First slide

Functions (XII)

Question

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

Easy

A

-1
B

$\frac{1}{2}$
C

-2
D

0

Solution

$\begin{array}{l} f (x) = \cos (\log x) \\ or f (x) f (y) - \frac{1}{2} [f (\frac{x}{y}) + f (x y)] \\ = \cos (\log x) \cos (\log y) - \frac{1}{2} [\cos (\log x - \log y)] + \cos (\log x + \log y)] \\ = \cos (\log x) \cos (\log y) - \frac{1}{2} [2 \cos (\log x) \cos (\log y)] \\ = 0 \end{array}$

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