02 NSS TM 5A (E)-2P-KJ.indd 85
x2 - 100x + 1 600 G 0
(x - 20)(x - 80) G 0
`
20 G x G 80
` The maximum length of the field along the
river will be 80 m.
22. (a)
Total surface area of the cylinder
= [2rr2 + 2rr(r - 2)] cm2
= (4rr2 - 4rr) cm2
(b) According to the question, we have
4rr2 - 4rr 1 48r
r2 - r 1 12
r2 - r - 12 1 0
(r + 3)(r - 4) 1 0
`
-3 1 r 1 4
a r 2 0 and r - 2 2 0, i.e. r 2 2
` 21r14
2010/7/20 4:40:35 PM
86
Solutions
23. (a)
According to the question, we have
m(36 - m) 2 320
36m - m2 - 320 2 0
m2 - 36m + 320 1 0
(m - 16)(m - 20) 1 0
`
16 1 m 1 20
` The possible values of m are 17, 18
(b) Solving 17 - 4x 1 x + 2:
15 1 5x
`
x 2 3 ............................. (i)
Solving 23x - 3 H 5x:
18x H 3
x H 61 ........................... (ii)
`
According to the question, x must satisfy
(i) and (ii).
Therefore x 2 3.
and 19.
(b) When m = 17,
the number of seats in the hall
= 17(36 - 17)
= 323
When m = 18,
the number of seats in the hall
= 18(36 - 18)
= 324
When m = 19,
the number of seats in the hall
= 19(36 - 19)
= 323
` The maximum number of seats in the
Graphical representation:
0
2. (a)
3
Rewrite the compound inequality as
2x - 2 G 6 and 6 G 1 - 5x
2x G 8 and 5x G -5
x G 4 and x G -1
` x G -1
Graphical representation:
hall is 324.
–1
Supp. Exercise 2 (P.95)
1. (a)
Solving 5x H 16 - 3x:
8x H 16
`
x H 2 ....................................... (i)
Solving 9x G 35 + 4x:
5x G 35
`
x G 7 ...................................... (ii)
According to the question, x must satisfy
(i) and (ii).
Therefore 2 G x G 7.
Graphical representation:
0
02 NSS TM 5A (E)-2P-KJ.indd 86
2
7
0
(b) Rewrite the compound inequality as
25 - 7x 1 3 + x and 3 + x G 12 - 8x
22 1 8x
and
9x G 9
x 2 11
4
and
xG1
` The compound inequality has no
solutions.
3. (a)
7 + 2x H -3
2x H -10
x H -5 ................................. (i)
and 8x - 9 1 7
8x 1 16
x 1 2 ................................... (ii)
According to the question, x must satisfy
(i) and (ii).
Therefore -5 G x 1 2.
2010/7/20 4:40:37 PM
Chapter 2: Inequalities in One Unknown
Graphical representation:
(c)
Solving 3x - 4 H 15 - 2x:
5x H 19
19
x H 5 ............................ (i)
`
–5
0
2
Solving
2x + 11 1 -1
2x 1 -12
x 1 -6 ................................ (i)
(b)
and
4x + 3
25
2
4x + 3 2 10
4x 2 7
87
x-5
1 3:
2
x-516
`
x 1 11 .............................. (ii)
According to the question, x must satisfy
(i) or (ii).
Therefore x can be any real number.
Graphical representation:
7
x 2 4 ................................. (ii)
[No values of x can satisfy both (i) and (ii).]
` This compound inequality has no
(d) Solving 2x + 7 2 10 - x:
solutions.
4. (a)
Solving 3x - 2 1 4:
3x 1 6
`
x 1 2 ............................... (i)
Solving -2x 1 -20:
2x 2 20
`
x 2 10 ................................. (ii)
According to the question, x must satisfy
(i) or (ii).
Therefore x 1 2 or x 2 10.
Graphical representation:
0
–4 –3 –2 –1 0 1 2 3 4
2
3x 2 3
x 2 1 ............................... (i)
`
x
Solving 3 - 1 G x:
-1 G 23 x
`
3
x H - 2 .......................... (ii)
According to the question, x must satisfy
(i) or (ii).
3
Therefore x H - 2 .
Graphical representation:
10
–3
2
0
(b) Solving 5 + 3x G -7:
`
3x G -12
x G -4 ............................ (i)
Solving 13x - 8 G 11x:
2x G 8
`
x G 4 ............................ (ii)
According to the question, x must satisfy
(i) or (ii).
Therefore x G 4.
