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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
Solution
(c) 0
Given : f(x) = cos (log x)
Then, \[f\left( x \right) f\left( 4 \right) - \frac{1}{2} \left\{ f\left( \frac{x}{4} \right) + f\left( 4x \right) \right\}\]
\[= \cos (\log x)\cos(\log 4) - \frac{1}{2}\left\{ \cos \left( \log\frac{x}{4} \right) + \cos\left( \log4x \right) \right\}\]
\[ = \frac{1}{2}\left[ \cos\left( \log x + \log 4 \right) + \cos \left( \log x - \log4 \right) \right] - \frac{1}{2}\left\{ \cos \left( \log\frac{x}{4} \right) + \cos\left( \log4x \right) \right\}\]
\[ = \frac{1}{2}\left\{ \cos (\log 4x) + \cos \left( \log \frac{x}{4} \right) - \cos \left( \log \frac{x}{4} \right) - \cos \left( \log 4x \right) \right\}\]
\[ = \frac{1}{2} \times 0 = 0\]
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