Chapter 7 Full Solutions
6 Inequalities
Review Exercise 6 (p. 6.6)
1. (a) [pic]
(b) [pic]
(c) [pic]
(d) [pic]
(e) [pic]
(f) [pic]
(e) [pic]
2. (a) x > 1
(b) x ( (1
(c) [pic]
(d) [pic]
3. (a) [pic]
(b) [pic]
(c) [pic]
(d) [pic]
4. (a) ∴ [pic]
Graphical representation:
[pic]
(b) ∴ [pic]
Graphical representation:
[pic]
(c) ∴ [pic]
Graphical representation:
[pic]
(d) ∴ [pic]
Graphical representation:
[pic]
(e) ∴ [pic]
Graphical representation:
[pic]
5. (a) [pic]
(b) Using the quadratic formula,
[pic]
∵ [pic]is not a real number.
∴ The equation has no real roots.
6. (a) [pic]
(b) When y = 0,
[pic]
∴ The x-intercepts are (3 and 2.
7. (a) (i) Since the coefficient of x2 is (3(< 0), the graph opens downwards.
(ii) When y = 0,
[pic]
∴ The x-intercepts are (1 and 3.
When x = 0,
[pic]
∴ The y-intercept is 9.
(b) (i) Since the coefficient of x2 is 8 (> 0), the graph opens upwards.
(ii) When y = 0,
[pic]
∴ The x-intercept is 2.5.
When x = 0,
[pic]
∴ The y-intercept is 50.
(c) (i) [pic]
Since the coefficient of x2 is (2(< 0), the graph opens downwards.
(ii) When y = 0,
[pic]
[pic]
∴ The graph has no x-intercepts.
When x = 0,
[pic]
∴ The y-intercept is –3.
Activity
Activity 6.1 (p. 6.28)
1. x + 3
[pic]
x – 2
[pic]
2. (x + 3)(x – 2)
[pic]
3. (a) (3 < x < 2
(b) x < (3 or x > 2
(c) x ( (3 or x ( 2
Classwork
Classwork (p. 6.10)
|Compound |Graphical representations |
|inequality |of the two inequalities |
|[pic] |[pic] |
|[pic] | |
| |[pic] |
|[pic] | |
| |[pic] |
|[pic] |[pic] |
|Compound |Graphical representation of the |Solutions of |
|inequality |solutions (if any) |the inequality|
|[pic] | |x > 4 |
| |[pic] | |
|[pic] | |[pic] |
| |[pic] | |
|[pic] |nil |no solutions |
|[pic] | |[pic] |
| |[pic] | |
Classwork (p. 6.15)
|Compound |Graphical representations |
|inequality |of the two inequalities |
|[pic] | |
| |[pic] |
|[pic] | |
| |[pic] |
|[pic] | |
| |[pic] |
|[pic] | |
| |[pic] |
|Compound |Graphical representation of the |Solutions of the |
|inequality |solutions |inequality |
|[pic] | |x > 3 |
| |[pic] | |
|[pic] | |[pic] |
| |[pic] | |
|[pic] | |all real values of|
| |[pic] |x |
|[pic] | |[pic] |
| |[pic] | |
Classwork (p. 6.20)
(a) [pic]
(b) [pic]
(c) [pic]
(d) [pic]
Classwork (p. 6.26)
1. [pic]
2. [pic]
3. [pic]
Classwork (p. 6.29)
(a) x < (6 or x > 1
(b) (6 < x < 1
(c) (6 ( x ( 1
Quick Practice
Quick Practice 6.1 (p. 6.11)
Rewrite the compound inequality as:
3x + 2 ( 5 and 8 – 5x ( –2
Solving 3x + 2 ( 5, we have
[pic]
Solving 8 – 5x ( –2, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are x ( 2.
Graphical representation:
[pic]
Quick Practice 6.2 (p. 6.11)
Rewrite the compound inequality as:
[pic]
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are –6 < x ( 2.
Graphical representation:
[pic]
Alternative Solution
[pic]
i.e. [pic]
∴ The solutions of the compound inequality are –6 < x ( 2.
Graphical representation:
[pic]
Quick Practice 6.3 (p. 6.16)
Solving 2x + 21 ( 5, we have
[pic]
Solving 10 – 5x > 55, we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x < –9 or
x ( –8.
Graphical representation:
[pic]
Quick Practice 6.4 (p. 6.16)
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x < –8 or [pic].
Graphical representation:
[pic]
Quick Practice 6.5 (p. 6.21)
Consider the corresponding quadratic function y = x2 – 3x – 10.
Since the coefficient of x2 is 1 (> 0), the graph of
y = x2 – 3x – 10 opens upwards.
When y = 0,
[pic]
Sketch the graph of y = x2 – 3x – 10, we have:
[pic]
From the graph, the solutions of x2 – 3x – 10 < 0 are [pic].
Quick Practice 6.6 (p. 6.22)
Consider the corresponding quadratic function y = –x2 – 5x – 6.
Since the coefficient of x2 is –1 (< 0), the graph of
y = –x2 – 5x – 6 opens downwards.
When y = 0,
[pic]
Sketch the graph of y = –x2 – 5x – 6, we have:
[pic]
From the graph, the solutions of –x2 – 5x – 6 ( 0 are
–3 ( x ( –2.
Alternative Solution
[pic]
Consider the quadratic function [pic].
Since the coefficient of x2 is 1 (> 0), the graph of [pic] opens upwards.
When y = 0, x = –3 or x = –2.
Sketch the graph of [pic], we have:
[pic]
From the graph, the solutions of [pic]
(i.e.[pic]) are –3 ( x ( –2.
Quick Practice 6.7 (p. 6.23)
[pic]
Consider the corresponding quadratic function y = x2 + 2x – 1.
Since the coefficient of x2 is 1 (> 0), the graph of
y = x2 + 2x – 1 opens upwards.
When y = 0,
[pic]
Sketch the graph of y = x2 + 2x – 1, we have:
[pic]
From the graph, the solutions of x2 + 2x – 1 ( 0
(i.e. [pic] are [pic] or [pic].
Quick Practice 6.8 (p. 6.24)
Consider the corresponding quadratic function y = –x2 + x – 1.
Since the coefficient of x2 is –1 (< 0), the graph opens downwards.
Since ( = 12 – 4(–1)(–1) = –3 < 0, the graph has no x-intercepts.
Sketch the graph of y = –x2 + x – 1, we have:
[pic]
From the graph, there are no solutions for –x2 + x – 1 > 0.
Quick Practice 6.9 (p. 6.27)
[pic]
∴ The solutions of x2 – x – 42 ( 0 are –6 ( x ( 7.
