A) \[2g(x).g(y)\]
B) \[2g(x+y).g(x-y)\]
C) \[2g(x)\]
D) none of these
Correct Answer: A
Solution :
| \[f(x)=\underset{even}{\mathop{g(x)}}\,+\underset{odd}{\mathop{h(x)}}\,\] |
| \[\Rightarrow g(x)\,\,=\,\,\frac{{{a}^{x}}+{{a}^{-x}}}{2}\] |
| \[g\,\,(x+y)+g\,\,(x-y)\] |
| \[=\frac{{{a}^{x+y}}+{{a}^{-y+x}}}{2}+\frac{{{a}^{x-y}}+{{a}^{y-x}}}{2}\] |
| \[=\frac{1}{2}[{{a}^{x}}.{{a}^{y}}+{{a}^{x}}.{{a}^{-y}}+{{a}^{y}}.{{a}^{-x}}+{{a}^{-x}}{{a}^{-y}}]\] |
| \[=\frac{1}{2}[{{a}^{x}}({{a}^{y}}+{{a}^{-y}})+{{a}^{-x}}({{a}^{y}}+{{a}^{-y}})]\] |
| \[=\frac{1}{2}({{a}^{y}}+{{a}^{-y}})({{a}^{x}}+{{a}^{-x}})\] |
| \[=2g(x).\,\,g(y)\] |
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