WBJEE · Maths · Application of Derivatives
The minimum value of \(\cos \theta+\sin \theta+\frac{2}{\sin 2 \theta}\)
for \(\theta \in(0, \pi / 2),\) is
- A \(2+\sqrt{2}\)
- B 2
- C \(1+\sqrt{2}\)
- D \(2 \sqrt{2}\)
Answer & Solution
Correct Answer
(A) \(2+\sqrt{2}\)
Step-by-step Solution
Detailed explanation
Here, \(\cos \theta+\sin \theta+\frac{2}{\sin 2 \theta}, \theta \in\left(0, \frac{\pi}{2}\right)\) For minimum value, \(\sin 2 \theta\) must be maximum \(\therefore \quad 2 \theta=\frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4}\) Hence,…
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