JEE Main 2019MathematicsLimitsEasyMCQ

JEE Main 2019 — Limits Question with Solution

JEE Main 2019 (08 Apr Shift 2)

Question

Let f:R→R be a differentiable function satisfying f'3+f'2=0. Then limx→01+f3+x-f31+f2-x-f21x is equal to

Choose an option

Show full solutionCorrect option: A
Correct answer
A1

Step-by-step explanation

This limit is 1∞ form

∴limx→01+f3+x-f31+f2-x-f21x=elimx→01x1+f3+x-f31+f2-x-f2-1

=elimx→01xf3+x-f3-f2-x+f21+f2-x-f200 form

=elimx→0 f'3+x+f'2-xx-f'2-x+1+f2-x-f2

=e0=1.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Limits 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 2019, covering the Limits 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.