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If $f(x)=\frac{2^{x}+2^{-x}}{2}$, then $f(x+y) \cdot f(x-y)$ is
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Verified Answer
The correct answer is:
$\frac{1}{2}[f(2 x)+f(2 y)]$
Given $f(x)=\frac{2^{x}+2^{-x}}{2}$
Now, $f(x+y)=\frac{2^{x+y}+2^{-x-y}}{2}$
and $f(x-y)=\frac{2^{x-y}+2^{-x+y}}{2}$
$\therefore f(x+y) \cdot f(x-y)$
$=\frac{\left(2^{x+y}+2^{-x-y}\right)}{2} \cdot \frac{\left(2^{x-y}+2^{-x+y}\right)}{2}$
$=\frac{2^{2 x}+2^{-2 y}+2^{2 y}+2^{-2 x}}{4}$
$=\frac{1}{2}\left\{\left(\frac{2^{2 x}+2^{-2 x}}{2}\right)+\left(\frac{2^{2 y}+2^{-2 y}}{2}\right)\right\}$
$=\frac{1}{2}\{f(2 x)+f(2 y)\}$
Now, $f(x+y)=\frac{2^{x+y}+2^{-x-y}}{2}$
and $f(x-y)=\frac{2^{x-y}+2^{-x+y}}{2}$
$\therefore f(x+y) \cdot f(x-y)$
$=\frac{\left(2^{x+y}+2^{-x-y}\right)}{2} \cdot \frac{\left(2^{x-y}+2^{-x+y}\right)}{2}$
$=\frac{2^{2 x}+2^{-2 y}+2^{2 y}+2^{-2 x}}{4}$
$=\frac{1}{2}\left\{\left(\frac{2^{2 x}+2^{-2 x}}{2}\right)+\left(\frac{2^{2 y}+2^{-2 y}}{2}\right)\right\}$
$=\frac{1}{2}\{f(2 x)+f(2 y)\}$
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