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Limits Continuity and Differentiability question

2023 · 15 Apr · Shift 1 · Q29
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  5. /2023 · 15 Apr · Shift 1 · Q29

Limits Continuity and Differentiability question

2023 · 15 Apr · Shift 1 · Q29

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let [x][x][x] denote the greatest integer function and f(x)=max⁡{1+x+[x],2+x,x+2[x]},0≤x≤2f(x)=\max \{1+x+[x], 2+x, x+2[x]\}, 0 \leq x \leq 2f(x)=max{1+x+[x],2+x,x+2[x]},0≤x≤2. Let mmm be the number of points in [0,2][0,2][0,2], where fff is not continuous and nnn be the number of points in (0,2)(0,2)(0,2), where fff is not differentiable. Then (m+n)2+2(m+n)^{2}+2(m+n)2+2 is equal to :
  1. A
    3
  2. B
    6
  3. C
    2
  4. D
    11
View written solutionFree

Correct answer: A

  1. Given function

We have

f(x)=max⁡{1+x+[x],  2+x,  x+2[x]},0≤x≤2.f(x)=\max\{1+x+[x],\;2+x,\;x+2[x]\}, \qquad 0\le x\le 2.f(x)=max{1+x+[x],2+x,x+2[x]},0≤x≤2.

Let

A(x)=1+x+[x],B(x)=2+x,C(x)=x+2[x].A(x)=1+x+[x],\quad B(x)=2+x,\quad C(x)=x+2[x].A(x)=1+x+[x],B(x)=2+x,C(x)=x+2[x].

Since [x][x][x] changes at integers, we analyze on intervals:

  • 0≤x<10\le x<10≤x<1
  • 1≤x<21\le x<21≤x<2
  • x=2x=2x=2
  1. For 0≤x<10\le x<10≤x<1

Here [x]=0[x]=0[x]=0. So

A(x)=1+x,B(x)=2+x,C(x)=x.A(x)=1+x,\quad B(x)=2+x,\quad C(x)=x.A(x)=1+x,B(x)=2+x,C(x)=x.

Clearly,

2+x>1+x>x.2+x > 1+x > x.2+x>1+x>x.

Hence,

f(x)=2+x(0≤x<1).f(x)=2+x \qquad (0\le x<1).f(x)=2+x(0≤x<1).
  1. For 1≤x<21\le x<21≤x<2

Here [x]=1[x]=1[x]=1. So

A(x)=1+x+1=x+2,A(x)=1+x+1=x+2,A(x)=1+x+1=x+2, B(x)=x+2,B(x)=x+2,B(x)=x+2, C(x)=x+2.C(x)=x+2.C(x)=x+2.

Thus all three are equal, and

f(x)=x+2(1≤x<2).f(x)=x+2 \qquad (1\le x<2).f(x)=x+2(1≤x<2).
  1. At x=2x=2x=2

Here [2]=2[2]=2[2]=2. So

A(2)=1+2+2=5,A(2)=1+2+2=5,A(2)=1+2+2=5, B(2)=2+2=4,B(2)=2+2=4,B(2)=2+2=4, C(2)=2+2⋅2=6.C(2)=2+2\cdot 2=6.C(2)=2+2⋅2=6.

Therefore,

f(2)=6.f(2)=6.f(2)=6.

But for x<2x<2x<2 close to 222, we had f(x)=x+2f(x)=x+2f(x)=x+2, so

lim⁡x→2−f(x)=4.\lim_{x\to 2^-} f(x)=4.x→2−lim​f(x)=4.

Since f(2)=6≠4f(2)=6\ne 4f(2)=6=4, fff is not continuous at x=2x=2x=2.

  1. Check continuity at x=1x=1x=1

For x<1x<1x<1, f(x)=x+2f(x)=x+2f(x)=x+2, so

lim⁡x→1−f(x)=3.\lim_{x\to 1^-} f(x)=3.x→1−lim​f(x)=3.

For x≥1x\ge 1x≥1, f(x)=x+2f(x)=x+2f(x)=x+2, so

lim⁡x→1+f(x)=3,\lim_{x\to 1^+} f(x)=3,x→1+lim​f(x)=3,

and

f(1)=3.f(1)=3.f(1)=3.

Thus fff is continuous at x=1x=1x=1.

Also on (0,1)(0,1)(0,1) and (1,2)(1,2)(1,2), f(x)=x+2f(x)=x+2f(x)=x+2, which is continuous.

Hence the number of points in [0,2][0,2][0,2] where fff is not continuous is

m=1.m=1.m=1.
  1. Check differentiability in (0,2)(0,2)(0,2)

On both intervals (0,1)(0,1)(0,1) and (1,2)(1,2)(1,2),

f(x)=x+2,f(x)=x+2,f(x)=x+2,

so

f′(x)=1.f'(x)=1.f′(x)=1.

At x=1x=1x=1,

  • left derivative =1=1=1
  • right derivative =1=1=1

So fff is differentiable at x=1x=1x=1 as well.

Thus there is no point in (0,2)(0,2)(0,2) where fff is not differentiable. Therefore,

n=0.n=0.n=0.
  1. Compute required value
(m+n)2+2=(1+0)2+2=1+2=3.(m+n)^2+2=(1+0)^2+2=1+2=3.(m+n)2+2=(1+0)2+2=1+2=3.

So the correct option is A.\boxed{\text{A}}.A​.

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