Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:
\(f(x)=\sin x+\cos 2x,(x>0)\) has minima for \(x=\)
Options:
\(\frac{n\pi}{2}\)
\(\frac{3}{2}(n+1)\pi\)
\(\frac{\pi}{2}(2n+1)\)
\(n\pi\)
Correct Answer:
\(\frac{\pi}{2}(2n+1)\)
Explanation:
\(\begin{aligned}f^{\prime}(x)&=0\\ \cos x-4\cos x\sin x&=0\\ \cos x&=0\Rightarrow x=(2n+1)\frac{\pi}{2}\\ 1-4\sin x&=0\Rightarrow \sin x=\frac{1}{4}\end{aligned}\hspace{5cm}\) Minimum value at \(\frac{\pi}{2}(2n+1)\)