CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    If\[f(x)=\cos (\log x),\]then \[f\left( \frac{1}{x} \right)f\left( \frac{1}{y} \right)-\frac{1}{2}\left[ f\left( \frac{x}{y} \right)+f(xy) \right]\]to:

    A)  \[\cos (x-y)\]   

    B)         \[\log [\cos (x+y)]\]

    C)  \[1\]                    

    D)         \[0\]

    E)  \[\cos (x+y)\]

    Correct Answer: D

    Solution :

    Now, \[f\left( \frac{1}{x} \right)f\left( \frac{1}{y} \right)-\frac{1}{2}\left[ f\left( \frac{x}{y} \right)+f(xy) \right]\] \[=\cos \left( \log \frac{1}{x} \right)\cos \left( \log \frac{1}{y} \right)\]                 \[-\frac{1}{2}\left[ \cos \left( \log \frac{x}{y} \right)+\cos (\log xy) \right]\] \[=\cos (\log x)\cos (\log y)\]                          \[-\frac{1}{2}[2\cos (\log x)\cos (\log y)]\] \[=\cos (\log x)\cos (\log y)-\cos (\log x)\]                                                 \[\cos (\log y)\] \[=0\]


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