Question 15 of 82intermediate🔧 ApplyNumerical4 marks

Find the maximum and minimum values of x + sin 2x on [0, 2π].

Correct Answer

Maximum value = 2π at x = 2π; Minimum value = 0 at x = 0

Exercise: EXERCISE 6.3 | Q: 12 | (Chapter: 30)
For More Understanding

Explanation

This question requires finding absolute maximum and minimum values on a closed interval [0, 2π]. Following the method shown in Example 27 of the textbook, we need to: (1) find critical points by setting derivative to zero, (2) evaluate the function at critical points and endpoints, and (3) compare all values to determine absolute extrema.

Solution Steps

  1. Step 1: Let f(x) = x + sin 2x. Find the derivative: f'(x) = 1 + 2cos 2x

  2. Step 2: Find critical points by setting f'(x) = 0: 1 + 2cos 2x = 0, which gives cos 2x = -1/2

  3. Step 3: Solving cos 2x = -1/2 on [0, 2π]: We get 2x = 2π/3, 4π/3, 8π/3, 10π/3, so x = π/3, 2π/3, 4π/3, 5π/3

  4. Step 4: Evaluate f(x) at all critical points and endpoints: f(0) = 0 + sin 0 = 0; f(π/3) = π/3 + 3\sqrt{3}/2; f(2π/3) = 2π/3 - 3\sqrt{3}/2; f(4π/3) = 4π/3 + 3\sqrt{3}/2; f(5π/3) = 5π/3 - 3\sqrt{3}/2; f(2π) = 2π + sin 4π = 2π

  5. Step 5: Comparing all values: f(0) = 0 is the smallest, and f(2π) = 2π is the largest. Therefore, absolute minimum value = 0 at x = 0 and absolute maximum value = 2π at x = 2π

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