text: # MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers
  
  Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Linear Inequalities Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 6 Linear Inequalities Objective Questions.
  
  ## Linear Inequalities Class 11 MCQs Questions with Answers
  
  Students are advised to solve the Linear Inequalities Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Linear Inequalities Class 11 with answers will boost your confidence thereby helping you score well in the exam.
  
  Explore numerous MCQ Questions of Linear Inequalities Class 11 with answers provided with detailed solutions by looking below.
  
  Question 1.
  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)
  
  Given, -2 < 2x – 1 < 2
  ⇒ -2 + 1 < 2x < 2 + 1
  ⇒ -1 < 2x < 3
  ⇒ -1/2 < x < 3/2
  ⇒ x ∈ (-1/2, 3/2)
  
  Question 2.
  If x² < -4 then the value of x is
  (a) (-2, 2)
  (b) (2, ∞)
  (c) (-2, ∞)
  (d) No solution
  
  Given, x² < -4
  ⇒ x² + 4 < 0
  Which is not possible.
  So, there is no solution.
  
  Question 3.
  If |x| < -5 then the value of x lies in the interval
  (a) (-∞, -5)
  (b) (∞, 5)
  (c) (-5, ∞)
  (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
  
  Question 4.
  The graph of the inequations x ≤ 0 , y ≤ 0, and 2x + y + 6 ≥ 0 is
  (a) exterior of a triangle
  (b) a triangular region in the 3rd quadrant
  (d) none of these
  
  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.
  
  Question 5.
  The graph of the inequalities x ≥ 0, y ≥ 0, 2x + y + 6 ≤ 0 is
  (a) a square
  (b) a triangle
  (c) { }
  (d) none of these
  
  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.
  
  Question 6.
  Solve: 2x + 1 > 3
  (a) [-1, ∞]
  (b) (1, ∞)
  (c) (∞, ∞)
  (d) (∞, 1)
  
  Given, 2x + 1 > 3
  ⇒ 2x > 3 – 1
  ⇒ 2x > 2
  ⇒ x > 1
  ⇒ x ∈ (1, ∞)
  
  This program is a multi purpose inequality solver.
  
  Question 7.
  The solution of the inequality 3(x – 2)/5 ≥ 5(2 – x)/3 is
  (a) x ∈ (2, ∞)
  (b) x ∈ [-2, ∞)
  (c) x ∈ [∞, 2)
  (d) x ∈ [2, ∞)
  
  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, ∞)
  
  Question 8.
  Solve: 1 ≤ |x – 1| ≤ 3
  (a) [-2, 0]
  (b) [2, 4]
  (c) [-2, 0] ∪ [2, 4]
  (d) None of these
  
  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]
  
  Question 9.
  Solve: -1/(|x| – 2) ≥ 1 where x ∈ R, x ≠ ±2
  (a) (-2, -1)
  (b) (-2, 2)
  (c) (-2, -1) ∪ (1, 2)
  (d) None of these
  
  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)
  
  Question 10.
  If x² < 4 then the value of x is
  (a) (0, 2)
  (b) (-2, 2)
  (c) (-2, 0)
  (d) None of these
  
  Given, x² < 4
  ⇒ x² – 4 < 0
  ⇒ (x – 2) × (x + 2) < 0
  ⇒ -2 < x < 2
  ⇒ x ∈ (-2, 2)
  
  Question 11.
  Solve: 2x + 1 > 3
  (a) [1, 1)
  (b) (1, ∞)
  (c) (∞, ∞)
  (d) (∞, 1)
  
  Given, 2x + 1 > 3
  ⇒ 2x > 3 – 1
  ⇒ 2x > 2
  ⇒ x > 1
  ⇒ x ∈ (1, ∞)
  
  Question 12.
  If a is an irrational number which is divisible by b then the number b
  (a) must be rational
  (b) must be irrational
  (c) may be rational or irrational
  (d) None of these
  
  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)
  
  Question 13.
  Sum of two rational numbers is ______ number.
  (a) rational
  (b) irrational
  (c) Integer
  (d) Both 1, 2 and 3
  
  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 14.
  If |x| = -5 then the value of x lies in the interval
  (a) (-5, ∞)
  (b) (5, ∞)
  (c) (∞, -5)
  (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.
  
  Question 15.
  The value of x for which |x + 1| + √(x – 1) = 0
  (a) 0
  (b) 1
  (c) -1
  (d) No value of x
  
  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.
  
  Question 16.
  If x² < -4 then the value of x is
  (a) (-2, 2)
  (b) (2, ∞)
  (c) (-2, ∞)
  (d) No solution
  
  Given, x² < -4
  ⇒ x² + 4 < 0
  Which is not possible.
  So, there is no solution.
  
  Question 17.
  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: (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)
  
  Question 18.
  The solution of the function f(x) = |x| > 0 is
  (a) R
  (b) R – {0}
  (c) R – {1}
  (d) R – {-1}
  
  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
  
  Question 19.
  Solve: |x – 1| ≤ 5, |x| ≥ 2
  (a) [2, 6]
  (b) [-4, -2]
  (c) [-4, -2] ∪ [2, 6]
  (d) None of these
  
  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]
  
  Question 20.
  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: (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
  
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