Question

If f(x) = cos (log_e x), then $f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)-\frac{1}{2}\left\{f\left(xy\right)+f\left(\frac{x}{y}\right)\right\}$is equal to
(a) cos (x − y)
(b) log (cos (x − y))
(c) 1
(d) cos (x + y)

Answer 1

Given:
$f\left(x\right)=\cos \left({\log }_{e}x\right)$
$$\begin{gathered} \Rightarrow f\left(\frac{1}{x}\right)=\cos \left({\log }_e\left(\frac{1}{x}\right)\right) \\ \Rightarrow f\left(\frac{1}{x}\right)=\cos \left(-{\log }_e\left(x\right)\right) \\ \Rightarrow f\left(\frac{1}{x}\right)=\cos \left({\log }_e\left(x\right)\right) \end{gathered}$$

Similarly,
$f\left(\frac{1}{y}\right)=\cos \left({\log }_{e}y\right)$

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

$$\begin{gathered} \Rightarrow f\left(\frac{x}{y}\right)+f\left(xy\right)=\cos \left({\log }_{e}x-{\log }_{e}y\right)+\cos \left({\log }_{e}x+{\log }_{e}y\right) \\ \Rightarrow f\left(\frac{x}{y}\right)+f\left(xy\right)=2\cos \left({\log }_{e}x\right)\cos \left({\log }_{e}y\right) \\ \Rightarrow \frac{1}{2}\left[f\left(\frac{x}{y}\right)+f\left(xy\right)\right]=\cos \left({\log }_{e}x\right)\cos \left({\log }_{e}y\right) \end{gathered}$$
$\Rightarrow f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)-\frac{1}{2}\left\{f\left(xy\right)+f\left(\frac{x}{y}\right)\right\}=\cos \left({\log }_{e}x\right)\cos \left({\log }_{e}y\right)-\cos \left({\log }_{e}x\right)\cos \left({\log }_{e}y\right)=0$
Disclaimer: The question in the book has some error, so none of the options are matching with the solution. The solution is created according to the question given in the book.