Question

The function $f(x) = x(x + 3)e^{- 1/2x}$ satisfies all the condition of Rolle's theorem in [– 3, 0]. The value of cis

• A. 0
• B. 1
• C. – 2
• D. – 3

Answer 1

Answer: (C)

– 2

Answer 2

To determine 'c' in Rolle's theorem, $f^{'}(c) = 0$

Here $f^{'}(x) = (x^{2} + 3x)e^{- (1/2)x}.\left( - \frac{1}{2} \right) + (2x + 3)e^{- (1/2)x}$

= $e^{- (1/2)x}\left\{ - \frac{1}{2}(x^{2} + 3x) + 2x + 3 \right\}$ = $- \frac{1}{2}e^{- (x/2)}\left\{ x^{2} - x - 6 \right\}$

$\therefore$ $f^{'}(c) = 0$ ⇒ $c^{2} - c - 6 = 0$ ⇒ $c = 3, - 2$.

But $c = 3 \notin \lbrack - 3,0\rbrack$, Hence c = –2.