JEE Main 2018MathematicsDefinite IntegrationHardMCQ

JEE Main 2018 — Definite Integration Question with Solution

JEE Main 2018 (16 Apr Online)

Question

If fx=∫0xtsinx-sintdt, then

Choose an option

Show full solutionCorrect option: D
Correct answer
Df'''x+f'x=cosx-2xsinx

Step-by-step explanation

Given fx=∫0xtsinx-sintdt

⇒fx=sinx∫0xtdt-∫0xtsintdt

On differentiating, we get

⇒f'x=sinxx+cosx∫0xtdt-xsinx

⇒f'x=cosx∫0xtdt

Differentiating the above equation again, we get

⇒f''x=cosxx-sinx∫0xtdt

⇒f'''x=x-sinx+cosx-sinxx-cosx∫0xtdt

⇒f'''x+f'x=cosx-2xsinx.

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About this question

This is a previous-year question from JEE Main 2018, 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.