Question

lf $f\left(x\right)={\int }_0^x{e}^{t^2}(t-2)(t-3)dt$ for all $x\in (0,\infty )$, then

• A. $f$ has a local maximum at $x=2$
• B. $f$ is decreasing on $(2,3)$
• C. there exists some $c\in (O,\infty )$ such that $f^{\prime \prime }\left(c\right)=0$
• D. $f$ has a local minimum at $x=3$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $f$ has a local maximum at $x=2$
B $f$ is decreasing on $(2,3)$
C there exists some $c\in (O,\infty )$ such that $f^{\prime \prime }\left(c\right)=0$
D $f$ has a local minimum at $x=3$

$f\left(x\right)={\int }_0^x{e}^{t^2}(t-2)(t-3)dt\forall xϵ(0,\infty )$
$f^{\prime }\left(x\right)$ has maxima at $x=2(\because f^{\prime }(x)$ changes sign from +ve to -ve)
$f^{\prime }\left(x\right)$ has minima at $x=3(\because f^{\prime }(x)$ changes sign from -ve to +ve).
Also f(x) is decreasing in $(2,3)$ $[\because f^{\prime }(x)<0]$
$f^{\prime }\left(x\right)=0$ for $x=2andx=3$.
So, by Rolle's theorem, there exists
$cϵ(2,3)$ for which $f"\left(c\right)=0$.
Option A, B, C and D are correct.