Graphical representation:
0
02 NSS TM 5A (E)-2P-KJ.indd 87
4
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88
Solutions
5. (a)
The corresponding quadratic function is
y = (x - 2)(x + 8).
When y = 0, we have
(x - 2)(x + 8) = 0
` The x-intercepts of the graph of the
quadratic function are 2 and -8.
Since the coefficient of x2 is 1 (2 0), the
parabola opens upward.
Sketch:
y
From the sketch,
y 1 0 when -15 1 x 1 0.
` The required solutions are
-15 1 x 1 0.
Graphical representation:
–15
(c)
0
3x2 G 18x
3x - 18x G 0
x2 - 6x G 0
The corresponding quadratic function is
y = x2 - 6x.
When y = 0, we have
x2 - 6x = 0
i.e. x(x - 6) = 0
` The x-intercepts of the graph of the
quadratic function are 0 and 6.
Since the coefficient of x2 is 1 (2 0), the
parabola opens upward.
Sketch:
2
y = (x – 2)(x + 8)
x
0
–8
2
From the sketch,
y 2 0 when x 1 -8 or x 2 2.
` The required solutions are x 1 -8 or
x 2 2.
Graphical representation:
y
y = x2 – 6x
–8
0 2
x
0
6
(b) The corresponding quadratic function is
y = x(x + 15).
When y = 0, we have
x(x + 15) = 0
` The x-intercepts of the graph of the
quadratic function are 0 and -15.
Since the coefficient of x2 is 1 (2 0), the
parabola opens upward.
Sketch:
From the sketch,
y G 0 when 0 G x G 6.
` The required solutions are 0 G x G 6.
Graphical representation:
y
y = x(x + 15)
–15
02 NSS TM 5A (E)-2P-KJ.indd 88
0
0
6
x
2010/7/20 4:40:42 PM
89
Chapter 2: Inequalities in One Unknown
(d) The corresponding quadratic function is
Sketch:
2
y = 2x - 7x - 15.
When y = 0, we have
2x2 - 7x - 15 = 0
i.e. (2x + 3)(x - 5) = 0
y
y = x2– x – 1
` The x-intercepts of the graph of the
3
1– 5
2
quadratic function are - 2 and 5.
x
0
1+ 5
2
Since the coefficient of x2 is 2 (2 0), the
parabola opens upward.
Sketch:
From the sketch,
y
y 2 0 when x 1
y = 2x2 – 7x – 15
1- 5
1+ 5
or x 2
.
2
2
` The required solutions are x 1
–3
2
x
0
or x 2
5
1- 5
2
1+ 5
.
2
Graphical representation:
From the sketch,
1– 5
2
3
y H 0 when x G - 2 or x H 5.
3
` The required solutions are x G - 2 or
x H 5.
Graphical representation:
(b)
0
1+ 5
2
2x2 - 3x G 1
2x2 - 3x - 1 G 0
The corresponding quadratic function is
y = 2x2 - 3x - 1.
When y = 0, we have
2x2 - 3x - 1 = 0
2
–3 0
5
2
x=
=
6. (a)
The corresponding quadratic function is
y = x2 - x - 1.
When y = 0, we have
x2 - x - 1 = 0
2
-(-1) ! (-1) - 4 (1)(-1)
x=
2 (1)
=
-(-3) ! (-3) - 4 (2)(-1)
2 (2)
3 ! 17
4
` The x-intercepts of the graph of the
quadratic function are
3 ! 17
.
4
Since the coefficient of x2 is 2 (2 0), the
parabola opens upward.
1! 5
2
` The x-intercepts of the graph of the
quadratic function are
1! 5
.
2
Since the coefficient of x2 is 1 (2 0), the
parabola opens upward.
02 NSS TM 5A (E)-2P-KJ.indd 89
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90
Solutions
a y 2 0 for all values of x.
` x can be any real number.
Sketch:
y
Graphical representation:
y = 2 x2– 3 x – 1
3 – 17
4
–4 –3 –2 –1 0 1 2 3 4
x
0
3 + 17
4
(b) The corresponding quadratic function is
y = x2 - 6x + 10.
When y = 0, we have
x2 - 6x + 10 = 0
Since the discriminant of x2 - 6x + 10 = 0
is (-6)2 - 4(1)(10) = -4 1 0,
y = x2 - 6x + 10 has no x-intercepts.
Since the coefficient of x2 is 1 (2 0), the
parabola opens upward.
Sketch:
From the sketch,
y G 0 when
3 - 17
3 + 17
GxG
.
4
4
` The required solutions are
3 - 17
3 + 17
GxG
.
