JEE Main 2021MathematicsContinuity and DifferentiabilityHardNumerical

JEE Main 2021 — Continuity and Differentiability Question with Solution

JEE Main 2021 (20 Jul Shift 2)

Question

Let a function g:0,4→R be defined as

gx=max t3-6t2+9t-30≤t≤x,0≤x≤34-x,3<x≤4

then the number of points in the interval 0, 4 where g(x) is NOT differentiable, is _________.

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Step-by-step explanation

We have,

gx=max t3-6t2+9t-30≤t≤x,0≤x≤34-x,3<x≤4

Let

fx=x3-6x2+9x-3

⇒f'(x)=3x2-12x+9

⇒f'(x)=3(x-1)(x-3)

For critical points:

f'x=0

⇒x=1, 3

And,

f"x=6x-12

Then,

f"1=6-12=-6<0(maxima)

f"3=6>0(minima)

Hence,

gx=f(x),0≤x≤11,1≤x≤34-x,3<x≤4

Hence, gx is continuous function.

g'x=3(x-1)(x-3),0<x<10,1<x<3-1,3<x<4

Clearly, g(x) is non-differentiable at x=3.

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About this question

This is a previous-year question from JEE Main 2021, 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.