Question

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

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

Answer 1

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

Given that,
$f\left(x\right)={\int }_{x^2}^{x^2+1}{e}^{-t^2}\mathrm{dt}\therefore f\text{'}\left(x\right)=e^{-(x^2+1{)}^2}\cdot 2x-e^{-x^4}\cdot 2x\Rightarrow f\text{'}\left(x\right)=2x\left(e^{-(x^2+1{)}^2}-e^{-x^4}\right]\because {\left(x^2+1\right)}^2>x^4\mathrm{or},e^{(x^2+1{)}^2}>e^{x^4}\mathrm{or},e^{-(x^2+1{)}^2}<e^{-x^4}\mathrm{or},e^{-(x^2+1{)}^2}-e^{-x^4}<0\therefore f\text{'}\left(x\right)>0\mathrm{for}\forall x<0\mathrm{Therefore},f\left(x\right)\mathrm{is}\mathrm{increasing}\mathrm{for}x<0.$