Question

If $f\left(x\right)=\cos \left(\log x\right)$, then $f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)-\frac{1}{2}\{f\left(\frac{x}{y}\right)+f(xy\left)\right\}=$

• A. $\log x$
• B. $\cos \left(\log x\right)$
• C. $\sin \left(\log x\right)$
• D. $0$

Answer 1

The correct option is D $0$

Given $f\left(x\right)=\cos \left(\log x\right)$

Consider, $f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)-\frac{1}{2}\{f\left(\frac{x}{y}\right)+f(xy\left)\right\}$

$=\left(\cos \left(\log \frac{1}{x}\right)\right)\left(\cos \left(\log \frac{1}{y}\right)\right)-\frac{1}{2}\{\left(\cos \left(\log \frac{x}{y}\right)\right)+\left(\cos \left(\log xy\right)\right)\}$

$=\left(\cos \right(\log x\left)\right)\left(\cos \right(\log y\left)\right)-\frac{1}{2}\left\{\cos \right(\log x-\log y)+\cos (\log x+\log y\left)\right\}$

$=\left(\cos \right(\log x\left)\right)\left(\cos \right(\log y\left)\right)-\frac{1}{2}\times 2\left(\cos \right(\log x\left)\right)\left(\cos \right(\log y\left)\right)$
$=0$