If f(x)=
Question:

If $f(x)=\frac{2^{x}+2^{-x}}{2}$, then $f(x+y) f(x-y)$ is equal to

(a) $\frac{1}{2}[f(2 x)+f(2 y)]$

(b) $\frac{1}{2}[f(2 x)-f(2 y)]$

(c) $\frac{1}{4}[f(2 x)+f(2 y)]$

(d) $\frac{1}{4}[f(2 x)-f(2 y)]$

Solution:

(a) $\frac{1}{2}[f(2 x)+f(2 y)]$

Given:

$f(x)=\frac{2^{x}+2^{-x}}{2}$

Now,

$f(x+y) f(x-y)=\left(\frac{2^{x+y}+2^{-x-y}}{2}\right)\left(\frac{2^{x-y}+2^{-x+y}}{2}\right)$

$\Rightarrow f(x+y) f(x-y)=\frac{1}{4}\left(2^{2 x}+2^{-2 y}+2^{2 y}+2^{-2 x}\right)$

$\Rightarrow f(x+y) f(x-y)=\frac{1}{2}\left(\frac{2^{2 x}+2^{-2 x}}{2}+\frac{2^{2 y}+2^{-2 y}}{2}\right)$

$\Rightarrow f(x+y) f(x-y)=\frac{1}{2}[f(2 x)+f(2 y)]$

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