Question

If $f(x)=cos(lo{g}_{e}x)$, 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

• A. $1$
• B. $1/2$
• C. $-2$
• D. $0$

Answer 1

Answer: (D)

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