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6. Application of derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths6. Application of derivatives1 Mark
MCQ
The maximum value of ${x^4}{e^{ - {x^2}}}$ is
  • A
    ${e^2}$
  • B
    ${e^{ - 2}}$
  • C
    $12{e^{ - 2}}$
  • ✓
    $4{e^{ - 2}}$
✓

Answer

Correct option: D.
$4{e^{ - 2}}$
d
(d) $f(x) = {x^4}{e^{ - {x^2}}}$ ==> $f'(x) = 4{x^3}{e^{ - {x^2}}} + {x^4}{e^{ - {x^2}}}( - 2x)$

For max., $f'(x) = 0$ ==> $4{x^3}{e^{ - {x^2}}} - 2{x^5}{e^{ - {x^2}}} = 0$

==> ${x^2} = 2 \Rightarrow x = \pm \sqrt 2 $

$f '' (x) = 12 x^2 e^{-x{^2}} + 4x^3  e^{-x{^2}} (-2x) -10 x^4  e^{-x{^2}} -2x^5  e^{-x{^2}} (-2x)$

==> $f''(\sqrt 2 ) = 24{e^{ - 2}} - 32{e^{ - 2}} - 40{e^{ - 2}} + 32{e^{ - 2}}$ = -ve

Hence, $f(x)$ is max. at $x = \sqrt 2 $

$\therefore$ Maximum value = $4{e^{ - 2}}$.

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