Question

The point(s) of extremum of $F\left(x\right)={\int }_1^x{e}^{-t^{2/2}}(1-t^2)dt$ is/are?

• A. $x=1$
• B. $x=-1$
• C. $x=2$
• D. $x=-2$

Answer 1

Answer: (A), (B)

The correct options are
A $x=1$
B $x=-1$

For points of extremum $F^{\prime }\left(x\right)=0$.

Now $F^{\prime }\left(x\right)=e^{-x^2/2}(1-x^2)-e^{-}1(1-1)$

$=>e^{-x^2/2}(1-x^2)=0$

$=>(1-x^2)=0$

$x=+1$ or $x=-1$