Q116 / 180 · KCET 2012 · Mathematics
The maximum value of $x \mathrm{e}^{-\mathrm{x}}$ is
- A$\mathrm{e}$
- Correct answerB$\frac{1}{\mathrm{e}}$
- C$-\mathrm{e}$
- D$-\frac{1}{e}$
Solution
Let $\quad y=x e^{-x}$
On differentiating w.r.t. ' $x$ ', we get
$$
\begin{aligned}
&\frac{d y}{d x}=x e^{-x}(-1)+e^{-x} \\
&\frac{d y}{d x}=e^{-x}(1-x)
\end{aligned}
$$
For maximum or minimum,
$$
\frac{\mathrm{dy}}{\mathrm{dx}}=0
$$
$$
\begin{array}{lrr}
\Rightarrow & \mathrm{e}^{-\mathrm{x}}(1-\mathrm{x})=0 & \\
\Rightarrow & 1-\mathrm{x}=0 & \left(\because \mathrm{e}^{-\mathrm{x}} \neq 0\right) \\
\Rightarrow & \mathrm{x}=1 &
\end{array}
$$
From Eq. (ii), we get
$$
\begin{gathered}
\begin{aligned}
\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}} &=\mathrm{e}^{-\mathrm{x}}(-1)+(1-\mathrm{x}) \mathrm{e}^{-\mathrm{x}}(-1) \\
&=-\mathrm{e}^{-\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}+\mathrm{x} \mathrm{} \mathrm{e}^{-\mathrm{x}} \\
&=-2 \mathrm{e}^{-\mathrm{x}}+\mathrm{xe}^{-\mathrm{x}} \\
\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}} &=\mathrm{e}^{-\mathrm{x}}(\mathrm{x}-2) \\
&=\mathrm{e}^{-1}(1-2)=\frac{1}{\mathrm{e}}(-1) \\
\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}}\right)_{\mathrm{at} \mathrm{} \mathrm{x}=1}
\end{aligned} \\
\therefore \text { At } \mathrm{x}=1, \mathrm{y} \text { is maximum and } \\
\text { maximum value }=1 \mathrm{e}^{-1}=\frac{1}{\mathrm{e}} .
\end{gathered}
$$
Hence, the correct answer is option: 'B'.