# MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers

Students are advised to solve the Linear Inequalities Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Linear Inequalities Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Linear Inequalities Class 11 with answers provided with detailed solutions by looking below.

Question 1.

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)

Given, -2 < 2x – 1 < 2

⇒ -2 + 1 < 2x < 2 + 1

⇒ -1 < 2x < 3

⇒ -1/2 < x < 3/2

⇒ x ∈ (-1/2, 3/2)

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

Given, x² < -4

⇒ x² + 4 < 0

Which is not possible.

So, there is no solution.

Question 3.

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

(a) (-∞, -5)

(b) (∞, 5)

(c) (-5, ∞)

(d) No Solution

## Answer

Answer: (d) No Solution

Given, |x| < -5

Now, LHS ≥ 0 and RHS < 0

Since LHS is non-negative and RHS is negative

So, |x| < -5 does not posses any solution

Question 4.

The graph of the inequations x ≤ 0 , y ≤ 0, and 2x + y + 6 ≥ 0 is

(a) exterior of a triangle

(b) a triangular region in the 3rd quadrant

(c) in the 1st quadrant

(d) none of these

## Answer

Answer: (b) a triangular region in the 3rd quadrant

Given inequalities x ≥ 0 , y ≥ 0 , 2x + y + 6 ≥ 0

Now take x = 0, y = 0 and 2x + y + 6 = 0

when x = 0, y = -6

when y = 0, x = -3

So, the points are A(0, 0), B(0, -6) and C(-3, 0)

So, the graph of the inequations x ≤ 0 , y ≤ 0 , and 2x + y + 6 ≥ 0 is a triangular region in the 3rd quadrant.

Question 5.

The graph of the inequalities x ≥ 0, y ≥ 0, 2x + y + 6 ≤ 0 is

(a) a square

(b) a triangle

(c) { }

(d) none of these

## Answer

Answer: (c) { }

Given inequalities x ≥ 0, y ≥ 0, 2x + y + 6 ≤ 0

Now take x = 0, y = 0 and 2x + y + 6 = 0

when x = 0, y = -6

when y = 0, x = -3

So, the points are A(0, 0), B(0, -6) and C(-3, 0)

Since region is outside from the line 2x + y + 6 = 0

So, it does not represent any figure.

Question 6.

Solve: 2x + 1 > 3

(a) [-1, ∞]

(b) (1, ∞)

(c) (∞, ∞)

(d) (∞, 1)

## Answer

Answer: (b) (1, ∞)

Given, 2x + 1 > 3

⇒ 2x > 3 – 1

⇒ 2x > 2

⇒ x > 1

⇒ x ∈ (1, ∞)

Question 7.

The solution of the inequality 3(x – 2)/5 ≥ 5(2 – x)/3 is

(a) x ∈ (2, ∞)

(b) x ∈ [-2, ∞)

(c) x ∈ [∞, 2)

(d) x ∈ [2, ∞)

## Answer

Answer: (d) x ∈ [2, ∞)

Given, 3(x – 2)/5 ≥ 5(2 – x)/3

⇒ 3(x – 2) × 3 ≥ 5(2 – x) × 5

⇒ 9(x – 2) ≥ 25(2 – x)

⇒ 9x – 18 ≥ 50 – 25x

⇒ 9x – 18 + 25x ≥ 50

⇒ 34x – 18 ≥ 50

⇒ 34x ≥ 50 + 18

⇒ 34x ≥ 68

⇒ x ≥ 68/34

⇒ x ≥ 2

⇒ x ∈ [2, ∞)

Question 8.

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]

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]

Question 9.

