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JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    \[\frac{d}{dx}\left\{ {{\sin }^{2}}x\left( {{\cot }^{-1}}\sqrt{\frac{1-x}{1+x}} \right) \right\}\]equals

    A) \[-1\]                                    

    B) \[1/2\]

    C)  \[-1/2\]                              

    D)  \[1\]

    Correct Answer: B

    Solution :

    Let                 \[y={{\sin }^{2}}\left\{ {{\cot }^{-1}}\sqrt{\frac{1-x}{1+x}} \right\}\]                 \[=\frac{1}{\cos e{{c}^{2}}\left\{ {{\cot }^{-1}}\sqrt{\frac{1-x}{1+x}} \right\}}\]                 \[=\frac{1}{1+{{\cot }^{2}}\left\{ {{\cot }^{-1}}\sqrt{\frac{1-x}{1+x}} \right\}}=\frac{1}{1+\frac{1-x}{1+x}}\]                 \[=\frac{1}{\frac{2}{1\_x}}=\frac{1+x}{2}\] \[\therefore \]  \[\frac{dy}{dx}=\frac{1}{2}\]


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