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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