JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f (x) = ⎩ ⎨ ⎧ 3 x, min {1 + x + [x], x + 2 [x]}, 5, x < 0 0 \leq x \leq 2 x > 2,$ where [.] denotes greatest integer function. If $α$ and $β$ are the number of points, where f is not continuous and is not differentiable, respectively, then $α + β$ equals

Q: Why can't we just differentiate the function directly without checking continuity?

That is a classic trap! Differentiability is a stricter condition than continuity. A function must be continuous to even be considered for differentiability. If there is a jump or a break at a point, the derivative simply does not exist there. Always check continuity first.

Q: Is $x=0$ really non-differentiable? It looks continuous!

You are absolutely right that it is continuous at $x=0$. However, continuity is not enough for differentiability. At $x=0$, the slope from the left is $3$ and the slope from the right is $1$. Because the slopes (derivatives) are not equal, there is a sharp corner, which makes the function non-differentiable.

Q: How do I handle the $\min$ operator inside the piecewise function?

The $\min$ operator is just a way of saying 'choose the smaller value.' By breaking the function into intervals where $[x]$ is constant, the expressions inside the $\min$ become simple linear functions. You just compare them in each interval to see which one is smaller.

Enter Numerical Value:

Visualized Solution

Analyzing the Piecewise Function

The function $f (x)$ is defined in three parts.
For $x < 0$ , $f (x) = 3 x$ .
For $0 \leq x \leq 2$ , $f (x) = min {1 + x + [x], x + 2 [x]}$ .
For $x > 2$ , $f (x) = 5$ .
Goal: Find $α$ (points of discontinuity) and $β$ (points of non-differentiability).

Simplifying the Interval $[0, 1)$

Consider the interval $0 \leq x < 1$ .
Here, the greatest integer function $[x] = 0$ .
Substitute $[x] = 0$ into the middle definition:
$f (x) = min {1 + x + 0, x + 2 (0)}$
$f (x) = min {1 + x, x}$
Since $x < 1 + x$ for all $x$ , $f (x) = x$ .

Simplifying the Interval $[1, 2)$

Consider the interval $1 \leq x < 2$ .
Here, the greatest integer function $[x] = 1$ .
Substitute $[x] = 1$ into the expression:
$f (x) = min {1 + x + 1, x + 2 (1)}$
$f (x) = min {x + 2, x + 2}$
Both terms are identical, so $f (x) = x + 2$ .

Evaluating at $x = 2$ and $x > 2$

At exactly $x = 2$ , $[x] = 2$ .
$f (2) = min {1 + 2 + 2, 2 + 2 (2)} = min {5, 6} = 5$ .
For $x > 2$ , the problem directly states $f (x) = 5$ .
Combining these, for $x \geq 2$ , $f (x) = 5$ .

The Fully Simplified Piecewise Function

We can now write the fully simplified function:
$f (x) = 3 x$ for $x < 0$
$f (x) = x$ for $0 \leq x < 1$
$f (x) = x + 2$ for $1 \leq x < 2$
$f (x) = 5$ for $x \geq 2$
Now we can easily analyze continuity and differentiability.

Checking Continuity at $x = 0$

Let's check the junction point $x = 0$ .
Left Hand Limit (LHL): $lim_{x \to 0^{-}} 3 x = 0$
Right Hand Limit (RHL): $lim_{x \to 0^{+}} x = 0$
Function value: $f (0) = 0$
Since $L H L = R H L = f (0)$ , the function is continuous at $x = 0$ .

Checking Continuity at $x = 1$

Let's check the junction point $x = 1$ .
Left Hand Limit (LHL): $lim_{x \to 1^{-}} x = 1$
Right Hand Limit (RHL): $lim_{x \to 1^{+}} (x + 2) = 3$
Since $L H L \neq = R H L$ , there is a jump in the graph.
The function is discontinuous at $x = 1$ .

Checking Continuity at $x = 2$

Let's check the final junction point $x = 2$ .
Left Hand Limit (LHL): $lim_{x \to 2^{-}} (x + 2) = 4$
Right Hand Limit (RHL): $lim_{x \to 2^{+}} 5 = 5$
Since $L H L \neq = R H L$ , there is another jump.
The function is discontinuous at $x = 2$ .
Total points of discontinuity: $α = 2$ .

Checking Differentiability at $x = 0$

A function must be continuous to be differentiable.
At $x = 1$ and $x = 2$ , it is discontinuous, so it is not differentiable there.
At $x = 0$ , it is continuous. Let's check the derivatives.
Left Hand Derivative (LHD): $\frac{d}{d x} (3 x) = 3$
Right Hand Derivative (RHD): $\frac{d}{d x} (x) = 1$
Since $L H D \neq = R H D$ , there is a sharp corner.
The function is not differentiable at $x = 0$ .

Final Calculation of $α + β$

Points of discontinuity: $x = 1, 2$ .
Therefore, $α = 2$ .
Points of non-differentiability: $x = 0, 1, 2$ .
Therefore, $β = 3$ .
We need to find $α + β$ .
$α + β = 2 + 3 = 5$ .

00:00 / 00:00

The Sigma Insight: Relationship Between Continuity and Differentiability

Solution Diagram

Analyzing the Setup

The function

f (x)

is defined piecewise, involving the greatest integer function

[x]

and the

min

operator. To understand its behavior, we must peel back the layers of the function interval by interval.