Quick Practice 6.10 (p. 6.27)
[pic]
∴ The solutions of –x2 + x + 20 < 0 are x < –4 or x > 5.
Quick Practice 6.11 (p. 6.28)
[pic]
∴ The solutions of 4x2 + 4x + 1 ( 0 are all real values of x.
Quick Practice 6.12 (p. 6.30)
[pic]
|Range of x |x < –3 |x = –3 |–3 < x < 8 |x = 8 |x > 8 |
|x – 8 |– |– |– |0 |+ |
|(x − 8)(x + 3) |+ |0 |– |0 |+ |
From the table, the solutions of [pic] are [pic] or [pic].
Quick Practice 6.13 (p. 6.30)
[pic]
|Range of x |x 1 |
|x – 1 |– |– |– |0 |+ |
|(2x + 1)(x – 1)|+ |0 |– |0 |+ |
From the table, the solutions of [pic] are [pic].
Quick Practice 6.14 (p. 6.31)
[pic]
|Range of x |x < –3 |x = –3 |x > –3 |
|x + 3 |– |0 |+ |
|(x + 3)2 |+ |0 |+ |
From the table, the solutions of x2 + 6x + 9 > 0 are all real values of x except x = –3.
Quick Practice 6.15 (p. 6.32)
Let x be the smaller negative integer, then the larger negative integer is x + 1.
[pic]
∴ The largest possible value of x is –16.
Hence, the largest possible values of these two consecutive integers are –16 and –15.
Quick Practice 6.16 (p. 6.33)
∵ 6x2 + (k – 1)x + (k2 – 13k + 6) ( 0 for all real values of x.
∴ 6x2 + (k – 1)x + (k2 – 13k + 6) = 0 has at most one real root.
[pic]
Hence, the range of possible values of k is [pic] or k ( 13.
Quick Practice 6.17 (p. 6.34)
(a) [pic]
(b) For Matthew’s claim to be correct,
[pic]
[pic]
Since x > 11, the area of the path cannot be greater than that of the garden.
Hence, Matthew’s claim is incorrect.
Further Practice
Further Practice (p. 6.12)
1. (a)
[pic]
(b)
[pic]
(c)
[pic]
2. (a) Solving 5x – 9 < 2, we have
[pic]
Solving 4 – 7x < 11 – 9x, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
(b) Rewrite the compound inequality as:
[pic] and [pic]
Solving x – 2(x – 1) < 6, we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (3) and (4).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
3. (a) [pic]
i.e. [pic]
∴ The solutions of the compound inequality are
–8 < x < –3.
Graphical representation:
[pic]
(b) Rewrite the compound inequality as:
1 + 3x ( x – 5 and x – 5 ( 2x + 3
Solving 1 + 3x ( x – 5, we have
[pic]
Solving x – 5 ( 2x + 3, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ There are no solutions.
Further Practice (p. 6.17)
1. (a)
[pic]
(b)
[pic]
2. (a) Solving 2x – 8 < 6, we have
[pic]
Solving 3(x + 3) > 2(x – 2), we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are all real values of x.
Graphical representation:
[pic]
(b) Solving 2x – 4 < 5x + 3, we have
[pic]
Solving 1 – 6x < 2 – 7x, we have
[pic]
∵ x must satisfy (3) or (4).
∴ The solutions of the compound inequality are all real values of x.
Graphical representation:
[pic]
3. (a) Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
(b) Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (3) or (4).
∴ The solutions of the compound inequality are
x ( 2 or x ( 8.
Graphical representation:
[pic]
Further Practice (p. 6.28)
1. (a) 18x2 – 15x + 2 = [pic]
(b) [pic]
∴ The solutions of [pic] are [pic].
2. (a) [pic]
∴ The solutions of y2 + 4y – 32 > 0 are y < –8 or
y > 4.
(b) Let y = x – 2.
[pic]
From (a), x – 2 < –8 or x – 2 > 4
[pic]
∴ The solutions of [pic] are [pic].
Further Practice (p. 6.31)
1. (a) 15x2 + 16x – 7 = [pic]
|(b) |Range of x |[pic] |[pic] |[pic] |
| |3x – 1 |– |– |– |
| |5x + 7 |– |0 |+ |
| |(3x – 1)(5x + 7) |+ |0 |– |
| |Range of x |[pic] |[pic] |
| |3x – 1 |0 |+ |
| |5x + 7 |+ |+ |
| |(3x – 1)(5x + 7) |0 |+ |
From the table, the solutions of 15x2 + 16x – 7 ( 0 are [pic].
2. (a) [pic]
|Range of x |[pic] |[pic] |[pic] |
|5x – 2 |– |0 |+ |
|(5x – 2)2 |+ |0 |+ |
From the table, the solutions of −25x2 + 20x − 4 < 0 are all real values of x except [pic].
(b) [pic]
|Range of x |[pic] |[pic] |[pic] |
|2x – 7 |– |0 |+ |
|(2x – 7)2 |+ |0 |+ |
From the table, the solution of
(2x – 5)2 – 4(2x – 5) ( –4 is [pic].
Further Practice (p. 6.34)
1. ∵ [pic] has no real roots.
∴ [pic]
Hence, the range of possible values of k is [pic].
2. [pic]
Since [pic],
[pic]
∴ The largest pentagonal number is 12th pentagonal number.
Exercise
Exercise 6A (p. 6.12)
Level 1
1. [pic]
2. [pic]
3. [pic]
4. [pic]
5. –2 ( x < 3
6. –7 ( x ( –1
7. Solving 3x ( 12, we have
[pic]
Solving 2x – 5 ( 9, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are x ( 7.
Graphical representation:
[pic]
8. [pic]
i.e. [pic]
∴ The solutions of the compound inequality are
2 < x ( 4.
Graphical representation:
[pic]
9. Solving 3x – 5 < 2x + 7, we have
[pic]
Solving x + 10 < 5x – 2, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are
3 < x < 12.
Graphical representation:
[pic]
10. Solving 2x + 8 > 7x – 2, we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
11. Rewrite the compound inequality as:
2x + 8 < 3x – 3 and 4x – 1 ( 2x + 1
Solving 2x + 8 < 3x – 3, we have
[pic]
Solving 4x – 1 ( 2x + 1, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are x > 11.
Graphical representation:
[pic]
12. Rewrite the compound inequality as:
[pic] and [pic]
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
13. Rewrite the compound inequality as:
–5 ≤ 2x + 3 and 2x + 3 < x + 11
Solving –5 ≤ 2x + 3, we have
[pic]
Solving 2x + 3 < x + 11, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are
–4 ( x < 8.
