JEE Main 2021MathematicsDifferentiationHardMCQ

JEE Main 2021 — Differentiation Question with Solution

JEE Main 2021 (16 Mar Shift 2)

Question

Let f:S→S where S=0,∞ be a twice differentiable function such that fx+1=xfx. If g:S→R be defined as gx=logefx, then the value of g''5-g''1 is equal to :

Choose an option

Show full solutionCorrect option: A
Correct answer
A205144

Step-by-step explanation

lnfx+1=lnxfx

lnfx+1=lnx+lnfx

⇒gx+1=lnx+gx

⇒gx+1-g(x)=lnx

⇒  g''x+1-g''x=-1x2

Put x=1,2,3,4

g''2-g''1=-112 ...1

g''3-g''2=-122 ...2

g''4-g''3=-132 ...3

g''5-g''4=-142 ...4

Add all the equations we get

g''5-g''1=-112-122-132-142

g"5-g"1=205144

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Differentiation 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 Differentiation 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.