Q. If $f(x) = \int_{x^{2}}^{x^{2} + 1}{e^{- t^{2}}dt}$, then $f(x)$ increases in
A) (2, 2)
B) No value of $x$
C) $(0,\ \infty)$
D) $( - \infty,\ 0)$
Explanation
(d)
Given, $f(x) = \int_{x^{2}}^{x^{2} + 1}{e^{- t^{2}}dt}$
$\Rightarrow f^{'}(x) = e^{- \left( x^{2} + 1 \right)^{2}}2x - e^{- \left( x^{2} \right)^{2}}.2x$
$= 2xe^{- (x^{4} + 2x^{2} + 1)}\left\{ 1 - e^{2x^{2} + 1} \right\}$
Here, $e^{2x^{2 + 1}} > 1$
And $e^{- (x^{4} + 2x^{2}1)} > 0$for all $x$
For $f^{'}(x) > 0,x < 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.