A) 1
B) \[\frac{1}{2}\]
C) \[-2\]
D) 0
E) \[-1\]
Correct Answer: D
Solution :
\[\because \]\[f(x)=\cos ({{\log }_{e}}x)\] \[\therefore \]\[f(x)f(y)-\frac{1}{2}\left[ f\left( \frac{y}{x} \right)+f(xy) \right]\] \[=\cos ({{\log }_{e}}x)\cos ({{\log }_{e}}y)\] \[-\frac{1}{2}\left[ \cos {{\log }_{e}}\left( \frac{y}{x} \right)+\cos {{\log }_{e}}(xy) \right]\]\[=\cos ({{\log }_{e}}x)\cos ({{\log }_{e}}y)-\frac{2}{2}\cos ({{\log }_{e}}x)\] \[\times \cos ({{\log }_{e}}y)\] \[=0\]
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