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JCECE Engineering JCECE Engineering Solved Paper-2004

  • question_answer
    If\[f(x)=\log \left( \frac{1+x}{1-x} \right)\], then\[f\left( \frac{2x}{1+{{x}^{2}}} \right)\]will be equal to:

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

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

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

    D) \[2f(x)\]

    Correct Answer: D

    Solution :

    Given that,\[f(x)=\log \left( \frac{1+x}{1-x} \right)\] \[\therefore \]\[f\left( \frac{2x}{1+{{x}^{2}}} \right)=\log \left( \frac{1+\frac{2x}{1+{{x}^{2}}}}{1-\frac{2x}{1+{{x}^{2}}}} \right)\]                 \[=\log {{\left( \frac{1+x}{1-x} \right)}^{2}}\]                 \[=2\log \left( \frac{1+x}{1-x} \right)=2f(x)\]


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