Question

If $f(x)=\cos ({\log }_{e}x),$ then $f(x)f(y)-\frac{1}{2}\left[f\left(\frac{y}{x}\right)+f(xy)\right]$ has the value:

• A. 1
• B. $\frac{1}{2}$
• C. $-2$
• D. 0
• E. $-1$

Answer 1

Answer: (D)

$\because$ $f(x)=\cos ({\log }_{e}x)$
$\therefore$ $f(x)f(y)-\frac{1}{2}\left[f\left(\frac{y}{x}\right)+f(xy)\right]$
$=\cos ({\log }_{e}x)\cos ({\log }_{e}y)$
$-\frac{1}{2}\left[\cos {\log }_e\left(\frac{y}{x}\right)+\cos {\log }_e(xy)\right]$
$=\cos ({\log }_{e}x)\cos ({\log }_{e}y)-\frac{2}{2}\cos ({\log }_{e}x)$ $\times \cos ({\log }_{e}y)$ $=0$