Let a ∈ ℤ and [t] be the greatest integer ≤ t. Then the number of points, where the function f(x)=[a+13 sin x], x ∈(0, π) is not differentiable, is __________.

Solution

Maths sir · 5 steps

Pull the integer a out of the bracket, then count where 13 sin x hits an integer on (0, π).

1. Remove a. Since a is an integer, [a + 13 sin x] = a + [13 sin x]. Adding a constant does not change where a function breaks, so a plays no part in the count.
2. Where it jumps. For x ∈ (0, π), sin x takes values in (0, 1], so 13 sin x takes values in (0, 13]. The greatest integer function jumps exactly where 13 sin x equals an integer k, which can be k = 1, 2, …, 13.
3. Levels 1 to 12. For each k from 1 to 12, sin x = k/13 < 1 has two solutions in (0, π): some x, and also π - x. At each one [13 sin x] jumps, so f is discontinuous there. That gives 12 × 2 = 24 points.
4. Level 13. sin x = 1 only at x = π/2. There f = a + 13, but on both sides 13 sin x is just below 13, so f = a + 12. The function is discontinuous at π/2, which adds 1 more point.
5. So. Every jump is a discontinuity, so f is not differentiable there. The total is 24 + 1 = 25.

The common slip: 24. Leaving out x = π/2, where 13 sin x reaches its peak value 13, gives 24 instead of 25.

The idea to keep. For [g(x)], mark each integer level that g reaches and count every x where g hits it. At a peak value that is an integer, [g] drops on both sides, so that point counts too.

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