Question

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

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

Answer 1

Answer: (D)

Given $f(x)=\cos (\log \,x)$
$\therefore \, f(x) \cdot f(y)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right]$
$=\cos (\log \,x) \cdot \cos (\log \,y)$
$-\frac{1}{2}\left[\cos \,\log \left(\frac{x}{y}\right)+\cos\, \log (x y)\right]$
$=\cos (\log x) \cos (\log y)-\frac{1}{2} \cdot 2[\cos (\log x)\times \cos (\log y)]$
$=\cos (\log x) \cos (\log y)$
$=0$
$\therefore \, f(x) \cdot f(y)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right]=0$