4
4
Graphical representation:
y
y = x2 – 6x + 10
3 – 17 0
3 + 17
4
4
7. (a)
The corresponding quadratic function is
y = x2 + 4x + 6.
When y = 0, we have
x2 + 4x + 6 = 0
Since the discriminant of x2 + 4x + 6 = 0
is 42 - 4(1)(6) = -8 1 0,
2
y = x + 4x + 6 has no x-intercepts.
Since the coefficient of x2 is 1 (2 0), the
parabola opens upward.
Sketch:
y
y = x2 + 4x + 6
0
02 NSS TM 5A (E)-2P-KJ.indd 90
0
x
a y G 0 is not true for all values of x.
` The inequality has no solutions.
(c)
4x2 + 11 H 12x
4x2 - 12x + 11 H 0
The corresponding quadratic function is
y = 4x2 - 12x + 11.
When y = 0, we have
4x2 - 12x + 11 = 0
Since the discriminant of 4x2 - 12x + 11 =
0 is (-12)2 - 4(4)(11) = -32 1 0,
y = 4x2 - 12x + 11 has no x-intercepts.
Since the coefficient of x2 is 4 (2 0), the
parabola opens upward.
x
2010/7/20 4:40:48 PM
Chapter 2: Inequalities in One Unknown
91
Graphical representation:
Sketch:
y
y = 4x2 – 12x + 11
0
(c)
5x2 - 17x + 6 1 0
(x - 3)(5x - 2) 1 0
` x - 3 2 0 and 5x - 2 1 0, i.e. x 2 3
2
and x 1 5 , this compound inequality
x
0
16
has no solutions.
a y H 0 for all values of x.
` x can be any real number.
or x - 3 1 0 and 5x - 2 2 0, i.e. x 1 3
2
and x 2 5 , the solutions of this
Graphical representation:
compound inequality are 25 1 x 1 3.
` The solutions of the quadratic inequality
–4 –3 –2 –1 0 1 2 3 4
8. (a)
(x + 3)(x - 6) G 0
` x + 3 H 0 and x - 6 G 0, i.e.
x H -3 and x G 6, the solutions of
this compound inequality are
-3 G x G 6.
or x + 3 G 0 and x - 6 H 0, i.e.
x G -3 and x H 6, this compound
inequality has no solutions.
` The solutions of the quadratic
inequality (x + 3)(x - 6) G 0 are
-3 G x G 6.
Graphical representation:
5x2 - 17x + 6 1 0 are 25 1 x 1 3.
Graphical representation:
0 2
3
5
(d)
4(3x - 1) G x(3x - 1)
0 G (3x - 1)(x - 4)
(3x - 1)(x - 4) H 0
` 3x - 1 G 0 and x - 4 G 0, i.e. x G
1
and x G 4, the solutions of this
3
compound inequality are x G 31 .
or 3x - 1 H 0 and x - 4 H 0, i.e.
–3
0
6
2
(b) 2x - 32x 2 0
2
x - 16x 2 0
x(x - 16) 2 0
` x 1 0 and x - 16 1 0, i.e. x 1 0
and x 1 16, the solutions of this
compound inequality are x 1 0.
or x 2 0 and x - 16 2 0, i.e. x 2 0
and x 2 16, the solutions of this
compound inequality are x 2 16.
` The solutions of the quadratic inequality
2x2 - 32x 2 0 are x 1 0 or x 2 16.
02 NSS TM 5A (E)-2P-KJ.indd 91
x H 31 and x H 4, the solutions of
this compound inequality are x H 4.
` The solutions of the quadratic
inequality 4(3x - 1) G x(3x - 1) are
x G 31 or x H 4.
Graphical representation:
01
3
4
2010/7/20 4:40:52 PM
92
Solutions
9. (a)
x2 + 8 H 2x
x2 - 2x + 8 H 0
(x - 1)2 + 7 H 0
a For all real values of x,
(x - 1)2 + 7 H 0 holds.
` x can be any real number.
12. According to the question, we have
15t - 5t2 2 10
-5t + 15t - 10 2 0
t2 - 3t + 2 1 0
(t - 1)(t - 2) 1 0
`
11t12
` The range of values of t is 1 1 t 1 2
2
Graphical representation:
such that the height of the ball from the
ground is more than 10 m.
–4 –3 –2 –1 0 1 2 3 4
13.
