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Question

If y=sin1(1+x)+(1x)2, then dydx=

A
1(1x2)
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B
1(1x2)
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C
12(1x2)
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D
None of these
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Solution

The correct option is C 12(1x2)
y=sin11+x+1x2Putx=cosθy=sin11+cosθ+1cosθ2=sin12cos2(θ2)+2sin2(θ2)2=sin12(cosθ2+sinθ2)2=sin1(12cosθ2+12sinθ2)=sin1(sinπ4cosθ2+cosπ4sinθ2)=sin1sin(π4+θ2)=π4+θ2y=π4+12cos1xdydx=1211x2=121x2

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