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Limits Continuity and Differentiability question

2025 · 4 Apr · Shift 1 · Q47
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Limits Continuity and Differentiability question

2025 · 4 Apr · Shift 1 · Q47

JEE MainMathematicsLimits Continuity and DifferentiabilityNumerical+4 / −1
Let mmm and nnn be the number of points at which the function f(x)=max⁡{x,x3,x5,…x21},x∈Rf(x)=\max \left\{x, x^3, x^5, \ldots x^{21}\right\}, x \in \mathbb{R}f(x)=max{x,x3,x5,…x21},x∈R, is not differentiable and not continuous, respectively. Then m+nm+nm+n is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 3

  1. Given function

We have

f(x)=max⁡{x,x3,x5,…,x21},x∈R.f(x)=\max\{x,x^3,x^5,\dots,x^{21}\},\qquad x\in\mathbb R.f(x)=max{x,x3,x5,…,x21},x∈R.

Let

  • mmm = number of points where fff is not differentiable,
  • nnn = number of points where fff is not continuous.

We must find m+nm+nm+n.

  1. Continuity of fff

Each function x,x3,x5,…,x21x, x^3, x^5, \dots, x^{21}x,x3,x5,…,x21 is a polynomial, hence continuous on R\mathbb RR.

The maximum of finitely many continuous functions is also continuous. Therefore, f(x) is continuous for all x∈R.f(x) \text{ is continuous for all } x\in\mathbb R.f(x) is continuous for all x∈R.

So, n=0.n=0.n=0.

  1. Determine which power is maximum in different regions

We compare the odd powers for different values of xxx.

Case 1: x>1x>1x>1

For x>1x>1x>1, higher powers are larger: x<x3<x5<⋯<x21.x<x^3<x^5<\cdots <x^{21}.x<x3<x5<⋯<x21. Hence, f(x)=x21(x>1).f(x)=x^{21} \qquad (x>1).f(x)=x21(x>1).

Case 2: 0<x<10<x<10<x<1

For 0<x<10<x<10<x<1, higher powers are smaller: x>x3>x5>⋯>x21.x>x^3>x^5>\cdots >x^{21}.x>x3>x5>⋯>x21. Hence, f(x)=x(0<x<1).f(x)=x \qquad (0<x<1).f(x)=x(0<x<1).

Case 3: −1<x<0-1<x<0−1<x<0

Let x=−ax=-ax=−a where 0<a<10<a<10<a<1. Then x2k+1=−a2k+1.x^{2k+1}=-a^{2k+1}.x2k+1=−a2k+1. Since a2k+1a^{2k+1}a2k+1 decreases as power increases, −a<−a3<−a5<⋯<−a21.-a < -a^3 < -a^5 < \cdots < -a^{21}.−a<−a3<−a5<⋯<−a21. So the maximum is the least negative term, i.e. f(x)=x21(−1<x<0).f(x)=x^{21} \qquad (-1<x<0).f(x)=x21(−1<x<0).

Case 4: x<−1x<-1x<−1

Let x=−ax=-ax=−a where a>1a>1a>1. Then x=−a, x3=−a3,…x=-a,\ x^3=-a^3,\dotsx=−a, x3=−a3,… and since a2k+1a^{2k+1}a2k+1 increases with power, −a>−a3>−a5>⋯>−a21.-a > -a^3 > -a^5 > \cdots > -a^{21}.−a>−a3>−a5>⋯>−a21. Thus, f(x)=x(x<−1).f(x)=x \qquad (x<-1).f(x)=x(x<−1).

  1. Check boundary points

The switching can occur only at points where two or more expressions are equal. Clearly, this happens at x=−1, 0, 1.x=-1,\ 0,\ 1.x=−1, 0, 1.

Indeed:

  • At x=1x=1x=1, all odd powers equal 111.
  • At x=−1x=-1x=−1, all odd powers equal −1-1−1.
  • At x=0x=0x=0, all odd powers equal 000.

Thus,

f(x)={x,x<−1,x21,−1<x<0,x,0<x<1,x21,x>1.f(x)= \begin{cases} x, & x<-1,\\[4pt] x^{21}, & -1<x<0,\\[4pt] x, & 0<x<1,\\[4pt] x^{21}, & x>1. \end{cases}f(x)=⎩⎨⎧​x,x21,x,x21,​x<−1,−1<x<0,0<x<1,x>1.​

And continuity already holds at x=−1,0,1x=-1,0,1x=−1,0,1.

  1. Check differentiability at the switching points

We compare left-hand and right-hand derivatives.

At x=−1x=-1x=−1

  • Left piece: f(x)=x⇒f−′(−1)=1f(x)=x \Rightarrow f'_-( -1)=1f(x)=x⇒f−′​(−1)=1
  • Right piece: f(x)=x21⇒f+′(−1)=21(−1)20=21f(x)=x^{21} \Rightarrow f'_+( -1)=21(-1)^{20}=21f(x)=x21⇒f+′​(−1)=21(−1)20=21

Since 1≠211\ne 211=21, fff is not differentiable at x=−1x=-1x=−1.

At x=0x=0x=0

  • Left piece: f(x)=x21⇒f−′(0)=0f(x)=x^{21} \Rightarrow f'_-(0)=0f(x)=x21⇒f−′​(0)=0
  • Right piece: f(x)=x⇒f+′(0)=1f(x)=x \Rightarrow f'_+(0)=1f(x)=x⇒f+′​(0)=1

Since 0≠10\ne 10=1, fff is not differentiable at x=0x=0x=0.

At x=1x=1x=1

  • Left piece: f(x)=x⇒f−′(1)=1f(x)=x \Rightarrow f'_-(1)=1f(x)=x⇒f−′​(1)=1
  • Right piece: f(x)=x21⇒f+′(1)=21f(x)=x^{21} \Rightarrow f'_+(1)=21f(x)=x21⇒f+′​(1)=21

Since 1≠211\ne 211=21, fff is not differentiable at x=1x=1x=1.

Hence, m=3.m=3.m=3.

  1. Compute m+nm+nm+n

m+n=3+0=3.m+n=3+0=3.m+n=3+0=3.

  1. Comparison with stored answer

Stored correct answer = 333.

Our derived answer also is 333, so they agree.

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