If $1(x)=\cos (\log x)$, then value of $f(x) f(4)-\frac{1}{2}\left\{f\left(\frac{x}{4}\right)+f(4 x)\right\}$ is
(a) 1
(b) −1
(c) 0
(d) ±1
(c) 0
Given : f(x) = cos (log x)
Then, $f(x) f(4)-\frac{1}{2}\left\{f\left(\frac{x}{4}\right)+f(4 x)\right\}$
$=\cos (\log x) \cos (\log 4)-\frac{1}{2}\left\{\cos \left(\log \frac{x}{4}\right)+\cos (\log 4 x)\right\}$
$=\frac{1}{2}[\cos (\log x+\log 4)+\cos (\log x-\log 4)]-\frac{1}{2}\left\{\cos \left(\log \frac{x}{4}\right)+\cos (\log 4 x)\right\}$
$=\frac{1}{2}\left\{\cos (\log 4 x)+\cos \left(\log \frac{x}{4}\right)-\cos \left(\log \frac{x}{4}\right)-\cos (\log 4 x)\right\}$
$=\frac{1}{2} \times 0=0$
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