Question

Statement I If $f(x)=x(x+3) \cdot e^{\frac{-x}{2}}$, then Rolle's theorem applies for $f(x)$ in $[-3,0]$.
Statement II Lagrange mean value theorem is applied in $f(x)=x(x+3) e^{-x / 2}$ in $[-3,0]$.

• A. Statement I is true; statement II is true; statement II is a correct explanation for statement I
• B. Statement I is true; statement II is true; statement II is not a correct explanation for statement I
• C. Statement I is true; statement II is false
• D. Statement I is false, statement II is true

Answer 1

Answer: (B)

We have, $x(x+3)$ and $e^{-x / 2}$ are continuous and differentiable every where, so $x(x+3) e^{-x / 2}$ is continuous and differentiable and $f(-3)=f(0)=0$
and $f(x)=\left(x^2+3 x\right) \cdot e^{-x / 2}$
$f^{\prime}(x) =(2 x+3) e^{-x / 2}+\left(x^2+3 x\right)\left(-\frac{1}{2}\right) e^{-x / 2} $
$ =\frac{e^{-x / 2}}{2}\left[2(2 x+3)-\left(x^2+3 x\right)\right] $
$ =\frac{e^{-x / 2}}{2}\left[4 x+6-x^2-3 x\right]$
$ =\frac{e^{-x / 2}}{2}\left[-x^2+x+6\right] $
$ =-\frac{1}{2} e^{-x / 2}\left(x^2-x-6\right)$
$ =-\frac{1}{2} e^{-x / 2}(x-3)(x+2) $
$\therefore f^{\prime} (c)=0 \Rightarrow -\frac{1}{2} e^{-c / 2}(c-3)(c+2) $
$\Rightarrow =3,-2 \Rightarrow 3 \notin[-3,0] $
$\therefore c=-2 \in[-3,0]$
$\therefore$ Rolle's theorem is verified.
LMVT is also applied.
$\because f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$
i.e., Rolle's theorem is a special case of LMVT
$\because f(a)=f(b) \Rightarrow f^{\prime}(c)=0$