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Question

Exhaustive values of x satisfying the equation |x4x212|=|x49||x2+3| is -

A
x [1,)
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B
x (,2][2,)
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C
x [2,2]
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D
x (,1][1,)
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Solution

The correct option is B x (,2][2,)
|x24||x2+3|=|x23||x2+3||x2+3|
|x24|=|x23|1
|x24|+1=|x23|
a=x24,b=1
|a|+|b|=|a+b|
ab=|a||b|
[By squaring both sides]
ab0
(x24)0
x24
So x ϵ (,2][2,)

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