Question

${\sin }^p\theta {\cos }^q\theta$ attains a maximum value when $\theta ={\tan }^{-1}\sqrt{\frac{p}{q}}.$

• A. True
• B. False

Answer 1

The correct option is A True

$y=si{n}^p\theta co{s}^q\theta \left(\frac{dy}{dx}\right)=psi{n}^{p-1}\theta cos\theta co{s}^q\theta -qsi{n}^p\theta co{s}^{q-1}\theta sin\theta =si{n}^{p-1}\theta co{s}^{q-1}\theta [pco{s}^2\theta -qsi{n}^2\theta ]\left(\frac{dy}{dx}\right)=0\left(forcriticalpoints\right)\therefore pco{s}^2\theta =qsi{n}^2\theta \therefore ta{n}^2\theta =\left(\frac{p}{q}\right)\therefore \theta =ta{n}^{-1}\sqrt{\left(\frac{p}{q}\right)}$