Question

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

• A. $-1$
• B. $2$
• C. $-2$
• D. $0$

Answer 1

Answer: (D)

$0$

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

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

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

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

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

$[\because \cos A+\cos B=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}]$

$\cos \left(\log x\right)\cos \left(\log y\right)-\cos \left(\log x\right)\cos \left(\log y\right)$

$=0$