Unit S7 Solutions 1

Unit S7 Ine qualities in One a − c = 0 and b − d = 1 − 2 = −1

Unkno wn ∴ a−c>b−d
∴ II may not be true.
Exercise (P. S7-4)
For III: Let a = −2, b = −1, c = 0 and d = 1.
Then a < b and c < d.
1. B
ac = 0 and bd = −1
For I: Let x = 1 and y = −2, then x > 0 > y.
∴ ac > bd
x2 = 1, y2 = (−2)2 = 4
∴ III may not be true.
∴ x2 < y2
∴ The answer is A.
∴ I may not be true.
For II: ∵ x>0>y
1 1 4. D
∴ >0>
x y Let a = 1 and b = −2.
∴ II must be true. b
For A: =−2<0
For III: Let x = 1 and y = −1, then x > 0 > y. a
x y −1
1 1 1 1 ∴ It may not be true.
  = ,   =  =3
3 
3 3 3
    For B: a + b = −1 < 0
x y
1 1 ∴ It may not be true.
∴   < 
3 3 For C: a2b3 = −8 < 0
∴ III may not be true. ∴ It may not be true.
∴ The answer is B.
For D: b – a = b + (−a)
∵ a>0
2. B ∴ −a < 0
For I and II: ∵ xy < 0 and b < 0
i.e. (x > 0 and y < 0) or (x < 0 and y > 0) ∴ b + (−a) < 0
x i.e. b − a < 0
∴ <0
y
∴ It must be true.
∴ I and II must be true.
∴ The answer is D.
For III: Let x = 2 and y = −1, then xy < 0.
x + y = 2 + (−1) = 1 > 0
∴ III may not be true. 5. D
4
∴ The answer is B. 3( x − 2)  ( x + 1)
3
9( x − 2)  4( x + 1)
3. A 9 x − 18  4 x + 4
For I: ∵ a < b and c<d 5 x  22
22
∴ a – b < 0 and c – d < 0 x
5
(a + c) − (b + d) = (a − b) + (c – d) < 0
i.e. a + c < b + d
∴ I is true.
For II: Let a = 0, b = 1, c = 0 and d = 2. ∴ The answer is D.

Then a < b and c < d.

2 Mathematics Today for HKDSE – Multiple Choice Practice

6. 12. A
A
3x + 1 > 2(2x + 1) x−2
x− >4 or 5 > 3 − x
3x + 1 > 4x + 2 3
3x − x + 2 > 12 or x > −2
−x > 1
2x > 10 or x > −2
x < −1
x>5 or x > −2
∴ −2 is a solution of the inequality.
∴ x > −2
7. D
1
x −1  2( x − 3)
2 13. D
x−2 4( x − 3)
x−2 4 x − 12
−3x  −10
10
x
3
7
∴ is not a solution of the inequality.
2

8. A
A From the graph,
2x − 4 < 6 < −3x the required solution is x  a or x  b.
2x − 4 < 6 and 6 < −3x
2x < 10 and 3x < −6
14. A
x < 5 and x < −2 Since the coefficient of x2 is −1(< 0), the parabola
∴ x < −2
opens downward.
Sketch:
9. B

24 − 3x > −6x and 5x + 15 > 0

3x > −24 and 5x > −15
x > −8 and x > −3
∴ x > −3
From the graph,
the required solution is x = −5.
10. C
1 15. D
− x2 or −x + 2  1
2
x2  9
x  −4 or −x  −1
x2 − 9  0
∴ x  −4 or x 1
(x + 3)(x − 3)  0
∴ x  −3 or x  3
11. A
2x + 3  1 or 8 − 7x  29
16. C
2x  −2 or −7x  21
x(x − 4)  0
x  −1 or x  −3
∴ x0 or x  4
∴ x  −1
 Unit S7 Solutions 3

17. C 23. B
(x − 1) > 0
2 2x + 1 4x − 1
>5>
3 7
∴ The required solution is all real values of x
except 1. 2x + 1 > 15 and 35 > 4x – 1
i.e. x < 1 or x > 1 2x > 14 and 36 > 4x
x>7 and x<9
18. C ∴ 7<x<9

2 + 3x − 2x2 > 0 As x is an integer, x = 8

2x2 − 3x − 2 < 0
(2x + 1)(x − 2) < 0 24. C
1 x > 0, y < 0, i.e. x is positive and y is negative.
∴ − <x<2
2
For I, (+)(−) = (−)

19. D For II, let x = 2 and y = −1.

(3x + 2)2  (3x + 2)(2x − 1) x + y = 2 + (−1) = 1 > 0

(+)
(3x + 2)(3x + 2 − 2x + 1)  0 For III, = ( −)
( −)
(3x + 2)(x + 3)  0
∴ Only I and III must be negative.
2
−3  x  −
3
2 25. D
∴ The maximum value of x is − .
3
An obtuse angle is greater than 90° but less than 180°.

