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  4. If f(x)=cos(loge x), then f (x) f (y)-(1/2)[f ((y/x))+f (xy)] has the value

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Q. If $f(x)=cos(log_{e} x)$, then $f \left(x\right) f \left(y\right)-\frac{1}{2}\left[f \left(\frac{y}{x}\right)+f \left(xy\right)\right]$ has the value

Relations and Functions

Solution:

$f \left(x\right)=cos\left(log_{e} x\right), f \left(\frac{y}{x}\right)=cos \left(log_{e} \frac{y}{x}\right)$,
$f \left(xy\right)=cos\left(log_{e} xy\right)$
$\therefore f\left(x\right)f\left(y\right)-\frac{1}{2} \left[f \left(\frac{y}{x}\right)+f \left(xy\right)\right]$
$=cos \left(log_{e} x\right)cos \left(log_{e} y\right)$
$-\frac{1}{2}\left[cos\left(log_{e} \frac{y}{x}\right)+cos\left(log_{e} xy\right)\right]$
$=cos\left(log_{e}\, x \right)cos\left(log_{e} y\right)$
$-\frac{1}{2}2cos \frac{log_{e} \frac{y}{x}+log_{e} xy}{2}cos\frac{log_{e} \frac{y}{x}-log_{e} xy}{2}$
$=cos \left(log_{e} x\right)cos\left(log_{e}y\right)-cos\left(log_{e} y\right)cos\left(log_{e} x\right)=0$