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JEE Main & Advanced JEE Main Paper (Held On 10 April 2015)

  • question_answer
    If \[f(x)=2ta{{n}^{-1}}x+{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right),x>1,\]then f(5) is equal to : JEE Main Online Paper (Held On 10 April 2015)

    A) \[\frac{\pi }{2}\]                                              

    B) \[\pi \]

    C) \[{{\tan }^{-1}}\left( \frac{65}{156} \right)\]                        

    D) \[4{{\tan }^{-1}}(5)\]

    Correct Answer: B

    Solution :

                    \[f(x)=2{{\tan }^{-1}}x+\pi -2{{\tan }^{-1}}x\] \[f\left( x \right)=\pi \] \[f\left( 5 \right)=\pi \]


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