Question

If f(x) = cos (log x), then the value of f(x²) f(y²) −$\frac{1}{2}\left\{f\left(\frac{x^2}{y^2}\right)+f\left(x^2{y}^2\right)\right\}$is
(a) −2
(b) −1
(c) 1/2
(d) None of these

Answer 1

(d) None of these

Given:
$f\left(x\right)=\cos \left(\log x\right)$
$$\begin{gathered} \Rightarrow f\left(x^2\right)=\cos \left(\log \left(x^2\right)\right) \\ \Rightarrow f\left(x^2\right)=\cos \left(2\log \left(x\right)\right) \end{gathered}$$

Similarly,
$f\left(y^2\right)=\cos \left(2\log \left(y\right)\right)$

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

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