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If $f(x)=\int_{0}^{x} e^{t^{2}}(t-2)(t-3) d t$ for all $x \in(0, \infty)$, then
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$f$ has a local maximum at $x=2$, $f$ is decreasing on $(2,3)$, there exists some $c \in(0, \infty)$, such that $f^{\prime \prime}(c)=0$, $f$ has a local minimum at $x=3$
$f(x)=\int_{0}^{x} e^{t^{2}}(t-2)(t-3) d t$
$\Rightarrow f^{\prime}(x)=e^{x^{2}} \cdot(x-2)(x-3)$
Put $f^{\prime}(x)=0 \Rightarrow x=2,3$
$f^{\prime \prime}(x)=e^{x^{2}} \cdot 2 x\left(x^{2}-5 x+6\right)+e^{x^{2}}(2 x-5)$
$f^{\prime \prime}(2)=-$ ve and $f^{\prime \prime}(3)=+$ ve
$\therefore \quad x=2$ is a point of local maxima.
and $x=3$ is a point of local minima.
Also for $x \in(2,3), f^{\prime}(x) < 0$
$\Rightarrow \quad f$ is decreasing on $(2,3)$.
Also we observe $f^{\prime \prime}(0) < 0$ and $f^{\prime \prime}(1)>0$
$\therefore \quad$ There exists some $C \in(0,1)$ such that $f^{\prime \prime}(C)=0$
Hence all the options are correct.
$\Rightarrow f^{\prime}(x)=e^{x^{2}} \cdot(x-2)(x-3)$
Put $f^{\prime}(x)=0 \Rightarrow x=2,3$
$f^{\prime \prime}(x)=e^{x^{2}} \cdot 2 x\left(x^{2}-5 x+6\right)+e^{x^{2}}(2 x-5)$
$f^{\prime \prime}(2)=-$ ve and $f^{\prime \prime}(3)=+$ ve
$\therefore \quad x=2$ is a point of local maxima.
and $x=3$ is a point of local minima.
Also for $x \in(2,3), f^{\prime}(x) < 0$
$\Rightarrow \quad f$ is decreasing on $(2,3)$.
Also we observe $f^{\prime \prime}(0) < 0$ and $f^{\prime \prime}(1)>0$
$\therefore \quad$ There exists some $C \in(0,1)$ such that $f^{\prime \prime}(C)=0$
Hence all the options are correct.
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