Question

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

• A. $cos(x-y)$
• B. $log\left(cos\right(x-y\left)\right)$
• C. $1$
• D. $cos(x+y)$

Answer 1

Given:
$f\left(x\right)=cos\left(lo{g}_{e}x\right)\Rightarrow f\left(\frac{1}{x}\right)=cos\left(lo{g}_e\left(\frac{1}{x}\right)\right)\Rightarrow f(\frac{1}{x}=cos(-lo{g}_e\left(x\right)\left)\right)\Rightarrow f\left(\frac{1}{x}\right)=cos\left(lo{g}_e\right(x\left)\right)$
Similarly,
$f\left(\frac{1}{y}\right)=cos\left(lo{g}_{e}y\right)$
Now,
$f\left(xy\right)=cos\left(lo{g}_{e}xy\right)=cos(lo{g}_{e}x+lo{g}_{e}y)$
and
$f\left(\frac{x}{y}\right)=cos\left(lo{g}_e\frac{x}{y}\right)=cos(lo{g}_{e}x-lo{g}_{e}y)\Rightarrow f\left(\frac{x}{y}\right)+cos\left(xy\right)=cos(lo{g}_{e}x-lo{g}_{e}y)+cos(lo{g}_{e}x+lo{g}_{e}y)\Rightarrow f\left(\frac{x}{y}\right)+f\left(xy\right)=cos\left(lo{g}_{e}x\right)cos\left(lo{g}_{e}y\right)\Rightarrow \frac{1}{2}[f\left(\frac{x}{y}\right)+f(xy\left)\right]=cos\left(lo{g}_{e}x\right)cos\left(lo{g}_{e}y\right)\Rightarrow f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)-\frac{1}{2}\left\{f\right(xy)+f\left(\frac{x}{y}\right)\}=cos\left(lo{g}_{e}x\right)cos\left(lo{g}_{e}y\right)-cos\left(lo{g}_{e}x\right)cos\left(lo{g}_{e}y\right)=0$