JEE Main 2021MathematicsLimitsMediumMCQ

JEE Main 2021 — Limits Question with Solution

JEE Main 2021 (26 Feb Shift 2)

Question

Let fx be a differentiable function at x=a with f'a=2 and fa=4. Then limx→axfa-afxx-a equals:

Choose an option

Show full solutionCorrect option: C
Correct answer
C4-2a

Step-by-step explanation

f'a=2,fa=4

limx→axfa-afxx-a

⇒limx→afa-af'x1 (L'Hospitals rule)

=fa-af'a

=4-2a

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About this question

This is a previous-year question from JEE Main 2021, 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.