X)=x\,\,ln\,\,x\] \[f'(x)=\FRAC{x}{x}+\] ln \[x=1+ln\,\,x\] Put \[f(x)=0\Rightarrow 1+ln\,\,x=0\] \[\Rightarrow \] ln \[x=-1\Rightarrow x={{e}^{-1}}\] Now, \[f''(x)=\frac{1}{x}\] \[f''(x)\left| _{x=e}-1=\frac{1}{{{e}^{-1}}}=e>0 \right.\] Hence, \[f(x)\] ATTAINS MINIMUM value at\[x={{e}^{-1}}\].

"> X)=x\,\,ln\,\,x\] \[f'(x)=\FRAC{x}{x}+\] ln \[x=1+ln\,\,x\] Put \[f(x)=0\Rightarrow 1+ln\,\,x=0\] \[\Rightarrow \] ln \[x=-1\Rightarrow x={{e}^{-1}}\] Now, \[f''(x)=\frac{1}{x}\] \[f''(x)\left| _{x=e}-1=\frac{1}{{{e}^{-1}}}=e>0 \right.\] Hence, \[f(x)\] ATTAINS MINIMUM value at\[x={{e}^{-1}}\].

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If\[f(x)=x\ell nx\], then \[f(x)\] attains minimum value at which one of the following points?

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

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[c] Let \[f(X)=x\,\,ln\,\,x\] \[f'(x)=\FRAC{x}{x}+\] ln \[x=1+ln\,\,x\] Put \[f(x)=0\Rightarrow 1+ln\,\,x=0\] \[\Rightarrow \] ln \[x=-1\Rightarrow x={{e}^{-1}}\] Now, \[f''(x)=\frac{1}{x}\] \[f''(x)\left| _{x=e}-1=\frac{1}{{{e}^{-1}}}=e>0 \right.\] Hence, \[f(x)\] ATTAINS MINIMUM value at\[x={{e}^{-1}}\].

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Posted on 09 Aug 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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