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Free download in PDF Class 11 Maths Chapter 6 Linear Inequalities Multiple Choice Questions and Answers for Board, JEE, NEET, AIIMS, JIPMER, IIT-JEE, AIEE and other competitive exams. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. These short solved questions or quizzes are provided by Gkseries./p>
(1)
Solve: |x – 3| < 5
[A]
(2, 8)
[B]
(-2, 8)
[C]
(8, 2)
[D]
(8, -2)
(2)
If x² = 4 then the value of x is
[A]
-2
[B]
2
[C]
-2, 2
[D]
None of these
(3)
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: (-∞, -2) ∪ [-1, 1] ∪ (2, ∞)
(4)
If x² = -4 then the value of x is
[A]
(-2, 2)
[B]
(-2, ∞)
[C]
(2, ∞)
[D]
No solution
(5)
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: 2 < x < 3 and x < 1
(6)
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
(7)
If x² = -4 then the value of x is
[A]
(-2, 2)
[B]
(-2, ∞)
[C]
(2, ∞)
[D]
No solution
(8)
f(x) = {(x – 1)×(2 – x)}/(x – 3) ≥ 0
[A]
(-∞, 1] ∪ (2, ∞)
[B]
(-∞, 1] ∪ (2, 3)
[C]
(-∞, 1] ∪ (3, ∞)
[D]
None of these
(9)
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)
(10)
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: interior of a triangle including the points on the sides
(11)
If |x| < 5 then the value of x lies in the interval
[A]
(-∞, -5)
[B]
(∞, 5)
[C]
(-5, ∞)
[D]
(-5, 5)
(12)
Sum of two rational numbers is ______ number.
[A]
rational
[B]
irrational
[C]
Integer
[D]
Both 1, 2 and 3
(13)
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
(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
(15)
If | x − 1| > 5, then
[A]
x∈(−∞, −4)∪(6, ∞]
[B]
x∈[6, ∞)
[C]
x∈(6, ∞)
[D]
x∈(−∞, −4)∪(6, ∞)
Answer: x∈(−∞, −4)∪(6, ∞)
(16)
1 ≤ |x – 1| ≤ 3
[A]
[-2, 0]
[B]
[2, 4]
[C]
[-2, 0] ∪ [2, 4]
[D]
None of these
(17)
Solve: (x + 1)² + (x² + 3x + 2)² = 0
[A]
x = -1, -2
[B]
x = -1
[C]
x = -2
[D]
None of these
(18)
If (x + 3)/(x – 2) > 1/2 then x lies in the interval
[A]
(-8, ∞)
[B]
(8, ∞)
[C]
(∞, -8)
[D]
(∞, 8)
(19)
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)
(20)
The solution of the inequality |x – 1| < 2 is
[A]
(1, ∞)
[B]
(-1, 3)
[C]
(1, -3)
[D]
(∞, 1)
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