The set of values of $x$ for which $sin x {cos}^{3} x > cos x {sin}^{3} x, 0 \leq x \leq π$ is

(0, π)

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(0, \frac{π}{4})

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(\frac{π}{4}, π)

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(0, \frac{π}{2})

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Solution

The correct option is B $(0, \frac{π}{4})$
since $sin x > 0$
we can cancel it from both sides
${cos}^{3} x > cos x {sin}^{2} x$
$cos x ({cos}^{2} x - {sin}^{2} x) > 0$
$cos x cos 2 x > 0$
which holds only if $0 \leq x < \frac{π}{4}$

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f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x + a \sqrt{2} sin x, & 0 \leq x < π / 4 2 x cot x + b, & π / 4 \leq x \leq π / 2 a cos 2 x - b sin x, & π / 2 < x \leq π \end{matrix}

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f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} x + a^{2} \sqrt{2} s i n x, & 0 \leq x < \frac{π}{4} x c o t x + b, & \frac{π}{4} \leq x < \frac{π}{2} b s i n 2 x - a c o s 2 x, & \frac{π}{2} \leq x \leq π \end{matrix}

[0, π]

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