# If f(x) = cos (log x), then the value of

Question:

If $f(x)=\cos (\log x)$, then the value of $f\left(x^{2}\right) f\left(y^{2}\right)-\frac{1}{2}\left\{f\left(\frac{x^{2}}{y^{2}}\right)+f\left(x^{2} y^{2}\right)\right\}$ is

(a) −2

(b) −1

(c) 1/2

(d) None of these

Solution:

(d) None of these

Given:

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

$\Rightarrow f\left(x^{2}\right)=\cos \left(\log \left(x^{2}\right)\right)$

$\Rightarrow f\left(x^{2}\right)=\cos (2 \log (x))$

Similarly,

$f\left(y^{2}\right)=\cos (2 \log (y))$

Now,

$f\left(\frac{x^{2}}{y^{2}}\right)=\cos \left(\log \left(\frac{x^{2}}{y^{2}}\right)\right)=\cos \left(\log x^{2}-\log y^{2}\right)$

and,

$f\left(x^{2} y^{2}\right)=\cos \left(\log x^{2} y^{2}\right)=\cos \left(\log x^{2}+\log y^{2}\right)$

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

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

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

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