JEE Main 2023 (10 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : (- 2, 2) \to R$ be defined by $f (x) = {x [x] (x - 1) [x], - 2 < x < 0, 0 \leq x < 2$ where $[x]$ denotes the greatest integer function. If $m$ and $n$ respectively are the number of points in $(- 2, 2)$ at which $y = ∣ f (x) ∣$ is not continuous and not differentiable, then $m + n$ is equal to ________.

Q: Wait, why is $x = -1$ a point of discontinuity? The function is defined there.

Great question! Continuity at a point $a$ requires that the limit as $x \to a$ exists and equals the function value $f(a)$. At $x = -1$, the left-hand limit (approaching from $(-2, -1)$) is $2$, while the right-hand limit (approaching from $[-1, 0)$) is $1$. Since $2 \ne 1$, the limit does not exist, creating a jump discontinuity.

Q: Why is $x = 0$ non-differentiable even though the function is continuous there?

Continuity is necessary but not sufficient for differentiability. At $x = 0$, the function is continuous because the LHL, RHL, and $f(0)$ all equal $0$. However, for differentiability, the slope must be consistent. The left-hand derivative is the slope of $-x$, which is $-1$. The right-hand derivative is the slope of $0$, which is $0$. Since $-1 \ne 0$, there is a sharp corner, making it non-differentiable.

Q: Do I need to worry about the absolute value sign changing the function?

In this specific case, no. If you analyze the function $f(x)$ in each interval, you will find that $f(x) \ge 0$ for all $x \in (-2, 2)$. Since $f(x)$ is always non-negative, $|f(x)|$ is identical to $f(x)$. This simplifies our task significantly, as we only need to analyze the original piecewise function.

Enter Numerical Value:

Progressive Hints

Hint 1

The greatest integer function

[x]

is the engine of this problem. It changes its value at every integer. In the interval

(- 2, 2)

, the critical points where the behavior of

f (x)

might change are

x = - 1, 0, 1

. Visualize the function as a series of distinct linear segments between these integers.

Hint 2

The absolute value function

∣ f (x) ∣

is a powerful tool. First, determine the sign of

f (x)

in each interval. If

f (x) \geq 0

, then

∣ f (x) ∣ = f (x)

. If

f (x) < 0

, then

∣ f (x) ∣ = - f (x)

. Check if the function ever dips below the x-axis in the given domain.

Hint 3

Continuity is a prerequisite for differentiability. Check the limits at the critical points

x = - 1, 0, 1

. If the left-hand limit (LHL) and right-hand limit (RHL) do not match, the function is discontinuous. If they match, check the slopes (derivatives) from both sides to see if there is a sharp corner, which would make the function non-differentiable.

Visualized Solution

Defining the Piecewise Function $f (x)$

Given function: $f (x) = {x [x] (x - 1) [x], - 2 < x < 0, 0 \leq x < 2$
We need to find $m$ (points of discontinuity) and $n$ (points of non-differentiability) for $y = ∣ f (x) ∣$ .
The function involves $[x]$ , which changes value at integers $x = - 1, 0, 1$ .

Interval Analysis: $(- 2, - 1)$

For $x \in (- 2, - 1)$ , $[x] = - 2$ .
Substituting this: $f (x) = x (- 2) = - 2 x$ .
As $x \to - 1^{-}$ , $f (x) \to 2$ .

Interval Analysis: $[- 1, 0)$

For $x \in [- 1, 0)$ , $[x] = - 1$ .
Substituting this: $f (x) = x (- 1) = - x$ .
At $x = - 1$ , $f (- 1) = 1$ .
As $x \to 0^{-}$ , $f (x) \to 0$ .

Interval Analysis: $[0, 1)$

For $x \in [0, 1)$ , $[x] = 0$ .
Substituting into $(x - 1) [x]$ : $f (x) = (x - 1) (0) = 0$ .
At $x = 0$ , $f (0) = 0$ .

