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If $f(x)=\cos (\log x)$, then the value of $f(x) \cdot f(4)-\frac{1}{2}\left[f\left(\frac{x}{4}\right)+f(4 x)\right]$
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Verified Answer
The correct answer is:
$0$
$f(x)=\cos (\log x)$
Now let
$y=f(x) . f(4)-\frac{1}{2}\left[f\left(\frac{x}{4}\right)+f(4 x)\right]$
$\begin{aligned}
\Rightarrow \quad y= & \cos (\log x) \cdot \cos (\log 4)-\frac{1}{2}\left[\cos \log \left(\frac{x}{4}\right)+\cos (\log 4 x)\right] \\
\Rightarrow y= & \cos (\log x) \cos (\log 4) -\frac{1}{2}[\cos (\log x-\log 4)+\cos (\log x+\log 4)] \\
\Rightarrow y= & \cos (\log x) \cos (\log 4)-\frac{1}{2}[2 \cos (\log x) \cos (\log 4)] \\
\Rightarrow y= & 0 .
\end{aligned}$
Now let
$y=f(x) . f(4)-\frac{1}{2}\left[f\left(\frac{x}{4}\right)+f(4 x)\right]$
$\begin{aligned}
\Rightarrow \quad y= & \cos (\log x) \cdot \cos (\log 4)-\frac{1}{2}\left[\cos \log \left(\frac{x}{4}\right)+\cos (\log 4 x)\right] \\
\Rightarrow y= & \cos (\log x) \cos (\log 4) -\frac{1}{2}[\cos (\log x-\log 4)+\cos (\log x+\log 4)] \\
\Rightarrow y= & \cos (\log x) \cos (\log 4)-\frac{1}{2}[2 \cos (\log x) \cos (\log 4)] \\
\Rightarrow y= & 0 .
\end{aligned}$
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