Question

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

• A. $-1$
• B. $\frac{1}{2}$
• C. $-2$
• D. $0$

Answer 1

The correct option is D

$0$

Explanation for the correct option.

Find the value of the given expression.

For the function $f\left(x\right)=\cos \left(\log x\right)$, the expression $f\left(x\right)·f\left(y\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f\left(xy\right)\right]$ can be written as:

$\begin{array}{rcl}f\left(x\right)·f\left(y\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f\left(xy\right)\right]& =& \cos \left(\log x\right)\cos \left(\log y\right)-\frac{1}{2}\left[\cos \left(\log \frac{x}{y}\right)+\cos \left(\log xy\right)\right]\\ & =& \cos \left(\log x\right)\cos \left(\log y\right)-\frac{1}{2}\left[\cos \left(\log x-\log y\right)+\cos \left(\log x+\log y\right)\right]\\ & =& \cos \left(\log x\right)\cos \left(\log y\right)-\frac{1}{2}\left[2\cos \left(\log x\right)\cos \left(\log y\right)\right]\left[\cos \left(A-B\right)+\cos \left(A+B\right)=2\cos A\cos B\right]\\ & =& \cos \left(\log x\right)\cos \left(\log y\right)-\cos \left(\log x\right)\cos \left(\log y\right)\\ & =& 0\end{array}$

Hence, the correct option is D.