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

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)

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

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

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.

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.

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

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, ∞)

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]

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)

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

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

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)

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.

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.

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.

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

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)

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

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]

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