Q. If $f(x) = \cos{(\log x)},$ then $f(x)f(y) - \frac{1}{2}\left\lbrack f\left( \frac{x}{y} \right) + f(xy) \right\rbrack$ has the value

A) -1
B) 1/2
C) -2
D) 0
Explanation

(d)
$f(x) = f(y) - \frac{1}{2}\left\lbrack f\left( \frac{x}{y} \right) \right\rbrack + f(x\ y)$
$\ \ \ \ \ \ = \cos{\ \left( \log x \right).\cos{\ \left( \log y \right) - \frac{1}{2}\left\lbrack \cos{\left( \log\left( \frac{x}{y} \right) \right) + \cos{\ (\log{xy})}} \right\rbrack}}$
$\ \ \ \ \ = \cos{\ \left( \log x \right)\cos{\ \left( \log y \right) - \frac{1}{2} \times 2\cos{\ \left( \log x \right)\cos{\ \left( \log y \right)}}}}$
$\ \ \ \ = \cos{\ \left( \log x \right)\cos{\ \left( \log y \right) - \cos{\ \left( \log x \right)\cos{\ (\log y)}}}}$
$\ \ \ \ = 0$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.