Question

If $f\left(x\right)={\int }_{x^2}^{x^2+1}{e}^{-t^2}dt$

then $f\left(x\right)$ increases in?

• A. $(0,\infty )$
• B. $(-\infty ,0)$
• C. $(2,2)$
• D. no value of x

Answer 1

The correct option is B $(-\infty ,0)$

$f\left(x\right)={\int }_{x^2}^{x^2+1}{e}^{-t^2}dt$
On differentiating both sides
$f^{\prime }\left(x\right)=e^{-{(x^2+1)}^2}2x-e^{-{\left(x^2\right)}^2}2x=2x{e}^{-(x^4+2x^2+1)}(1-e^{2x^2+1})$
As $e^{2x^2+1}>1\forall x$ and $e^{-(x^4+2x^2+1)}>0\forall x$
Then
$f^{\prime }\left(x\right)>0\Rightarrow x<0\Rightarrow xϵ(-\infty ,0)$