**CBSE Class 11 Maths – MCQ and Online Tests – Unit 6 – Linear Inequalities**

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**CBSE Class 11 Maths – MCQ and Online Tests – Unit 6 – Linear Inequalities**

Question 1.

If x² = -4 then the value of x is

(a) (-2, 2)

(b) (-2, 8)

(c) (2, 8)

(d) No solution

## Answer

Answer: (d) No solution

Hint:

Given, x² = -4

Since LHS = 0 and RHS < 0

So, No solution is possible.

Question 2.

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 3.

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 4.

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 5.

If (x + 3)/(x – 2) > 1/2 then x lies in the interval

(a) (-8, 8)

(b) (8, 8)

(c) (8, -8)

(d) (8, 8)

## Answer

Answer: (a) (-8, 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, 8)

Question 6.

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 7.

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 8.

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 9.

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 10.

If x² = -4 then the value of x is

(a) (-2, 2)

(b) (-2, 8)

(c) (2, 8)

(d) No solution

## Answer

Answer: (d) No solution

Hint:

Given, x² = -4

Since LHS = 0 and RHS < 0

So, No solution is possible.

Question 11.

The solution of |2/(x – 4)| > 1 where x ? 4 is

(a) (2, 6)

(b) (2, 4) ? (4, 6)

(c) (2, 4) ? (4, 8)

(d) (-8, 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 12.

If (|x| – 1)/(|x| – 2) ?= 0, x ? R, x ?± 2 then the interval of x is

(a) (-8, -2) ? [-1, 1]

(b) [-1, 1] ? (2, 8)

(c) (-8, -2) ? (2, 8)

(d) (-8, -2) ? [-1, 1] ? (2, 8)

## Answer

Answer: (d) (-8, -2) ? [-1, 1] ? (2, 8)

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] ? (-8, -2) ? (2, 8)

Hence the solution set is:

x ? (-8, -2) ? [-1, 1] ? (2, 8)

Question 13.

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 14.

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 15.

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]

Question 16.

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 17.

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

(a) (-8, -5)

(b) (8, 5)

(c) (-5, 8)

(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 18.

Solve: f(x) = {(x – 1)×(2 – x)}/(x – 3) = 0

(a) (-8, 1] ? (2, 8)

(b) (-8, 1] ? (2, 3)

(c) (-8, 1] ? (3, 8)

(d) None of these

## Answer

Answer: (b) (-8, 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 ? (-8, 1] ? (2, 3)

Question 19.

The solution of the inequality |x – 1| < 2 is

(a) (1, 8)

(b) (-1, 3)

(c) (1, -3)

(d) (8, 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 20.

If | x – 1| > 5, then

(a) x?(-8, -4)?(6, 8]

(b) x?[6, 8)

(c) x?(6, 8)

(d) x?(-8, -4)?(6, 8)

## Answer

Answer: (d) x?(-8, -4)?(6, 8)

Hint:

Given |x-1| >5

Case 1:

(x – 1) > 5

? x > 6

? x ? (6,8)

Case 2:

-(x – 1) > 5

? -x + 1 > 5

? -x > 4

? x < -4

? x ? (-8, -4)

So the range of x is (-8, -4)?(6, 8)

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