Q. If $f(x) = be^{ax} + ae^{bx}$, then $f^{''}(0)$ is equal to
A) 0
B) $2ab$
C) $ab(a + b)$
D) $ab$
Explanation
(c)
Given, $f(x) = be^{ax} + ae^{bx}$
$\Rightarrow \ \ \ f^{'}(x) = \ abe^{ax} + abe^{bx}\$
$\Rightarrow \ \ \ f^{''}(x) = a^{2}be^{ax} + ab^{2}e^{bx}$
$\Rightarrow \ \ \ {\ f}^{''}(0) = a^{2}b + ab^{2} = ab(a + b)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.