JEE Main 2019MathematicsDefinite IntegrationMediumMCQ

JEE Main 2019 — Definite Integration Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

If fx=2-xcosx2+xcosx and g(x)=loge⁡x, then the value of the integral ∫-π4π4gfxdx is

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Show full solutionCorrect option: C
Correct answer
Cloge⁡1

Step-by-step explanation

Given,

fx=2-xcosx2+xcosx, g(x)=loge⁡x

⇒gfx=loge2-xcosx2+xcosx

⇒gf-x=⁡loge2-(-x)cos⁡(-x)2+(-x)cos⁡(-x)

⇒gf-x=loge2+xcosx2-xcosx

⇒gf-x=-log2-xcosx2+xcosx

⇒gf-x=–g(fx)

Hence, g(fx) is an odd function.

By using the property of definite integration, ∫-aafxdx=2∫0afxdx,  f-x=fx0,f-x=-fx, we can write
∫-π4π4g(f(x))dx=0=loge⁡1

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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.