JEE Mains · Maths · STD 12 - 6. Application of derivatives
\(\max_{0 \leq x \leq \pi}\left(16\sin\left(\dfrac{x}{2}\right)\cos^3\left(\dfrac{x}{2}\right)\right)\) is equal to:
- A \(\dfrac{3\sqrt{3}}{2}\)
- B \(3\sqrt{3}\)
- C \(4\sqrt{3}\)
- D \(6\sqrt{3}\)
Answer & Solution
Correct Answer
(B) \(3\sqrt{3}\)
Step-by-step Solution
Detailed explanation
Let \(y = \dfrac{x}{2}\). Since \(0 \leq x \leq \pi\), we have \(0 \leq y \leq \dfrac{\pi}{2}\). The given expression becomes \(f(y) = 16\sin y \cos^3 y\). Differentiating with respect to \(y\): \(f'(y) = 16(\cos y \cdot \cos^3 y + \sin y \cdot 3\cos^2 y(-\sin y))\)…
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