Let $f(x)=x^2 e^{-2 x}, x>0$. The maximum value of $f(x)$ is

• A

  0
• Correct answer
  B

  $\frac{1}{e^2}$
• C

  $\frac{1}{4 e^2}$
• D

  $\frac{1}{2 e}$

Solution

$$ \begin{aligned} & \text { } y=x^2 \cdot e^{-2 x} \quad x>0 \\ & \frac{d y}{d x}=x^2 \cdot(-2) e^{-2 x}+e^{-2 x} \cdot 2 x \end{aligned} $$ $$ \frac{d y}{d x}=0 \Rightarrow 2 x e^{-2 x}\{1-x\}=0 \Rightarrow x=0,1 $$ By first derivative method Minima at $x=0$ and Maxima at $x=1$ So, maximum value $=y_{x=1}=(1)^2 e^{-2}$ $$ \Rightarrow \quad \frac{1}{e^2} $$

Hence, the correct answer is option: 'B'.