Graphical representation:
[pic]
14. Rewrite the compound inequality as:
–2x + 3 < 3x + 5 and 3x + 5 ( 5x – 1
Solving –2x + 3 < 3x + 5, we have
[pic]
Solving 3x + 5 ( 5x – 1, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are x ( 3.
Graphical representation:
[pic]
15. From the question,
2n + 12 < 4n – 18 and 3n – 5 > 4n – 23
Solving 2n + 12 < 4n – 18, we have
[pic]
Solving 3n – 5 > 4n – 23, we have
[pic]
∵ n must satisfy both (1) and (2).
∴ The solutions of the compound inequality are
15 < n < 18.
Hence, the possible values of n are 16 and 17.
16. Let n be the larger integer, then the smaller integer is
n – 1.
From the question,
[pic]
Solving n + (n – 1) < 15, we have
[pic]
∵ n must satisfy both (1) and (2).
∴ The solutions of the compound inequality are n < 8.
∴ The largest possible value of n is 7.
Hence, the largest possible sum of the two consecutive integers is 7 + 6 = 13.
Level 2
17. Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are x ( 0.
Graphical representation:
[pic]
18. Rewrite the compound inequality as:
[pic]
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ There are no solutions.
19. Solving 4(x + 2) ( 3(2x – 1), we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
20. Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
21. Rewrite the compound inequality as:
[pic]
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
22. Solving [pic], we have
[pic]
Solving 2 – 5x > k, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are x > 4 and [pic].
Suppose the solutions are in the form a < x < b. Then,
[pic]
∴ [pic]
[pic]
Thus k can be any value less than –18.
∴ [pic] (or any other values less than –18)
23. Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1), (2) and (3).
∴ The solutions of the compound inequality are
4 ( x < 5.
24. Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1), (2) and (3).
∴ The solutions of the compound inequality are x < –3.
25. (a) Number of girls = 1200 – x
∴ 1200 – x ( x – 50 and 1200 – x ( x – 100
(or any other equivalent inequalities)
(b) (i) Solving 1200 – x ( x – 50, we have
[pic]
Solving 1200 – x ( x – 100, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are 625 ( x ( 650.
(ii) Minimum number of girls = 1200 – 650 = 550
26. Let x cm be the length of each equal side of the triangle.
Perimeter of the triangle = (x + x + 16) cm
= (2x + 16) cm
From the question,
[pic]
Minimum height of the triangle [pic]
Minimum area of the triangle [pic]
Maximum height of the triangle [pic]
Maximum area of the triangle
= [pic]
Exercise 6B (p. 6.17)
Level 1
1. [pic]
2. [pic]
3. x < –4 or x ( 4
4. [pic] or [pic]
5. Solving 1 ( 2x – 5, we have
[pic]
Solving 2x – 5 ( 9, we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x ( 3.
Graphical representation:
[pic]
6. Solving 7 – 2x > –1, we have
[pic]
Solving 2x < 3, we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x < 4.
Graphical representation:
[pic]
7. Solving 4x + 2 < 2x + 4, we have
[pic]
Solving 2x – 3 ( 4x – 9, we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x < 1 or x ( 3.
Graphical representation:
[pic]
8. Solving –2(x + 2) > –12, we have
[pic]
Solving 4 – 3x ( 9 + 2x, we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x < 4.
Graphical representation:
[pic]
9. Solving [pic], we have
[pic]
Solving 4x + 3 > 7x – 6, we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x < 3.
Graphical representation:
[pic]
10. Solving 3x – 2 > x + 3, we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
11. Solving [pic], we have
[pic]
Solving 2x – 3 > 3x + 4, we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
12. Solving [pic], we have
[pic]
Solving 2(3x – 1) < x + 3, we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x < 1 or [pic].
Graphical representation:
[pic]
13. From the question,
2n + 5 > 5n – 13 or 5n + 18 < 4n + 13
Solving 2n + 5 > 5n – 13, we have
[pic]
Solving 5n + 18 < 4n + 13, we have
[pic]
∵ n must satisfy (1) or (2).
∴ The solutions of the compound inequality are n < 6.
Since n is a positive integer, the possible values of n are 1, 2, 3, 4 and 5.
14. From the question,
n + (2n + 5) < 20 or 2n + 5 < 10
Solving n + (2n + 5) < 20, we have
[pic]
Solving 2n + 5 < 10, we have
[pic]
∵ n must satisfy (1) or (2).
∴ The solutions of the compound inequality are n < 5.
∴ The largest possible value of n is 4.
Hence, the largest possible values of the two integers n and 2n + 5 are 4 and 13 respectively.
Level 2
15. Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
16. Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are [pic] or [pic].
Graphical representation:
[pic]
17. Consider [pic]. ……(*)
Rewrite (*) as:
x + 5 ( 3x – 3 and 3x – 1 < 2x + 4
Solving x + 5 ( 3x – 3, we have
[pic]
Solving 3x – 1 < 2x + 4, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of (*) are 4 ( x < 5. …… (3)
Solving x – 2 > 3(1 – x), we have
[pic]
∵ x must satisfy (3) or (4).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
18. Consider 2x + 1 < 3x – 3 ( x + 11. ……(*)
Rewrite (*) as:
2x + 1 < 3x – 3 and 3x – 3 ( x + 11
Solving 2x + 1 < 3x – 3, we have
[pic]
Solving 3x – 3 ( x + 11, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of (*) are 4 < x ( 7. …… (3)
Solving [pic], we have
[pic]
∵ x must satisfy (3) or (4).
∴ The solutions of the compound inequality are x > 4.
Graphical representation:
[pic]
19. Consider 2x – 3 ( 4x – 6 ( x + 18. ……(*)
Rewrite (*) as:
2x – 3 ( 4x – 6 and 4x – 6 ( x + 18
Solving 2x – 3 ( 4x – 6, we have
[pic]
Solving 4x – 6 ( x + 18, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of (*) are [pic]
Solving [pic], we have
[pic]
∵ x must satisfy (3) or (4).
∴ The solutions of the compound inequality are [pic] or [pic].
Graphical representation:
[pic]
20. Solving [pic], we have
[pic]
Solving 5 – 6x < k, we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are [pic] or [pic].
Suppose the solutions are in the form ‘x > a or x < b’, where a > b. Then,
[pic]
Thus k can be any integer less than or equal to –9.
∴ k = –9 or –10 (or any other integers less than or equal to –9)
21. Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1), (2) or (3).
∴ The solutions of the compound inequality are all real values of x.
22. Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
Solving 2(7x – 6) – 3(4x – 3) < x + 2, we have
[pic]
∵ x must satisfy (1), (2) or (3).
∴ The solutions of the compound inequality are all real values of x.
23. (a) Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are [pic].
(b) From (a), the solutions of ‘[pic]’ are [pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (3) and (4).
∴ The solutions of the compound inequality are [pic].
24. (a) Number of 10-dollar coins = 100 – x
Gabriel’s claim is x > 100 – x
Cecilia’s claim is 5x + 10(100 – x) < 800
∴ The required compound inequality is
x > 100 – x or 5x + 10(100 – x) < 800
(or any other equivalent inequalities)
(b) (i) Solving x > 100 – x, we have
[pic]
Solving 5x + 10(100 – x) < 800, we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x > 40.
(ii) To maximize the total amount of money, the number of 5-dollar coins must be minimized.
∵ The smallest possible value of x is 41.
i.e. The minimum number of 5-dollar coins is 41.
∴ Maximum total amount
= $5 × 41 + $10 × (100 – 41) = $795
25. Let the speed of Peter’s car be x km/h
Distance travelled by Peter = [pic]
Distance travelled by Mary = [pic]
Speed of Mary’s car = [pic]
From the question,
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x ( 80.
∴ Maximum speed of Peter’s car = 80 km/h
∴ Maximum possible distance travelled by Peter
= [pic]
Exercise 6C (p. 6.24)
Level 1
1. Consider the corresponding quadratic function y = x(x + 2).
Since the coefficient of x2 is 1 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are 0 and –2.
Sketch the graph of y = x(x + 2).
[pic]
From the graph, the solutions of x(x + 2) < 0 are
–2 < x < 0.
2. Consider the corresponding quadratic function
y = (x – 3)(x + 8).
Since the coefficient of x2 is 1 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are 3 and –8.
Sketch the graph of y = (x – 3)(x + 8).
[pic]
From the graph, the solutions of (x – 3)(x + 8) ( 0 are
x ( –8 or x ( 3.
3. Consider the corresponding quadratic function
y = (x + 1)(x + 7).
Since the coefficient of x2 is 1 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are –1 and –7.
Sketch the graph of y = (x + 1)(x + 7).
[pic]
From the graph, the solutions of (x + 1)(x + 7) ( 0 are
–7 ( x ( –1.
4. Consider the corresponding quadratic function
y = (x – 1)(5 – x).
Since the coefficient of x2 is –1 (< 0), the graph opens downwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are 1 and 5.
Sketch the graph of y = (x – 1)(5 – x).
[pic]
From the graph, the solutions of (x – 1)(5 – x) < 0 are
x < 1 or x > 5.
5. Consider the corresponding quadratic function y = x2 – 2x.
Since the coefficient of x2 is 1 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are 0 and 2.
Sketch the graph of y = x2 – 2x.
[pic]
From the graph, the solutions of x2 – 2x > 0 are x < 0 or
x > 2.
6. Consider the corresponding quadratic function y = 16 – x2.
Since the coefficient of x2 is –1 (< 0), the graph opens downwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are –4 and 4.
Sketch the graph of y = 16 – x2.
[pic]
From the graph, the solutions of 16 – x2 > 0 are –4 < x < 4.
7. Consider the corresponding quadratic function
y = x2 – 6x + 5.
Since the coefficient of x2 is 1 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are 1 and 5.
Sketch the graph of y = x2 – 6x + 5.
[pic]
From the graph, the solutions of x2 – 6x + 5 ( 0 are
1 ( x ( 5.
8. Consider the corresponding quadratic function
y = 2x2 + 7x – 4.
Since the coefficient of x2 is 2 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are –4 and [pic].
Sketch the graph of y = 2x2 + 7x – 4.
[pic]
From the graph, the solutions of 2x2 + 7x – 4 ( 0 are
x ( –4 or [pic].
9. Consider the corresponding quadratic function
y = 12 + x – x2.
Since the coefficient of x2 is –1 (< 0), the graph opens downwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are –3 and 4.
Sketch the graph of y = 12 + x – x2.
[pic]
From the graph, the solutions of 12 + x – x2 < 0 are x < –3 or x > 4.
10. Consider the corresponding quadratic function
y = –6x2 + x + 5.
Since the coefficient of x2 is –6 (< 0), the graph opens downwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are [pic] and 1.
Sketch the graph of y = –6x2 + x + 5.
[pic]
From the graph, the solutions of –6x2 + x + 5 ( 0 are [pic].
Level 2
11. Consider the corresponding quadratic function
y = 6x2 – 7x – 20.
Since the coefficient of x2 is 6 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are [pic] and [pic].
Sketch the graph of y = 6x2 – 7x – 20.
[pic]
From the graph, the solutions of 6x2 – 7x – 20 ( 0 are [pic].
12. Consider the corresponding quadratic function
y = –12x2 + 28x – 15.
Since the coefficient of x2 is –12 (< 0), the graph opens downwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are [pic] and [pic].
Sketch the graph of y = –12x2 + 28x – 15.
[pic]
From the graph, the solutions of –12x2 + 28x – 15 > 0 are [pic].
13. Consider the corresponding quadratic function
y = 2x2 + 4x – 1.
Since the coefficient of x2 is 2 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are [pic] and [pic].
Sketch the graph of y = 2x2 + 4x – 1.
[pic]
From the graph, the solutions of 2x2 + 4x – 1 > 0 are [pic].
14. Consider the corresponding quadratic function
y = –x2 + 6x – 3.
Since the coefficient of x2 is –1 (< 0), the graph opens downwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are [pic] and [pic].
Sketch the graph of y = –x2 + 6x – 3.
[pic]
From the graph, the solutions of –x2 + 6x – 3 ( 0 are [pic].
15. [pic]
Consider the corresponding quadratic function y = x2 – 3x.
Since the coefficient of x2 is 1 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are 0 and 3.
Sketch the graph of y = x2 – 3x.
[pic]
From the graph, the solutions of x2 –3x ( 0 (i.e. x2 ( 3x) are 0 ( x ( 3.
16. [pic]
Consider the corresponding quadratic function [pic].
Since the coefficient of x2 is 1 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are –6 and 4.
Sketch the graph of [pic].
[pic]
From the graph, the solutions of [pic]
(i.e. (x + 1)2 ( 25) are x ( –6 or x ( 4.
17. [pic]
Consider the corresponding quadratic function y = 4x2 – 9.
Since the coefficient of x2 is 4 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are [pic] and [pic].
Sketch the graph of y = 4x2 – 9.
