JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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 ________.

Q: Why does the 'winning' function change at $x=1$ and $x=-1$?

Think of it as a transition of power. At $x=1$, for $x > 1$, $x^{21}$ is clearly larger than $x$. But for $x < 1$, $x$ is larger than $x^{21}$. The intersection points are where the curves cross, and that is exactly where the 'winner' of the max function switches, creating a potential sharp corner.

Q: Why is $x=0$ a point of non-differentiability?

Let's look at the slopes. For $x$ just slightly less than $0$, the function is $x^{21}$, and its derivative is $21x^{20}$, which approaches $0$ as $x \to 0$. For $x$ just slightly greater than $0$, the function is $x$, and its derivative is $1$. Since $0 \neq 1$, the slopes do not match, creating a sharp turn at the origin.

Enter Numerical Value:

Visualized Solution

Understanding the Function $f (x)$

Given function: $f (x) = max {x, x^{3}, x^{5}, \dots, x^{21}}$ for $x \in R$ .
The set contains odd powers of $x$ : $g_{k} (x) = x^{k}$ where $k \in {1, 3, 5, \dots, 21}$ .
We need to find $m$ (points of non-differentiability) and $n$ (points of non-continuity).

Analyzing Continuity ( $n$ )

Each function $g_{k} (x) = x^{k}$ is a polynomial and thus continuous on $R$ .
The maximum of a finite set of continuous functions is always continuous.
Therefore, $f (x)$ is continuous for all $x \in R$ .
Conclusion: $n = 0$ .

Finding Critical Intersection Points

Intersection points occur where $x^{k} = x^{j}$ for $k \neq = j$ .
Solving $x = x^{3} = x^{5} = \dots$ gives $x \in {- 1, 0, 1}$ .
These are the potential points of non-differentiability where the 'winning' function changes.

Region 1: $x > 1$

For $x > 1$ , higher powers grow faster.
$x^{21} > x^{19} > \dots > x^{3} > x$ .
Thus, $f (x) = x^{21}$ for $x \in (1, \infty)$ .

Region 2: $0 < x < 1$

For $0 < x < 1$ , higher powers result in smaller values.
$x > x^{3} > x^{5} > \dots > x^{21}$ .
Thus, $f (x) = x$ for $x \in (0, 1)$ .

Region 3: $- 1 < x < 0$

For $- 1 < x < 0$ , all $x^{k}$ are negative.
The value closest to $0$ is the maximum.
Since $∣ x^{21} ∣ < ∣ x^{19} ∣ < \dots < ∣ x ∣$ , then $x^{21} > x^{19} > \dots > x$ .
Thus, $f (x) = x^{21}$ for $x \in (- 1, 0)$ .

Region 4: $x < - 1$

For $x < - 1$ , all $x^{k}$ are negative and $∣ x ∣ > 1$ .
Higher powers result in larger magnitudes, making them 'more negative'.
$x > x^{3} > x^{5} > \dots > x^{21}$ .
Thus, $f (x) = x$ for $x \in (- \infty, - 1)$ .

Piecewise Definition Summary

The piecewise definition of $f (x)$ is:
$f (x) = ⎩ ⎨ ⎧ x x^{21} x x^{21} x \leq - 1 - 1 < x \leq 0 0 < x \leq 1 x > 1$

Checking Differentiability at $x = 1$ and $x = - 1$

At $x = 1$ : $L H D = \frac{d}{d x} (x) = 1$ , $R H D = \frac{d}{d x} (x^{21}) = 21 (1)^{20} = 21$ .
At $x = - 1$ : $L H D = \frac{d}{d x} (x) = 1$ , $R H D = \frac{d}{d x} (x^{21}) = 21 (- 1)^{20} = 21$ .
Since $L H D \neq = R H D$ at both points, $f (x)$ is not differentiable at $x = 1$ and $x = - 1$ .

Checking Differentiability at $x = 0$

At $x = 0$ : $L H D = \frac{d}{d x} (x^{21}) ∣_{x = 0} = 0$ .
At $x = 0$ : $R H D = \frac{d}{d x} (x) ∣_{x = 0} = 1$ .
Since $0 \neq = 1$ , $f (x)$ is not differentiable at $x = 0$ .

Final Calculation

Number of points of non-continuity ( $n$ ) = $0$ .
Number of points of non-differentiability ( $m$ ) = $3$ (at $x = - 1, 0, 1$ ).
Final sum: $m + n = 3 + 0 = 3$ .
Final Answer: 3

00:00 / 00:00

The Sigma Insight: Relationship Between Continuity and Differentiability

Solution Diagram

Analyzing the Function Structure

We are examining the function defined as

f (x) = max {x, x^{3}, x^{5}, \dots, x^{21}}

. This function represents the upper envelope of a finite set of power functions.

Because each individual component

g_{k} (x) = x^{k}

is a polynomial, every component is continuous for all

x \in R

. A fundamental theorem in calculus states that the maximum of a finite set of continuous functions is itself continuous.

