Question

Let f(x) = min {1, 1 + x sin x}, 0 ≤ x ≤ 2π. If m is the number of points, where f is not differentiable, and n is the number of points, where f is not continuous, then the ordered pair (m, n) is equal to

• A. (2, 0)
• B. (1, 0)
• C. (1, 1)
• D. (2, 1)

Answer 1

The Correct Option is B

 To determine the number of points where the function \(f(x) = \min \{1, 1 + x \sin x\}\) is not differentiable and not continuous, we need to analyze the function carefully within the interval \(0 \le x \le 2\pi\).

1. First, consider the two expressions inside the minimum function:
   • \(1\): This is a constant function and is both continuous and differentiable everywhere in the domain.
   • \(1 + x \sin x\): A function composed of continuous and differentiable functions (i.e., linear and trigonometric functions), hence it is continuous and differentiable for all \(x\) in the domain.
2. The function \(f(x) = \min \{1, 1 + x \sin x\}\) will switch between the two expressions at points where they are equal, i.e., where: \(1 = 1 + x \sin x\)
   • Solving \(x \sin x = 0\), we find these points correspond to \(x = 0, \pi, 2\pi\) (where \(\sin x = 0\)).
3. Analyze continuity:
   • At \(x = 0\): \(f(x) = 1 = 1 + x \sin x\), hence \(f(x)\) is continuous.
   • At \(x = \pi\): \(f(x)\) changes from \(1\) to \(1 + \pi \sin \pi = 1\), so \(f(x)\) is continuous.
   • At \(x = 2\pi\): \(f(x)\) is the same \(1\), thus continuous.
4. Analyze differentiability:
   • At \(x = \pi\), the function switches between two different expressions. Hence, it is not differentiable at this point.
   • At \(x = 0\) and \(x = 2\pi\): Since \(f(x)\) does not switch, it remains differentiable.

Therefore, the function \(f(x)\) is differentiable everywhere except at \(x = \pi\). It is continuous everywhere. Thus, the ordered pair \((m, n)\) is (1, 0).