JEE Main 2021MathematicsDefinite IntegrationNewton Lebnitz Rule Of DifferentiationmediumMCQ

JEE Main 2021Definite Integration Question with Solution

From: JEE Main 2021 (Online) 27th July Evening Shift

Question

Let f : (a, b) R be twice differentiable function such that for a differentiable function g(x). If f(x) = 0 has exactly five distinct roots in (a, b), then g(x)g'(x) = 0 has at least :

Choose an option

Show full solutionCorrect option: C
Correct answer
Cseven roots in (a, b)

Step-by-step explanation



f′(x) = g(x)

f′'(x) = g'(x)

Given, g(x).g'(x) = 0

f′(x).f′'(x) = 0

Also given f(x) has exactly 5 roots.

So from Rolle's theorem we can say,

f′(x) has 4 roots and f′'(x) has 3 roots.

f′(x).f′'(x) = 0 has 4 + 3 = 7 roots.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Definite Integration chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2021, covering the Definite Integration chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.