[pic]
From the graph, the solutions of 4x2 – 9 > 0
(i.e. 4x(x – 1) > 9 – 4x) are [pic] or [pic].
18. [pic]
Consider the corresponding quadratic function y = x2 + 9x.
Since the coefficient of x2 is 1 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are –9 and 0.
Sketch the graph of y = x2 + 9x.
[pic]
From the graph, the solutions of [pic]
(i.e. (x + 3)(x + 6) < 18) are –9 < x < 0.
19. [pic]
Consider the corresponding quadratic function
y = –x(5x – 4).
Since the coefficient of x2 is –5 (< 0), the graph opens downwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are 0 and [pic].
Sketch the graph of y = –x(5x – 4).
[pic]
From the graph, the solutions of –x(5x – 4) < 0
(i.e. x(5x – 4) < 2x(5x – 4)) are x < 0 or [pic].
20. [pic]
Consider the corresponding quadratic function
y = 2x2 + 4x + 3.
Since the coefficient of x2 is 2 (> 0), the graph opens upwards.
Since ( = 42 – 4(2)(3) = –8 < 0, the graph has no x-intercepts.
Sketch the graph of y = 2x2 + 4x + 3.
[pic]
From the graph, there are no solutions for [pic] (i.e. [pic]).
21. [pic]
Consider the corresponding quadratic function
y = 2x2 + x + 2.
Since the coefficient of x2 is 2 (> 0), the graph opens upwards.
Since ( = 12 – 4(2)(2) = –15 < 0, the graph has no x-intercepts.
Sketch the graph of y = 2x2 + x + 2.
[pic]
From the graph, the solutions of 2x2 + x + 2 > 0
(i.e. (2x + 1)2 > 2x – 3) are all real values of x.
22. (a) [pic]
Consider the corresponding quadratic function
y = x2 + x – 10.
Since the coefficient of x2 is 1 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are [pic] and [pic].
Sketch the graph of y = x2 + x – 10.
[pic]
From the graph, the solutions of x2 + x – 10 ( 0
(i.e. [pic]) are
[pic].
(b) (i) From (a), [pic]
∴ Minimum value of k = –3
(ii) From (a), [pic]
∴ Maximum value of k = 2
23. (a) Consider the corresponding quadratic function
y = 2x2 – 5x – 12.
Since the coefficient of x2 is 2 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are [pic] and 4.
Sketch the graph of y = 2x2 – 5x – 12.
[pic]
From the graph, the solutions of 2x2 – 5x – 12 ( 0 are [pic].
(b) (i) [pic]
From (a), we have
[pic]
(ii) [pic], –1, 0 and 1 can satisfy the inequality.
24. Consider the corresponding quadratic function [pic].
Since the coefficient of x2 is 1 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercept of the graph is –1.
Sketch the graph of y = x2 + 2x + 1.
[pic]
(a) From the graph, the solutions of x2 + 2x + 1 > 0 are all real values of x except x = –1.
(b) From the graph, the solutions of x2 + 2x + 1 ( 0 are all real values of x.
(c) From the graph, there are no solutions for
x2 + 2x + 1 < 0.
(d) From the graph, the solution of x2 + 2x + 1 ( 0 is
x = –1.
25. (a) (i) Consider the corresponding quadratic function
y = x2 – 2x – 63.
Since the coefficient of x2 is 1 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are –7 and 9.
Sketch the graph of y = x2 – 2x – 63.
[pic]
From the graph, the solutions of x2 – 2x – 63 < 0 are –7 < x < 9. …… (1)
(ii) [pic]
(b) ∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are
x > –7.
26. (a) (i) Consider the corresponding quadratic function
y = 4x2 + 4x – 15.
Since the coefficient of x2 is 4 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are [pic] and [pic].
Sketch the graph of y = 4x2 + 4x – 15.
[pic]
From the graph, the solutions of
4x2 + 4x – 15 ( 0 are
[pic]
(ii) [pic]
(b) ∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are [pic].
Exercise 6D (p. 6.31)
Level 1
1. [pic]
∴ The solutions of (x + 1)(x – 12) ( 0 are x ( –1 or
x ( 12.
2. [pic]
∴ The solutions of –(x + 5)(x – 8) > 0 are –5 < x < 8.
|3. |Range of x |x < –7 |x = –7 |–7 < x < 1 |x = 1 |x > 1 |
| |x + 7 |– |0 |+ |+ |+ |
| |(x – 1)( x + |+ |0 |– |0 |+ |
| |7) | | | | | |
From the table, the solutions of (x – 1)(x + 7) > 0 are
x < –7 or x > 1.
4. [pic]
|Range of x |x < 0 |x = 0 |0 < x < 5 |x = 5 |x > 5 |
|x – 5 |– |– |– |0 |+ |
|x(x – 5) |+ |0 |– |0 |+ |
From the table, the solutions of x(x – 5) ( 0
(i.e.[pic]) are 0 ( x ( 5.
5. [pic]
∴ The solutions of 2x2 – 7x + 6 < 0 are [pic].
6. [pic]
∴ The solutions of –x2 – 7x + 120 < 0 are x < –15 or
x > 8.
7. [pic]
∴ The solutions of x(23 – x) ( 132 are 11 ( x ( 12.
8. [pic]
∴ The solutions of (x – 1)2 > 4(x – 1) are x < 1 or x > 5.
9. [pic]
|Range of x |[pic] |[pic] |[pic] |x = 3 |x > 3 |
|x – 3 |– |– |– |0 |+ |
|(2x + 3)(x – 3)|+ |0 |– |0 |+ |
From the table, the solutions of [pic]
(i.e. 2x2 – 3x – 9 < 0) are [pic].
10. [pic]
|Range of x |[pic] |[pic] |[pic] |x = 5 |x > 5 |
|x – 5 |– |– |– |0 |+ |
|(3x + 13)(x – 5)|+ |0 |– |0 |+ |
From the table, the solutions of [pic]
(i.e. 65 + 2x – 3x2 ( 0) are [pic] or x ( 5.
11. [pic]
|Range of x |[pic] |[pic] |[pic] |x = 2 |x > 2 |
|x – 2 |– |– |– |0 |+ |
|(6x + 5)(x – 2)|+ |0 |– |0 |+ |
From the table, the solutions of [pic]
(i.e. 7x + 4 > 6(x – 1)(x + 1)) are [pic].
12. [pic]
|Range of x |[pic] |[pic] |[pic] |x = 0 |x > 0 |
|x |– |– |– |0 |+ |
|(3x + 5)x |+ |0 |– |0 |+ |
From the table, the solutions of [pic]
(i.e. (3x + 2)2 + (3x + 2) – 6 > 0) are [pic] or x > 0.
