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Question

Solve the irrational inequality 4x2+x2x0

A
[3,0)(0,2]
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B
[2,2]
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C
[2,4][5,13]
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D
ϕ
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Solution

The correct option is A [3,0)(0,2]
Given 4x2+x2x0 will be true
if 4x20(x2)(x+2)0,x0
x[2,2]{0}(1)

Case 1: when x0
4x2x2xx2x<0x2x>0 x>0(1)

Case 2: when x<0
4x2x2x4x21,x0x230,x0(x3)(x+3)0,x0
x[3,0)(3)

Hence from (1),(2) and (3)
x[3,2]{0}

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