JEE Main 2019MathematicsFunctionsMediumMCQ

JEE Main 2019 — Functions Question with Solution

JEE Main 2019 (08 Apr Shift 2)

Question

Let fx=ax (a>0) be written as fx=f1x+f2x,  where f1(x) is an even function and f2(x) is an odd function. Then f1x+y+f1(x-y) equals:

Choose an option

Show full solutionCorrect option: A
Correct answer
A2f1xf1y

Step-by-step explanation

fx=ax, a>0

fx=ax+a-x+ax-a-x2
⇒f1x=ax+a-x2
f2x=ax-a-x2

⇒f1x+y+f1x-y

=ax+y+a-x-y2+ax-y+a-x+y2
=ax+a-x2ay+a-y

=f1x×2f1y

=2f1xf1y

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About this question

This is a previous-year question from JEE Main 2019, covering the Functions 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.