JEE Main 2022 (29 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $[t]$ denotes the greatest integer $\leq t$ , then number of points, at which the function $f (x) = 4∣2 x + 3∣ + 9 [x + 1/2] - 12 [x + 20]$ is not differentiable in the open interval $(- 20, 20)$ , is ______.

Q: I'm confused about the interval. The question says $(-20, 20)$. Do I include -20 and 20 in my count?

Great catch! The interval is open, denoted by parentheses $(-20, 20)$. This means we strictly exclude the endpoints $-20$ and $20$. When you are counting your points, ensure that your smallest and largest values are strictly greater than $-20$ and strictly less than $20$. If you include the boundaries, you will be off by one or two points, which is a classic trap in JEE Advanced problems.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: Visualize the function as a superposition of three distinct behaviors. The modulus term

∣2 x + 3∣

creates a sharp 'V' corner, while the greatest integer terms

[x + \frac{1}{2}]

and

[x + 20]

create a series of discrete 'staircase' jumps. Non-differentiability occurs precisely at these geometric singularities.

Hint 2

Hint 2: For the modulus term, the point of non-differentiability is where the argument vanishes. For the greatest integer terms, the function is non-differentiable wherever the argument inside the brackets becomes an integer. Set the arguments equal to an arbitrary integer

I

and solve for

x

to find the set of critical points.

Hint 3

Hint 3: Once you have identified the two sets of points—the half-integers from the first floor function and the integers from the second—carefully count them within the open interval

(- 20, 20)

. Finally, check for any overlap between these sets and the root of the modulus function to ensure you do not double-count.

Visualized Solution

Analyzing the Function $f (x)$

The function is $f (x) = 4∣2 x + 3∣ + 9 [x + \frac{1}{2}] - 12 [x + 20]$
We need to find the number of points of non-differentiability in $x \in (- 20, 20)$

Conditions for Non-Differentiability

Non-differentiability occurs at:
1. Sharp corners of the modulus function.
2. Jump discontinuities of the greatest integer functions.

The Modulus Term $4∣2 x + 3∣$

Term 1: $4∣2 x + 3∣$
A modulus function $∣ g (x) ∣$ is non-differentiable where $g (x) = 0$
Set $2 x + 3 = 0 ⟹ x = - 1.5$
Since $- 1.5 \in (- 20, 20)$ , this is a point of non-differentiability.

The First Floor Term $9 [x + \frac{1}{2}]$

Term 2: $9 [x + \frac{1}{2}]$
$[g (x)]$ is non-differentiable at points where $g (x)$ is an integer.
Set $x + \frac{1}{2} = I$ (where $I$ is an integer)
$x = I - 0.5$ (These are half-integers)

The Second Floor Term $12 [x + 20]$

Term 3: $12 [x + 20]$
Set $x + 20 = m$ (where $m$ is an integer)
$x = m - 20$
Since $m$ and $20$ are integers, $x$ must be an integer.

Counting Points for $[x + \frac{1}{2}]$

Points from $[x + \frac{1}{2}]$ in $(- 20, 20)$ :
$x \in {- 19.5, - 18.5, \dots, 19.5}$
Number of points = $19.5 - (- 19.5) + 1$
Total = $40$ points.

Counting Points for $[x + 20]$

Points from $[x + 20]$ in $(- 20, 20)$ :
$x \in {- 19, - 18, \dots, 19}$
Number of points = $19 - (- 19) + 1$
Total = $39$ points.

Checking for Overlaps

Set 1 (Half-integers): $40$ points.
Set 2 (Integers): $39$ points.
Intersection: The sets are disjoint because an integer cannot be a half-integer.
The modulus point $x = - 1.5$ is already included in Set 1.

Final Calculation and Summary

Total points of non-differentiability = $40 + 39 = 79$
Key Takeaway:
1. Modulus functions are non-diff at their roots.
2. Greatest Integer functions are non-diff at integer values of their arguments.
Final Answer: $79$

00:00 / 00:00

The Sigma Insight: Relationship Between Continuity and Differentiability

Solution Diagram

Analyzing the Setup

We are tasked with finding the number of points where the function

f (x) = 4∣2 x + 3∣ + 9 [x + \frac{1}{2}] - 12 [x + 20]

fails to be differentiable within the open interval

(- 20, 20)

The function is composed of a modulus term and two greatest integer (floor) functions. We will analyze the non-differentiability of each component individually.

The Modulus—The Sharp Turn

Consider the term

4∣2 x + 3∣

. A modulus function

∣ g (x) ∣

is non-differentiable at points where

g (x) = 0

, provided

g (x)

is a linear function.

Setting the argument to zero:

2 x + 3 = 0 ⟹ x = - 1.5

Since

- 1.5

lies within the interval

(- 20, 20)

, this is our first point of non-differentiability. Note that this point is a "half-integer."

The Floor Functions—The Staircase of Discontinuity

The greatest integer function

[u]

is non-differentiable whenever its argument

u

is an integer.

