Q. The function $f(x)$ which satisfies $f(x) = f( - x) = \frac{f^{'}(x)}{x}$, is given by
A) $f(x) = \frac{1}{2}e^{x^{2}}$
B) $f(x) = \frac{1}{2}e^{- x^{2}}$
C) $f(x) = x^{2}e^{x^{2}/2}$
D) $f(x) = e^{x^{2}/2}$
Explanation
(d)
Let $f(x) = e^{x^{2}/2}$
$\therefore\ \ \ \ f( - x) = e^{( - x)^{2}} = {e^{x}}^{2/2}$
And $\frac{f^{'}(x)}{x} = \frac{1}{x}\left( e^{x^{2}/2}.\frac{2x}{2} \right) = e^{x^{2}/2}$
$\Rightarrow \ \ \ \ f(x) = f( - x)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{f^{'}(x)}{x}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.