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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    \[5{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)+7{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\]\[-4{{\tan }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)-{{\tan }^{-1}}x=5\pi ,\]then\[x\]is equal to

    A)  3                            

    B)         \[-\sqrt{3}\]

    C)  \[\sqrt{2}\]                       

    D)         \[2\]

    E)  \[\sqrt{3}\]

    Correct Answer: E

    Solution :

    Given, \[5{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)+7{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] \[-4{{\tan }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)-{{\tan }^{-1}}x=5\pi \] \[\therefore \] \[5(2{{\tan }^{-1}}x)+7(2{{\tan }^{-1}}x)\]                                 \[-4(2{{\tan }^{-1}}x)-{{\tan }^{-1}}x=5\pi \] \[\Rightarrow \] \[15{{\tan }^{-1}}x=5\pi \] \[\Rightarrow \] \[{{\tan }^{-1}}x=\frac{\pi }{3}\] \[\Rightarrow \] \[x=\sqrt{3}\]


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