Learn

Saturday, 31 October

JEE Main 2021 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 25 February 2021, Shift 1, Maths Q65

Question 65 of 90 in this shift, Maths question 5 of 30, Section A.

If Rolle's theorem holds for the function f(x)=x3−ax2+bx−4f(x) = x^3 - ax^2 + bx - 4, x∈[1,2]x \in [1, 2] with f′(43)=0f'\left(\frac{4}{3}\right) = 0, then ordered pair (a, b) is equal to :
  1. (1)(5, 8)Official answer
  2. (2)(5, −8-8)
  3. (3)(−5-5, 8)
  4. (4)(−5-5, −8-8)

Official answer

Option 1

NTA final key.

Same topic in other shifts

All Rolle's Theorem questions
  1. 4 Apr 2024, Shift 2 · Q25Let f:R→Rf:\mathbb{R}\to\mathbb{R} be a thrice differentiable function such that f(0)=0,f(1)=1,f(2)=−1,f(3)=2f(0)=0, f(1)=1, f(2)=-1, f(3)=2 and f(4)=−2f(4)=-2. Then, the…HardNumerical value
  2. 30 Jan 2024, Shift 1 · Q9Let g:R→Rg:\mathbf{R}\to\mathbf{R} be a non constant twice differentiable function such that…HardSingle correct
  3. 28 Jul 2022, Shift 1 · Q23Let f:[0,1]→Rf:[0,1]\to\mathbf{R} be a twice differentiable function in (0,1)(0,1) such that f(0)=3f(0)=3 and f(1)=5f(1)=5. If the line y=2x+3y=2x+3 intersects…MediumNumerical value
  4. 24 Jun 2022, Shift 2 · Q11The number of distinct real roots of the equation x7−7x−2=0x^7 - 7x - 2 = 0 isMediumSingle correct
  5. 29 Jun 2022, Shift 2 · Q26Let ff and gg be twice differentiable even functions on (−2,2)(-2, 2) such that f(14)=0f\left(\frac{1}{4}\right) = 0,…HardNumerical value
  6. 31 Aug 2021, Shift 2 · Q68Let ff be any continuous function on [0,2][0, 2] and twice differentiable on (0,2)(0, 2). If f(0)=0f(0) = 0, f(1)=1f(1) = 1 and f(2)=2f(2) = 2, then :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.