Check the below NCERT MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Linear Inequalities Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

## Linear Inequalities Class 11 MCQs Questions with Answers

Question 1.

Sum of two rational numbers is ______ number.

(a) rational

(b) irrational

(c) Integer

(d) Both 1, 2 and 3

## Answer

Answer: (a) rational

Hint:

The sum of two rational numbers is a rational number.

Ex: Let two rational numbers are 1/2 and 1/3

Now, 1/2 + 1/3 = 5/6 which is a rational number.

Question 2.

If x² = -4 then the value of x is

(a) (-2, 2)

(b) (-2, ∞)

(c) (2, ∞)

(d) No solution

## Answer

Answer: (d) No solution

Hint:

Given, x² = -4

Since LHS ≥ 0 and RHS < 0

So, No solution is possible.

Question 3.

Solve: (x + 1)² + (x² + 3x + 2)² = 0

(a) x = -1, -2

(b) x = -1

(c) x = -2

(d) None of these

## Answer

Answer: (b) x = -1

Hint:

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

Question 4.

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

(a) (-8, ∞)

(b) (8, ∞)

(c) (∞, -8)

(d) (∞, 8)

## Answer

Answer: (a) (-8, ∞)

Hint:

Given,

(x + 3)/(x – 2) > 1/2

⇒ 2(x + 3) > x – 2

⇒ 2x + 6 > x – 2

⇒ 2x – x > -2 – 6

⇒ x > -8

⇒ x ∈ (-8, ∞)

Question 5.

The region of the XOY-plane represented by the inequalities x ≥ 6, y ≥ 2, 2x + y ≤ 10 is

(a) unbounded

(b) a polygon

(c) none of these

(d) exterior of a triangle

## Answer

Answer: (c) none of these

Hint:

Given inequalities x ≥ 6, y ≥ 2, 2x + y ≤ 10

Now take x = 6, y = 2 and 2x + y = 10

when x = 0, y = 10

when y = 0, x = 5

So, the points are A(6, 2), B(0, 10) and C(5, 0)

So, the region of the XOY-plane represented by the inequalities x ≥ 6, y ≥ 2, 2x + y ≤ 10 is not defined.

Question 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

## Answer

Answer: (d) 2 < x < 3 and x < 1

Hint:

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

Question 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)

## Answer

Answer: (b) (-1/2, 3/2)

Hint:

Given, -2 < 2x – 1 < 2

⇒ -2 + 1 < 2x < 2 + 1

⇒ -1 < 2x < 3

⇒ -1/2 < x < 3/2

⇒ x ∈(-1/2, 3/2)

Question 8.

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

(a) (1, ∞)

(b) (-1, 3)

(c) (1, -3)

(d) (∞, 1)

## Answer

Answer: (b) (-1, 3)

Hint:

Given, |x – 1| < 2

⇒ -2 < x – 1 < 2

⇒ -2 + 1 < x < 2 + 1

⇒ -1 < x < 3

⇒ x ∈ (-1, 3)

Question 9.

If | x − 1| > 5, then

(a) x∈(−∞, −4)∪(6, ∞]

(b) x∈[6, ∞)

(c) x∈(6, ∞)

(d) x∈(−∞, −4)∪(6, ∞)

## Answer

Answer: (d) x∈(−∞, −4)∪(6, ∞)

Hint:

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, ∞)

Question 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)

## Answer

Answer: (b) (2, 4) ∪ (4, 6)

Hint:

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)

Question 11.

