If f(x)=cos (log_e x),then the value of f(x)f(y)-$\frac{1}{2}$$\left[f\left(\frac{x}{y}\right)+f\left(xy\right)\right]$is

Given,f(x)=cos(log_ex)

$\therefore$f(x)f(y)-$\frac{1}{2}${f($\frac{x}{y}$)+f(xy)}

=cos (logx)cos (logy)-$\frac{1}{2}${cos log $\left|\frac{x}{y}\right|$+cos log (xy)}

=cos (logx)cos( logy)-$\frac{1}{2}${cos (logx-logy)+cos (logx+log y)

=cos (logx)cos (logy)-$\frac{1}{2}${2cos (logx )cos (log y)}=0