MCQ
The maximum value of ${x^4}{e^{ - {x^2}}}$ is
- A$e^2$
- B$e^{-2}$
- C$12e^{-2}$
- ✓$4e^{-2}$
✓
Answer
Correct option: D.
$4e^{-2}$
d
$f(x)=x^{4} e^{-x^{2}}$ or $f^{\prime}(x)=4 x^{3} e^{-x^{2}}+x^{4} e^{-x^{2}}(-2 x)$
$f(x)=x^{4} e^{-x^{2}}$ or $f^{\prime}(x)=4 x^{3} e^{-x^{2}}+x^{4} e^{-x^{2}}(-2 x)$
$2 x^{3} e^{-x^{2}}\left(2-x^{2}\right)$
Sign scheme of $f^{\prime}(x)$ is as follows:
Hence, $f(x)$ is maximum at $x=\pm \sqrt{2}.$
Thus, maximum value $=4 \mathrm{e}^{-2}.$
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