Continuity & DifferentiabilitymediumPYQ · JEE Main · 2 Apr 2026 · Shift 2 (Afternoon)Free

Greatest Integer Function Discontinuities on [2,4]: 10 Points | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
The number of points in the interval [2,4][2,4], at which the function f(x)=[x2−x−12]f(x)=\left[x^2-x-\dfrac12\right], where [⋅][\cdot] denotes the greatest integer function, is discontinuous, is
Solution
Answer: 10 (± 0.01)
Step 1: On x∈[2,4]x\in[2,4], let g(x)=x2−x−12g(x)=x^2-x-\dfrac12. It is increasing there, with
g(2)=4−2−12=1.5,g(4)=16−4−12=11.5,g(2)=4-2-\tfrac12=1.5,\qquad g(4)=16-4-\tfrac12=11.5,
 so the range of gg is [1.5, 11.5][1.5,\ 11.5]. Step 2: The greatest integer function [g(x)][g(x)] is discontinuous exactly where g(x)g(x) takes an integer value. Step 3: The integers in the interval [1.5,11.5][1.5,11.5] are
2,3,4,5,6,7,8,9,10,11,2,3,4,5,6,7,8,9,10,11,
 which is 1010 values. Since gg is strictly increasing, each is attained exactly once. Step 4: Therefore the number of points of discontinuity is 1010. Correct answer: 10
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Continuity & Differentiability · hard
The number of points, at which the function f(x)=max⁡{6x, 2+3x2}+∣x−1∣cos⁡∣x2−14∣f(x)=\max\{6x,\ 2+3x^2\}+|x-1|\cos\left|x^2-\dfrac14\right|, x∈(−π,π)x\in(-\pi,\pi), is not differentiable, is
Continuity & Differentiability · hard
Let f(x)={ex−1,x<0x2−5x+6,x≥0f(x)=\begin{cases}e^{x-1}, & x<0\\ x^2-5x+6, & x\ge0\end{cases} and g(x)=f(∣x∣)+∣f(x)∣g(x)=f(|x|)+|f(x)|. If the number of points where gg is not continuous and is not differentiable are α\alpha and β\beta respectively, then α+β\alpha+\beta is equal to

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath