Q. If $a < 0$, the function $(e^{ax} + e^{- ax})$ is a decreasing function for all values of $x$, where
A) $x < 0$
B) $x > 0$
C) $x < 1$
D) $x > 1$
Explanation
(a)
Let $f(x) = e^{ax} + e^{- ax},\ \ a < 0$
Now, $f^{'}(x) = ae^{ax} - ae^{- ax}$
For decreasing, $f^{'}(x) < 0\ \ \ \Rightarrow \ \ \ \ a\left( e^{ax} - e^{- ax} \right) < 0$
$\Rightarrow \ \ \ \ \ \ a\left( e^{2ax} - 1 \right)$<0
$\Rightarrow \ \ \ \ \ e^{2ax} > 1\ \ \ \Rightarrow \ \ \ \ x < 0$ [$\because\ \ \ a < 0$]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.