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KVPY Sample Paper KVPY Stream-SX Model Paper-31

  • question_answer
    If \[f(x)=lo{{g}_{e}}\left( \frac{1-x}{1+x} \right),\] then \[f\left( \frac{2x}{1+{{x}^{2}}} \right)\] is equal to:

    A) \[f(x)\]                          

    B) \[2f(x)\]

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

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

    Correct Answer: B

    Solution :

    \[f(x)=\ell n\left( \frac{1-x}{1+x} \right)\]
    \[f\left( \frac{2x}{1+{{x}^{2}}} \right)=\ell n\left( \frac{1-\frac{2x}{1+{{x}^{2}}}}{1+\frac{2x}{1+{{x}^{2}}}} \right)=\ell n\left( \frac{1+{{x}^{2}}-2x}{1+{{x}^{2}}+2x} \right)\]
    \[=\ell n\left( {{\left( \frac{1-x}{1+x} \right)}^{2}} \right)=2\ell n\left( \frac{1-x}{1+x} \right)=2f(x).\]


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