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)
$f^{'}(x) = 2xe^{- \left( x^{2} + 1 \right)^{2}} - 2xe^{- x^{4}}$
$= 2x\left( e^{- \left( x^{2} + 1 \right)^{2}} - e^{- x^{4}} \right)\ \ \ \ \ \left\lbrack here\ e^{- x^{4}} > e^{- \left( x^{2} + 1 \right)^{2}} \right\rbrack$
So, $f^{'}(x) > 0,\ when\ x < 0\ \ \ \ \left\lbrack as\ x^{4} < \left( x^{2} + 1 \right)^{2} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.