A) \[2g(x)\,g(y)\]
B) \[2g(x+y)\,g(x-y)\]
C) \[2g(x)\]
D) none of these
Correct Answer: A
Solution :
| Clearly, g(x) \[=\frac{1}{2}({{a}^{x}}+{{a}^{-x}})\] and \[h(x)\,=\frac{1}{2}\left( {{a}^{x}}-{{a}^{-x}} \right)\] |
| Now \[h\left( x+y \right)+g\left( x-y \right)\]\[=\frac{1}{2}({{a}^{x+y}}+{{a}^{-(x+y)}}+\frac{1}{2}({{a}^{x-y}}+{{a}^{-x+y}})\] |
| =\[\frac{1}{2}({{a}^{x}}{{a}^{y}}+{{a}^{x}}{{a}^{-y}}+{{a}^{-x}}{{a}^{y}}+{{a}^{-x}}{{a}^{-y}})\]\[=\frac{1}{2}({{a}^{x}}({{a}^{y}}+{{a}^{-y}})+{{a}^{-x}}({{a}^{y}}+{{a}^{-y}}))\] |
| =\[2\left( \frac{1}{2}({{a}^{x}}+{{a}^{-x}}) \right)\left( \frac{1}{2}({{a}^{y}}+{{a}^{-y}}) \right)\]\[=2g(x)g(y)\] |
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