JEE Main 2022MathematicsDefinite IntegrationMediumMCQ

JEE Main 2022 — Definite Integration Question with Solution

JEE Main 2022 (27 Jun Shift 2)

Question

Let f be a differentiable function in 0,π2. If ∫cosx1t2ftdt=sin3x+cosx, then 13f'13 is equal to

Choose an option

Show full solutionCorrect option: C
Correct answer
C6-92

Step-by-step explanation

∫cosx1t2ftdt=sin3x+cosx

On differentiating, we get

fcosxsinx·cos2x=3sin2xcosx-sinx

fcosx=3tanx-sec2x

Again differentiating, we get

-sinxf'cosx=3sec2x-2sec2xtanx

When cosx=13 then secx=3, tanx=2 & sinx=23

Then -23f'13=3×3-2×32

⇒13f'13=6-92

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About this question

This is a previous-year question from JEE Main 2022, 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.