JEE Main 2020MathematicsContinuity and DifferentiabilityMediumMCQ

JEE Main 2020 — Continuity and Differentiability Question with Solution

JEE Main 2020 (04 Sep Shift 2)

Question

Let f:(0,∞)→(0,∞) be a differentiable function such that f(1)=e and limt→xt2f2(x)-x2f2(t)t-x=0. If f(x)=1, then x is equal to:

Choose an option

Show full solutionCorrect option: A
Correct answer
A1e

Step-by-step explanation

limt→xt2f2(x)-x2f2(t)t-x=0
Using L'Hospital

limt→x2tf2(x)-x22f(t)f'(t)1=0
⇒2xf2(x)-x22f(x)f'(x)=0
⇒2 x f(x)[ f(x)-xf'(x)]=0
But f(x)≠0

So, xf'(x)=f(x)
⇒xdydx=y
⇒1ydy=1xdx
Integration gives

lny=lnx+lnc
⇒y=cx⇒f(x)=cx
Now

f(1)=c=e (given)

So, f(x)=e x
Now if f(x)=1, then ex=1⇒x=1e

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About this question

This is a previous-year question from JEE Main 2020, covering the Continuity and Differentiability 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.