JEE Main 2021 — Application Of Derivatives Question with Solution
From: JEE Main 2021 (Online) 24th February Evening Shift
Question
Let be defined as
f(x) = \left\{ {\matrix{ { - 55x,} & {if\,x < - 5} \cr {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr } } \right.
Let A = {x R : f is increasing}. Then A is equal to :
f(x) = \left\{ {\matrix{ { - 55x,} & {if\,x < - 5} \cr {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr } } \right.
Let A = {x R : f is increasing}. Then A is equal to :
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
f(x) = \left\{ {\matrix{
{ - 55x,} & {if\,x < - 5} \cr
{2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr
{2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr
} } \right.
Now, f'(x) = \left\{ {\matrix{ { - 55} & ; & {x < - 5} \cr {6({x^2} - x - 20)} & ; & { - 5 < x < 4} \cr {6({x^2} - x - 6)} & ; & {x > 4} \cr } } \right.
f'(x) = \left\{ {\matrix{ { - 55} & ; & {x < - 5} \cr {6(x - 5)(x + 4)} & ; & { - 5 < x < 4} \cr {6(x - 3)(x + 2)} & ; & {x > 4} \cr } } \right.
Hence, f(x) is monotonically increasing in interval
Now, f'(x) = \left\{ {\matrix{ { - 55} & ; & {x < - 5} \cr {6({x^2} - x - 20)} & ; & { - 5 < x < 4} \cr {6({x^2} - x - 6)} & ; & {x > 4} \cr } } \right.
f'(x) = \left\{ {\matrix{ { - 55} & ; & {x < - 5} \cr {6(x - 5)(x + 4)} & ; & { - 5 < x < 4} \cr {6(x - 3)(x + 2)} & ; & {x > 4} \cr } } \right.
Hence, f(x) is monotonically increasing in interval
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Application Of Derivatives chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2021, 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.