JEE Main 2023MathematicsFunctionsMediumMCQ

JEE Main 2023 — Functions Question with Solution

JEE Main 2023 (29 Jan Shift 2)

Question

Consider a function f:ℕ→ℝ, satisfying f1+2f2+3f3+…+xfx=xx+1fx ;x≥2 with f1=1. Then 1f2022+1f2028 is equal to

 

Choose an option

Show full solutionCorrect option: D
Correct answer
D8100

Step-by-step explanation

Given:

f1+2f2+3f3+…+xfx=xx+1fx

⇒f1+2f2+3f3+…+x-1fx-1=xx-1fx-1

Now when x=2, then

f1+2f2=6f2

⇒1=4f2

⇒f2=14

When x=3, 

f1+2f2=9f3⇒f3=16

When x=4, f1+2f2+3f3=16f4⇒f4=18

So,

fx=12x

Hence, 

1f2022+1f2028=4044+4056=8100

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About this question

This is a previous-year question from JEE Main 2023, 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.