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

Linear Inequalities Class 11 Maths MCQs Questions with Answers

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.

Class 11 Maths Chapter 6 Quiz

Class 11 Maths Chapter 6 MCQ Online Test

You can refer to NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Linear Inequalities Class 11 Maths MCQ online test

Q1.	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)

Ans: (-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)

Q2.	If x² < -4 then the value of x is
	A.(-2, 2)
	B.(2, ∞)
	C.(-2, ∞)
	D.No solution

Ans: No solution
Given, x² < -4
⇒ x² + 4 < 0
Which is not possible.
So, there is no solution.

Q3.	If \|x\| < -5 then the value of x lies in the interval
	A.(-∞, -5)
	B.(∞, 5)
	C.(-5, ∞)
	D.No Solution

Ans: 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

Q4.	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

Ans: 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)
MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers 1

MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers 1

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

Q5.	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

Ans: { }
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)
MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers 2

MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers 2

Since region is outside from the line 2x + y + 6 = 0
So, it does not represent any figure.

Q6.	Solve: 2x + 1 > 3
	A.[-1, ∞]
	B.(1, ∞)
	C.(∞, ∞)
	D.(∞, 1)

Ans: (1, ∞)
Given, 2x + 1 > 3
⇒ 2x > 3 – 1
⇒ 2x > 2
⇒ x > 1
⇒ x ∈ (1, ∞)

Q7.	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, ∞)

Ans: 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, ∞)

Q8.	Solve: 1 ≤ \|x – 1\| ≤ 3
	A.[-2, 0]
	B.[2, 4]
	C.[-2, 0] ∪ [2, 4]
	D.None of these

Ans: [-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]

Q9.	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

Ans: (-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
MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers 3

Using number line rule:
1 ≤ |x| < 2
⇒ x ∈ (-2, -1) ∪ (1, 2)

Q10.	If x² < 4 then the value of x is
	A.(0, 2)
	B.(-2, 2)
	C.(-2, 0)
	D.None of these

Ans: (-2, 2)
Given, x² < 4
⇒ x² – 4 < 0
⇒ (x – 2) × (x + 2) < 0
⇒ -2 < x < 2
⇒ x ∈ (-2, 2)

Q11.	Solve: 2x + 1 > 3
	A.[1, 1)
	B.(1, ∞)
	C.(∞, ∞)
	D.(∞, 1)

Ans: (1, ∞)
Given, 2x + 1 > 3
⇒ 2x > 3 – 1
⇒ 2x > 2
⇒ x > 1
⇒ x ∈ (1, ∞)

Q12.	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

Ans: 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)

Q13.	Sum of two rational numbers is ______ number.
	A.rational
	B.irrational
	C.Integer
	D.Both 1, 2 and 3

Ans: 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.

Q14.	If \|x\| = -5 then the value of x lies in the interval
	A.(-5, ∞)
	B.(5, ∞)
	C.(∞, -5)
	D.No solution

Ans: No solution
Given, |x| = -5
Since |x| is always positive or zero
So, it can not be negative
Hence, given inequality has no solution.

Q15.	The value of x for which \|x + 1\| + √(x – 1) = 0
	A.0
	B.1
	C.-1
	D.No value of x

Ans: 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.

Q16.	If x² < -4 then the value of x is
	A.(-2, 2)
	B.(2, ∞)
	C.(-2, ∞)
	D.No solution

Ans: No solution
Given, x² < -4
⇒ x² + 4 < 0
Which is not possible.
So, there is no solution.

Q17.	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)

Ans: (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)

Q18.	The solution of the function f(x) = \|x\| > 0 is
	A.R
	B.R – {0}
	C.R – {1}
	D.R – {-1}

Ans: 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

Q19.	Solve: \|x – 1\| ≤ 5, \|x\| ≥ 2
	A.[2, 6]
	B.[-4, -2]
	C.[-4, -2] ∪ [2, 6]
	D.None of these

Ans: [-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]

Q20.	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

Ans: 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