JEE Main 2016MathematicsDefinite IntegrationHardMCQ

JEE Main 2016 — Definite Integration Question with Solution

JEE Main 2016 (10 Apr Online)

Question

For x∈R, x≠0, if yx is a differentiable function such that x∫1xytdt=x+1 ∫1xtytdt, then yx equals (where C is a constant)

Choose an option

Show full solutionCorrect option: D
Correct answer
DCx3 e-1x

Step-by-step explanation

x∫1xytdt=x∫1xtytdt+ ∫1xtytdt

Differentiate w.r.t. x

∫1xy(t)dt+xy(x)=∫1xty(t)dt+x[xy(x)]+xy(x)

∫1xy(t)dt=∫1xty(t)dt+x2y(x) 

Differentiate again w.r.t. x

y(x)=xy(x)+2x y(x)+x2y'(x) 

1-3xyx=x2y'x

y'xyx=1-3xx2

1ydydx=1-3xx2

Integrating on both sides

⇒lny=-1x-3lnx+C

⇒ln(yx3)=-1x+C

⇒yx3= e-1x+C

⇒y=e-1x+Cx3

⇒y=Cx3e-1x

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About this question

This is a previous-year question from JEE Main 2016, 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.