Question

The points of extremum of the function
$\varphi \left(x\right)={\int }_1^x{e}^{-\frac{t^2}{2}}(1-t^2)dt$ are

• A. $x=0$
• B. $x=1$
• C. $x=\frac{1}{2}$
• D. $x=-2$

Answer 1

The correct option is B $x=1$

$\varphi \left(x\right)={\int }_1^x{e}^{-\frac{t^2}{2}}(1-t^2)$
$\Rightarrow {\varphi }^{\prime }\left(x\right)=e^{-\frac{x^2}{2}}(1-x^2)=0$ for extrema.
$\Rightarrow x^2-1=0\Rightarrow x=\pm 1$