(b)
2
4x 1 4x - 5
2
4x - 4x + 5 1 0
2
4(x - x) + 5 1 0
2x2 + (k - 4)x = 2(k - 1)
2x + (k - 4)x - 2(k - 1) = 0
For the quadratic equation to have two distinct
real roots, the discriminant of the equation
= (k - 4)2 - 4(2)[-2(k - 1)] 2 0.
k2 - 8k + 16 + 16k - 16 2 0
k2 + 8k 2 0
k(k + 8) 2 0
` k 1 -8 or k 2 0
2
2
4 a x - 21 k + 4 1 0
a For all real values of x,
2
4 a x - 21 k + 4 1 0 does not hold.
` The inequality has no solutions.
10. When the length of the rectangle is x cm, its
width is (x - 6) cm and its perimeter is
[2(x + x - 6)] cm.
` 60 G 2(x + x - 6) G 80
i.e. 30 G 2x - 6 and 2x - 6 G 40
36 G 2x
and
2x G 46
x H 18
and
x G 23
` The possible range of values of x is
18 G x G 23.
14. (a)
Solving 3x + 4 H -1 - 2x:
5x H -5
`
x H -1 ............................ (i)
Solving 8 - x 1 4x - 2:
10 1 5x
`
x 2 2 ................................ (ii)
According to the question, x must satisfy
(i) and (ii).
Therefore x 2 2.
Graphical representation:
11. According to the question, we have
9
9
c + 32 1 -37.9 or 5 c + 32 2 674.1
5
9
c 1 -69.9 or
5
9
c 2 642.1
5
c 1 -38.8 or
c 2 356.7,
cor. to 1 d.p.
` The possible range of the temperature in
0
(b) Rewrite the compound inequality as
3
- 6x G 2x + 1 and 2x + 1 G 9
2
Celsius scale is below -38.8cC or above
356.7cC when mercury is not in liquid
state.
`
02 NSS TM 5A (E)-2P-KJ.indd 92
2
1
G 8x
2
and
2x G 8
1
x H 16
and
xG4
1
GxG4
16
2010/7/20 4:40:55 PM
Chapter 2: Inequalities in One Unknown
and 3x + 4 2 x - 2
Graphical representation:
0 1
5x - 9 2 x - 1
4x 2 8
x 2 2 ..................................... (i)
(c)
2x 2 -6
x 2 -3 ................................ (ii)
According to the question, x must satisfy
(i) and (ii).
Therefore -3 1 x G 2.
4
16
93
(b) If x is a positive integer, the possible
values of x are 1 and 2.
8x
and 6x + 2 G 3 - 2
16. (a)
10x
G -4
3
Rewrite the compound inequality as
x+22
6
x G - 5 .............................. (ii)
4x + 8 2 3x + 2
x 2 -6 ................................ (i)
[No values of x can satisfy both (i) and (ii).]
` This compound inequality has no
and
solutions.
x H 6x - 60
60 H 5x
x G 12 ........................................... (i)
and 3x + 4 G 4(x - 2)
3x + 4 G 4x - 8
12 G x
x H 12 ................................. (ii)
According to the question, x must satisfy
(i) and (ii).
Therefore x = 12.
Graphical representation:
0
15. (a)
3x + 2
H 2x + 1
4
3x + 2 H 8x + 4
-2 H 5x
x
H x - 10
6
(d)
12
2x - 1
3x
G4- 2
3
2(2x - 1) G 24 - 3(3x)
4x - 2 G 24 - 9x
13x G 26
x G 2 ................................ (i)
x G - 25 .............................. (ii)
According to the question, x must satisfy
(i) and (ii).
Therefore -6 1 x G - 25 .
(b) The maximum value of x = - 1
The minimum value of x = - 5
17. (a)
Solving 2(x + 8) 2 30 - 5x:
2x + 16 2 30 - 5x
7x 2 14
`
x 2 2 ............................ (i)
Solving
2x + 5
1 x:
3
2x + 5 1 3x
51x
`
x 2 5 .............................. (ii)
According to the question, x must satisfy
(i) or (ii).
Therefore x 2 2.
Graphical representation:
0
02 NSS TM 5A (E)-2P-KJ.indd 93
3x + 2
4
2
2010/7/20 4:40:59 PM
94
Solutions
(b) Solving 3(3x - 1) 2 4:
Graphical representation:
9x - 3 2 4
9x 2 7
`
7
x 2 9 ......................... (i)
3x
Solving 2x - 21 G 2 :
–5 0
18. (a)
x
G 21
2
`
x G 1 ............................. (ii)
According to the question, x must satisfy
(i) or (ii).
Therefore x can be any real number.
