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 $f(x)=\left(x^2+3x\right)\cdot e^{-x/2}$
$f^{\prime }(x)=(2x+3)e^{-x/2}+\left(x^2+3x\right)\left(-\frac{1}{2}\right)e^{-x/2}$
$=\frac{e^{-x/2}}{2}\left[2(2x+3)-\left(x^2+3x\right)\right]$
$=\frac{e^{-x/2}}{2}\left[4x+6-x^2-3x\right]$
$=\frac{e^{-x/2}}{2}\left[-x^2+x+6\right]$
$=-\frac{1}{2}{e}^{-x/2}\left(x^2-x-6\right)$
$=-\frac{1}{2}{e}^{-x/2}(x-3)(x+2)$
$\therefore f^{\prime }(c)=0\Rightarrow -\frac{1}{2}{e}^{-c/2}(c-3)(c+2)$
$\Rightarrow =3,-2\Rightarrow 3\notin [-3,0]$
$\therefore c=-2\in [-3,0]$
$\therefore$ 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
$\because f(a)=f(b)\Rightarrow f^{\prime }(c)=0$