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Question

If |x1|+|x+1|=2, then x belongs to

A
(0,)
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B
[1,1]
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C
{1,1}
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D
(,0]
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Solution

The correct option is B [1,1]
|x1|+|x+1|=2

Case 1: When x1
x1+x+1=2
x=1[1,)

Case 2: When 1<x<1
(x1)+x+1=2
x+1+x+1=2
2=2
x can have any value as it will satisfy the equation, but here the interval is (1,1)
i.e. 1<x<1 this whole internal would satisfy case 2.

Case 3: When x1
(x1)(x+1)=2
x+1x1=2
2x=2
x=1(,1]
From 123
x[1,1]

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