Question

The value of $x$ for which the function $f\left(x\right)={\int }_0^x(1-t^2)e^{-t^2/2}dt$ has an extremum is

• A. $0$
• B. $1$
• C. $-1$
• D. none of these

Answer 1

Answer: (B), (C)

The correct options are
B $1$
C $-1$

$f\left(x\right)={\int }_0^x(1-t^2)e^{-t^2/2}$
$f^{\prime }\left(x\right)=(1-x^2)e^{-x^2/2}$
For extreme value of $f\left(x\right)$
$f^{\prime }\left(x\right)=0=(1-x^2)e^{-x^2/2}$
$\Rightarrow x=\pm 1$