JEE Main 2019MathematicsDefinite IntegrationEasyMCQ

JEE Main 2019 — Definite Integration Question with Solution

JEE Main 2019 (12 Apr Shift 1)

Question

Let f:R→R be a continuous and differentiable function such that f2=6 and f'2=148. If ∫6f(x)4t3dt=x-2gx, then limx→2gx is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
B18

Step-by-step explanation

limx→2gx=limx→2∫6fx4t3dtx-2      00 form.

By using L'Hospital rule
=limx→24.f3(x)⋅f'x1
=4f32f'2=4×6×6×6×148=18.

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About this question

This is a previous-year question from JEE Main 2019, 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.