Solve: -1/(|x| – 2) ≥ 1 where x ∈ R, x ≠ ±2

(a) (-2, -1)

(b) (-2, 2)

(c) (-2, -1) ∪ (1, 2)

(d) None of these

## Answer

Answer: (c) (-2, -1) ∪ (1, 2)

Given, -1/(|x| – 2) ≥ 1

⇒ -1/(|x| – 2) – 1 ≥ 0

⇒ {-1 – (|x| – 2)}/(|x| – 2) ≥ 0

⇒ {1 – |x|}/(|x| – 2) ≥ 0

⇒ -(|x| – 1)/(|x| – 2) ≥ 0

Using number line rule:

1 ≤ |x| < 2

⇒ x ∈ (-2, -1) ∪ (1, 2)

Question 10.

If x² < 4 then the value of x is

(a) (0, 2)

(b) (-2, 2)

(c) (-2, 0)

(d) None of these

## Answer

Answer: (b) (-2, 2)

Given, x² < 4

⇒ x² – 4 < 0

⇒ (x – 2) × (x + 2) < 0

⇒ -2 < x < 2

⇒ x ∈ (-2, 2)

Question 11.

Solve: 2x + 1 > 3

(a) [1, 1)

(b) (1, ∞)

(c) (∞, ∞)

(d) (∞, 1)

## Answer

Answer: (b) (1, ∞)

Given, 2x + 1 > 3

⇒ 2x > 3 – 1

⇒ 2x > 2

⇒ x > 1

⇒ x ∈ (1, ∞)

Question 12.

If a is an irrational number which is divisible by b then the number b

(a) must be rational

(b) must be irrational

(c) may be rational or irrational

(d) None of these

## Answer

Answer: (b) must be irrational

If a is an irrational number which is divisible by b then the number b must be irrational.

Ex: Let the two irrational numbers are √2 and √3

Now, √2/√3 = √(2/3)

Question 13.

Sum of two rational numbers is ______ number.

(a) rational

(b) irrational

(c) Integer

(d) Both 1, 2 and 3

## Answer

Answer: (a) rational

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 14.

If |x| = -5 then the value of x lies in the interval

(a) (-5, ∞)

(b) (5, ∞)

(c) (∞, -5)

(d) No solution

## Answer

Answer: (d) No solution

Given, |x| = -5

Since |x| is always positive or zero

So, it can not be negative

Hence, given inequality has no solution.

Question 15.

The value of x for which |x + 1| + √(x – 1) = 0

(a) 0

(b) 1

(c) -1

(d) No value of x

## Answer

Answer: (d) No value of x

Given, |x + 1| + √(x – 1) = 0, where each term is non-negative.

So, |x + 1| = 0 and √(x – 1) = 0 should be zero simultaneously.

i.e. x = -1 and x = 1, which is not possible.

So, there is no value of x for which each term is zero simultaneously.

Question 16.

If x² < -4 then the value of x is

(a) (-2, 2)

(b) (2, ∞)

(c) (-2, ∞)

(d) No solution

## Answer

Answer: (d) No solution

Given, x² < -4

⇒ x² + 4 < 0

Which is not possible.

So, there is no solution.

Question 17.

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)

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 18.

The solution of the function f(x) = |x| > 0 is

(a) R

(b) R – {0}

(c) R – {1}

(d) R – {-1}

## Answer

Answer: (b) R – {0}

Given, f(x) = |x| > 0

We know that modulus is non negative quantity.

So, x ∈ R except that x = 0

⇒ x ∈ R – {0}

This is the required solution

Question 19.

Solve: |x – 1| ≤ 5, |x| ≥ 2

(a) [2, 6]

(b) [-4, -2]

(c) [-4, -2] ∪ [2, 6]

(d) None of these

## Answer

Answer: (c) [-4, -2] ∪ [2, 6]

Given, |x – 1| ≤ 5, |x| ≥ 2

⇒ -(5 ≤ (x – 1) ≤ 5), (x ≤ -2 or x ≥ 2)

⇒ -(4 ≤ x ≤ 6), (x ≤ -2 or x ≥ 2)

Now, required solution is

x ∈ [-4, -2] ∪ [2, 6]

Question 20.

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

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

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