A) \[{{a}_{0}}=1\]
B) \[{{a}_{0}}=e\]
C) \[{{a}_{0}}={{e}^{e}}\]
D) \[{{a}_{0}}={{e}^{2}}\]
E) \[\frac{\pi }{2}\]
Correct Answer: B
Solution :
\[({{e}^{{{e}^{x}}}})={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+....\] Now, \[{{e}^{{{e}^{x}}}}=1+\frac{{{e}^{x}}}{1!}+\frac{{{e}^{2x}}}{2!}+....\] \[=1+\frac{1}{1!}\left( 1+x+\frac{{{x}^{2}}}{2!}+..... \right)\] \[+\frac{1}{2!}\left( 1+\frac{2x}{1!}+\frac{{{(2x)}^{2}}}{2!}..... \right)\] \[\therefore \]\[{{e}^{{{e}^{x}}}}=\left( 1+1+\frac{1}{2!}+\frac{1}{3!}+.... \right)\] \[+x\left( \frac{1}{1!}+\frac{1}{2!1!}+.... \right)....\] \[\therefore \] \[{{a}_{0}}=e\]
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