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B
−1√(1−x2)
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C
−12√(1−x2)
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D
None of these
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Solution
The correct option is C−12√(1−x2) y=sin−1√1+x+√1−x2Putx=cosθ∴y=sin−1√1+cosθ+√1−cosθ2=sin−1√2cos2(θ2)+√2sin2(θ2)2=sin−1√2(cosθ2+sinθ2)2=sin−1(1√2cosθ2+1√2sinθ2)=sin−1(sinπ4cosθ2+cosπ4sinθ2)=sin−1sin(π4+θ2)=π4+θ2∴y=π4+12cos−1x∴dydx=12−1√1−x2=−12√1−x2