Question

Class 11 Maths MCQ Questions of Linear Inequalities with Answers?

Answer 1

Practice the Important Class 11 Maths MCQ Questions of Linear Inequalities, which are given here. Go through the page given below to get the important MCQ Questions of class 11 Maths. In this chapter, students can learn how to derive the solution for a linear inequality, representation of the solution in the number line, and its graphical representation. 

Explore MCQ Questions of class 11 Maths with answers furnished with detailed solutions by looking underneath.

Practice MCQ Questions for class 11 Maths Chapter-Wise

1. Sum of two rational numbers is ______ number.

(a) rational
(b) irrational
(c) Integer
(d) Both 1, 2 and 3

2. Solve: (x + 1)² + (x² + 3x + 2)² = 0

(a) x = -1, -2
(b) x = -1
(c) x = -2
(d) None of these

3. If ∣x−3∣<2x+9, then x lies in the interval

(a) (−∞,−2)
(b) (−2,0)
(c) (−2,∞)
(d) (2,∞)

4. If (x + 3)/(x – 2) > 1/2 then x lies in the interval

(a) (-8, ∞)
(b) (8, ∞)
(c) (∞, -8)
(d) (∞, 8)

5. The region represented by the inequalities x≥6,y≥2,2x+y≤10,x≥0,y≥0 is

(a) unbounded
(b) a polygon
(c) a triangle
(d) None of these

6.The interval in which f(x) = (x – 1) × (x – 2) × (x – 3) is negative is

(a) x > 2
(b) 2 < x and x < 1
(c) 2 < x < 1 and x < 3
(d) 2 < x < 3 and x < 1

7. If -2 < 2x – 1 < 2 then the value of x lies in the interval

(a) (1/2, 3/2)
(b) (-1/2, 3/2)
(c) (3/2, 1/2)
(d) (3/2, -1/2)

8. The solution of the inequality |x – 1| < 2 is

(a) (1, ∞)
(b) (-1, 3)
(c) (1, -3)
(d) (∞, 1)

9. If | x − 1| > 5, then

(a) x∈(−∞, −4)∪(6, ∞]
(b) x∈[6, ∞)
(c) x∈(6, ∞)
(d) x∈(−∞, −4)∪(6, ∞)

10. The solution of |2/(x – 4)| > 1 where x ≠ 4 is

(a) (2, 6)
(b) (2, 4) ∪ (4, 6)
(c) (2, 4) ∪ (4, ∞)
(d) (-∞, 4) ∪ (4, 6)

11. The solution of the -12 < (4 -3x)/(-5) < 2 is

(a) 56/3 < x < 14/3
(b) -56/3 < x < -14/3
(c) 56/3 < x < -14/3
(d) -56/3 < x < 14/3

12. Solve: |x – 3| < 5

(a) (2, 8)
(b) (-2, 8)
(c) (8, 2)
(d) (8, -2)

13. The graph of the inequations x ≥ 0, y ≥ 0, 3x + 4y ≤ 12 is

(a) none of these
(b) interior of a triangle including the points on the sides
(c) in the 2nd quadrant
(d) exterior of a triangle

14. If |x| < 5 then the value of x lies in the interval

(a) (-∞, -5)
(b) (∞, 5)
(c) (-5, ∞)
(d) (-5, 5)

15. If x² = 4 then the value of x is

(a) -2
(b) 2
(c) -2, 2
(d) None of these

16. Solve: 1 ≤ |x – 1| ≤ 3

(a) [-2, 0]
(b) [2, 4]
(c) [-2, 0] ∪ [2, 4]
(d) None of these

17. The function f(x)=∣x∣ at x=0 is:

(a) continuous but non-differentiable
(b) discontinuous and differentiable
(c) discontinuous and non-differentiable
(d) continuous and differentiable

18. If x is real number and |x| < 3, then

(a) x ≥ 3
(b) –3 < x < 3
(c) x ≤ -3
(d) -3 ≤ x ≤ 3

19. The integral value of x which satisfies the inequality x⁴−3x³−x+3<0 is

(a) 0
(b) 1
(c) 2
(d) 3

20. The length of a rectangle is three times the breadth. If the minimum perimeter of the rectangle is 160 cm, then

(a) breadth > 20 cm
(b) length < 20 cm
(c) breadth x ≥ 20 cm 
(d) length ≤ 20 cm

Answer 2

1. Answer: (a) rational

Explanation: adding two rationals is the same as adding two such fractions, which will result in another fraction of this same form since integers are closed under addition and multiplication. Thus, adding two rational numbers produces another rational number.

2. Answer: (b) x = -1

Explanation: Given, (x + 1)² + (x² + 3x + 2)² = 0

This is true when each term is equal to zero simultaneously,

So, (x + 1)² = 0 and (x² + 3x + 2)² = 0

⇒ x + 1 = 0 and x² + 3x + 2 = 0

⇒ x = -1, and x = -1, -2

Now, the common solution is x = -1

So, solution of the equation is x = -1

3. Answer: (c) (−2,∞)

Explanation: Given, ∣x−3∣<2x+9

⇒−(2x+9)<x−3<2x+9

⇒−2x−9<x−3 and x−3<2x+9

⇒−3x<6 and −x<12

⇒x>−2 and x>−12

⇒x>−2

Therefore, x∈(−2,∞)

4. Answer: (a) (-8, ∞)

Explanation: (x + 3)/(x – 2) > 1/2

⇒ 2(x + 3) > x – 2

⇒ 2x + 6 > x – 2

⇒ 2x – x > -2 – 6

⇒ x > -8

⇒ x ∈ (-8, ∞)

5. Answer: (d) None of these

Explanation:

Let l₁: x=6

l₂: y=2

l₃ :2x+y=10

l₄: x=0

and l₅: y=0

The graph of the inequalities 2x+y≤10,

y≥2,x≥0,y≥0 is

the region bounded by ΔBCD. This region has no point in common with the region x≥6,y≥2 as is clear from the figure. Hence, the region of the given inequalities is an empty set.

