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## 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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