NCERT Class 12 · Ch 6 Exercise 6.3 · Q12
NCERT Class 12 Maths · Chapter 6 Applications Of Derivatives · Exercise 6.3 Question 12
Find maximum and minimum of x+sin 2x on [0,2π].
💡 Concept Hints
- Identify the domain and the inner and outer functions, or the relevant one-sided limits.
- Apply the appropriate differentiation or limit rule, keep all chain factors, and check exceptional points.
✅ Step-by-Step Solution
Final Answer: Minimum 0 at x=0; maximum 2π at x=2π.
- Differentiate: f′=\(2 \cos{\left(2 x \right)} + 1\). The interior stationary points in this interval are \(\frac{\pi}{3}\), \(\frac{2 \pi}{3}\), \(\frac{4 \pi}{3}\), \(\frac{5 \pi}{3}\).
- Evaluate every candidate: f(\(0\))=\(0\); f(\(\frac{\pi}{3}\))=\(\frac{3 \sqrt{3} + 2 \pi}{6}\); f(\(\frac{2 \pi}{3}\))=\(\frac{- 3 \sqrt{3} + 4 \pi}{6}\); f(\(\frac{4 \pi}{3}\))=\(\frac{3 \sqrt{3} + 8 \pi}{6}\); f(\(\frac{5 \pi}{3}\))=\(\frac{- 3 \sqrt{3} + 10 \pi}{6}\); f(\(2 \pi\))=\(2 \pi\).
- Comparing these values gives the absolute maximum and minimum; endpoints must be included.
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