Question

$si{n}^p\theta co{s}^q\theta$ attains a maximum, when $\theta =$

• A. $ta{n}^{-1}\sqrt{\frac{p}{q}}$
• B. $ta{n}^{-1}\left(\frac{p}{q}\right)$
• C. $ta{n}^{-1}q$
• D. $ta{n}^{-1}\left(\frac{q}{p}\right)$

Answer 1

Answer: (A)

Let $y=si{n}^p\theta co{s}^q\theta$. Then,
$\frac{dy}{d\theta }=psi{n}^{p-1}\theta co{s}^q\theta +si{n}^p\theta qco{s}^{q-1}\theta \left(-sin\theta \right)$
$\Rightarrow \frac{dy}{d\theta }=psi{n}^{p-1}\theta co{s}^{q+1}\theta -qsi{n}^{p+1}\theta co{s}^{q-1}\theta$
$\Rightarrow \frac{dy}{d\theta }=si{n}^{p-1}\theta co{s}^{q-1}\theta \left(pco{s}^2\theta -qsi{n}^2\theta \right)$
$\Rightarrow \frac{dy}{d\theta }=si{n}^p\theta co{s}^q\theta \left(\frac{pco{s}^2\theta -qsi{n}^2\theta }{sin\theta cos\theta }\right)$
$\Rightarrow \frac{dy}{d\theta }=si{n}^p\theta co{s}^q\theta \left(pcot\theta -qtan\theta \right)$
For maximum or minimum, we must have
$\frac{dy}{d\theta }=0$
$\Rightarrow si{n}^p\theta co{s}^q\theta \left(pcot\theta -qtan\theta \right)=0$
$\Rightarrow si{n}^p\theta =0$ or
$co{s}^q\theta =0$ or
$pcot\theta -qtan\theta =0$
$\Rightarrow si{n}^p\theta =0$ or
$co{s}^q\theta =0$ or,
$tan\theta =\sqrt{\frac{p}{q}}=\alpha$
$\Rightarrow \theta =0$ or,
$\theta =\frac{\pi }{2}$ or,
$\theta =ta{n}^{-1}\sqrt{\frac{p}{q}}=\alpha \left(say\right)$
Now,
$\frac{dy}{d\theta }=si{n}^p\theta co{s}^q\theta \left(pcot\theta -qtan\theta \right)$
$=y\left(pcot\theta -qtan\theta \right)$
$\Rightarrow \frac{d^{2}y}{d{\theta }^2}=\frac{dy}{d\theta }\left(pcot\theta -qtan\theta \right)+y\left(-pcose{c}^2\theta -qse{c}^2\theta \right)$
$\Rightarrow {\left(\frac{d^{2}y}{d{\theta }^2}\right)}_{\theta =\alpha }={\left(\frac{dy}{d\theta }\right)}_{\theta =\alpha }\left(p\sqrt{\frac{q}{p}-q\sqrt{\frac{p}{q}}}\right)$
$+si{n}^p\theta co{s}^q\theta \left[-pcose{c}^2\theta -qse{c}^2\theta \right]$
$\Rightarrow {\left(\frac{d^{2}y}{d{\theta }^2}\right)}_{\theta =\alpha }=0-si{n}^p\theta co{s}^q\theta \left(pcose{c}^2\theta -qse{c}^2\theta \right)$
Hence, $y$ is maximum when $\theta =\alpha =ta{n}^{-1}\sqrt{\frac{p}{q}}$.