If \(f(x)=\int_{0}^{x} e^{t^{2}}(t-2)(t-3) d t\) for all \(x \in(0, \infty)\), then

\(f(x)=\int_{0}^{x} e^{t^{2}}(t-2)(t-3) d t\)

\(\Rightarrow f^{\prime}(x)=e^{x^{2}} \cdot(x-2)(x-3)\)

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.