JEE Main 2020 — Functions Question with Solution
From: JEE Main 2020 (Online) 2nd September Evening Slot
Question
Let f : R R be a function which satisfies
f(x + y) = f(x) + f(y) x, y R. If f(1) = 2 and
g(n) = , n N then the value of n, for which g(n) = 20, is :
f(x + y) = f(x) + f(y) x, y R. If f(1) = 2 and
g(n) = , n N then the value of n, for which g(n) = 20, is :
Choose an option
Show full solutionCorrect option: C
Correct answer
C5
Step-by-step explanation
Given f(1) = 2 ;
f(x + y) = f(x) + f(y)
When x = y = 1 f(2) = 2 + 2 = 4
When x = 2, y = 1 f(3) = 4 + 2 = 6
g(n) =
= f(1) + f(2) +.........+ f(n - 1)
= 2 + 4 + 6 + ......+ 2(n - 1)
= 2
= n2 - n
Given g(n) = 20
n2– n = 20
n2 – n – 20 = 0
n = 5
f(x + y) = f(x) + f(y)
When x = y = 1 f(2) = 2 + 2 = 4
When x = 2, y = 1 f(3) = 4 + 2 = 6
g(n) =
= f(1) + f(2) +.........+ f(n - 1)
= 2 + 4 + 6 + ......+ 2(n - 1)
= 2
= n2 - n
Given g(n) = 20
n2– n = 20
n2 – n – 20 = 0
n = 5
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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.