CBSE (Science) Class 12CBSE
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

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

Login
Create free account


      Forgot password?

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?
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.
S
View in app×