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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2002

  • question_answer
    If\[f(x)=\log \frac{1+x}{1-x},\]then\[f(x)\]is:

    A)  even

    B)                         \[f({{x}_{1}})f({{x}_{2}})=f({{x}_{1}}+{{x}_{2}})\]

    C)                         \[\frac{f({{x}_{1}})}{f({{x}_{2}})}=f({{x}_{1}}-{{x}_{2}})\]

    D)                         odd

    E)                         neither even nor odd

    Correct Answer: D

    Solution :

    \[\because \]\[f(-x)=\log \left( \frac{1-x}{1+x} \right)\] \[=-\log \left( \frac{1+x}{1-x} \right)\] \[\Rightarrow \]\[f(-x)=-f(x)\] \[\therefore \]Given function is an odd function.


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