Solving 3(3x + 2) 1 5(x - 2):
9x + 6 1 5x - 10
4x 1 -16
`
x 1 -4 ...................... (ii)
According to the question, x must satisfy
(i) or (ii).
Therefore x can be any real number except
-4.
1
4
(b) 0, - 2 and 5 cannot satisfy the compound
inequality in (a).
19. (a)
Solving 12 a 81 x - 1k 1 -x - 22:
3
x - 12 1 -x - 22
2
5
x 1 -10
2
x 1 -4 .................... (i)
5
1
Solving 24
x+ 1 2 4:
0
(d) Solving 2(3x - 4) G 2x - 13:
6x - 8 G 2x - 13
4x G -5
`
Therefore x 1 - 45 or x 2 1.
`
Graphical representation:
–4
x 1 - 45 ...................... (i)
Solving 3(x + 2) 2 13 - 4x:
3x + 6 2 13 - 4x
7x 2 7
`
x 2 1 ............................ (ii)
According to the question, x must satisfy
(i) or (ii).
–4 –3 –2 –1 0 1 2 3 4
Solving 4x - (3x - 1) 2 -3:
4x - 3x + 1 2 -3
`
x 2 -4 ................ (i)
Solving 2(1 - 2x) 2 6 + x:
2 - 4x 2 6 + x
-4 2 5x
`
Graphical representation:
(c)
4
4
5
x G - 4 ...................... (i)
x
6
Solving 6 - a x + 5 k 1 5 :
30 - 5x - 6 1 x
24 1 6x
`
x 2 4 .................... (ii)
According to the question, x must satisfy
(i) or (ii).
x + 24 2 30
`
x 2 6 ........................... (ii)
According to the question, x must satisfy
(i) or (ii).
Therefore x 1 -4 or x 2 6.
Graphical representation:
–4
0
6
(b) From (a), x 1 -4 or x 2 6.
` -4 G t G 6
5
Therefore x G - 4 or x 2 4.
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95
Chapter 2: Inequalities in One Unknown
20. (a)
Rewrite the compound inequality as
3x - 2 G 5x + 6 and 5x + 6 G 2x + 15
-8 G 2x
and
3x G 9
x H -4
and
xG3
` -4 G x G 3 ..................................... (i)
or
1-x
20
2
1-x20
12x
x 1 1 ........................................ (ii)
According to the question, x must satisfy
(i) or (ii).
Therefore x G 3.
(b)
4x H 3 - 4x2
4x2 + 4x - 3 H 0
The corresponding quadratic function is
y = 4x2 + 4x - 3.
When y = 0, we have
4x2 + 4x - 3 = 0
i.e. (2x - 1)(2x + 3) = 0
` The x-intercepts of the graph of the
3
quadratic function are 21 and - 2 .
Since the coefficient of x2 is 4 (2 0), the
parabola opens upward.
Sketch:
(b) Graphical representation of the solutions
y
in (a):
y = 4x2 + 4x – 3
0
3
–3
x
0
1
2
2
21. (a)
The corresponding quadratic function is
y = 11x - 24 - x2.
When y = 0, we have
11x - 24 - x2 = 0
i.e. -(x - 3)(x - 8) = 0
` The x-intercepts of the graph of the
quadratic function are 3 and 8.
Since the coefficient of x2 is -1 (1 0),
the parabola opens downward.
Sketch:
From the sketch,
3
y H 0 when x G - 2 or x H 21 .
` The required solutions are
3
x G - 2 or x H 21 .
Graphical representation:
y
x
0
8
3
–3
2
0
1
2
y = 11x – 24 – x2
From the sketch,
y 2 0 when 3 1 x 1 8.
` The required solutions are 3 1 x 1 8.
Graphical representation:
0
02 NSS TM 5A (E)-2P-KJ.indd 95
3
8
2010/7/20 4:41:08 PM
96
Solutions
22. (a)
(c)
The corresponding quadratic function is
y = 9x2 - 12x + 4.
When y = 0, we have
9x2 - 12x + 4 = 0
(3x - 2)2 = 0
From the sketch in (a),
y = 0 when x = 23 .
` The solution of 9x2 - 12x + 4 G 0 is
x = 23 .
x = 23 (repeated)
Graphical representation:
Since the coefficient of x2 is 9 (2 0), the
parabola opens upward.
Sketch:
0
y
y = 9x2 – 12x + 4
2
3
(d) From the sketch in (a),
a y 1 0 is not true for all values of x.
` 9x2 - 12x + 4 1 0 has no solutions.
23.
x
0
2
3
From the sketch,
y H 0 when x H 23 or x G 23 .
