Question

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

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

Answer 1

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

Given $f\left(x\right)={\int }_{x^2}^{x^2+1}{e}^{{-t}^2}dt$
On differentiating both sides using Newton's Leibnitz formula ,we get
$f^{\prime }\left(x\right)=e^{-{(x^2+1)}^2}\left\{\frac{d}{dx}(x^2+1)\right\}-e^{{-\left(x^2\right)}^2}\left\{\frac{d}{dx}{\left(x\right)}^2\right\}=e^{-{(x^2+1)}^2}.2x-e^{-{\left(x^2\right)}^2}.2x={2xe}^{-(x^4+{2x}^2+1)}(1-e^{{2x}^2}+1)$
(where $e^{{2x}^2+1}>1,\forall x$ and $e^{-(x^4+{2x}^2+1)}>0\forall x$)
$\therefore f^{\prime }\left(x\right)>0$ which shows $2x<0$ or $x<0$
$\Rightarrow x\in (-\infty ,0)$