A) \[{{e}^{2}}\]
B) \[{{e}^{-2}}\]
C) \[12{{e}^{-2}}\]
D) \[4{{e}^{-2}}\]
Correct Answer: D
Solution :
\[f(x)={{x}^{4}}{{e}^{-{{x}^{2}}}}\] Þ \[{f}'(x)=4{{x}^{3}}{{e}^{-{{x}^{2}}}}+{{x}^{4}}{{e}^{-{{x}^{2}}}}(-2x)\] For max., \[{f}'(x)=0\] Þ \[4{{x}^{3}}{{e}^{-{{x}^{2}}}}-2{{x}^{5}}{{e}^{-{{x}^{2}}}}=0\] Þ \[{{x}^{2}}=2\Rightarrow x=\pm \sqrt{2}\] \[f''(x)=12{{x}^{2}}{{e}^{-{{x}^{2}}}}+4{{x}^{3}}{{e}^{-{{x}^{2}}}}(-2x)-10{{x}^{4}}{{e}^{-{{x}^{2}}}}-2{{x}^{5}}{{e}^{-{{x}^{2}}}}(-2x)\] Þ \[{f}''(\sqrt{2})=24{{e}^{-2}}-32{{e}^{-2}}-40{{e}^{-2}}+32{{e}^{-2}}\] = ?ve Hence, \[f(x)\] is max. at \[x=\sqrt{2}\] \ Maximum value = \[4{{e}^{-2}}\].
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