` The solutions of 9x2 - 12x + 4 H 0
can be any real number.
Graphical representation:
–4 –3 –2 –1 0 1 2 3 4
12x(x - 1) G x + 4
12x2 - 12x G x + 4
12x2 - 13x - 4 G 0
The corresponding quadratic function is
y = 12x2 - 13x - 4.
When y = 0, we have
12x2 - 13x - 4 = 0
i.e. (3x - 4)(4x + 1) = 0
` The x-intercepts of the graph of the
quadratic function are 43 and - 41 .
Since the coefficient of x2 is 12 (2 0), the
parabola opens upward.
Sketch:
(b) From the sketch in (a),
y
y 2 0 when x 2 23 or x 1 23 .
y = 12x2 – 13x – 4
` x can be any real number except 23 .
Graphical representation:
0
2
3
–1
4
0
x
4
3
From the sketch,
y G 0 when - 41 G x G 43 ,
and x is an integer.
` The required solutions are 0 and 1.
02 NSS TM 5A (E)-2P-KJ.indd 96
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Chapter 2: Inequalities in One Unknown
24. (a)
(2x - 1)(x + 2) G (3x + 2)(x - 1)
2x2 + 3x - 2 G 3x2 - x - 2
0 G x2 - 4x
2
x - 4x H 0
The corresponding quadratic function is
y = x2 - 4x.
When y = 0, we have
x2 - 4x = 0
i.e. x(x - 4) = 0
` The x-intercepts of the graph of the
quadratic function are 0 and 4.
Since the coefficient of x2 is 1 (2 0), the
parabola opens upward.
Sketch:
(b)
9x2 + 16 2 24x
9x2 - 24x + 16 2 0
(3x - 4)2 2 0
` 3x - 4 2 0, i.e. x 2 43 ,
4
or 3x - 4 1 0, i.e. x 1 3 .
` The required solutions are
any real number except 43 .
Graphical representation:
0
4
3
y
y = x2 – 4x
0
From the sketch,
y H 0 when x G 0 or x H 4.
` The required solutions are x G 0 or
x H 4.
(b) From the sketch in (a),
when 0 1 x 1 4, y = x2 - 4x 1 0, i.e.
(2x - 1)(x + 2) 2 (3x + 2)(x - 1).
` The required solutions are 0 1 x 1 4.
25. (a)
2
x + 10x + 25 G 0
2
(x + 5) G 0
` x + 5 H 0 and x + 5 G 0, i.e.
x H -5 and x G -5, the solution of
this compound inequality is x = -5.
` The required solution is x = -5.
Graphical representation:
–5
02 NSS TM 5A (E)-2P-KJ.indd 97
(c)
4x2 + 49 1 28x
4x2 - 28x + 49 1 0
(2x - 7)2 1 0
a For all real values of x, (2x - 7)2 1 0
does not hold.
` The inequality has no solutions.
26. (a)
(x - 2)(3x + 5) + (2 - x)(4x + 7) 1 0
(x - 2)(3x + 5) - (x - 2)(4x + 7) 1 0
(x - 2)[(3x + 5) - (4x + 7)] 1 0
(x - 2)(-x - 2) 1 0
(x - 2)(x + 2) 2 0
` x - 2 2 0 and x + 2 2 0, i.e. x 2 2
and x 2 -2, the solutions of this
compound inequality are x 2 2.
or x - 2 1 0 and x + 2 1 0, i.e. x 1 2
and x 1 -2, the solutions of this
compound inequality are x 1 -2.
` The required solutions are x 2 2 or
x
4
97
x 1 -2.
Graphical representation:
–2
0
2
0
2010/7/20 4:41:14 PM
98
Solutions
(b) From (a),
30. (a)
the required solutions are -2 G x G 2.
Graphical representation:
–2
27. (a)
0
2
x(x + 1) G 3(x + 1)
(x + 1)(x - 3) G 0
`
-1 G x G 3 ........................ (i)
and
4x2 + 4x - 3 1 0
(2x - 1)(2x + 3) 1 0
`
3
- 2 1 x 1 21 .......... (ii)
According to the question, we have
15 + 0.005 85(x - 20) 1 15(1 + 0.5%)
i.e.
0.005 85x 1 0.192
` x 1 33, cor. to the nearest integer .. (i)
or 15 + 0.005 85(x - 20) 2 15(1 + 2%)
i.e.
0.005 85x 2 0.417
` x 2 71, cor. to the nearest integer.. (ii)
According to the question, x must satisfy
(i) or (ii), and must satisfy x 2 20 at the
same time.
i.e. 20 1 x 1 33 or x 2 71.
