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5. Continuity and Differentiation

medium

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

A

$0$

B

$-1$

C

$-2$

D

$-3$

Solution

(c) 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)}}\{ {x^2} – x – 6\} $

$\therefore f'(c) = 0 \Rightarrow {c^2} – c – 6 = 0 \Rightarrow c = 3,\, – 2,$

But $c = 3 \notin [ – 3,\,0].$

Std 12

Mathematics

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