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Saturday, 31 October

JEE Main 2021 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 3 August 2021, Shift 1, Maths Q66

Question 66 of 90 in this shift, Maths question 6 of 30, Section A.

Suppose that the function f(x)={x2−[x]+2a;x<−14x+5−b;−1≤x<23x2+6[x]−2a;x≥2f(x) = \begin{cases} x^2 - [x] + 2a; & x < -1 \\ 4x + 5 - b; & -1 \leq x < 2 \\ 3x^2 + 6[x] - 2a; & x \geq 2 \end{cases} is continuous on (−2,3)(-2, 3) for some values of aa and bb where [x][x] denotes the greatest integer function. Then the value of 8ab8ab is
  1. (1)135
  2. (2)−117-117Official answer
  3. (3)−104-104
  4. (4)99

Official answer

Option 2

NTA final key.

Chapter
Continuity and Differentiability
Topic
Continuity and Differentiability
Idea tested
Finding Unknowns using Continuity

Same idea in other shifts

Asked 3× in all
  1. 8 Apr 2026, Shift 2 · Q19Let f(x)={13,x≤π/2b(1−sin⁡x)(π−2x)2,x>π/2f(x) = \begin{cases} \frac{1}{3}, & x \le \pi/2 \\ \frac{b(1 - \sin x)}{(\pi - 2x)^2}, & x > \pi/2 \end{cases}. If ff is continuous…MediumSingle correct
  2. 3 Apr 2025, Shift 1 · Q17Let f(x)={(1+ax)1/x,x<01+b,x=0(x+4)1/2−2(x+c)1/3−2,x>0f(x) = \begin{cases} (1+ax)^{1/x}, & x < 0 \\ 1+b, & x = 0 \\ \dfrac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2}, & x > 0 \end{cases} be…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.