Question

If $f\left(x\right)=\cos \left(\log x\right)$, then $f\left(x\right)f\left(y\right)-\frac{1}{2}[f\left(\frac{x}{y}\right)+f(xy\left)\right]$ is equal to :

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

Answer 1

The correct option is D $0$

Substituting the values we get $\cos \left(\log \left(x\right)\right).\cos \left(\log \left(y\right)\right)-\frac{1}{2}[\cos \left(\log \left(\frac{x}{y}\right)\right)+\cos \left(\log \left(xy\right)\right)]$

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

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