For the term

9 [x + \frac{1}{2}]

, the function is non-differentiable when:

x + \frac{1}{2} = I, where I \in Z

x = I - 0.5

In the interval

(- 20, 20)

, the possible values for

x

are

{- 19.5, - 18.5, \dots, 18.5, 19.5}

. The number of such points is calculated as:

\frac{19.5 - ( - 19.5 )}{1} + 1 = 40 points

For the term

12 [x + 20]

, the function is non-differentiable when:

x + 20 = m, where m \in Z

x = m - 20

In the interval

(- 20, 20)

, the possible values for

x

are

{- 19, - 18, \dots, 18, 19}

. The number of such points is:

\frac{19 - ( - 19 )}{1} + 1 = 39 points

The Final Synthesis

To find the total number of points, we must sum the unique points of non-differentiability. We have identified three sets of points: the modulus root, the half-integers, and the integers.

First, we observe that the modulus root

x = - 1.5

is a half-integer. Therefore, it is already contained within the set of 40 points generated by

9 [x + \frac{1}{2}]

Second, we observe that the set of integers and the set of half-integers are disjoint. An integer can never equal a half-integer.

Thus, the total number of points is simply the sum of the points from the two floor functions:

Total Points = 40 + 39 = 79

The function

f (x)

fails to be differentiable at 79 points within the interval

(- 20, 20)

Common Traps & FAQs

Wait, why do we only look for where the argument is an integer for the greatest integer function? Can't it be non-differentiable elsewhere?

That is a fantastic question! The greatest integer function

[u]

is a step function. It is constant between any two consecutive integers, meaning its derivative is zero in those intervals. However, at the exact moment the input

u

hits an integer, the value jumps. A function cannot be differentiable at a jump discontinuity because the limit of the difference quotient does not exist. So, yes, the only points of non-differentiability for

[u]

are exactly where

u

is an integer.

I'm confused about the interval. The question says

(- 20, 20)

. Do I include -20 and 20 in my count?

Great catch! The interval is open, denoted by parentheses

(- 20, 20)

. This means we strictly exclude the endpoints

- 20

and

20

. When you are counting your points, ensure that your smallest and largest values are strictly greater than

- 20

and strictly less than

20

. If you include the boundaries, you will be off by one or two points, which is a classic trap in JEE Advanced problems.

Why didn't we have to subtract the modulus point from the total? I thought we had to check for overlaps?

We absolutely must check for overlaps! In this specific problem, the modulus term

∣2 x + 3∣

gives us the point

x = - 1.5

. When we look at the set of points generated by

[x + \frac{1}{2}]

, we see that

x = - 1.5

is actually a member of that set (since

- 1.5 + 0.5 = - 1

, which is an integer). Because it is already included in the count of the floor function's discontinuities, we don't need to add it again. It is already 'accounted for' in the total count of 79.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $[t]$ denotes the greatest integer $\leq t$ , then number of points, at which the function $f (x) = 4∣2 x + 3∣ + 9 [x + 1/2] - 12 [x + 20]$ is not differentiable in the open interval $(- 20, 20)$ , is ______.

Enter Numerical Value:

Visualized Solution

Analyzing the Function $f (x)$

Conditions for Non-Differentiability

The Modulus Term $4∣2 x + 3∣$

The First Floor Term $9 [x + \frac{1}{2}]$

The Second Floor Term $12 [x + 20]$

Counting Points for $[x + \frac{1}{2}]$

Counting Points for $[x + 20]$

Checking for Overlaps

Final Calculation and Summary

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Modulus—The Sharp Turn

The Floor Functions—The Staircase of Discontinuity

The Final Synthesis

Similar Questions

The number of points, at which the function $f (x) = ∣2 x + 1∣ - 3∣ x + 2∣ + ∣ x^{2} + x - 2∣$ , $x \in R$ is not differentiable, is

A function $f$ is defined on $[- 3, 3]$ as $f (x) = {min {∣ x ∣, 2 - x^{2}}, [∣ x ∣], - 2 \leq x \leq 2 2 < ∣ x ∣ \leq 3$ where $[x]$ denotes the greatest integer $\leq x$ . The number of points, where $f$ is not differentiable in $(- 3, 3)$ 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 $m$ 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 :

Enter Numerical Value:

Visualized Solution

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

Conditions for Non-Differentiability

The Modulus Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4∣2x+3∣

The First Floor Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 9[x+21​]

The Second Floor Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 12[x+20]

Counting Points for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x+21​]

Counting Points for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x+20]

Checking for Overlaps

Final Calculation and Summary

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Modulus—The Sharp Turn

The Floor Functions—The Staircase of Discontinuity

The Final Synthesis

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of points, at which the function f(x)=∣2x+1∣−3∣x+2∣+∣x2+x−2∣, x∈R is not differentiable, is

.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 :

Analyzing the Function $f (x)$

The Modulus Term $4∣2 x + 3∣$

The First Floor Term $9 [x + \frac{1}{2}]$

The Second Floor Term $12 [x + 20]$

Counting Points for $[x + \frac{1}{2}]$

Counting Points for $[x + 20]$

The number of points, at which the function $f (x) = ∣2 x + 1∣ - 3∣ x + 2∣ + ∣ x^{2} + x - 2∣$ , $x \in R$ is not differentiable, is

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 :