JEE Main & Advanced Sample Paper JEE Main - Mock Test - 7

  • question_answer
    If \[y={{\tan }^{-1}}\left( \frac{{{\log }_{e}}(e/{{x}^{2}})}{{{\log }_{e}}(e{{x}^{2}})} \right)+{{\tan }^{-1}}\left( \frac{3+2{{\log }_{e}}x}{1-6{{\log }_{e}}x} \right)\], then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\]is

    A) \[2\]                             

    B) \[1\]

    C) \[0\]                             

    D) \[-1\]

    Correct Answer: C

    Solution :

    \[y={{\tan }^{-1}}\left( \frac{{{\log }_{e}}(e/{{x}^{2}})}{{{\log }_{e}}(e{{x}^{2}})} \right)+{{\tan }^{-1}}\left( \frac{3+2{{\log }_{e}}x}{1-6{{\log }_{e}}x} \right)\]
    \[={{\tan }^{-1}}\left( \frac{1-2{{\log }_{e}}x}{1+2{{\log }_{e}}x} \right)+{{\tan }^{-1}}\left( \frac{3+2{{\log }_{e}}x}{1-3.2\,lo{{g}_{e}}x} \right)\]
    \[={{\tan }^{-1}}(a)-{{\tan }^{-1}}(2{{\log }_{e}}x)\]\[+{{\tan }^{-1}}(c)+ta{{n}^{-1}}(2{{\log }_{e}}x)\]
    \[={{\tan }^{-1}}(a)+{{\tan }^{-1}}(c)\]
    \[\therefore \,\,\frac{dy}{dx}=0\]


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