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The value of c in Rolle's theorem for the function f ( x ) = x ( x + 1 ) e x defined on [−1, 0] is

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Question

The value of c in Rolle's theorem for the function \[f\left( x \right) = \frac{x\left( x + 1 \right)}{e^x}\] defined on [−1, 0] is

Options

  • 0.5

  • \[\frac{1 + \sqrt{5}}{2}\]

  • \[\frac{1 - \sqrt{5}}{2}\] 

  • −0.5

MCQ
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Solution

 \[\frac{1 - \sqrt{5}}{2}\]

Given:

\[f\left( x \right) = \frac{x\left( x + 1 \right)}{e^x}\]
Differentiating the given function with respect to x, we get 

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

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

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

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

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

\[ \therefore f'\left( c \right) = 0 \]

\[ \Rightarrow \frac{- c^2 + c + 1}{e^c} = 0\]

\[ \Rightarrow c^2 - c - 1 = 0\]

\[ \Rightarrow c = \frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\]

\[ \therefore c = \frac{1 - \sqrt{5}}{2} \in \left( - 1, 0 \right)\]

Hence, the required value of c is \[\frac{1 - \sqrt{5}}{2}\] .

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Chapter 14: Mean Value Theorems - Exercise 15.4 [Page 20]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 14 Mean Value Theorems
Exercise 15.4 | Q 8 | Page 20

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