JEE Main 2018MathematicsApplication of DerivativesMediumMCQ

JEE Main 2018 — Application of Derivatives Question with Solution

JEE Main 2018 (16 Apr Online)

Question

Let M and m be respectively the absolute maximum and the absolute minimum values of the function, fx=2x3-9x2+12x+5 in the interval [0,3] . Then M-m is equal to

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Show full solutionCorrect option: A
Correct answer
A9

Step-by-step explanation

Given fx=2x3-9x2+12x+5, x∈0,3

Differentiating w.r.t x,

⇒f′x=6x2-18x+12

⇒f′x=6x-1x-2

For maxima/minima f'x=0

⇒6x-1x-2=0⇒x=1,2

Now, f"x=12x-18

So at x=1, f"x<0

and at x=2, f"x>0

So, for absolute maxima/minima

⇒f1=10

⇒f2=9

⇒f0=5

⇒f3=14

⇒M=14

⇒m=5

∴ M-m=9

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About this question

This is a previous-year question from JEE Main 2018, 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.