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Paper 1

1. –1 < x , 3Answer: D

2. – 4 , x < 2Answer: B

3. x + 1 < 4 x < 4 – 1 x < 3x is a positive integer.Hence, x = 1, 2, 3Answer: D

4. 2x – 1 , 9 2x , 9 + 1 2x , 10

x , 10—–2

x , 5x is a positive integer.Hence, x = 1, 2, 3, 4Answer: C

5.

–1 0 1 2 3

x = –1, 0, 1, 2Answer: B

6. 3m – 1 . 11 3m . 11 + 1 3m . 12

m . 12–––3

m . 4Answer: A

7. p—2 . 3

p . 2 × 3 p . 6Answer: C

8. 5 – h < 2 5 < 2 + h 5 – 2 < h 3 < h h > 3

Answer: A

Paper 1

1. 2 – x > – 4 and 2x + 1 . 3 2 + 4 > x 2x . 3 – 1 6 > x 2x . 2 x . 1

0 1 2 3 4 5 6 7 8–1–2

Answer: C

2. 2x – 1 > 15 and 4x–––3 , 16

4x , 16 × 3

x , 48–––4

x , 12

2x > 15 + 1

x > 16–––2

x > 8

8 9 10 11 12

x = 8, 9, 10, 11

Answer: C

3. 3 < x , 8 and 1 – y , 5 1 – 5 , y –4 , y

Maximum value of x – y= Maximum value of x – Minimum value of y= 7 – (–3)= 10

Answer: C

CHAPTER

5 Linear InequalitiesCHAPTER

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Mathematics SPM Chapter 5

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4. 4 – 3x , 16 4 , 16 + 3x 4 – 16 , 3x –12 , 3x

– 12–––3 , x

–4 , x x = –3, –2, –1, …

Answer: A

5. n––2 – 1 > 3n + 4

–1 – 4 > 3n – n––2

–5 > 6n–––2 – n––

2

–5 > 5n–––2

–5 × 2 > 5n

–10––––5 > n

–2 > n

Answer: D

6. 12

x , 2 and 2 – 5x < 7 2 – 7 < 5x –5 < 5x –1 < x

x , 4

–1 10 2 3 4

The integers are –1, 0, 1, 2, and 3.

Answer: C

7. 3(x + 5) < –2x 3x + 15 < –2x 3x + 2x < –15 5x < –15

x < – 155

x < –3

Answer: D

8. 23

x > 4 and 2 + x , 10

2x > 12 x , 10 – 2 x > 6 x , 8

6 87

x = 6, 7

Answer: B

9. h – 2–4 . 3

h – 2 . –12 h . –12 + 2 h . –10

Answer: C

Paper 1

1. 2x + 3 > –3 and 2 – x > 0 2x > –3 – 3 2 > x

x > –6–––2

x > –3

–1 1 20–3 –2

x = –3, –2, –1, 0, 1, 2

Answer: A

2. 8 – e < 6 and 3(e – 6) , 12 – 2e 3e – 18 , 12 – 2e 3e + 2e , 12 + 18 5e , 30

e , 30–––5

e , 6

8 – 6 < e 2 < e

4 652 3

e = 2, 3, 4, 5

Answer: C

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Mathematics SPM Chapter 5

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3. 2k < 8 + k and k––2 > 1

k > 1 × 2 k > 2

2k – k < 8 k < 8

Hence, 2 < k < 8

Answer: D

4. Given 4 , 3x – 11 < 16then 4 , 3x – 11 and 3x – 11 < 16

3x < 16 + 11

x < 27–––3

x < 9

4 + 11 , 3x

15–––3 , x

5 , x

5 6 7 8 9

x = 6, 7, 8, 9

Answer: A

5. Given –1 , 1 – x—2 < 2

then –1 , 1 – x—2 and 1 – x—

2 < 2

1 – 2 < x—2

–1 < x—2

–2 < x

x—2 – 1 , 1

x—2 , 1 + 1

x , 4

–2 –1 0 2 31 4

x = –2, –1, 0, 1, 2, 3

Answer: C

6. 7 – w < 2 and 3w – 27 , 0 3w , 27

w , 27–––3

w , 9

7 – 2 < w 5 < w

Hence, 5 < w , 9

Answer: B

7. 2 + t , 9 and 7 – 4t < 3 7 – 3 < 4t 4 < 4t 1 < t

t , 9 – 2 t , 7

Hence, 1 < t , 7

Answer: D

8. Given 3 > h . 1 – h—3 ,

then 3 > h and h . 1 – h—3

h + h—3 . 1

4h–––3 . 1

4h . 1 × 3

h . 3—4

0 1 2 334–

h = 1, 2, 3

Answer: A

9. 2m – 1 , 2 and m > –3 2m , 2 + 1 2m , 3

m , 3—2

–3 –2 –1 0 1 232–

m = –3, –2, –1, 0, 1

Answer: C

10. 1—3 x < 2 and 5 – 3x , 8

5 – 8 , 3x –3 , 3x

–3–––3 , x

–1 , x

x < 2 × 3 x < 6

Hence, –1 , x < 6

Answer: C

4

Mathematics SPM Chapter 5

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11. 2—5 x . –2 and 3 + 4x < 15

4x < 15 – 3

x < 12–––4

x < 3

2x . –2 × 5

x . – 10–––2

x . –5

Hence, –5 , x < 3

Answer: D

12. 7 – 2p . – 4 and 2p–––3

. 2

2p . 2 × 3

p . 6—2

p . 3

7 + 4 . 2p

11–––2 . p

5 1—2 . p

3 4 5 65 12–

p = 4, 5

Answer: A

13. x2 + y2 = 22 + 02

= 4

Answer: B

14. Given 1 + y < 9 – y , 2y + 6

then 1 + y < 9 – y and 9 – y , 2y + 6 9 – 6 , 2y + y 3 , 3y

3—3 , y

1 , y

y + y < 9 – 1 2y < 8 y < 8—

2 y < 4

Hence, 1 , y < 4Compare with p , y < q.Therefore, p = 1, q = 4

Answer: C