Question
${x^4}{e^{ - {x^2}}}$ का अधिकतम मान है
अधिकतम मान के लिए, $f'(x) = 0$ ==> $4{x^3}{e^{ - {x^2}}} - 2{x^5}{e^{ - {x^2}}} = 0$
==> ${x^2} = 2 \Rightarrow x = \pm \sqrt 2 $
$f'' (x) = 12 x^2 e^{-x{^2}} + 4x^3 e^{-x{^2}} (-2x) -10 x^4 e^{-x{^2}} -2x^5 e^{-x{^2}} (-2x)$
==> $f''(\sqrt 2 ) = 24{e^{ - 2}} - 32{e^{ - 2}} - 40{e^{ - 2}} + 32{e^{ - 2}} =$ ऋणात्मक
अत: $f(x)$, $x = \sqrt 2 $ पर अधिकतम होगा
$\therefore$ अधिकतम मान $= 4{e^{ - 2}}$.
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