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Question

The equation sin4x2cos2x+a2=0 is solvable if

A
3a3
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B
2a2
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C
1a1
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D
None of these
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Solution

The correct option is D 2a2
sin4x2sin2x+a22=0
If sin2x=y, then the equation reduce to y2+2y+a22=0
Now, Δ=44(a22)03a20a23y=sin2x=1±3a2 ...(1)

Now, y=sin2x is positive
0y101±3a211±3a22
13a242a21a220 and a2+10 ...(2)

Using (1) and (2), we have
a220(a+2)(a2)02a2

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