JEE Main 2020 — Functions Question with Solution
From: JEE Main 2020 (Online) 7th January Morning Slot
Question
If g(x) = x2 + x - 1 and
(goƒ) (x) = 4x2 - 10x + 5, then ƒ is equal to:
(goƒ) (x) = 4x2 - 10x + 5, then ƒ is equal to:
Choose an option
Show full solutionCorrect option: C
Correct answer
C-
Step-by-step explanation
Given, (goƒ) (x) = 4x2 - 10x + 5
g(f(x)) = 4x2 - 10x + 5
g(f()) = = ...(1)
Also given, g(x) = x2 + x - 1
g(f(x)) = f2(x) + f(x) –1
g(f()) = f2() + f() –1 ....(2)
from (1) & (2)
f2() + f() –1 =
f2() + f() + = 0
(f() + )2 = 0
f() = -
g(f(x)) = 4x2 - 10x + 5
g(f()) = = ...(1)
Also given, g(x) = x2 + x - 1
g(f(x)) = f2(x) + f(x) –1
g(f()) = f2() + f() –1 ....(2)
from (1) & (2)
f2() + f() –1 =
f2() + f() + = 0
(f() + )2 = 0
f() = -
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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.