Q. If $e^{e^{x}} = a_{0} + a_{1}x + a_{2}x^{2} + \ldots$, then
A) $a_{0} = 1$
B) $a_{o} = e$
C) $a_{0} = e^{e}$
D) $a_{0} = e^{2}$
Explanation
(b)
$e^{e^{x}} = 1 + \frac{e^{x}}{1!} + \frac{e^{2^{x}}}{2!} + ...$
$= 1 + \frac{1}{1!}\left( 1 + x + \frac{x^{2}}{2!} + ... \right) + \frac{1}{2!}\left( 1 + \frac{2x}{1!} + \frac{(2x)^{2}}{2!} + ... \right) + ...$
$\therefore{e^{e}}^{x} = \left( 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... \right) + x\left( \frac{1}{1!} + \frac{2}{2!1!} + ... \right) + ...$
$\therefore\ a_{0} = e$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.