Therefore,

f (x)

is continuous everywhere on the real line. This implies that the number of points of discontinuity,

n

, is equal to

0

Identifying Critical Junctions

The "winner" of the competition—the function that defines

f (x)

—changes only at the intersection points of these curves. We find these points by solving the equality:

x = x^{3} = x^{5} = \dots = x^{21}

Solving this equation yields the critical junctions at

x = - 1

x = 0

, and

x = 1

. These are the only candidates for points where the function may fail to be differentiable.

Defining the Piecewise Behavior

To determine the behavior of

f (x)

, we analyze the magnitude of the powers in different intervals:

For

x > 1

, higher powers grow faster, so

f (x) = x^{21}

For

0 < x < 1

, raising a fraction to a higher power results in a smaller value, so

f (x) = x

For

- 1 < x < 0

, all powers are negative. The value closest to

0

is the maximum; since

∣ x^{21} ∣ < ∣ x ∣

, we have

f (x) = x^{21}

For

x < - 1

, the magnitude of

x^{21}

is larger than

x

, but since both are negative, the value closer to

0

is the maximum. Thus,

f (x) = x

Testing Differentiability

We now check the points of intersection for sharp turns:

x = 1

, the function switches from

x

x^{21}

. The left-hand derivative is

1

, while the right-hand derivative is:

\frac{d}{d x} (x^{21})_{x = 1} = 21 (1)^{20} = 21

Since $1 eq 21$, the function is not differentiable at

x = 1

x = - 1

, the function switches from

x^{21}

x

. Similar to the previous case, the derivatives do not match, confirming non-differentiability.

x = 0

, the function switches from

x^{21}

x

. The left-hand derivative is

21 (0)^{20} = 0

, and the right-hand derivative is

1

. Since $0 eq 1$, the function is not differentiable at

x = 0

Final Calculation

We have identified exactly

3

points of non-differentiability (

m = 3

) and

0

points of discontinuity (

n = 0

The final result is $3$ .

Common Traps & FAQs

Wait, why is

n = 0

? Isn't the max function usually discontinuous?

That is a great intuition, but remember the theorem: the maximum of a finite set of continuous functions is always continuous. Since each

g_{k} (x) = x^{k}

is a polynomial, and polynomials are continuous everywhere on

R

, their maximum must also be continuous. There are no jumps or breaks here!

Why does the 'winning' function change at

x = 1

and

x = - 1

Think of it as a transition of power. At

x = 1

, for

x > 1

x^{21}

is clearly larger than

x

. But for

x < 1

x

is larger than

x^{21}

. The intersection points are where the curves cross, and that is exactly where the 'winner' of the max function switches, creating a potential sharp corner.

Why is

x = 0

a point of non-differentiability?

Let's look at the slopes. For

x

just slightly less than

0

, the function is

x^{21}

, and its derivative is

21 x^{20}

, which approaches

0

x \to 0

. For

x

just slightly greater than

0

, the function is

x

, and its derivative is

1

. Since

0 \neq = 1

, the slopes do not match, creating a sharp turn at the origin.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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 ________.

Enter Numerical Value:

Visualized Solution

Understanding the Function $f (x)$

Analyzing Continuity ( $n$ )

Finding Critical Intersection Points

Region 1: $x > 1$

Region 2: $0 < x < 1$

Region 3: $- 1 < x < 0$

Region 4: $x < - 1$

Piecewise Definition Summary

Checking Differentiability at $x = 1$ and $x = - 1$

Checking Differentiability at $x = 0$

Final Calculation

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Function Structure

Identifying Critical Junctions

Defining the Piecewise Behavior

Testing Differentiability

Final Calculation

Similar Questions

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 :

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

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 $f (x) = ⎩ ⎨ ⎧ 3 x, min {1 + x + [x], x + 2 [x]}, 5, x < 0 0 \leq x \leq 2 x > 2,$ 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

Let $f (x) = min {1, 1 + x sin x}, 0 \leq x \leq 2 π$ . If $m$ is the number of points, where $f$ is not differentiable and $n$ is the number of points, where $f$ is not continuous, then the ordered pair $(m, n)$ is equal to

Enter Numerical Value:

Visualized Solution

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

Analyzing Continuity ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n)

Finding Critical Intersection Points

Region 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>1

Region 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0<x<1

Region 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −1<x<0

Region 4: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<−1

Piecewise Definition Summary

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

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

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Function Structure

Identifying Critical Junctions

Defining the Piecewise Behavior

Testing Differentiability

Final Calculation

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 :

Understanding the Function $f (x)$

Analyzing Continuity ( $n$ )

Region 1: $x > 1$

Region 2: $0 < x < 1$

Region 3: $- 1 < x < 0$

Region 4: $x < - 1$

Checking Differentiability at $x = 1$ and $x = - 1$

Checking Differentiability at $x = 0$

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 :