Q. $\frac{\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \ldots}{1 + \frac{1}{3!} + \frac{1}{5!} + \ldots}$ equals
A) $e + 1$
B) $\frac{e - 1}{e + 1}$
C) $e - 1$
D) None of these
Explanation
(b)
Given, $\frac{\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + ...}{1 + \frac{1}{3!} + \frac{1}{5!} + ...} = \frac{\frac{e + e^{- 1}}{2} - 1}{\frac{e - e^{- 1}}{2}}$
$= \frac{e^{2} + 1 - 2e}{e^{2} - 1} = \frac{e - 1}{e + 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.