JEE Main 2021 · 25 July · morning · Single correct · +4 / −1
Let f:[0,∞ ) → [0,∞ ) be defined as f(x) = ∫₀ˣ [y]dy
where [x] is the greatest integer less than or equal to x. Which of the following is true?
- (a) f is continuous at every point in [0,∞ ) and differentiable except at the integer points.
- (b) f is both continuous and differentiable except at the integer points in [0,∞ ).
- (c) f is continuous everywhere except at the integer points in [0,∞ ).
- (d) f is differentiable at every point in [0,∞ ).
Answer and solution
Answer (a) f is continuous at every point in [0,∞ ) and differentiable except at the…
Solution
Maths sir · 4 steps
Area under a step graph never jumps, but its slope copies the step and jumps at every integer.
- Add the areas. On [k, k+1) the value of [y] is k, so each full unit strip adds area k. For x ∈ [n, n+1), add the full strips from 0 to n and then the partial strip from n to x.
f(x) = (0 + 1 + … + (n-1)) + n(x-n) = n(n-1)/2 + n(x-n), n ≤ x < n+1
- Continuity. As x → (n+1)⁻, the formula gives n(n-1)/2 + n = n(n+1)/2, which is exactly f(n+1). The pieces join with no gap, so f is continuous at every point of [0,∞).
- Differentiability. Each piece is a straight line, so f'(x) = [x] between integers. At a positive integer n the slope on the left is n-1 and on the right is n. These do not match, so the graph has a corner there.
- So. f is continuous everywhere on [0,∞) and fails to be differentiable only at the integer points. That is option (a).
The idea to keep. Integrating a function that jumps makes it one level smoother: the integral is continuous everywhere, and it has corners exactly where the integrand jumps.
Answer from the NTA final answer key. Solution by PrepWiser.