If (|x| – 1)/(|x| – 2) ≥ 0, x ∈ R, x ± 2 then the interval of x is

(a) (-∞, -2) ∪ [-1, 1]

(b) [-1, 1] ∪ (2, ∞)

(c) (-∞, -2) ∪ (2, ∞)

(d) (-∞, -2) ∪ [-1, 1] ∪ (2, ∞)

## Answer

Answer: (d) (-∞, -2) ∪ [-1, 1] ∪ (2, ∞)

Hint:

Given, (|x| – 1)/(|x| – 2) ≥ 0

Let y = |x|

So, (y – 1)/(y – 2) ≥ 0

⇒ y ≤ 1 or y > 2

⇒ |x| ≤ 1 or |x| > 2

⇒ (-1 ≤ x ≤ 1) or (x < -2 or x > 2)

⇒ x ∈ [-1, 1] ∪ (-∞, -2) ∪ (2, ∞)

Hence the solution set is:

x ∈ (-∞, -2) ∪ [-1, 1] ∪ (2, ∞)

Question 12.

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

## Answer

Answer: (d) -56/3 < x < 14/3

Hint:

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

Question 13.

If x² = -4 then the value of x is

(a) (-2, 2)

(b) (-2, ∞)

(c) (2, ∞)

(d) No solution

## Answer

Answer: (d) No solution

Hint:

Given, x² = -4

Since LHS ≥ 0 and RHS < 0

So, No solution is possible.

Question 14.

Solve: |x – 3| < 5

(a) (2, 8)

(b) (-2, 8)

(c) (8, 2)

(d) (8, -2)

## Answer

Answer: (b) (-2, 8)

Hint:

Given, |x – 3| < 5

⇒ -5 < (x – 3) < 5

⇒ -5 + 3 < x < 5 + 3

⇒ -2 < x < 8

⇒ x ∈ (-2, 8)

Question 15.

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

## Answer

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

Hint:

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.

Question 16.

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

(a) (-∞, -5)

(b) (∞, 5)

(c) (-5, ∞)

(d) (-5, 5)

## Answer

Answer: (d) (-5, 5)

Hint:

Given, |x| < 5

It means that x is the number which is at distance less than 5 from 0

Hence, -5 < x < 5

⇒ x ∈ (-5, 5)

Question 17.

Solve: f(x) = {(x – 1)×(2 – x)}/(x – 3) ≥ 0

(a) (-∞, 1] ∪ (2, ∞)

(b) (-∞, 1] ∪ (2, 3)

(c) (-∞, 1] ∪ (3, ∞)

(d) None of these

## Answer

Answer: (b) (-∞, 1] ∪ (2, 3)

Hint:

Given, f(x) = {(x – 1)×(2 – x)}/(x – 3) ≥ 0

or f(x) = -{(x – 1)×(2 – x)}/(x – 3)

which gives x – 3 ≠ 0

⇒ x ≠ 3

Using number line rule as shown in the figure,

which gives f(x) ≥ 0 when x ≤ 1 or 2 ≤ x < 3

i.e. x ∈ (-∞, 1] ∪ (2, 3)

Question 18.

If x² = 4 then the value of x is

(a) -2

(b) 2

(c) -2, 2

(d) None of these

## Answer

Answer: (c) -2, 2

Hint:

Given, x² = 4

⇒ x² – 4 = 0

⇒ (x – 2)×(x + 2) = 0

⇒ x = -2, 2

Question 19.

The solution of the 15 < 3(x – 2)/5 < 0 is

(a) 27 < x < 2

(b) 27 < x < -2

(c) -27 < x < 2

(d) -27 < x < -2

## Answer

Answer: (a) 27 < x < 2

Hint:

Given inequality is:

15 < 3(x-2)/5 < 0

⇒ 15 × 5 < 3(x-2) < 0 × 5

⇒ 75 < 3(x-2) < 0

⇒ 75/3 < x-2 < 0

⇒ 25 < x-2 < 0

⇒ 25 +2 < x <0+2

⇒ 27 < x < 2

Question 20.

Solve: 1 ≤ |x – 1| ≤ 3

(a) [-2, 0]

(b) [2, 4]

(c) [-2, 0] ∪ [2, 4]

(d) None of these

## Answer

Answer: (c) [-2, 0] ∪ [2, 4]

Hint:

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]

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