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Question

If |x3||x+1|=12, then x=

A
3
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B
5
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C
3
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D
5
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Solution

The correct options are
A 3
B 5
|x3||x+1|=12
|x22x3|=12

Case 1: x22x30 i.e., x(,1][3,)
Then x22x3=12
x22x15=0
(x5)(x+3)=0
x=3,5

Case 2: x22x3<0 i.e., x(1,3)
Then x22x3=12
x22x+9=0
(x1)2+8=0 (Not possible)


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