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Question

The solution set of inequality |x -1| + |x + 1| < 4 is


Your Answer
A
(,2)
Correct Answer
B
(2,2)
Your Answer
C
(2,)
Your Answer
D
(,)

Solution

The correct option is C (2,2)
Best way to solve such problem is by checking at different values of x.
At x = -3, |x -1| + |x + 1| = 6 is not less than 4.
At x = 3, |x -1| + |x + 1| = 6 is not less than 4.
Hence, option (a), (c), and (d) can be eliminated.
Option (b) is correct answer.

Alternatively:
The solution will consist of three intervals
x<1,1x<1 and x1If x<1,then |x1|=(x1) and |x+1|=(x+1)We get (x1)(x+1)<4x>2 We have x<1 and x>22<x<1xϵ(2,1)If 1x<1, then |x1|=(x1) and |x+1|=x+1.
We get, (x1)+(x+1)<42<4, which is valid 
Inequality holds true for all x satisfying 
1x<1xϵ[1,1)If x1,then |x1|=(x1) and |x+1|=x+1we get x1+x+1<4x<2.We have x1 andx<21x<2xϵ[1,2)
Combining the solutions of the above three cases, we get
xϵ(2,1)[1,1)[1,2)(2,2)

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