Question

The function $f(x)=x(x+3) e^{-(1 / 2) x}$ satisfies the conditions of Rolle's theorem in $[-3,0]$ The value of $c$ is

(A) 0

(B) -1

(C) -2

(D) -3

Answer 1

$\begin{aligned}f^{\prime}(x) & =\left(x^2+3 x\right) e^{(-1 / 2) x} \cdot\left(-\frac{1}{2}\right)+(2 x+3) e^{-(\alpha / 2) x} \\& =-\frac{1}{2} e^{-(1 / 2) x}\left\{x^2-x-6\right\}\end{aligned}$

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).