Q107 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics
If Rolle's theorem is applicable for the function $f(x)=x(x+3) e^{-\frac{x}{2}}$ on $[-3,0]$. then the value of $c$ is
- A3
- B3 and -2
- Correct answerC-2
- D-1
Solution
Given, $f(x)=x(x+3) e^{-\frac{x}{2}},[-3,0]$
$\because f^{\prime}(x)=\frac{d}{d x}\left(\left(x^2+3 x\right) e^{\frac{-x}{2}}\right)=\frac{1}{2}\left(-x^2+x+6\right) e^{\frac{-x}{2}}$
Since, $f(x)$ is satisfy Rolle's theorem so, $c \in[-3,0]$ such that $f^{\prime}(c)=0$
$\begin{aligned}
& \Rightarrow \frac{1}{2}\left(-c^2+c+6\right) e^{\frac{-c}{2}}=0 \Rightarrow c^2-c-6=0 \\
& \Rightarrow c=3,-2
\end{aligned}$
Since, $c=3 \notin[-3,0]$. So, $c=-2$.
Hence, the correct answer is option: 'C'.