Level 2
13. (a) [pic]
(b) [pic]
∵ (x + 2.5)2 ( 0 for all real values of x
∴ The solutions of x2 + 5x + 7 > 0 are all real values of x.
14. (a) [pic]
(b) [pic]
∵ (x – 1)2 ( 0 for all real values of x
∴ There are no solutions for –3x2 + 6x – 9 ( 0.
15. [pic]
∵ [pic]for all real values of x
∴ The solution of x2 – 12x + 36 ( 0 is x = 6.
16. [pic]
Consider the corresponding quadratic function
y = x2 + x – 1.
When y = 0,
[pic]
Let [pic] and [pic].
∴ [pic]
|Range of x |x < ( |x = ( |( < x < ( |x = ( |x > ( |
|x – ( |– |– |– |0 |+ |
|(x − ()(x – ()|+ |0 |– |0 |+ |
From the table, the solutions of [pic]
(i.e. x2 + x < 1) are [pic].
17. [pic]
∴ The solutions of 6x2 + 7x – 1 > (2x + 1)2 are x < –2 or [pic].
18. [pic]
|Range of x |x < –3 |x = –3 |–3 < x < –1 |x = –1 |x > –1 |
|x + 1 |– |– |– |0 |+ |
|(x + 3)(x + 1) |+ |0 |– |0 |+ |
From the table, the solutions of [pic]
(i.e. (x + 1)(2x − 1) – (3x – 1)(x + 2) ( 4) are –3 ( x ( –1.
Exercise 6E (p. 6.34)
Level 1
1. ∵ x2 – kx + 16 = 0 has no real roots.
∴ [pic]
Hence, the range of possible values of k is –8 < k < 8.
2. ∵ x2 + (k + 1)x – 2(k + 1) = 0 has real roots.
∴ [pic]
Hence, the range of possible values of k is k ( –9 or
k ( –1.
3. Let x be one of the positive integers, then the other positive integer is 36 – x.
[pic]
∴ The greatest possible value of x is 25.
Hence, the greatest possible value of the larger integer is 25.
4. Let x be the smaller positive even integer, then the larger integer is x + 2.
[pic]
Since x is a positive even integer, the possible values of x are 2, 4 and 6.
Hence, the greatest possible value of the smaller integer is 6.
5. From the question,
[pic]
Hence, the required range of values of t is 2 ( t ( 4.
6. From the question,
[pic]
Hence, the minimum number of cakes to be made is 41.
7. From the question,
[pic]
∴ The first term that exceeds 3003 = the 78th term
=[pic]
=[pic]
8. Amount after two years = $10 000 × (1 + r%)2
Total interest = $[10 000 × (1 + r%)2 – 10 000]
[pic]
Hence, the range of possible values of r is r ( 5.
Level 2
9. (a) ∵ The graph of the quadratic function is always
below the x-axis.
∴ –x2 + kx – 2k + 3 < 0 for all real values of x
∴ –x2 + kx – 2k + 3 = 0 has no real roots.
[pic]
Hence, the range of possible values of k is 2 < k < 6.
(b) When k = 6, y = –x2 + 6x – 2(6) + 3
= –x2 + 6x – 9
∴ By the result of (a), the graph of y = –x2 + 6x – 9 is not always below the x-axis.
10. (a) Let x m be the length of the rectangle, then the width is (5 – x) m.
[pic]
In addition, since the length must be greater than or equal to the width,
[pic]
∴ The range of possible values of x is 2.5 ( x ( 3.
Hence, the maximum length of the rectangle is 3 m.
(b) When x = 2.5,
area enclosed [pic]
When x = 3,
area enclosed [pic]
∴ The length found in (a) does not give the maximum possible enclosed area.
11. (a) [pic]
(b) For Judy’s assertion to be correct,
[pic]
∴ The range of possible values of x is
x >[pic].
Hence, Judy’s assertion is correct.
12. Since the equation has real roots,
[pic]
Since the roots are negative,
[pic]
Since k must satisfy (1) and (2), the range of possible values of k is k ( –49.
13. ∵ 3x2 + (2 + k)x + 3k > 18 for all real values of x
i.e. 3x2 + (2 + k)x + 3k − 18 > 0
∴ 3x2 + (2 + k)x + 3k – 18 = 0 has no real roots.
[pic]
Hence, the range of possible values of k is 10 < k < 22.
Revision Exercise 6 (p. 6.38)
Level 1
1. Solving 4x + 6 ( 1, we have
[pic]
Solving 3x – 2 > 5, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
2. Solving 3 ( 4x – 3, we have
[pic]
Solving 4x – 3 > 7, we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
3. Rewrite the compound inequality as:
3 – 2x > 2 – x and 2 – x ( 4 – 3x
Solving 3 – 2x > 2 – x, we have
[pic]
Solving 2 – x ( 4 – 3x, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ There are no solutions.
4. Solving 3(x + 1) – 5 ( 7, we have
[pic]
Solving 8 – 5x ( –2, we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x ( 3.
Graphical representation:
[pic]
5. Solving [pic], we have
[pic]
Solving 2x + 7 > x + 4, we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are all real values of x.
Graphical representation:
[pic]
6. Solving 2x – 3 ( 5, we have
[pic]
Solving 3x – 2 > 4, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are
2 < x ( 4.
Graphical representation:
[pic]
7. Solving 5x – 3 > x + 1, we have
[pic]
Solving 4x + 1 > 2x – 5, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are x > 1.
Graphical representation:
[pic]
8. Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x > –4.
Graphical representation:
[pic]
9. Consider the corresponding quadratic function
y = x2 – 15x + 50.
Since the coefficient of x2 is 1 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are 5 and 10.
Sketch the graph of y = x2 – 15x + 50.
[pic]
From the graph, the solutions of x2 – 15x + 50 ( 0 are
x ( 5 or x ( 10.
10. Consider the corresponding quadratic function
y = –x2 + 3x – 2.
Since the coefficient of x2 is –1 (< 0), the graph opens downwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are 1 and 2.
Sketch the graph of y = –x2 + 3x – 2.
[pic]
From the graph, the solutions of –x2 + 3x – 2 > 0 are
1 < x < 2.
11. Consider the corresponding quadratic function
y = –x2 – 4x + 21.
Since the coefficient of x2 is –1 (< 0), the graph opens downwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are –7 and 3.
Sketch the graph of y = –x2 – 4x + 21.
[pic]
From the graph, the solutions of –x2 – 4x + 21 ( 0 are
x ( –7 or x ( 3.
