JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let [ ] denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function $f (x) = [x] + ∣ x - 2∣$ , $- 2 < x < 3,$ is not continuous and not differentiable. Then $m + n$ is equal to:

Q: Wait, why do we count $x=2$ as a point of discontinuity? Isn't the modulus function continuous there?

You are absolutely right that $|x-2|$ is continuous at $x=2$. However, the function $f(x)$ is the sum of $[x]$ and $|x-2|$. Since $[x]$ has a jump discontinuity at every integer, including $x=2$, the sum of a discontinuous function and a continuous one is always discontinuous. The 'jump' from $[x]$ dominates the behavior at that point.

Q: Why is $n=4$ if the modulus function has a sharp corner at $x=2$?

That is a brilliant observation! The modulus function $|x-2|$ is indeed non-differentiable at $x=2$ because of the sharp corner. However, we already identified $x=2$ as a point of discontinuity. Since every point of discontinuity is automatically a point of non-differentiability, $x=2$ is already accounted for in our count of $n$. We don't double-count it!

Q: Can I just use the derivative formula to check for differentiability?

You could, but it would be a trap! The derivative formula assumes the function is continuous in a neighborhood. Since $[x]$ is a step function, its derivative is zero where it exists and undefined at the jumps. Using formal limits is safer, but recognizing the geometric properties of the functions is the 'JEE way' to save time and avoid errors.

Q: Does the interval $(-2, 3)$ include the endpoints?

Great catch! The notation $(-2, 3)$ denotes an open interval, meaning $x$ can be any value strictly between $-2$ and $3$. Therefore, we do not include $-2$ or $3$ in our set of points. Always double-check your boundaries—they are the most common places to lose marks in competitive exams.

Q: Is it possible for a function to be non-differentiable but continuous?

Yes, absolutely! The modulus function $|x-2|$ is the perfect example. It is continuous everywhere, but it is not differentiable at $x=2$ because the left-hand derivative is $-1$ and the right-hand derivative is $+1$. They don't match, creating that sharp 'V' shape.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the function as a sum of two distinct behaviors. The greatest integer function

[x]

creates a 'staircase' of jumps, while the modulus function

∣ x - 2∣

creates a 'V-shaped' valley. Where do these behaviors clash with the smoothness of a function?

Hint 2

Hint 2: Recall the fundamental hierarchy of calculus: a function must be continuous to be differentiable. If a function breaks its continuity at a point, can it possibly have a defined slope there? Check your list of integers in the domain

(- 2, 3)

carefully.

Hint 3

Hint 3: Start by identifying the set of integers

I = {- 1, 0, 1, 2}

within the interval

(- 2, 3)

. For each

x \in I

, evaluate the continuity of

f (x) = [x] + ∣ x - 2∣

. Then, check if there are any additional points where the derivative fails to exist, specifically looking at the vertex of the modulus function.

Visualized Solution

Defining the Function $f (x)$

Given function: $f (x) = [x] + ∣ x - 2∣$
Interval: $x \in (- 2, 3)$
Goal: Find $m$ (points of discontinuity) and $n$ (points of non-differentiability).

Discontinuity of $[x]$

The greatest integer function $[x]$ is discontinuous at all integer values.
Integers in the interval $(- 2, 3)$ are: $- 1, 0, 1, 2$ .

Continuity of $∣ x - 2∣$

The function $∣ x - 2∣$ is a modulus function.
Modulus functions are continuous for all $x \in R$ .

Calculating $m$ (Points of Discontinuity)

Property: $Discontinuous + Continuous = Discontinuous$ .
Therefore, $f (x)$ is discontinuous at $x = {- 1, 0, 1, 2}$ .
Total points of discontinuity: $m = 4$ .

Non-Differentiability: The Rule

Rule: If a function is discontinuous at $x = a$ , it is strictly non-differentiable at $x = a$ .
Points ${- 1, 0, 1, 2}$ are automatically points of non-differentiability.

Checking $x = 2$ for Modulus

Modulus functions have sharp corners where their inner expression is zero.
$∣ x - 2∣$ has a sharp corner at $x = 2$ .
Sharp corners imply non-differentiability.

Calculating $n$ (Points of Non-Differentiability)

The sharp corner at $x = 2$ is already included in our set ${- 1, 0, 1, 2}$ .
No other points of non-differentiability exist in $(- 2, 3)$ .
Total points of non-differentiability: $n = 4$ .

