JEE Main 2019MathematicsDefinite IntegrationEasyMCQ

JEE Main 2019 — Definite Integration Question with Solution

JEE Main 2019 (09 Apr Shift 2)

Question

If f:R→R is a differentiable function and f2=6, then limx→2∫6fx2tdtx-2 is:

Choose an option

Show full solutionCorrect option: D
Correct answer
D12f'2

Step-by-step explanation

The given limit can be written as

l=limx→2∫6fx2tdtx-2  

Applying L' Hospital’s rule i.e. if l=limx→agxhx and limx→agx→0 & limx→ahx→0, then l=limx→ag'xh'x

l=limx→2ddx∫6fx2tdt1

Now, applying Newton's Leibnitz rule i.e. ddx∫abgxdx=gb·b'-ga·a', we get

⇒l=limx→22fx·f'x-0

⇒l=2f2⋅f'2

Put the given value, to get

⇒l=2×6×f'2=12f'2.

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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.