On maximum point of ${\sin ^p}x\,{\cos ^q}x$ is

$x = {\tan ^{ - 1}}\left( {\sqrt {\frac{q}{p}} } \right)$

$x = {\tan ^{ - 1}}\left( {\frac{p}{q}} \right)$

$x = {\tan ^{ - 1}}\left( {\frac{q}{p}} \right)$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

On maximum point of

{\sin ^p}x\,{\cos ^q}x

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Correct option: A

x = {\tan ^{ - 1}}\left( {\sqrt {\frac{p}{q}} } \right)

Option A is correct because the maximum point is found by setting the derivative of the function to zero and solving for x.

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