SOLUTION —
$\begin{aligned}f^{\prime}(x) & =\left(x^2+3 x\right) e^{(-1 / 2) x} \cdot\left(-\frac{1}{2}\right)+(2 x+3) e^{-(\alpha / 2) x} \\& =-\frac{1}{2} e^{-(1 / 2) x}\left\{x^2-x-6\right\}\end{aligned}$
Since, $f(x)$ satisfies the Rolle's theorem
$\begin{array}{lrrl}\therefore & f^{\prime}(x) & =0 \\\Rightarrow & -\frac{1}{2} e^{-c / 2}\left(c^2-c-6\right) & =0 \\\Rightarrow & & c=3,-2\end{array}$
So, The correct option will be (C).