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Saturday, 31 October

JEE Main 2022 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 29 June 2022, Shift 1, Maths Q4

Question 4 of 90 in this shift, Maths question 4 of 30, Section A.

Let f:R→Rf: \mathbf{R} \to \mathbf{R} be a function defined by : f(x)={max⁡t≤x{t3−3t};  x≤2x2+2x−6;  2<x<3[x−3]+9;  3≤x≤52x+1;  x>5f(x) = \begin{cases} \max_{t \le x}\{t^3 - 3t\} & ; \; x \le 2 \\ x^2 + 2x - 6 & ; \; 2 < x < 3 \\ [x - 3] + 9 & ; \; 3 \le x \le 5 \\ 2x + 1 & ; \; x > 5 \end{cases} where [t][t] is the greatest integer less than or equal to tt. Let mm be the number of points where ff is not differentiable and I=∫−22f(x) dxI = \int_{-2}^{2} f(x)\,dx. Then the ordered pair (m,I)(m, I) is equal to :
  1. (1)(3,274)\left(3, \frac{27}{4}\right)
  2. (2)(3,234)\left(3, \frac{23}{4}\right)
  3. (3)(4,274)\left(4, \frac{27}{4}\right)Official answer
  4. (4)(4,234)\left(4, \frac{23}{4}\right)

Official answer

Option 3

NTA final key (2022 Session 1).

Chapter
Continuity and Differentiability
Topic
Continuity and Differentiability
Idea tested
Piecewise Differentiability

Same idea in other shifts

Asked 7× in all
  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. 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
  3. 27 Jul 2022, Shift 1 · Q20Let a function f:R→Rf : \mathbb{R} \rightarrow \mathbb{R} be defined as :…HardSingle correct
  4. 22 Jul 2021, Shift 2 · Q85Let f:R→Rf:\mathbf{R}\to\mathbf{R} be a function defined as…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. 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

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.