JEE Main 2022 · MathsNumerical answerNumerical valueHardMulti-step

JEE Main 29 June 2022, Shift 2, Maths Q26

Question 26 of 90 in this shift, Maths question 26 of 30, Section B.

Let

f

and

g

be twice differentiable even functions on

(-2, 2)

such that

f\left(\frac{1}{4}\right) = 0

f\left(\frac{1}{2}\right) = 0

f(1) = 1

and

g\left(\frac{3}{4}\right) = 0

g(1) = 2

Then, the minimum number of solutions of

f(x)g''(x) + f'(x)g'(x) = 0

(-2, 2)

is equal to___.

Official answer

NTA final key.

Chapter: Application of Derivatives
Topic: Rolle's Theorem
Idea tested: Auxiliary Functions: Product and Quotient Rules

Same topic in other shifts

All Rolle's Theorem questions

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24 Jun 2022, Shift 2 · Q11The number of distinct real roots of the equation $x^7 - 7x - 2 = 0$ isMediumSingle correct
31 Aug 2021, Shift 2 · Q68Let $f$ be any continuous function on $[0, 2]$ and twice differentiable on $(0, 2)$ . If $f(0) = 0$ , $f(1) = 1$ and $f(2) = 2$ , then :MediumSingle correct
25 Feb 2021, Shift 1 · Q65If Rolle's theorem holds for the function $f(x) = x^3 - ax^2 + bx - 4$ , $x \in [1, 2]$ with $f'\left(\frac{4}{3}\right) = 0$ , then ordered…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.