Explanation

(c)
Given, $f(x) = \left( x^{2} + 3x \right)e^{- (1/2)x}$
$\therefore\ \ \ \ \ \ f^{'}(x) = \left( x^{2} + 3x \right)\ e^{- (1/2)x}.\left( - \frac{1}{2} \right) + (2x + 3)e^{- (1/2)x}$
$= - \frac{1}{2}e^{- (1/2)x}\{ x^{2} - x - 6\}$
Since, $f(x)$ satisfies the Rolle’s theorem
$\therefore\ \ \ \ f^{'}(c) = 0\ \ \ \Rightarrow \ \ - \frac{1}{2}e^{- \left( \frac{c}{2} \right)}\left( c^{2} - c - 6 \right) = 0$
$\Rightarrow \ \ \ \ c = 3,\ - 2$
But $c = 3 \notin \lbrack - 3,\ 0\rbrack$
$\therefore\ c = - 2$