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JEE Main & Advanced JEE Main Solved Paper-2015

  • question_answer
    Let \[{{\tan }^{-1}}y={{\tan }^{-1}}x+{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right),\]where \[|x|<\frac{1}{\sqrt{3}}.\]Then a value of y is : [JEE Main Solved Paper-2015 ]

    A) \[\frac{3x-{{x}^{3}}}{1+3{{x}^{2}}}\]                       

    B) \[\frac{3x+{{x}^{3}}}{1+3{{x}^{2}}}\]

    C) \[\frac{3x-{{x}^{3}}}{1-3{{x}^{2}}}\]        

    D) \[\frac{3x+{{x}^{3}}}{1-3{{x}^{2}}}\]

    Correct Answer: C

    Solution :

    \[{{\tan }^{-1}}y={{\tan }^{-1}}x+2{{\tan }^{-1}}x\] \[=3{{\tan }^{-1}}x\] \[{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)\] \[y=\frac{3x-{{x}^{3}}}{1-3{{x}^{2}}}\]


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