Question

lf $f\left(\theta \right)={\sin }^{99}\theta {\cos }^{94}\theta ;\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$, attains a maximum at $\theta$ equals

• A. ${\tan }^{-1}\sqrt{\frac{94}{99}}$
• B. ${\tan }^{-1}\sqrt{\frac{99}{94}}$
• C. $\frac{\pi }{2}$
• D. $0$

Answer 1

The correct option is B ${\tan }^{-1}\sqrt{\frac{99}{94}}$

$f^{\prime }\left(\theta \right)=99{\sin }^{98}\theta {\cos }^{95}\theta -94{\sin }^{100}\theta {\cos }^{93}\theta$
$f^{\prime }\left(\theta \right)=0$ when ${\tan }^2\theta =\frac{99}{94}$
$\theta ={\tan }^{-1}\sqrt{\frac{99}{94}}$