Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:
One maximum point of \(\sin^{p}x\cos^{q}x\) is
Options:
\(x=\tan^{-1}\sqrt{\frac{p}{q}}\)
\(x=\tan^{-1}\sqrt{\frac{q}{p}}\)
\(x=\tan^{-1}\left(\frac{p}{q}\right)\)
\(x=\tan^{-1}\left(\frac{q}{p}\right)\)
Correct Answer:
\(x=\tan^{-1}\sqrt{\frac{p}{q}}\)
Explanation:
Let \(f(x)=\sin^{p}x\cos^{q}x\hspace{6cm}\) \(\begin{aligned}f^{\prime}(x)&=p\sin^{p-1}x\cos x\cos^{q}x-q\sin^{p}x\cos^{q-1}x\sin x\\ f^{\prime}(x)&=0\\ \tan^{2}x&=\frac{p}{q}\end{aligned}\)