MCQMediumJEE 2023Continuity

JEE Mathematics 2023 Question with Solution

Let [x][x] denote the greatest integer function and f(x)=max{1+x+[x],2+x,x+2[x]},0x2.f(x)=\max\{1+x+[x],\,2+x,\,x+2[x]\},\quad 0\le x\le 2. Let mm be the number of points in [0,2][0,2], where ff is not continuous and nn be the number of points in (0,2)(0,2), where ff is not differentiable. Then (m+n)2+2(m+n)^2+2 is equal to

  • A

    22

  • B

    33

  • C

    66

  • D

    1111

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: f(x)=max{1+x+[x],2+x,x+2[x]}f(x)=\max\{1+x+[x],\,2+x,\,x+2[x]\} for 0x20\le x\le 2.

Find: The value of (m+n)2+2(m+n)^2+2, where mm is the number of discontinuity points in [0,2][0,2] and nn is the number of non-differentiable points in (0,2)(0,2).

The greatest integer function [x][x] changes value at integers, so split the interval at x=1x=1 and check the endpoint values separately.

For 0x<10\le x<1, [x]=0[x]=0. Then

f(x)=max{1+x,2+x,x}=2+xf(x)=\max\{1+x,\,2+x,\,x\}=2+x

For 1x<21\le x<2, [x]=1[x]=1. Then

f(x)=max{2+x,2+x,x+2}=2+xf(x)=\max\{2+x,\,2+x,\,x+2\}=2+x

Thus, on both open intervals,

f(x)=2+xf(x)=2+x

Now check continuity at the junction points.

At x=0x=0,

limx0+f(x)=2,f(0)=max{1,2,0}=2\lim_{x\to 0^+} f(x)=2,\quad f(0)=\max\{1,2,0\}=2

So ff is continuous at x=0x=0.

At x=1x=1,

limx1f(x)=3,limx1+f(x)=3,f(1)=3\lim_{x\to 1^-} f(x)=3,\quad \lim_{x\to 1^+} f(x)=3,\quad f(1)=3

So ff is continuous at x=1x=1.

At x=2x=2,

limx2f(x)=4,f(2)=max{1+2+2,4,2+4}=6\lim_{x\to 2^-} f(x)=4,\quad f(2)=\max\{1+2+2,\,4,\,2+4\}=6

Since the left limit is not equal to the function value, ff is not continuous at x=2x=2.

Hence,

m=1m=1

For differentiability in (0,2)(0,2), the reduced form is f(x)=2+xf(x)=2+x on both sides of x=1x=1, so the derivative is the same on each side. Therefore ff is differentiable everywhere in (0,2)(0,2).

Hence,

n=0n=0

Now compute

(m+n)2+2=(1+0)2+2=3(m+n)^2+2=(1+0)^2+2=3

Therefore, the correct option is B.

Piecewise Reduction Trick

Given: The function contains both max\max and [x][x].

Find: The value of (m+n)2+2(m+n)^2+2.

The quickest approach is to first note that [x][x] is constant on [0,1)[0,1) and [1,2)[1,2).

  • On [0,1)[0,1), [x]=0[x]=0, so the three expressions become 1+x1+x, 2+x2+x, and xx. Their maximum is 2+x2+x.
  • On [1,2)[1,2), [x]=1[x]=1, so the three expressions become 2+x2+x, 2+x2+x, and x+2x+2. Again the maximum is 2+x2+x.

So inside (0,2)(0,2), the function is just

f(x)=2+xf(x)=2+x

This immediately shows there is no non-differentiable point in (0,2)(0,2), so n=0n=0.

Only endpoints and the jump point of [x][x] need checking for continuity: x=0,1,2x=0,1,2. At x=0x=0 and x=1x=1, the function matches the limit. At x=2x=2,

limx2f(x)=4,f(2)=6\lim_{x\to 2^-} f(x)=4,\quad f(2)=6

So only one discontinuity occurs, giving m=1m=1.

Therefore,

(m+n)2+2=(1+0)2+2=3(m+n)^2+2=(1+0)^2+2=3

So the correct option is B.](streamdown:incomplete-link)

Common mistakes

  • Checking only the open intervals and forgetting to test the endpoint x=2x=2. This is wrong because mm counts discontinuity points in [0,2][0,2], which includes endpoints. You must compare f(2)f(2) with the left-hand limit at x=2x=2.

  • Assuming that a greatest integer function automatically makes the whole function discontinuous at every integer. This is wrong because the surrounding max\max expression can simplify so that the final function remains continuous at some integers. First reduce the expression piecewise, then test continuity.

  • Concluding that ff is not differentiable at x=1x=1 only because [x][x] changes there. This is wrong because on both sides of x=1x=1 the simplified form is the same linear function 2+x2+x. Compare the left and right derivatives of the reduced function, not the unsimplified pieces alone.

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