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1 © Penerbitan Pelangi Sdn. Bhd.
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
2
Mathematics SPM Chapter 5
© Penerbitan Pelangi Sdn. Bhd.
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
3
Mathematics SPM Chapter 5
© Penerbitan Pelangi Sdn. Bhd.
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
© Penerbitan Pelangi Sdn. Bhd.
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