Question

If f(x) = cos (log x), then the value of f(x) f(y) −$\frac{1}{2}\left\{f\left(\frac{x}{y}\right)+f\left(xy\right)\right\}$ is
(a) −1
(b) 1/2
(c) −2
(d) None of these

Answer 1

(d) None of these

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

Now,
$f\left(\frac{x}{y}\right)=\cos \left(\log \left(\frac{x}{y}\right)\right)=\cos \left(\log x-\log y\right)$
and
$f\left(xy\right)=\cos \left(\log xy\right)=\cos \left(\log x+\log y\right)$

$$\begin{gathered} \Rightarrow f\left(\frac{x}{y}\right)+f\left(xy\right)=\cos \left(\log x-\log y\right)+\cos \left(\log x+\log y\right) \\ \Rightarrow f\left(\frac{x}{y}\right)+f\left(xy\right)=2\cos \left(\log x\right)\cos \left(\log y\right) \\ \Rightarrow \frac{1}{2}\left[f\left(\frac{x}{y}\right)+f\left(xy\right)\right]=\cos \left(\log x\right)\cos \left(\log y\right) \end{gathered}$$
$\Rightarrow f\left(x\right)f\left(y\right)-\frac{1}{2}\left\{f\left(xy\right)+f\left(\frac{x}{y}\right)\right\}=\cos \left(\log x\right)\cos \left(\log y\right)-\cos \left(\log x\right)\cos \left(\log y\right)=0$