Final Calculation: $m + n$

Value of $m = 4$
Value of $n = 4$
Final sum: $m + n = 4 + 4 = 8$

00:00 / 00:00

The Sigma Insight: Relationship Between Continuity and Differentiability

Solution Diagram

Analyzing the Setup

We are examining the function

f (x) = [x] + ∣ x - 2∣

within the open interval

x \in (- 2, 3)

. This function is a composite of the Greatest Integer Function, which is inherently discontinuous at integers, and the Modulus Function, which is continuous everywhere but possesses a sharp corner at its vertex.

The Staircase of Discontinuity

The Greatest Integer Function

[x]

exhibits jump discontinuities at every integer value. Within the interval

(- 2, 3)

, these integer points are

x = - 1, 0, 1, 2

Because the modulus function

∣ x - 2∣

is continuous, it cannot compensate for the jumps produced by

[x]

. Consequently, the sum

f (x)

remains discontinuous at these specific points. We identify the number of points of discontinuity as

m = 4

The Sharpness of the Modulus

A function is non-differentiable at any point of discontinuity. Therefore, the points

{- 1, 0, 1, 2}

are already confirmed as points where the derivative does not exist.

We must also investigate points where the function is continuous but possesses a "sharp corner." The term

∣ x - 2∣

has a vertex at

x = 2

, where the slope changes abruptly from

- 1

1

The Synthesis

We evaluate the set of points where the function fails to be differentiable. This set includes all points of discontinuity

{- 1, 0, 1, 2}

and any additional points of non-differentiability.

Since the sharp corner at

x = 2

is already included in the set of points of discontinuity, we do not count it as a separate, unique point of non-differentiability. Thus, the total number of points of non-differentiability is

n = 4

Final Calculation

Having determined the values for the points of discontinuity and non-differentiability, we perform the final summation:

m + n = 4 + 4 = 8

The final result is 8.

Common Traps & FAQs

Wait, why do we count

x = 2

as a point of discontinuity? Isn't the modulus function continuous there?

You are absolutely right that

∣ x - 2∣

is continuous at

x = 2

. However, the function

f (x)

is the sum of

[x]

and

∣ x - 2∣

. Since

[x]

has a jump discontinuity at every integer, including

x = 2

, the sum of a discontinuous function and a continuous one is always discontinuous. The 'jump' from

[x]

dominates the behavior at that point.

Why is

n = 4

if the modulus function has a sharp corner at

x = 2

That is a brilliant observation! The modulus function

∣ x - 2∣

is indeed non-differentiable at

x = 2

because of the sharp corner. However, we already identified

x = 2

as a point of discontinuity. Since every point of discontinuity is automatically a point of non-differentiability,

x = 2

is already accounted for in our count of

n

. We don't double-count it!

Can I just use the derivative formula to check for differentiability?

You could, but it would be a trap! The derivative formula assumes the function is continuous in a neighborhood. Since

[x]

is a step function, its derivative is zero where it exists and undefined at the jumps. Using formal limits is safer, but recognizing the geometric properties of the functions is the 'JEE way' to save time and avoid errors.

Does the interval

(- 2, 3)

include the endpoints?

Great catch! The notation

(- 2, 3)

denotes an open interval, meaning

x

can be any value strictly between

- 2

and

3

. Therefore, we do not include

- 2

3

in our set of points. Always double-check your boundaries—they are the most common places to lose marks in competitive exams.

Is it possible for a function to be non-differentiable but continuous?

Yes, absolutely! The modulus function

∣ x - 2∣

is the perfect example. It is continuous everywhere, but it is not differentiable at

x = 2

because the left-hand derivative is

- 1

and the right-hand derivative is

+ 1

. They don't match, creating that sharp 'V' shape.

Select Answer:

Visualized Solution

Defining the Function $f (x)$

Discontinuity of $[x]$

Continuity of $∣ x - 2∣$

Calculating $m$ (Points of Discontinuity)

Non-Differentiability: The Rule

Checking $x = 2$ for Modulus

Calculating $n$ (Points of Non-Differentiability)

Final Calculation: $m + n$

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Staircase of Discontinuity

The Sharpness of the Modulus

The Synthesis

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Discontinuity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x]

Continuity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣x−2∣

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m (Points of Discontinuity)

Non-Differentiability: The Rule

Checking .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2 for Modulus

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n (Points of Non-Differentiability)

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Staircase of Discontinuity

The Sharpness of the Modulus

The Synthesis

Final Calculation

Similar Questions

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Calculating $n$ (Points of Non-Differentiability)

Final Calculation: $m + n$

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