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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    If \[y={{\sin }^{-1}}(3x-4{{x}^{3}})+{{\cos }^{-1}}(4{{x}^{3}}-3x)\] \[+{{\tan }^{-1}}(E),\]then\[\frac{dy}{dx}\]is equal to

    A)  \[5\]                                    

    B)  \[0\]

    C)  \[\frac{2}{\sqrt{1+{{x}^{2}}}}\]

    D)         \[\frac{2}{\sqrt{1-{{x}^{2}}}}\]

    E)  \[\frac{2}{\sqrt{1-{{x}^{2}}}}\]

    Correct Answer: B

    Solution :

    Given, \[y={{\sin }^{-1}}{{(3x-4x)}^{2}}+{{\cos }^{-1}}(4{{x}^{3}}-3x)\]                                                 \[+{{\tan }^{-1}}(E)\] \[\Rightarrow \]\[y=3{{\sin }^{-1}}x+3{{\cos }^{-1}}x+{{\tan }^{-1}}(E)\] \[\Rightarrow \]               \[y=\frac{3\pi }{2}+{{\tan }^{-1}}(E)\] On differentiating w.r.t.\[x,\]we get                 \[\frac{dy}{dx}=0\]


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