JEE Main 2019 — Differential Equations Question with Solution
From: JEE Main 2019 (Online) 9th January Evening Slot
Question
Let f : [0,1] R be such that f(xy) = f(x).f(y), for all x, y [0, 1], and f(0) 0. If y = y(x) satiesfies the differential equation, = f(x) with y(0) = 1, then y + y is equal to :
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Show full solutionCorrect option: A
Correct answer
A3
Step-by-step explanation
If f(xy) = f(x) f(y) x, y R and f(0) 0
put x = y = 0
f(0) = [f(0)]2
f(0) = 1
put y = 0 f(0) = f(x) f(0)
f(x) = 1
given that = f(x)
= 1 y = x + k
given that y(0) = 1
k = 1
hence y = x + 1
y + y = + = 3
put x = y = 0
f(0) = [f(0)]2
f(0) = 1
put y = 0 f(0) = f(x) f(0)
f(x) = 1
given that = f(x)
= 1 y = x + k
given that y(0) = 1
k = 1
hence y = x + 1
y + y = + = 3
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This is a previous-year question from JEE Main 2019, covering the Differential Equations 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.