12. Consider the corresponding quadratic function
y = 2x2 + 5x – 3.
Since the coefficient of x2 is 2 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are –3 and [pic].
Sketch the graph of y = 2x2 + 5x – 3.
[pic]
From the graph, the solutions of 2x2 + 5x – 3 < 0 are [pic].
13. [pic]
∴ The solutions of x2 + 11x – 26 > 0 are x < –13 or
x > 2.
14. [pic]
∴ The solutions of –x2 – 23x + 140 ( 0 are –28 ( x ( 5.
15. [pic]
∴ The solutions of 2x2 – 5x ( –2 are [pic].
16. [pic]
∴ The solutions of (2x + 1)2 – 3(2x + 1) ( 0 are [pic].
17. [pic]
|Range of x |[pic] |[pic] |[pic] |
|5x + 12 |– |0 |+ |
|x – 3 |– |– |– |
|(5x + 12)(x – 3) |+ |0 |– |
|Range of x |x = 3 |x > 3 |
|5x + 12 |+ |+ |
|x – 3 |0 |+ |
|(5x + 12)(x – 3) |0 |+ |
From the table, the solutions of 5x2 – 3x – 36 < 0 are [pic].
18. [pic]
|Range of x |x < –9 |x = –9 |–9 < x < 16 |x = 16 |x > 16 |
|x – 16 |– |– |– |0 |+ |
|(x + 9)(x – 16)|+ |0 |– |0 |+ |
From the table, the solutions of –x2 + 7x + 144 < 0 are
x < –9 or x > 16.
19. [pic]
|Range of x |[pic] |[pic] |[pic] |[pic] |[pic] |
|5x – 4 |– |– |– |0 |+ |
|(5x – 3)(5x – 4)|+ |0 |– |0 |+ |
From the table, the solutions of (5x – 3)2 ( 5x – 3 are [pic].
20. [pic]
|Range of x |[pic] |[pic] |[pic] |[pic] |[pic] |
|2x – 3 |– |– |– |0 |+ |
|(4x – 3)(2x – 3) |+ |0 |– |0 |+ |
From the table, the solutions of –8x2 + 18x ( 9 are [pic].
21. (a) [pic]
(b) ∵ x must simultaneously satisfy x < 6 and
–8 ( x ( 8.
∴ The range of possible values of x is –8 ( x < 6.
22. (a) [pic]
(b) Since x is an odd integer, the possible values of x are 11 and 13.
23. (a) Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
(b) –5, –3, –2.5, 2.5 and 3
24. (a) [pic]
Consider the corresponding quadratic function [pic].
Since the coefficient of x2 is 1 (> 0), the graph opens upwards.
When y = 0,
[pic]
Sketch the graph of [pic].
[pic]
From the graph, the solutions of [pic] are x < –3 or x > 3.
(b) ∵ x is a positive integer.
∴ x > 3
∴ The least possible value of x is 4.
25. (a) Number of $10 coins = 120 – x
∴ [pic]
(b) Solving [pic], we have
[pic]
Solving 5x + 10(120 – x) < 1000, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are
40 < x < 80.
26. From the question,
(x + 25) – 4 ( 2(x – 4) and (x + 25) + 4 < 2(x + 4)
Solving (x + 25) – 4 ( 2(x – 4), we have
[pic]
Solving (x + 25) + 4 < 2(x + 4), we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are
21 < x ( 29.
Hence, the possible values of x are 22, 23, 24, 25, 26, 27, 28 and 29.
27. Let x be the smaller integer, then the larger integer is x + 5.
[pic]
Hence, the least possible value of the smaller integer is 16.
28. ∵ x2 – 2kx + k = 0 has two distinct real roots.
∴ [pic]
Hence, the range of possible values of k is k < 0 or k > 1.
29. ∵ kx2 + (2k + 1)x + (2k + 1) = 0 has no real roots.
∴ [pic]
Hence, the range of possible values of k is [pic].
30. From the question,
[pic]
Hence, the range of possible values of x is x > 4.
Level 2
31. Solving 2x – 1 < 3(x – 1) + 2, we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x < –12 or x > 0.
Graphical representation:
[pic]
32. Rewrite the compound inequality as:
[pic] and [pic]
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
33. Rewrite the compound inequality as:
[pic] and [pic]
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
34. Consider
‘[pic]’. ……(*)
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of (*) are [pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (3) and (4).
∴ The solutions of the compound inequality are [pic].
Graphical representation:
[pic]
35. Consider
‘[pic]’. ……(*)
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of (*) are x ( –1. …… (3)
Solving [pic], we have
[pic]
∵ x must satisfy (3) or (4).
∴ The solutions of the compound inequality are all real values of x.
Graphical representation:
[pic]
36. [pic]
Consider the corresponding quadratic function
[pic].
Since the coefficient of x2 is 9 (> 0), the graph opens upwards.
When y = 0,
[pic]
( The x-intercepts of the graph are [pic] and [pic].
Sketch the graph of [pic].
[pic]
From the graph, the solutions of [pic] are [pic].
37. [pic]
Consider the corresponding quadratic function
y = 5x2 – 7x + 2.
Since the coefficient of x2 is 5 (> 0), the graph opens upwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are [pic] and 1.
Sketch the graph of y = 5x2 – 7x + 2.
[pic]
From the graph, the solutions of
(3x – 2)2 – 1 < (2x – 1)2 – x are [pic].
38. [pic]
∴ The solutions of 6x2 – 13x – 63 ( 0 are [pic].
39. [pic]
∴ The solutions of 3(2x + 3)2 – (x – 1)2 > 2(2x + 1)2 are [pic].
40. Solving [pic], we have
[pic]
Solving (x + 2)2 ( 3(x + 8), we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are x ( –5 or x > –1.
41. Rewrite the compound inequality as:
–12 < x2 – 7x and x2 – 7x < –10
Solving –12 < x2 – 7x, we have
[pic]
Solving x2 – 7x < –10, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are
2 < x < 3 or 4 < x < 5.
42. (a) Rewrite the compound inequality as:
[pic]
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The range of possible values of x is [pic].
(b) The possible values of x are –1 and 0.
43. (a) Consider the corresponding quadratic function
y = –5x2 – 14x + 3.
Since the coefficient of x2 is –5 (< 0), the graph opens downwards.
When y = 0,
[pic]
∴ The x-intercepts of the graph are –3 and [pic].
Sketch the graph of y = –5x2 – 14x + 3.
[pic]
From the graph, the solutions of –5x2 – 14x + 3 > 0 are [pic].
