Question

If $f\left(x\right)=cos\left(logx\right)$, then $f\left(x\right)f\left(y\right)-\frac{1}{2}\{f\left(\frac{x}{y}\right)+f(xy\left)\right\}$ has the value

• A. $-1$
• B. $\frac{1}{2}$
• C. $-2$
• D. None of these

Answer 1

The correct option is D

None of these

Given :
$f\left(x\right)=cos\left(logx\right)$
$\therefore f\left(y\right)=cos\left(logy\right)$
Now,
$f\left(\frac{x}{y}\right)=cos\left(log\left(\frac{x}{y}\right)\right)=cos(logx-logy)$
and
$f\left(xy\right)=cos\left(logxy\right)=cos(logx+logy)\Rightarrow f\left(\frac{x}{y}\right)+f\left(xy\right)=cos(logx-logy)+cos(logx+logy)\Rightarrow f\left(\frac{x}{y}\right)+f\left(xy\right)=2cos\left(logx\right)cos\left(logy\right)\Rightarrow \frac{1}{2}[f\left(\frac{x}{y}\right)+f(xy\left)\right]=cos\left(logx\right)cos\left(logy\right)\Rightarrow f\left(x\right)f\left(y\right)-\frac{1}{2}\left[f\right(xy)+f\left(\frac{x}{y}\right)]$
$=cos\left(logx\right)cos\left(logy\right)-cos\left(logx\right)cos\left(logy\right)=0$