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The function $f(x)=\sin x(1+ \cos x)$ which is defined for all real values of $x$

  1. has a maximum at $x= \pi /3$
  2. has a maximum at $x= \pi$
  3. has a minimum at $x= \pi /3$
  4. has neither a maximum nor a minimum at $x=\pi/3$
in Calculus
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1 Answer

1 vote

$\underline{\textbf{Answer:}\Rightarrow\;\mathbf A}$

$\underline{\textbf{Explanation:}\Rightarrow }$

Given: $f(x) = \sin x(1+\cos x)$

On differentiating the above expression  w.r.t. $\mathrm x$, we get:

$f'(x) = \sin x (1+\cos x)+\sin x(-\sin x) = \cos x + \cos^2 x-\sin ^2x = \cos x + \cos 2x$

Equate the above expression with $0$:

$\implies \cos x = -\cos 2x$

This is equal when $x = \frac{\pi}{3}$

So, this is the critical point.

Now, for knowing whether its maxima or minima, further differentiate $f{''}(x)$, we get:

$f^{''}(x) = -\sin x - 2\sin 2x = -(\sin x + 2\sin 2x)$

$\implies f^{''}(\frac{\pi}{3}) = -(\frac{\sqrt 3}{2} + 2.\frac{\sqrt 3}{2}) \lt 0$

$\therefore$ It's a point of Maxima.

$\therefore \mathbf A$ is the correct option.


edited by
1
Answer is A actually. You solved it correctly though :)
0
Thanks.

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