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प्रश्न
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
पर्याय
0.5
\[\frac{1 + \sqrt{5}}{2}\]
\[\frac{1 - \sqrt{5}}{2}\]
−0.5
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उत्तर
\[\frac{1 - \sqrt{5}}{2}\]
Given:
\[f'\left( x \right) = \frac{e^x \left( 2x + 1 \right) - x\left( x + 1 \right) e^x}{\left( e^x \right)^2}\]
\[ \Rightarrow f'\left( x \right) = \frac{\left( 2x + 1 \right) - x\left( x + 1 \right)}{e^x}\]
\[ \Rightarrow f'\left( x \right) = \frac{2x + 1 - x^2 - x}{e^x} \]
\[ \Rightarrow f'\left( x \right) = \frac{- x^2 + x + 1}{e^x}\]
\[ \Rightarrow f'\left( c \right) = \frac{- c^2 + c + 1}{e^c}\]
\[ \therefore f'\left( c \right) = 0 \]
\[ \Rightarrow \frac{- c^2 + c + 1}{e^c} = 0\]
\[ \Rightarrow c^2 - c - 1 = 0\]
\[ \Rightarrow c = \frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\]
\[ \therefore c = \frac{1 - \sqrt{5}}{2} \in \left( - 1, 0 \right)\]
Hence, the required value of c is \[\frac{1 - \sqrt{5}}{2}\] .
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