Question

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

Answer 1

(c) 0

Given : f(x) = cos (log x)
Then, $f\left(x\right)f\left(4\right)-\frac{1}{2}\left\{f\left(\frac{x}{4}\right)+f\left(4x\right)\right\}$
$$\begin{gathered} =\cos (\log x)\cos (\log 4)-\frac{1}{2}\left\{\cos \left(\log \frac{x}{4}\right)+\cos \left(\log 4x\right)\right\} \\ =\frac{1}{2}\left[\cos \left(\log x+\log 4\right)+\cos \left(\log x-\log 4\right)\right]-\frac{1}{2}\left\{\cos \left(\log \frac{x}{4}\right)+\cos \left(\log 4x\right)\right\} \\ =\frac{1}{2}\left\{\cos (\log 4x)+\cos \left(\log \frac{x}{4}\right)-\cos \left(\log \frac{x}{4}\right)-\cos \left(\log 4x\right)\right\} \\ =\frac{1}{2}\times 0=0 \end{gathered}$$