JEE Main 2014MathematicsApplication of DerivativesEasyMCQ

JEE Main 2014 — Application of Derivatives Question with Solution

JEE Main 2014 (06 Apr)

Question

If x=-1 and x=2 are extreme points of fx=αlogx+βx2+x, then 

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Show full solutionCorrect option: A
Correct answer
A α = 2 ,  β = - 1 2

Step-by-step explanation

fx=αlogx+βx2+x

If x < 0

f⁡x=αlog-x+βx2+x

∴   f⁡'x=-α-x+2βx+1

If x > 0

f⁡x=αlogx+βx2+x

∴   f⁡'x=αx+2βx+1

f⁡ ′ - 1 = - α - 2 β + 1 = 0

f⁡ ′ 2 = α 2 + 4 β + 1 = 0

2 f⁡ ′ - 1 = - 2 α - 4 β + 2 = 0

and f'2=α2+4β+1=0

Adding,

- 3 α 2 + 3 = 0

∴    α = 2

∴    β = - 1 2

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About this question

This is a previous-year question from JEE Main 2014, 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.