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The Minimum Value of X Log E X is (A) E (B) 1/E (C) 1 (D) None of These

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प्रश्न

The minimum value of \[\frac{x}{\log_e x}\] is _____________ .

विकल्प

  • e

  • 1/e

  • 1

  • none of these

MCQ
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उत्तर

e

 

\[\text { Given }: f\left( x \right) = \frac{x}{\log_e x}\]

\[ \Rightarrow f'\left( x \right) = \frac{\log_e x - 1}{\left( \log_e x \right)^2}\]

\[\text { For a local maxima or a local minima, we must have } \]

\[f'\left( x \right) = 0\]

\[ \Rightarrow \frac{\log_e x - 1}{\left( \log_e x \right)^2} = 0\]

\[ \Rightarrow \log_e x - 1 = 0\]

\[ \Rightarrow \log_e x = 1\]

\[ \Rightarrow x = e\]

\[\text { Now,} \]

\[f''\left( x \right) = \frac{- 1}{x \left( \log_e x \right)^2} + \frac{2}{x \left( \log_e x \right)^3}\]

\[ \Rightarrow f''\left( e \right) = \frac{- 1}{e} + \frac{2}{e} = \frac{1}{e} > 0\]

\[\text { So, x = e is a local minima } . \]

\[ \therefore \text { Minimum value of } f\left( x \right) = \frac{e}{\log_e e} = e\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 17: Maxima and Minima - Exercise 18.7 [पृष्ठ ८१]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 17 Maxima and Minima
Exercise 18.7 | Q 3 | पृष्ठ ८१
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