Question

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

• A. $(-2,2)$
• B. no value of $x$
• C. $(1/2,\infty )$
• D. $(-\infty ,0)$

Answer 1

The correct option is C $(1/2,\infty )$

$f\left(x\right)={\int }_x^{x^2+1}{e}^{-2}dt,$
$f^{\prime }\left(x\right)=2x-e^{-2}-e^{-2}=\frac{1}{e^{-2}}=(2x-1)>0$
$\Rightarrow 2x-1>0c\Rightarrow 2x>1\Rightarrow x>\frac{1}{2}$
$\therefore f\left(x\right)$ is increasing in $(\frac{1}{2},\infty )$