Consider function f(x) = {e^{ - 2x}} sin, 2x over interval left( {0,{π over 2}} right) . A real num

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5. Continuity and Differentiation

easy

Consider the function $f(x) = {e^{ - 2x}}$ $sin\, 2x$ over the interval $\left( {0,{\pi \over 2}} \right)$. A real number $c \in \left( {0,{\pi \over 2}} \right)\,,$ as guaranteed by Rolle’s theorem, such that $f'\,(c) = 0$ is

A

$\pi /8$

B

$\pi /6$

C

$\pi /4$

D

$\pi /3$

Solution

(a) $f(x) = {e^{ – 2x}}\sin 2x$ ==> $f'(x) = 2{e^{ – 2x}}(\cos 2x – \sin 2x)$

Now, $f'(c) = 0$

==>$\cos 2c – \sin 2c = 0$==>$\tan 2c = 1$==>$c = \frac{\pi }{8}$.

Standard 12

Mathematics

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