JEE Main 2018MathematicsApplication of DerivativesMediumMCQ

JEE Main 2018 — Application of Derivatives Question with Solution

JEE Main 2018 (08 Apr)

Question

Let fx=x2+1x2 and gx=x-1x, x∈R--1, 0, 1. If hx=fxgx , then the local minimum value of hx is:
 

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Show full solutionCorrect option: A
Correct answer
A22

Step-by-step explanation

fx=x2+1x2=x-1x2+2

gx=x-1x

x∈R--1,0,1

Let x-1x=t

hx=fxgx=t2+2t=t+2t

t∈R-0

hx=t+2t

AM≥GM

Minimum of hx⇒ t+2t2 ≥ t×2t

t+2t≥22

Local minimum =22

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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.