(b) Graphical representation of the range in
(a):
According to the question, x must satisfy
(i) and (ii).
0
20
33
71
Therefore -1 G x 1 21 .
(b) The maximum value of x is 0.
28. Let the number of the $2 coins be x, then the
number of the $5 coins is 36 - x.
According to the question, we have
2x + 5(36 - x) H 90 and 2x + 5(36 - x) G 120
2x + 180 - 5x H 90 and 2x + 180 - 5x G 120
-3x H -90 and
-3x G -60
x G 30 and
x H 20
` The maximum number of the $2 coins is
30 and the minimum number is 20.
31. Let the larger odd number be 2x + 3, then the
smaller odd number is 2x + 1.
According to the question, we have
(2x + 1)2 + (2x + 3)2 1 100
2
4x + 4x + 1 + 4x2 + 12x + 9 1 100
8x2 + 16x + 10 1 100
8x2 + 16x - 90 1 0
4x2 + 8x - 45 1 0
(2x + 9)(2x - 5) 1 0
9
5
` - 2 1 x 1 2 ......................................... (i)
and 2x + 1 2 0
2x 2 -1
29. According to the question, we have
t
t
50: 60 2 45 and 50: 60 1 100 - 30
5
t 2 45 and
6
5
t 1 70
6
t 2 54 and
` 54 1 t 1 84
t 1 84
`
x 2 - 21 ...................................... (ii)
According to the question, x must satisfy (i) and
(ii).
5
` - 21 1 x 1 2
The possible values of x are 0, 1 and 2.
` The possible values of the larger odd
number are 3, 5 and 7.
02 NSS TM 5A (E)-2P-KJ.indd 98
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Chapter 2: Inequalities in One Unknown
32. (a)
If 2x2 - 4x + k H 0, the discriminant of
the quadratic equation 2x2 - 4x + k = 0
is:
D = (-4)2 - 4(2)(k) G 0
16 - 8k G 0
16 G 8k
kH2
` The minimum value of k is 2.
According to the question, x must satisfy
(i) and (ii).
Therefore -1 G x G 0 or 4 G x G 5.
35. (a)
3
= 21 x $ 2 x cm2
3
2
= 4 x cm2
(b) Total cost of the ornament
2
= $ b 0.5 # 12x + 2 # 4 x # 8 l
3
2
= $ (6x + 4 3 x )
(c)
According to the question, we have
*
2
6x + 4 3 x H 50
6x + 4 3 x2 G 300
6x + 4 3 x2 H 50
4 3 x2 + 6x - 50 H 0
2 3 x2 + 3x - 25 H 0
-2 G 2 cos i G 2
i.e. -2 G 3x - 1 G 2
-2 G 3x - 1 and 3x - 1 G 2
-1 G 3x
and
3x G 3
and
` xG
y2 - y - 6 G 0
(y - 3)(y + 2) G 0
`
-2 G y G 3
or x H
2
-2 G x - 4x - 2 G 3.
2
-2 G x - 4x - 2
2
x - 4x H 0
x(x - 4) H 0
`
x G 0 or x H 4 .......... (i)
and
02 NSS TM 5A (E)-2P-KJ.indd 99
-3 + 9 + 200 3
4 3
i.e. x H 2.3, cor. to 1 d.p. .............. (i)
6x + 4 3 x2 G 300
4 3 x + 6x - 300 G 0
2 3 x2 + 3x - 150 G 0
and
2
`
(b) From the result in (a), we have
-3 - 9 + 200 3
4 3
i.e. x G -3.2, cor. to 1 d.p.
xG1
` - 31 G x G 1
34. (a)
2
` Area of one triangular face of the
ornament
33. a -1 G cos i G 1
x H - 31
2
3
(b) If -x - 4x + k G 8,
Integrated Questions
(cross-chapters)
x - a 2x k cm
Height of the triangle =
= 2 x cm
2
then x2 + 4x - k + 8 H 0.
The discriminant of the quadratic equation
x2 + 4x - k + 8 = 0 is:
D = 42 - 4(1)(-k + 8) G 0
16 + 4k - 32 G 0
4k G 16
kG4
` The maximum value of k is 4.
99
-3 - 9 + 1 200 3
GxG
4 3
-3 + 9 + 1 200 3
4 3
i.e. -7.0 G x G 6.2, cor. to 1 d.p. ..... (ii)
a x must satisfy (i), (ii) and x 2 0.
