Question

If $f(x)=cos(logx)$, then $f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)-\frac{1}{2}\left(f\left(\frac{x}{y}\right)+f\left(xy\right)\right)$ is equal to

• A. $cos(x-y)$
• B. $log(cos(x+y))$
• C. $1$
• D. $0$

Answer 1

Answer: (D)

We have, $f\left(x\right)=cos\left(logx\right)$
$f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)-\frac{1}{2}\left(f\left(\frac{x}{y}\right)+f\left(xy\right)\right)$
$=cos\left(log\frac{1}{x}\right)cos\left(log\frac{1}{y}\right)-\frac{1}{2}\left(cos\left(log\frac{x}{y}\right)+cos\left(logxy\right)\right)$
$=cos\left(log\frac{1}{x}\right)cos\left(log\frac{1}{y}\right)$
$-\frac{1}{2}\times 2cos\left(\frac{log\frac{x}{y}+logxy}{2}\right)cos\left(\frac{log\frac{x}{y}-log\left(xy\right)}{2}\right)$
$=cos\left(log\frac{1}{x}\right)cos\left(log\frac{1}{y}\right)$
$-\left(cos\frac{\left(logx-logy+logx+logy\right)}{2}\right)\times$
$\left(cos\frac{logx-logy-logx-logy}{2}\right)$
$=cos\left(log\frac{1}{x}\right)cos\left(log\frac{1}{y}\right)-cos\left(logx\right)cos\left(logy\right)$
$=cos\left(log1-logx\right)cos\left(log1-logy\right)-cos\left(logx\right)cos\left(logy\right)$.
$=cos\left(logx\right)cos\left(logy\right)-cos\left(logx\right)cos\left(logy\right)$
$=0.\left(\because log1=0\right)$