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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    Let\[f\left( x+\frac{1}{x} \right)={{x}^{2}}+\frac{1}{{{x}^{2}}},x\ne 0,\]then\[f(x)\] equals to:

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

    B)         \[{{x}^{2}}-1\]

    C)  \[{{x}^{2}}-2\]                 

    D)         \[{{x}^{2}}+1\]

    E)  \[{{x}^{2}}+2\]

    Correct Answer: C

    Solution :

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


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