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Saturday, 31 October

JEE Main · Maths · Continuity and Differentiability

Piecewise Differentiability

JEE Main asked Piecewise Differentiability 7 times in 7 of the 108 shifts Lumi analysed, in 2020, 2021, 2022 and 2024. It belongs to Continuity and Differentiability (Maths). The longest wait between two years asking it was 2 years. In 2021 and 2022 it came in more than one shift of the same year.

Updated 2026-10-24

Times asked
7
in 7 shifts
Years
4 of 8
First asked
2020
last in 2024
Pattern
Comes in cycles
across the analysed shifts

Questions on this idea a shift, by year

2017–2026. Asked in 2020, 2021, 2022 and 2024. Years differ in how many shifts Lumi analysed, so each year is scaled to one shift of 25 Maths questions.

00.10.20’170.08’200.07’210.11’220’230.05’240’250’26

Raw count: 2017: 0 in 1 shift · 2020: 1 in 12 shifts · 2021: 2 in 24 shifts · 2022: 3 in 24 shifts · 2023: 0 in 12 shifts · 2024: 1 in 18 shifts · 2025: 0 in 8 shifts · 2026: 0 in 9 shifts.

Shift by shift

Each square is one analysed shift, in date order. Open a filled one for that paper.

  • 20260 of 9
  • 20250 of 8
  • 20241 of 18
  • 20230 of 12
  • 20223 of 24
  • 20212 of 24
  • 20201 of 12
  • 20170 of 1

Asked in that shift More than once Not asked

How JEE Main framed it

Single correct is the usual form. Skill tested: multi-step.

  • Continuity and differentiability of piecewise function
  • Non-differentiable points and integral of piecewise function

Difficulty

Easy 0Medium 3Hard 4

3 of the 7 questions had a numerical answer.

Every question

All 7 questions

From the official JEE Main shift papers, newest first. Open one for the full question, the options and the official answer.

  1. 30 Jan 2024, Shift 1 · Q30If the function f(x)={1∣x∣,∣x∣≥2ax2+2b,∣x∣<2f(x)=\begin{cases}\frac{1}{|x|}, & |x|\geq2\\ ax^2+2b, & |x|<2\end{cases} is differentiable on R\mathbf{R}, then…MediumNumerical value
  2. 29 Jun 2022, Shift 1 · Q4Let f:R→Rf: \mathbf{R} \to \mathbf{R} be a function defined by :…HardSingle correct
  3. 27 Jul 2022, Shift 1 · Q20Let a function f:R→Rf : \mathbb{R} \rightarrow \mathbb{R} be defined as :…HardSingle correct
  4. 25 Jul 2022, Shift 1 · Q26Let f(x)={∣ 4x2−8x+5 ∣,if 8x2−6x+1≥0[ 4x2−8x+5 ],if 8x2−6x+1<0,f(x)=\begin{cases}\left|\,4x^2-8x+5\,\right|, & \text{if } 8x^2-6x+1\ge0\\ [\,4x^2-8x+5\,], & \text{if } 8x^2-6x+1<0,\end{cases}…HardNumerical value
  5. 25 Jul 2021, Shift 2 · Q72If f(x)={∫0x(5+∣1−t∣) dt,x>25x+1,x≤2f(x)=\begin{cases}\int_0^x(5+|1-t|)\,dt, & x>2\\ 5x+1, & x\le 2\end{cases}, thenMediumSingle correct
  6. 22 Jul 2021, Shift 2 · Q85Let f:R→Rf:\mathbf{R}\to\mathbf{R} be a function defined as…HardNumerical value
  7. 4 Sep 2020, Shift 2 · Q59The function f(x)={π4+tan⁡−1x,∣x∣≤112(∣x∣−1),∣x∣>1f(x) = \begin{cases} \frac{\pi}{4} + \tan^{-1} x, & |x| \leq 1 \\ \frac{1}{2}(|x| - 1), & |x| > 1 \end{cases} is :MediumSingle correct

Shifts: 4 Sep 2020, Shift 2, 22 Jul 2021, Shift 2, 25 Jul 2021, Shift 2, 29 Jun 2022, Shift 1, 25 Jul 2022, Shift 1, 27 Jul 2022, Shift 1, 30 Jan 2024, Shift 1.