The length of the longest interval, in which $f (x) = 3 sin x - 4 {sin}^{2} x$ is increasing, is

\frac{π}{3}

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\frac{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 $\frac{π}{3}$
Let $f (x) = 3 sin x - 4 {sin}^{3} x = sin 3 x$
Since, $sin x$ is increasing in the interval $[- \frac{π}{2}, \frac{π}{2}]$
Thus $- \frac{π}{2} \leq 3 x \leq \frac{π}{2} \Rightarrow - \frac{π}{6} \leq x \leq \frac{π}{6}$
Hence, the length of interval $= [\frac{π}{6} - (- \frac{π}{6})] = \frac{π}{3}$ .

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