If fx= cos logx then fx2.fy2−12fx2y2+fx2y2=

fx=coslogxfx2=cos2logx, fy2=cos2logyfx2y2= cos2logxy=cos2logx+2logyfx2y2=cos2logxy=cos2logx−2logyfx2.fy2−12fx2y2+fx2y2=cos2logxcos2logy−12cos2logx+2logy+cos2logx−2logy=cos2logx.cos2logy−122cos2logx.cos2logy=0

If fx= cos logx then fx2.fy2−12fx2y2+fx2y2=

fx=coslogxfx2=cos2logx, fy2=cos2logyfx2y2= cos2logxy=cos2logx+2logyfx2y2=cos2logxy=cos2logx−2logyfx2.fy2−12fx2y2+fx2y2=cos2logxcos2logy−12cos2logx+2logy+cos2logx−2logy=cos2logx.cos2logy−122cos2logx.cos2logy=0

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$I f f (x) = \cos (\log x) t h e n f (x^{2}) . f (y^{2}) - \frac{1}{2} [f (x^{2} y^{2}) + f (\frac{x^{2}}{y^{2}})] =$

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answer is D.

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Detailed Solution

$\begin{array}{l} f (x) = \cos (\log x) \\ f (x^{2}) = \cos (2 \log x), f (y^{2}) = \cos (2 \log y) \\ f (x^{2} y^{2}) = \cos 2 (\log x y) = \cos (2 \log x + 2 \log y) \\ f (\frac{x^{2}}{y^{2}}) = \cos 2 \log (\frac{x}{y}) = \cos (2 \log x - 2 \log y) \\ f (x^{2}) . f (y^{2}) - \frac{1}{2} [f (x^{2} y^{2}) + f (\frac{x^{2}}{y^{2}})] = \cos (2 \log x) \cos (2 \log y) - \frac{1}{2} [\cos (2 \log x + 2 \log y) + \cos (2 \log x - 2 \log y)] \\ = \cos (2 \log x) . \cos (2 \log y) - \frac{1}{2} (2 \cos (2 \log x) . \cos (2 \log y)) \\ = 0 \end{array}$