Question

If $f\left(x\right)=cos\left(logx\right)$, then value of $f\left(x\right)$
$f\left(4\right)-\frac{1}{2}\{f\left(\frac{x}{4}\right)+f(4x\left)\right\}$ is

• A. $1$
• B. $-1$
• C. $0$
• D. $\pm 1$

Answer 1

The correct option is C

$0$

$f\left(x\right)=cos\left(logx\right)$
Then, $f\left(x\right)f\left(4\right)-\frac{1}{2}\{f\left(\frac{x}{4}\right)+f(4x\left)\right\}$
$=cos\left(logx\right)cos\left(log4\right)-\frac{1}{2}\{cos\left(log\frac{x}{4}\right)+cos(log4x\left)\right\}$
$=\frac{1}{2}\left[cos\right(logx+log4)+cos(logx-log4\left)\right]-\frac{1}{2}\{cos\left(log\frac{x}{4}\right)+cos(log4x\left)\right\}$
$=\frac{1}{2}\left\{cos\right(log4x)+cos\left(log\frac{x}{4}\right)-cos\left(log\frac{x}{4}\right)-cos(log4x\left)\right\}$
$=\frac{1}{2}\times 0=0$