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Question

The equation
2cos2x2sin2x=x2+x2;0<xπ2
has


A
No real solution 
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B
One real solution
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C
more than one solution 
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D
none of these
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Solution

The correct option is A No real solution 

The given equation is
2cos2(x2)sin2x=x2+1x2
where 0<xπ2
L.H.S.=2cos2(x2)sin2x=(1+cosx)sin2x1+cosx<2 and sin2x1 for 0<xπ2(1+cosx)sin2x<2
and R.H.S.=x2+1x22
Therefore, for 0<xπ2, the given equation is not possible for any real value of x.


Mathematics

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