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# Solution for The Value of C in Rolle'S Theorem for the Function F ( X ) = X ( X + 1 ) E X Defined on [−1, 0] is (A) 0.5 (B) 1 + √ 5 2 (C) 1 − √ 5 2 (D) −0.5 - CBSE (Science) Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### 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
(a) 0.5
(b) $\frac{1 + \sqrt{5}}{2}$

(c)$\frac{1 - \sqrt{5}}{2}$

(d) −0.5

#### Solution

(c) $\frac{1 - \sqrt{5}}{2}$

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 $\frac{1 - \sqrt{5}}{2}$ .

Is there an error in this question or solution?

#### APPEARS IN

Solution The Value of C in Rolle'S Theorem for the Function F ( X ) = X ( X + 1 ) E X Defined on [−1, 0] is (A) 0.5 (B) 1 + √ 5 2 (C) 1 − √ 5 2 (D) −0.5 Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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