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

Do you need some help in preparing for your upcoming Class 11 Maths exams? We’ve compiled a list of MCQ on Linear Inequalities Class 11 MCQs Questions with Answers to get you started with the subject. You can download NCERT MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers Pdf free download, and learn how smart students prepare well ahead with MCQ Questions for Class 11 Maths with Answers.

## Linear Inequalities Class 11 MCQs Questions with Answers

Solving the MCQ Questions of Linear Inequalities Class 11 with answers can help you understand the concepts better.

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)

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

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

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
(d) none of these

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

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)

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: (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: (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: (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

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)

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

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

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

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: (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

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: (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}

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: (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: (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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