max 0 x { x - 2 sin x cos x + 1 3 sin 3 x } = [2023]

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Solution

Q.
$\max_{0 \leq x \leq π} {x - 2 \sin x \cos x + \frac{1}{3} \sin 3 x} =$ [2023]

1 $π$

2 $\frac{π + 2 - 3 \sqrt{3}}{6}$

3 $\frac{5 π + 2 + 3 \sqrt{3}}{6}$

4 0

Ans.
(3)

Let $f (x) = x - 2 \sin x \cos x + \frac{1}{3} \sin 3 x, 0 \leq x \leq π$

$f (x) = x - \sin 2 x + \frac{1}{3} \sin 3 x$

$f' (x) = 1 - 2 \cos 2 x + \cos 3 x = 0$

$\Rightarrow 4 \cos^{3} x - 3 \cos x - 2 (2 \cos^{2} x - 1) + 1 = 0$

$\Rightarrow 4 \cos^{3} x - 4 \cos^{2} x - 3 \cos x + 3 = 0$

$\Rightarrow 4 \cos^{2} x (\cos x - 1) - 3 (\cos x - 1) = 0$

$\Rightarrow (\cos x - 1) (4 \cos^{2} x - 3) = 0$

$\Rightarrow \cos x = 1, \pm \frac{\sqrt{3}}{2} ∴ x = 0, \frac{π}{6}, \frac{5 π}{6}$

Now, $f'' (x) = 4 \sin 2 x - 3 \sin 3 x$

Now, $f'' (0) = 0$

$f'' (\frac{π}{6}) > 0 and f'' (\frac{5 π}{6}) < 0 \Rightarrow (\frac{5 π}{6}) is a point of maxima.$

Hence, $f (\frac{5 π}{6}) = \frac{5 π}{6} - 2 \sin \frac{5 π}{6} \cos \frac{5 π}{6} + \frac{1}{3} \sin \frac{5 π}{2}$

$= \frac{5 π}{6} - \sin 2 (\frac{5 π}{6}) + \frac{1}{3} = \frac{5 π}{2} + \frac{\sqrt{3}}{2} + \frac{1}{3}$ $= \frac{5 π + 3 \sqrt{3} + 2}{6}$

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