# If f(x) = cos

Question:

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

(a) cos (x − y)

(b) log (cos (x − y))

(c) 1

(d) cos (x + y)

Solution:

Given:

$f(x)=\cos \left(\log _{e} x\right)$

$\Rightarrow f\left(\frac{1}{x}\right)=\cos \left(\log _{e}\left(\frac{1}{x}\right)\right)$

$\Rightarrow f\left(\frac{1}{x}\right)=\cos \left(-\log _{e}(x)\right)$

$\Rightarrow f\left(\frac{1}{x}\right)=\cos \left(\log _{e}(x)\right)$

Similarly,

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

Now,

$f(x y)=\cos \left(\log _{e} x y\right)=\cos \left(\log _{e} x+\log _{e} y\right)$

and

$f\left(\frac{x}{y}\right)=\cos \left(\log _{e} \frac{x}{y}\right)=\cos \left(\log _{e} x-\log _{e} y\right)$

$\Rightarrow f\left(\frac{x}{y}\right)+f(x y)=\cos \left(\log _{e} x-\log _{e} y\right)+\cos \left(\log _{e} x+\log _{e} y\right)$

$\Rightarrow f\left(\frac{x}{y}\right)+f(x y)=2 \cos \left(\log _{e} x\right) \cos \left(\log _{e} y\right)$

$\Rightarrow \frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right]=\cos \left(\log _{e} x\right) \cos \left(\log _{e} y\right)$

$\Rightarrow f\left(\frac{1}{x}\right) f\left(\frac{1}{y}\right)-\frac{1}{2}\left\{f(x y)+f\left(\frac{x}{y}\right)\right\}=\cos \left(\log _{e} x\right) \cos \left(\log _{e} y\right)-\cos \left(\log _{e} x\right) \cos \left(\log _{e} y\right)=0$

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.