Phase 1

The Deconstruction

For the interval

0 \leq x < 1

, the greatest integer function

[x]

is always

0

. The function simplifies to:

f (x) = min {1 + x + 0, x + 2 (0)} = min {1 + x, x}

Since

x < 1 + x

for all

x

, the function simplifies to

f (x) = x

For the interval

1 \leq x < 2

, we have

[x] = 1

. The function becomes:

f (x) = min {1 + x + 1, x + 2 (1)} = min {x + 2, x + 2}

Because both expressions are identical, the function simplifies to

f (x) = x + 2

x = 2

, we have

[x] = 2

, yielding:

f (2) = min {1 + 2 + 2, 2 + 2 (2)} = min {5, 6} = 5

For

x > 2

, the function is defined as

5

. Thus, for

x \geq 2

f (x) = 5

Phase 2

The Junctions

Now we analyze the continuity at the junction points

x = 0

x = 1

, and

x = 2

x = 0

, the left-hand limit is

lim_{x \to 0^{-}} 3 x = 0

and the right-hand limit is

lim_{x \to 0^{+}} x = 0

. Since these match

f (0) = 0

, the function is continuous at

x = 0

x = 1

, the left-hand limit is

lim_{x \to 1^{-}} x = 1

, while the right-hand limit is

lim_{x \to 1^{+}} (x + 2) = 3

. Since $1 eq 3$, the function is discontinuous at

x = 1

x = 2

, the left-hand limit is

lim_{x \to 2^{-}} (x + 2) = 4

, while the right-hand limit is

lim_{x \to 2^{+}} 5 = 5

. Since $4 eq 5$, the function is discontinuous at

x = 2

Consequently, we have two points of discontinuity, so

α = 2

Phase 3

The Sharp Corners

Any point of discontinuity is automatically a point of non-differentiability. Therefore,

x = 1

and

x = 2

are already confirmed as non-differentiable.

We must check

x = 0

for differentiability. The left-hand derivative is:

\frac{d}{d x} (3 x) = 3

The right-hand derivative is:

\frac{d}{d x} (x) = 1

Since $3 eq 1$, there is a sharp corner at

x = 0

The function is not differentiable at

x = 0, 1, 2

, which gives us

β = 3

. The final sum is:

α + β = 2 + 3 = 5

Common Traps & FAQs

Why can't we just differentiate the function directly without checking continuity?

That is a classic trap! Differentiability is a stricter condition than continuity. A function must be continuous to even be considered for differentiability. If there is a jump or a break at a point, the derivative simply does not exist there. Always check continuity first.

x = 0

really non-differentiable? It looks continuous!

You are absolutely right that it is continuous at

x = 0

. However, continuity is not enough for differentiability. At

x = 0

, the slope from the left is

3

and the slope from the right is

1

. Because the slopes (derivatives) are not equal, there is a sharp corner, which makes the function non-differentiable.

How do I handle the

min

operator inside the piecewise function?

The

min

operator is just a way of saying 'choose the smaller value.' By breaking the function into intervals where

[x]

is constant, the expressions inside the

min

become simple linear functions. You just compare them in each interval to see which one is smaller.

Enter Numerical Value:

Visualized Solution

Analyzing the Piecewise Function

Simplifying the Interval $[0, 1)$

Simplifying the Interval $[1, 2)$

Evaluating at $x = 2$ and $x > 2$

The Fully Simplified Piecewise Function

Checking Continuity at $x = 0$

Checking Continuity at $x = 1$

Checking Continuity at $x = 2$

Checking Differentiability at $x = 0$

Final Calculation of $α + β$

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

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:

Let $m$ and $n$ be the number of points at which the function $f (x) = max {x, x^{3}, x^{5}, \dots, x^{21}}, x \in R$ , is not differentiable and not continuous, respectively. Then $m + n$ is equal to ________.

Let $f (x) = ∣2 x^{2} + 5∣ x ∣ - 3∣, x \in R$ . If $m$ and $n$ denote the number of points where $f$ is not continuous and not differentiable respectively, then $m + n$ is equal to :

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Consider the function $f : (0, \infty) \to R$ defined by $f (x) = e^{- ∣ l o g_{6} x ∣}$ . If $m$ and $n$ be respectively the number of points at which $f$ is not continuous and $f$ is not differentiable, then $m + n$ is

Enter Numerical Value:

Visualized Solution

Analyzing the Piecewise Function

Simplifying the Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [0,1)

Simplifying the Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [1,2)

Evaluating at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>2

The Fully Simplified Piecewise Function

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

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

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

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

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

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=∣2x2+5∣x∣−3∣,x∈R. If m and n denote the number of points where f is not continuous and not differentiable respectively, then m+n is equal to :

Simplifying the Interval $[0, 1)$

Simplifying the Interval $[1, 2)$

Evaluating at $x = 2$ and $x > 2$

Checking Continuity at $x = 0$

Checking Continuity at $x = 1$

Checking Continuity at $x = 2$

Checking Differentiability at $x = 0$

Final Calculation of $α + β$

Let $f (x) = ∣2 x^{2} + 5∣ x ∣ - 3∣, x \in R$ . If $m$ and $n$ denote the number of points where $f$ is not continuous and not differentiable respectively, then $m + n$ is equal to :