` 2.3 G x G 6.2
2
x - 4x - 2 G 3
2
x - 4x - 5 G 0
(x - 5)(x + 1) G 0
`
-1 G x G 5 ........ (ii)
2010/7/20 4:41:22 PM
100
Solutions
2
36. Let the length of BC be l cm.
`
It is given that the perimeter of ABCD is 20 cm,
and AB = AD, BC = DC.
10 (r + 2) - 100 (r + 2) - 4 (r + 2) $ 50 (r - 2)
2 (r + 2)
GlG
2
` AB + BC = 21 (20) cm
10 (r + 2) + 100 (r + 2) - 4 (r + 2) $ 50 (r - 2)
2 (r + 2)
` 1.3 G l G 8.7, cor. to 1 d.p.
` The maximum length of BC is 8.7 cm.
= 10 cm
` Length of AB = (10 - l ) cm
A
Public Exam Questions
D
B
37. 2 1 x 1 6
38. x 1 -3 or x 2 2
39. -8 1 x G -5
40. x 1 3; 0, 1, 2
C
Lively Maths Problem
Join AC.
a 3ABC , 3ADC (SSS)
` +ABC = +ADC
a +ABC + +ADC = 180c
` +ABC = +ADC = 90c
Length of AC =
2
BC + AB
2
41. (a)
Area of the kite = 21 x $ (76 - x) cm2
2
2
x
= c 38x - 2 m cm
2
2
= l + (10 - l ) cm
a +ABC = 90c
` AC is a diameter.
Area of the shaded region
2
2
l + (10 - l ) F
- 21 l (10 - l ) $ 2 3 cm2
= )r <
2
2
2
2
l + (100 - 20l + l ) F
- 10l + l 2 2 cm2
= (r <
4
(b) According to the question, we have
x
2
38x - 2 H 720
76x - x2 H 1 440
x - 76x + 1 440 G 0
(x - 36)(x - 40) G 0
` 36 G x G 40
2
2
rl - 10rl + 50r
- 10l + l 2 m cm2
=c
2
According to the question, we have
2
rl - 10rl + 50r
- 10l + l 2 G 50
2
rl 2 - 10rl + 50r - 20l + 2l 2 G 100
(r + 2)l 2 - (10r + 20)l + 50r - 100 G 0
(r + 2)l 2 - 10(r + 2)l + 50(r - 2) G 0
02 NSS TM 5A (E)-2P-KJ.indd 100
2010/7/20 4:41:26 PM
Chapter 2: Inequalities in One Unknown
Revision Test (P.105)
1.
5.
2
9x - 6 2 2x - 4x
2
0 2 2x - 13x + 6
2
2x - 13x + 6 1 0
(x - 6)(2x - 1) 1 0
x 2 - 21 ..................................... (i)
3x - 15
2 -4
7
and
3x - 15 2 -28
3x 2 -13
9x - 6 2 x(2x - 4)
C
-4x 1 2
B
101
`
6.
A
1
1x16
2
Let the length of BC be x cm, then
AB = (x + 4) cm
13
x 2 - 3 ........................... (ii)
2
2
2
2
According to the question, x must satisfy
(i) and (ii).
AC = x + (x + 4) cm
According to the question, we have
x20
Therefore x 2 - 21 .
and
x + (x + 4) 2 0
x2 + (x + 4) 2 1 106
x2 + (x + 4)2 1 106
2
2
x + x + 8x + 16 - 106 1 0
2x2 + 8x - 90 1 0
x2 + 4x - 45 1 0
(x + 9)(x - 5) 1 0
-9 1 x 1 5
` 01x15
i.e. 0 cm 1 length of BC 1 5 cm
and
2.
A
-2 G 13 - 7x
7x G 15
15
x G 7 ........................................ (i)
or
x G 7 - 12x
13x G 7
7
x G 13 ....................................... (ii)
According to the question, x must satisfy
(i) or (ii).
15
Therefore x G 7 .
3.
x 2 1 ........................................... (i)
D
HKCEE Questions
7. A
8. D
and 2x 1 3x - 4
41x
x 2 4 .......................................... (ii)
and 3x - 4 1 5
3x 1 9
x 1 3 .................................. (iii)
According to the question, x must satisfy
(i), (ii) and (iii).
Therefore the compound inequality has no
9. A
solutions.
4.
A
From the graph,
y = 0 when x = 3
and there does not exist other values of x
such that y 1 0.
` The solution of f(x) G 0 is x = 3.
02 NSS TM 5A (E)-2P-KJ.indd 101
2010/7/20 4:41:29 PM