Question

If f(x) = cos (log x), then value of \[f\left( x \right) f\left( 4 \right) - \frac{1}{2} \left\{ f\left( \frac{x}{4} \right) + f\left( 4x \right) \right\}\] is 

Options

• (a) 1

• (b) −1

• (c) 0

• (d) ±1

   

Answer 1

(c) 0

Given : f(x) = cos (log x)
Then, \[f\left( x \right) f\left( 4 \right) - \frac{1}{2} \left\{ f\left( \frac{x}{4} \right) + f\left( 4x \right) \right\}\]

\[= \cos (\log x)\cos(\log 4) - \frac{1}{2}\left\{ \cos \left( \log\frac{x}{4} \right) + \cos\left( \log4x \right) \right\}\]
\[ = \frac{1}{2}\left[ \cos\left( \log x + \log 4 \right) + \cos \left( \log x - \log4 \right) \right] - \frac{1}{2}\left\{ \cos \left( \log\frac{x}{4} \right) + \cos\left( \log4x \right) \right\}\]
\[ = \frac{1}{2}\left\{ \cos (\log 4x) + \cos \left( \log \frac{x}{4} \right) - \cos \left( \log \frac{x}{4} \right) - \cos \left( \log 4x \right) \right\}\]
\[ = \frac{1}{2} \times 0 = 0\]