Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. Can't be determined

Answer 1

The correct option is A $0$

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

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

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

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

$=0$

$\therefore 2\cos A\cos B=\cos (A+B)+\cos (A-B)$