The length of the longest interval in which the function $3 sin x - 4 {sin}^{3} x$ is increasing is

π / 3

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π / 2

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\frac{3 π}{2}

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π

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Solution

The correct option is A $π / 3$
We know that, $3 sin x - 4 {sin}^{3} x = sin 3 x$
We know that the interval of the sin function is $(- 1, + 1)$
i.e., $- 1 < sin θ < + 1$
Let $θ = 3 x$
$∴ \frac{- π}{2} < θ < \frac{π}{2}$

$∴ \frac{- π}{2} < 3 x < \frac{π}{2}$

$⟹ \frac{- π}{6} < θ < \frac{π}{6}$

$x ϵ (\frac{- π}{6}, \frac{π}{6})$

$\frac{- π}{6} < x < \frac{π}{6}$

$∴$ The longest interval is,

$\frac{π}{6} - (- \frac{π}{6})$

$\frac{2 π}{6}$

$\frac{π}{3}$
Option A

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