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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