(b) [pic]
From (a), we have
[pic]
44. (a) Number of cows = 2000 – x
∴ 500 ( x – (2000 – x) ( 600 and 2000 – x ( 720
(b) [pic]
Solving 2000 – x ( 720, we have
[pic]
∵ x must satisfy (1) and (2).
∴ The range of possible values of x is
1280 ( x ( 1300.
Maximum number of sheep = 1300
∴ Minimum number of cows = 2000 – 1300 = 700
45. Distance travelled by David = t(t + 400) m
Distance travelled by Paul = t(t + 300) m
∴ [pic]
Rewrite the compound inequality as:
3600 ( t2 + 350t and t2 + 350t ( 4000
Solving 3600 ( t2 + 350t, we have
[pic]
Solving t2 + 350t ( 4000, we have
[pic]
∵ t is non-negative and must satisfy both (1) and (2).
∴ The range of possible values of t is 10 ( t ( 11.1.
46. (a) (C = 180° – (A – (B (( sum of △)
= 180° – (2x° – 18°) – (x° + 30°)
= 168° – 3x°
(b) (i) Since (A, (B and (C must be positive, we have
(A > 0° and (B > 0° and (C > 0°
i.e. 2x – 18 > 0 and x + 30 > 0 and
168 – 3x > 0
(ii) Solving 2x – 18 > 0, we have
[pic]
Solving x + 30 > 0, we have
[pic]
Solving 168 – 3x > 0, we have
[pic]
∵ x must satisfy (1), (2) and (3).
∴ The range of possible values of x is
9 < x < 56. …… (4)
(c) (i) Since (ABC is an obtuse-angled triangle,
(A > 90° or (B > 90° or (C > 90°
i.e. 2x – 18 > 90 or x + 30 > 90 or
168 – 3x > 90
(ii) Consider ‘2x – 18 > 90 or x + 30 > 90 or 168 – 3x > 90’. ……(*)
Solving 2x – 18 > 90, we have
[pic]
Solving x + 30 > 90, we have
[pic]
Solving 168 – 3x > 90, we have
[pic]
∵ x must satisfy (5), (6) or (7).
∴ The solutions of (*) is
x < 26 or x > 54. …… (8)
∵ x must satisfy both (4) and (8).
∴ The range of possible values of x is
9 < x < 26 or 54 < x < 56.
47. (a) ∵ 2x2 + (k – 1)x – 2k > 0 for all real values of x
∴ 2x2 + (k – 1)x – 2k = 0 has no real roots.
[pic]
Hence, the range of possible values of k is [pic].
(b) From (a), we have
[pic]
Since k is an integer, the maximum possible value of k is –1.
48. (a) [pic]
∵ x2 + (2k + 1)x + (1 – 2k) > 0 for all real values of x
∴ x2 + (2k + 1)x + (1 – 2k) = 0 has no real roots.
[pic]
Hence, the range of possible values of k is [pic].
(b) From (a), we have
[pic]
Since k is an integer, the possible values of k are
–3, –2, –1 and 0.
49. ∵ [pic] and [pic] are two real numbers.
∴ x2 + 8x – 48 ( 0 and 6x2 + 7x – 24 ( 0
Solving x2 + 8x – 48 ( 0, we have
[pic]
Solving 6x2 + 7x – 24 ( 0, we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The range of possible values of x is x ( –12 or x ( 4.
50. (a) Number of beads required to construct the nth pattern
[pic]
(b) ∵ [pic]
∴ [pic]
i.e. [pic]
∵ n is a positive integer.
∴ [pic]
∴ The largest possible pattern is the 48th pattern.
51. (a) Total surface area of the cuboid
[pic]
(b) From the question,
[pic]
∵ r > 0
∴ The range of possible values of r is [pic].
52. (a) From the question,
[pic]
Hence, the range of possible values of x is x ( 4.
(b) Minimum length of the wire
= [4 × 4 + 2 × (6 + 5)] cm
= 38 cm
Multiple Choice Questions (p. 6.41)
1. Answer: C
Solving [pic], we have
[pic]
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are [pic].
2. Answer: D
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy (1) or (2).
∴ The solutions of the compound inequality are all real values of x.
3. Answer: C
For choice C,
solving [pic], we have
[pic]
solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of the compound inequality are [pic].
∴ The answer is C.
4. Answer: B
Consider ‘[pic]’. ……(*)
Solving [pic], we have
[pic]
Solving [pic], we have
[pic]
∵ x must satisfy both (1) and (2).
∴ The solutions of (*) are [pic]
Solving [pic], we have
[pic]
∵ x must satisfy (3) or (4).
∴ The solutions of the compound inequality are [pic].
5. Answer: D
From the graph of [pic], the solutions of [pic] are [pic] or [pic].
6. Answer: B
From the table, the solutions of [pic] are [pic].
7. Answer: A
[pic]
[pic] or [pic]
[pic] or [pic]
[pic] or no solutions
∴ The solutions of [pic] are [pic].
8. Answer: D
The length of the rectangle[pic]
Area of the rectangle[pic]
[pic]
∴ The range of possible values of x is [pic].
9. Answer: C
The larger number is a + 3.
[pic]
∴ The largest possible value of a is 8.
HKMO (p. 6.43)
1. [pic]
∴ The greatest value of x, i.e. h, is 8.
2. Consider the straight line L: 3x + 5y = 1.
∵ The straight line L passes through (2, (1) and its slope is [pic].
∴ The equation 3x + 5y = 1 can be rewritten as
[pic].
∴ Let [pic], where k is an integer
[pic]
[pic]
∴ The least possible value of k is 251.
[pic]
∴ The least possible value of S is 2011.
Investigation Corner (p. 6.44)
a) The number of beads (x) to construct pattern n is n2.
The number of sticks (y) to construct pattern n is
[n(n – 1) + (n – 1)n], i.e. 2n(n – 1).
|[pic] | | |
| |(∵ n ( 1) | |
| |(cor. to 2 d.p.) | |
| | |……(1) |
and [pic]
[pic](∵ n ( 1)
∵ n must satisfy both (1) and (2).
∴ The range of possible values of n is [pic].
∴ The largest pattern is pattern 26.
|(b) |[pic] | | |
| | |(∵ n ( 1) | |
| | |(cor. to 2 d.p.) | |
| | | |……(3) |
and [pic]
[pic](∵ n ( 1)
∵ n must satisfy both (3) and (4).
∴ The range of possible values of n is [pic].
∴ The largest pattern is pattern 16.
(c) Let X and Y be the number of beads and the number of sticks to construct pattern M.
∴ [pic] (∵ M ( 1)
∴[pic](∵ M ( 1)
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