If $f (x) = \cos (\log_e x)$, then $f(x)f(y)-\frac{1}{2}\left\{\begin{matrix}f(\frac{x}{y})+f(xy)\end{matrix}\right\}$ is equal to
Answer & explanation
Correct answer: option 1
We have, $f (x) = \cos (\log_e x)$
$∴f(x)f(y)-\frac{1}{2}\left\{\begin{matrix}f(\frac{x}{y})+f(xy)\end{matrix}\right\}$
$= \cos (\log x) \cos (\log y) -\frac{1}{2}\left\{\begin{matrix}\cos\log(\frac{x}{y})+\cos(\log(xy))\end{matrix}\right\}$
$= \cos (\log x) \cos (\log y) -\frac{1}{2}\left\{\begin{matrix}\cos (\log x - \log y)+\cos (\log x + \log y)\end{matrix}\right\}$
$= \cos (\log x) \cos (\log y) -\frac{1}{2}\left\{\begin{matrix}2 \cos (\log x) \cos (\log y)\end{matrix}\right\}$
$=0$