A) \[\frac{1}{2}[f(x+y)+f(x-y)]\]
B) \[\frac{1}{2}[f(2x)+f(2y)]\]
C) \[\frac{1}{2}[f(x+y).f(x-y)]\]
D) None of these
Correct Answer: B
Solution :
[b] \[f(x+y).f(x-y)\] \[=\frac{{{2}^{x+y}}+{{2}^{-x-y}}}{2}.\frac{{{2}^{x-y}}+{{2}^{-x+y}}}{2}\] \[=\frac{{{2}^{2x}}+{{2}^{2y}}+{{2}^{-2x}}+{{2}^{-2y}}}{2\times 2}\] \[=\frac{1}{2}\left[ \frac{{{2}^{2x}}+{{2}^{-2x}}}{2}+\frac{{{2}^{2y}}+{{2}^{-2y}}}{2} \right]\] \[=\frac{1}{2}[f(2x)+f(2y)]\]
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