Q103 / 160 · WBJEE 2011 · Mathematics

Let \(f(x)=x^3 e^{-3 x}, x > 0\). Then the maximum value of \(f(x)\) is

  • Correct answer
    A
    \(\mathrm{e}^{-3}\)
  • B
    \(3 \mathrm{e}^{-3}\)
  • C
    \(27 \mathrm{e}^{-9}\)
  • D
    \(\infty\)

Solution

\(\begin{aligned} & \text { Hints: } f(x)=x^3 \cdot e^{-3 x} \\ & =f^{\prime}(x)=3 x^2 e^{-3 x}+x^3 e^{-3 x}(-3) \\ & =x^2 3 e^{-3 x}[1-x]=0, x=1 \\ & \text { Maximum at } x=1 \\ & f(1)=e^{-3} \end{aligned}\)

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

Relevant Questions