20. B ∴ 90° < (2x − 10)° < 180°

100 < 2x < 190
3x2 – x − 4  0
50 < x < 95
(3x − 4)(x + 1)  0
4
−1  x 
3 26. A
The positive integer satisfying the inequality is 1. BC is the longest side.
∴ There is only 1 positive integer. ∴ x > 8 ………………………………..(i)
By Triangle Inequality Theorem,
21. D
x<6+8
x − 4 > 5 − 2x or 4 > 3x + 10
x < 14 …………………………………..(ii)
3x > 9 or −6 > 3x
Combining (i) with (ii), 8 < x < 14
x>3 or x < −2
∴ x < −2 or x > 3
27. C
22. C ∵ x2 + 6x + k > 0 is true for all real values of x.
2x − 4 ∴ x2 + 6x + k = 0 has no real roots.
−x < 3 or > −2
4 i.e. ∆<0
x > −3 or 2x − 4 > −8
6 − 4(1)(k) < 0
2
x > −3 or 2x > −4
36 < 4k
x > −3 or x > −2
∴ k>9
∴ x > −3
4 Mathematics Today for HKDSE – Multiple Choice Practice

28. 3. B
B
x log 0.01 < log 10 2x − 5  5 − 2x

x(−2) < 1 4x  10
5
∴ x > −0.5 x
2
x can be 1 or 2 , so the least value of x is 1.
29. D
Substituting x = −1 into x2 − 4x + c = 0,
4. D
(−1)2 − 4(−1) + c = 0
0 < 3x − 1 < 2x + 7
1+4+c=0
0 < 3x − 1 and 3x − 1 < 2x + 7
c = −5
1 < 3x and x<8
x2 + 4x − 5 < 0 1
∴ <x<8
(x + 5)(x − 1) < 0 3
∴ −5 < x < 1 S does not satisfy the inequality, since it is a point
greater than 8.

Test (P. S7-7)

5. B
1. C −7x < 18 − 4x and 5x − 6 > −x

For I: Let a = −2 and b = 1. −3x < 18 and 6x > 6

a2 = 4 > 1 = b2 x > −6 and x>1
∴ I may not be true. ∴ x>1
For II: a<b
a−k<b−k
6. C
∴ II must be true.
2x
For III: ∵ k2 > 0 and a < b − < 4 < 5x − 1
3
∴ ak2 < bk2 2x
− < 4 and 4 < 5x − 1
∴ III must be true. 3
∴ The answer is C. −2x < 12 and 5 < 5x
x > −6 and x>1
2. C ∴ x>1
For A: As x > 3, −x < −3.
∴ It is true.
7. D
For B: 2x > 6
x>3 7 − 2x > 3x − 8 or 9 − 4x < 1
∴ It is true. 15 > 5x or 8 < 4x
For C: Let x = 4 > 3, then x<3 or x>2
(x − 3)(x − 5) = (4 − 3)(4 − 5) = −1 < 0 ∴ The solution is all real numbers.
∴ It may not be true.
8. B
1 1
For D: < 5 1
x 3 −2x + 1 > 5 or 2x + < x− 2
2 2
x>3 3 9
−2x > 4 or x<−
∴ It is true. 2 2
∴ The answer is C. x < −2 or x < −3
 Unit S7 Solutions 5

∴ x < −2 14. A
9. A x < −5 or x>a
ax > 2( x − a ) x+5<0 or x − a > 0
ax > 2 x − 2 a
(x + 5)(x − a) > 0
( a − 2) x > − 2 a
x2 + 5x – ax – 5a > 0
− 2a
x< (As a < 2, a − 2 < 0.) x2 + (5 − a)x − 5a > 0
a−2
2a i.e. 5a = 5 and 5 − a = b
∴ x<
2−a ∴ a = 1 and b=4

10. B
15. C
(1 − 2x)(x + 1)  1 − 2x x2 + 2x − k2 = −2
(1 − 2x)(x + 1 − 1)  0 x2 + 2x − (k2 − 2) = 0
x(1 − 2x)  0 Since the equation has no real roots, ∆ < 0.
x(2x − 1)  0 i.e. 22 − 4(1)[−(k2 − 2)] < 0
1
∴ 0 x  4 + 4(k2 − 2) < 0
2
k2 < 1
11. C ∴ −1 < k < 1
(2x + 1)2 – 4(2x + 1) – 5 > 0
(2x + 1 − 5)(2x + 1 + 1) > 0
(2x − 4)(2x + 2) > 0
∴ x < −1 or x > 2

12. C
(x − 4)2 < 4
(x − 4)2 − 22 < 0
(x − 4 + 2)(x − 4 − 2) < 0
(x − 2)(x − 6) < 0
2<x<6
∴ The smallest integer satisfying the inequality is 3.

13. C
x2 − kx + 1 = x
x2 − (k + 1)x + 1 = 0
Since the equation has real roots, ∆  0.
i.e. [−(k + 1)]2 − 4(1)(1)  0
(k + 1)2 – 22  0
(k + 1 + 2)(k + 1 − 2)  0
(k + 3)(k − 1)  0
∴ k  −3 or k  1