JEE Main 2019 — Application Of Derivatives Question with Solution
From: JEE Main 2019 (Online) 11th January Morning Slot
Question
The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S = is :
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
S = {x R, x2 + 30 11x 0}
= {x R, 5 x 6}
Now f(x) = 3x3 18x2 + 27x 40
f '(x) = 9(x 1)(x 3),
which is positive in [5, 6]
f(x) increasing in [5, 6]
Hence maximum value = f(6) = 122
= {x R, 5 x 6}
Now f(x) = 3x3 18x2 + 27x 40
f '(x) = 9(x 1)(x 3),
which is positive in [5, 6]
f(x) increasing in [5, 6]
Hence maximum value = f(6) = 122
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This is a previous-year question from JEE Main 2019, covering the Application Of Derivatives 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.