E}^{x}}};f'(x)=\frac{2X.{{e}^{x}}-{{e}^{x}}.{{x}^{2}}}{{{\LEFT( {{e}^{x}} \right)}^{2}}}\] \[f'(x)=\frac{2x-{{x}^{2}}}{{{e}^{x}}}\] As \[{{e}^{x}}\] is always positive and for monotonically increasing; \[2x-{{x}^{2}}>0\] \[\Rightarrow {{x}^{2}}-2x<0\Rightarrow x(x-2)<0\Rightarrow x\in (0,2)\]

"> E}^{x}}};f'(x)=\frac{2X.{{e}^{x}}-{{e}^{x}}.{{x}^{2}}}{{{\LEFT( {{e}^{x}} \right)}^{2}}}\] \[f'(x)=\frac{2x-{{x}^{2}}}{{{e}^{x}}}\] As \[{{e}^{x}}\] is always positive and for monotonically increasing; \[2x-{{x}^{2}}>0\] \[\Rightarrow {{x}^{2}}-2x<0\Rightarrow x(x-2)<0\Rightarrow x\in (0,2)\]

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The function \[f(x)=\frac{{{x}^{2}}}{{{e}^{x}}}\] monotonically increasing if

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) . 11 months ago

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[c] \[f(x)=\frac{{{x}^{2}}}{{{E}^{x}}};f'(x)=\frac{2X.{{e}^{x}}-{{e}^{x}}.{{x}^{2}}}{{{\LEFT( {{e}^{x}} \right)}^{2}}}\] \[f'(x)=\frac{2x-{{x}^{2}}}{{{e}^{x}}}\] As \[{{e}^{x}}\] is always positive and for monotonically increasing; \[2x-{{x}^{2}}>0\] \[\Rightarrow {{x}^{2}}-2x<0\Rightarrow x(x-2)<0\Rightarrow x\in (0,2)\]

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Posted on 31 Jul 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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