$f (x) = sin x (1 + cos x)$ is maximum at:

x = \frac{π}{4}

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x = \frac{π}{6}

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x = \frac{π}{3}

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x = \frac{π}{2}

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Solution

The correct option is B $x = \frac{π}{3}$
$f (x) = s i n x (1 + c o s x)$
Therefore
$f (x) = s i n x + s i n x . c o s x$ .
Now for a maximum
$f^{'} (x) = 0$
Or
$\Rightarrow c o s x - s i n^{2} x + c o s^{2} x = 0$
$\Rightarrow c o s x - (1 - c o s^{2} x) + c o s^{2} x = 0$
$\Rightarrow 2 c o s^{2} x + c o s x - 1 = 0$
$\Rightarrow 2 c o s^{2} x + 2 c o s x - c o s x - 1 = 0$
$\Rightarrow 2 c o s x (1 + c o s x) - (1 + c o s x) = 0$
$\Rightarrow (1 + c o s x) (2 c o s x - 1) = 0$
$c o s x = - 1$ or $c o s x = \frac{1}{2}$ .
Hence
$x = π$ or $x = \frac{π}{3}, \frac{5 π}{3}$ .
Now
$f (π) = 0$ ....(not a maximum).
$f (\frac{π}{3}) = \frac{\sqrt{3}}{2} (\frac{3}{2})$
$= \frac{3 \sqrt{3}}{4}$ ...(ii)
And
$f (\frac{5 π}{3}) = \frac{- 3 \sqrt{3}}{4}$
Hence
$f (x)$ attains a maximum at $x = \frac{π}{3}$ .

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