The points of extremum of $\phi(x)=\int\limits_1^x e^{-t^2 / 2}\left(1-t^2\right) d t$ are
Answer & explanation
Correct answer: option 1
We have,
$\phi(x)=\int\limits_1^x e^{-t^2 / 2}\left(1-t^2\right) d t \Rightarrow \phi'(x)=e^{-x^2 / 2}\left(1-x^2\right)$
∴ $\phi'(x)=0 \Rightarrow x= \pm 1$
Hence, the points of extremum are $x= \pm 1$