JEE Main 2019MathematicsApplication of DerivativesMediumMCQ

JEE Main 2019 — Application of Derivatives Question with Solution

JEE Main 2019 (12 Jan Shift 2)

Question

If the function f given by fx=x3-3a-2x2+3ax+7, for some a∈R is increasing in 0, 1 and decreasing in 1, 5, then a root of the equation, fx-14x-12=0, x≠1 is :

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Show full solutionCorrect option: A
Correct answer
A7

Step-by-step explanation

At x=1, f′x=0

∴ f′x=3x2-6a-2x+3a

Now, f'1=0

⇒3-6 a-2+3a=0

⇒a=5

∴ fx-14x-12=0

⇒x3-9x2+15x-7x-12=0

⇒x-7x2-2x+1x-12=0

⇒x-7=0⇒x=7

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About this question

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.