Class 11 Maths Chapter 6 Linear Inequalities MCQs

Class 11 Maths Chapter 6 Linear Inequalities MCQs are given here to help students in their preparation for their exams. MCQs of Class 11 Maths Chapter 6 provided here contain correct options along with detailed explanations. All these multiple-choice questions are given as per the latest NCERT curriculum that covers all the concepts of Chapter 6 of Class 11 Maths.

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Students can practice all the objective type questions given for the chapter 6 Linear Inequalities, Class 11 Maths and verify their answers with the help of solutions provided.

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MCQs for Class 11 Maths Chapter 6 Linear Inequalities with Answers

1. The length of a rectangle is three times the breadth. If the minimum perimeter of the rectangle is 160 cm, then

(a) breadth > 20 cm

(b) length < 20 cm

(c) breadth x ≥ 20 cm

(d) length ≤ 20 cm

Correct option: (c) breadth x ≥ 20 cm

Solution:

Let x be the breadth of a rectangle.

So, length = 3x

Given that the minimum perimeter of a rectangle is 160 cm.

Thus, 2 (3x + x) ≥ 160

⇒ 4x ≥ 80

⇒ x ≥ 20

2. If – 3x + 17 < – 13, then

(a) x ∈ (10, ∞)

(b) x ∈ [10, ∞)

(c) x ∈ (– ∞, 10]

(d) x ∈ [– 10, 10)

Correct option: (a) x ∈ (10, ∞)

Solution:

Given,

-3x + 17 < -13

Subtracting 17 from both sides,

-3x + 17 – 17 < -13 – 17

⇒ -3x < -30

⇒ x > 10 {since the division by negative number inverts the inequality sign}

⇒ x ∈ (10, ∞)

3. The inequality representing the following graph is

Class 11 maths chapter 6 Linear Inequalities MCQ 3

(a) | x | < 5

(b) | x | ≤ 5

(c) | x | > 5

(d) | x | ≥ 5

Correct option: (a) | x | < 5

Solution:

The given graph shows the shaded region corresponding to x > – 5 and x < 5.

Therefore, by combining the above two inequalities, we get |x| < 5.

4. Observe the figure given below.

Class 11 maths chapter 6 Linear Inequalities MCQ 4

The interval at which the value of x lies is

(a) x ∈ (– ∞, – 2)

(b) x ∈ (– ∞, – 2]

(c) x ∈ (– 2, ∞]

(d) x ∈ [– 2, ∞)

Correct option: (b) x ∈ (– ∞, – 2]

Solution:

In the given figure, the circle is filled with dark colour at -2 which means -2 is included and the highlighted is towards the left of -2.

So, x ∈ (– ∞, – 2]

5. Given that x, y and b are real numbers and x < y, b < 0, then

(a) x/b < y/b

(b) x/b ≤ y/b

(c) x/b > y/b

(d) x/b ≥ y/b

Correct option: (a) x/b < y/b

Solution:

Given that x, y and b are real numbers and x < y, b < 0.

Consider, x < y

Divide both sides of the inequality by “b”

x/b < y/b {since b < 0}

6. If |x −1| > 5, then

(a) x ∈ (– 4, 6)

(b) x ∈ [– 4, 6]

(c) x ∈ (– ∞, – 4) U (6, ∞)

(d) x ∈ [– ∞, – 4) U [6, ∞)

Correct option: (c) x ∈ (– ∞, – 4) ∪ (6, ∞)

Solution:

|x – 1| > 5

x – 1 < – 5 and x – 1 > 5

x < -4 and x > 6

Therefore, x ∈ (-∞, -4) U (6, ∞)

7. If |x – 7|/(x – 7) ≥ 0, then

(a) x ∈ [7, ∞)

(b) x ∈ (7, ∞)

(c) x ∈ (– ∞, 7)

(d) x ∈ (– ∞, 7]

Correct option: (b) x ∈ (7, ∞)

Solution:

Given,

|x – 7|/(x – 7) ≥ 0

This is possible when x − 7 ≥ 0, and x – 7 ≠ 0.

Here, x ≥ 7 but x ≠ 7

Therefore, x > 7, i.e. x ∈ (7, ∞).

8. If |x + 3| ≥ 10, then

(a) x ∈ (– 13, 7]

(b) x ∈ (– 13, 7]

(c) x ∈ (– ∞, – 13] ∪ [7, ∞)

(d) x ∈ [– ∞, – 13] ∪ [7, ∞)

Correct option: (d) x ∈ [– ∞, – 13] ∪ [7, ∞)

Solution:

Given,

|x + 3| ≥ 10

⇒ x + 3 ≤ – 10 or x + 3 ≥ 10

⇒ x ≤ – 13 or x ≥ 7

⇒ x ∈ (– ∞, – 13] ∪ [7, ∞)

9. If 4x + 3 < 6x +7, then x belongs to the interval

(a) (2, ∞)

(b) (-2, ∞)

(c) (-∞, 2)

(d) (-4, ∞)

Correct option: (b) (-2, ∞)

Solution:

Given,

4x + 3 < 6x + 7

Subtracting 3 from both sides,

4x + 3 – 3 < 6x + 7 – 3

⇒ 4x < 6x + 4

Subtracting 6x from both sides,

4x – 6x < 6x + 4 – 6x

⇒ – 2x < 4 or

⇒ x > – 2 i.e., all the real numbers greater than –2, are the solutions of the given inequality.

Hence, the solution set is (–2, ∞), i.e. x ∈ (-2, ∞)

10. Solving – 8 ≤ 5x – 3 < 7, we get

(a) –1/2 ≤ x ≤ 2

(b) 1 ≤ x < 2

(c) –1 ≤ x < 2

(d) –1 < x ≤ 2

Correct option: (c) –1 ≤ x < 2

Solution:

Given,

– 8 ≤ 5x – 3 and 5x – 3 < 7

Let us solve these two inequalities simultaneously.

– 8 ≤ 5x – 3 and 5x – 3 < 7 can be written as:

– 8 ≤ 5x –3 < 7

Adding 3, we get

– 8 + 3 ≤ 5x – 3 + 3 < 7 + 3

–5 ≤ 5x < 10

Dividing by 5, we get

–1 ≤ x < 2

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