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

Solution

Maths sir · 5 steps

The min switches pieces only where x sin x changes sign, so check that one point for a corner.

1. Sign check. On [0, π] both x and sin x are ≥ 0, so x sin x ≥ 0. On [π, 2π], sin x ≤ 0, so x sin x ≤ 0.
2. Write f piecewise. On [0, π] we have 1 + x sin x ≥ 1, so the minimum is 1. On [π, 2π] we have 1 + x sin x ≤ 1, so the minimum is 1 + x sin x. Both pieces equal 1 at x = π.

   f(x) = 1, 0 ≤ x ≤ π ; 1 + x sin x, π < x ≤ 2π

3. Continuity. The minimum of two continuous functions is continuous, so f has no break anywhere and n = 0.
4. Check the join. At x = π the left derivative is 0. The right derivative is sin x + x cos x at x = π, which is 0 - π = -π. They differ, so x = π is a corner, and each piece is smooth everywhere else.
5. So. There is exactly one point of non-differentiability and none of discontinuity, so (m, n) = (1, 0), option (b).

The idea to keep. For min or max of two functions, find where they cross and rewrite piecewise. Corners can only appear at crossing points, so compare the one-sided derivatives there.

Answer from the NTA final answer key. Solution by PrepWiser.