JEE Main 2019 — Functions Question with Solution
From: JEE Main 2019 (Online) 8th April Evening Slot
Question
Let ƒ(x) = ax
(a > 0) be written as
ƒ(x) = ƒ1 (x) + ƒ2 (x), where ƒ1 (x) is an even function of ƒ2 (x) is an odd function.
Then ƒ1 (x + y) + ƒ1 (x – y) equals
ƒ(x) = ƒ1 (x) + ƒ2 (x), where ƒ1 (x) is an even function of ƒ2 (x) is an odd function.
Then ƒ1 (x + y) + ƒ1 (x – y) equals
Choose an option
Show full solutionCorrect option: A
Correct answer
A2ƒ1
(x)ƒ1
(y)
Step-by-step explanation
f(x) = ax
As f1(x) is even function then
f1(x) =
=
As f2(x) is odd function then
f2(x) =
=
Now,
ƒ1 (x + y) + ƒ1 (x – y)
=
Also ƒ1 (x)ƒ1 (y) =
=
=
ƒ1 (x + y) + ƒ1 (x – y) = 2ƒ1 (x)ƒ1 (y)
As f1(x) is even function then
f1(x) =
=
As f2(x) is odd function then
f2(x) =
=
Now,
ƒ1 (x + y) + ƒ1 (x – y)
=
Also ƒ1 (x)ƒ1 (y) =
=
=
ƒ1 (x + y) + ƒ1 (x – y) = 2ƒ1 (x)ƒ1 (y)
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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.