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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Morning)

  • question_answer
    If\[f(x)={{\log }_{e}}\left( \frac{1-x}{1+x} \right),|x|<1,\]then\[f\left( \frac{2x}{1+{{x}^{2}}} \right)\]is equal to: [JEE Main 8-4-2019 Morning]

    A) \[2f(x)\]

    B) \[2f({{x}^{2}})\]

    C) \[{{(f(x))}^{2}}\]                 

    D) \[-2f(x)\]

    Correct Answer: A

    Solution :

    \[f(x)=lo{{g}_{e}}\left( \frac{1-x}{1+x} \right),|x|<1\]           \[f\left( \frac{2x}{1+{{x}^{2}}} \right)=\ell n\left( \frac{1-\frac{2x}{1+2{{x}^{2}}}}{1+\frac{2x}{1+{{x}^{2}}}} \right)\]           \[=\ell n\left( \frac{{{(x-1)}^{2}}}{{{(x+1)}^{2}}} \right)=2\ell n\left| \frac{1-x}{1+x} \right|=2f(x)\]                      


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