6. Answer: (d) 2 < x < 3 and x < 1

Explanation: Given, f(x) = (x – 1) × (x – 2) × (x – 3) has all factors with odd powers. So, put them zero

i.e. x – 1 = 0, x – 2 = 0, x – 3 = 0

⇒ x = 1, 2, 3

Now, f(x) < 0 when 2 < x < 3 and x < 1

7. Answer: (b) (-1/2, 3/2)

Explanation: Given, -2 < 2x – 1 < 2

⇒ -2 + 1 < 2x < 2 + 1

⇒ -1 < 2x < 3

⇒ -1/2 < x < 3/2

⇒ x ∈(-1/2, 3/2)

8. Answer: (b) (-1, 3)

Explanation: Given, |x – 1| < 2

⇒ -2 < x – 1 < 2

⇒ -2 + 1 < x < 2 + 1

⇒ -1 < x < 3

⇒ x ∈ (-1, 3)

9. Answer: (d) x∈(−∞, −4)∪(6, ∞)

Explanation: Given |x−1| >5

Case 1:

(x – 1) > 5

⇒ x > 6

⇒ x ∈ (6,∞)

Case 2:

-(x – 1) > 5

⇒ -x + 1 > 5

⇒ -x > 4

⇒ x < -4

⇒ x ∈ (−∞, −4)

So the range of x is (−∞, −4)∪(6, ∞)

10. Answer: (b) (2, 4) ∪ (4, 6)

Explanation: Given, |2/(x – 4)| > 1

⇒ 2/|x – 4| > 1

⇒ 2 > |x – 4|

⇒ |x – 4| < 2

⇒ -2 < x – 4 < 2

⇒ -2 + 4 < x < 2 + 4

⇒ 2 < x < 6

⇒ x ∈ (2, 6) , where x ≠ 4

⇒ x ∈ (2, 4) ∪ (4, 6)

Answer 3

11. Answer: (a) 56/3 < x < 14/3

Explanation: Given inequality is :

-12 < (4 -3x)/(-5) < 2

⇒ -2 < (4-3x)/5 < 12

⇒ -2 × 5 < 4 – 3x < 12 × 5

⇒ -10 < 4 – 3x < 60

⇒ -10 – 4 < -3x < 60-4

⇒ -14 < -3x < 56

⇒ -56 < 3x < 14

⇒ -56/3 < x < 14/3

12. Answer: (b) (-2, 8)

Explanation: Given, |x – 3| < 5

⇒ -5 < (x – 3) < 5

⇒ -5 + 3 < x < 5 + 3

⇒ -2 < x < 8

⇒ x ∈ (-2, 8)

13. Answer: (b) interior of a triangle including the points on the sides

Explanation: Given inequalities x ≥ 0, y ≥ 0, 3x + 4y ≤ 12

Now take x = 0, y = 0 and 3x + 4y = 12

when x = 0, y = 3

when y = 0, x = 4

So, the points are A(0, 0), B(0, 3) and C(4, 0)

So, the graph of the inequations x ≥ 0, y ≥ 0, 3x + 4y ≤ 12 is interior of a triangle including the points on the sides.

14. Answer: (d) (-5, 5)

Explanation: It means that x is the number which is at distance less than 5 from 0. Hence, -5 < x < 5

⇒ x ∈ (-5, 5)

15. Answer: (c) -2, 2

Explanation: Given, x² = 4

⇒ x² – 4 = 0

⇒ (x – 2)×(x + 2) = 0

⇒ x = -2, 2

16. Answer: (c) [-2, 0] ∪ [2, 4]

Explanation: Given, 1 ≤ |x – 1| ≤ 3

⇒ -3 ≤ (x – 1) ≤ -1 or 1 ≤ (x – 1) ≤ 3

i.e. the distance covered is between 1 unit to 3 units

⇒ -2 ≤ x ≤ 0 or 2 ≤ x ≤ 4.

Hence, the solution set of the given inequality is x ∈ [-2, 0] ∪ [2, 4]

17. Answer: (a) continuous but non-differentiable

Explanation: Given, f(x)=∣x∣

As we can see the curve is continuous everywhere. But at x=0 the curve changes its direction abruptly so it is not differentiable.

18. Answer: (b) –3 < x < 3

Explanation: Given, |x|<3

⇒−3<x<3 

[∵|x|<a ⇒ −a<x<a]

19. Answer: (c) 2 

Explanation:  x⁴−3x³−x+3, we get

x⁴−3x³−x+3 = (x−1)(x−3)(x²+x+1)

So the inequality is equivalent to

(x−1)(x−3)(x²+x+1)<0

⇒(x−1)(x−3)<0 (∵x²+x+1>0)

∴1<x<3 so integral value of x = 2

20. Answer: (c) breadth x ≥ 20 cm 

Explanation: If x cm is the breadth, then

2 (3x + x) ≥ 160 

⇒ x ≥ 20

Click here to practice MCQ Questions for Linear Inequalities class 11