JEE Main 2020MathematicsFunctionsHardNumerical

JEE Main 2020 — Functions Question with Solution

JEE Main 2020 (04 Sep Shift 1)

Question

Suppose a differentiable function fx satisfies the identity fx+y=fx+fy+xy2+x2y, for all real x and y. If limx→0fxx=1, then f'3 is equal to :

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Show full solutionCorrect answer: 10
Correct answer
10

Step-by-step explanation

∵limx→0fxx=1 ⇒f'0=1

fx+y=fx+fy+xy2+x2y

Differentiate w.r.t. x keeping y constant

f'x+y=f'x+0+y2+2xy

put y=-x

f'0=f'x+x2−2x2

1=f'x−x2

f'x=1+x2

f'3=10.

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About this question

This is a previous-year question from JEE Main 2020, 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.