Question

If $f\left(x\right)=\cos \left(\log x\right)$ then the value of $f\left(x\right)\cdot f\left(y\right)-\frac{1}{2}[f\left(\frac{x}{y}\right)+f(xy\left)\right]$ equals

• A. $x^2$
• B. $x^2+2x+1$
• C. $0$
• D. None of these

Answer 1

Answer: (C)

$0$

Given $f\left(x\right)=\cos \left(\log x\right)$
$\therefore f\left(y\right)=\cos \left(\log y\right)$
Now
$f\left(x\right)f\left(y\right)-\frac{1}{2}[f\left(\frac{x}{y}\right)+f\left(xy\right)]$
$=\cos \left(\log x\right)\cos \left(\log y\right)-\frac{1}{2}[\cos \log \left(\frac{x}{y}\right)+\cos \log \left(xy\right)]$
$=\cos \left(\log x\right)\cos \left(\log y\right)-\frac{1}{2}[\cos (\log x-\log y)+\cos (\log x+\log y)]$
$=\cos A\cos B-\frac{1}{2}[\cos (A-B)+\cos (A+B)]$, where A$=\log x,B=\log y$
$=\cos A\cos B-\frac{1}{2}\left(2\cos A\cos B\right)=0$