Interval Analysis: $[1, 2)$

For $x \in [1, 2)$ , $[x] = 1$ .
Substituting into $(x - 1) [x]$ : $f (x) = (x - 1) (1) = x - 1$ .
At $x = 1$ , $f (1) = 0$ .

Consolidating $∣ f (x) ∣$

Since $f (x) \geq 0$ for all $x \in (- 2, 2)$ , $∣ f (x) ∣ = f (x)$ .
The consolidated function is:
$∣ f (x) ∣ = ⎩ ⎨ ⎧ - 2 x - x 0 x - 1, - 2 < x < - 1, - 1 \leq x < 0, 0 \leq x < 1, 1 \leq x < 2$

Checking Continuity at $x = - 1$

At $x = - 1$ :
LHL $= lim_{x \to - 1^{-}} (- 2 x) = 2$
RHL $= lim_{x \to - 1^{+}} (- x) = 1$
Since LHL $\neq =$ RHL, $∣ f (x) ∣$ is discontinuous at $x = - 1$ .

Checking Continuity at $x = 0$ and $x = 1$

At $x = 0$ : LHL $= 0$ , RHL $= 0$ , $f (0) = 0$ . (Continuous)
At $x = 1$ : LHL $= 0$ , RHL $= 0$ , $f (1) = 0$ . (Continuous)
Number of points of discontinuity, $m = 1$ (at $x = - 1$ ).

Differentiability at $x = 0$

At $x = 0$ :
LHD $= \frac{d}{d x} (- x) = - 1$
RHD $= \frac{d}{d x} (0) = 0$
LHD $\neq =$ RHD, so not differentiable at $x = 0$ .

Differentiability at $x = 1$

At $x = 1$ :
LHD $= \frac{d}{d x} (0) = 0$
RHD $= \frac{d}{d x} (x - 1) = 1$
LHD $\neq =$ RHD, so not differentiable at $x = 1$ .
Also, non-differentiable at $x = - 1$ due to discontinuity.

Final Calculation of $m + n$

Points of discontinuity: $x \in {- 1} ⟹ m = 1$
Points of non-differentiability: $x \in {- 1, 0, 1} ⟹ n = 3$
Final Sum: $m + n = 1 + 3 = 4$

00:00 / 00:00

The Sigma Insight: Relationship Between Continuity and Differentiability

Solution Diagram

Analyzing the Setup

The greatest integer function

[x]

acts as the heartbeat of this problem, changing its value at every integer. In our domain of

(- 2, 2)

, the critical points where the function behavior shifts are

x = - 1

x = 0

, and

x = 1

We define the function

f (x)

as:

f (x) = {x [x] (x - 1) [x], - 2 < x < 0, 0 \leq x < 2

Deconstructing the Staircase

Let us analyze the function interval by interval to identify points of discontinuity.

For

x \in (- 2, - 1)

[x] = - 2

, so

f (x) = - 2 x

. As

x \to - 1^{-}

f (x) \to 2

For

x \in [- 1, 0)

[x] = - 1

, so

f (x) = - x

. At

x = - 1

f (- 1) = 1

Because the Left-Hand Limit (LHL) is

2

and the Right-Hand Limit (RHL) is

1

x = - 1

, the function is discontinuous at

x = - 1

The Positive Territory

Now, let us examine the domain

x \in [0, 2)

For

x \in [0, 1)

[x] = 0

, so

f (x) = (x - 1) (0) = 0

. The function is constant and flat on the x-axis.

For

x \in [1, 2)

[x] = 1

, so

f (x) = (x - 1) (1) = x - 1

. This is a line with a positive slope starting from

0

x = 1

The Modulus and Sharp Corners

Since

f (x) \geq 0

for all

x \in (- 2, 2)

, the absolute value

∣ f (x) ∣

is identical to

f (x)

. A function is non-differentiable if it is discontinuous or if it possesses a sharp corner (kink).

We have already identified a discontinuity at

x = - 1

, which automatically makes it a point of non-differentiability.

Now, we check the points

x = 0

and

x = 1

for differentiability:

x = 0

: The Left-Hand Derivative (LHD) of

- x

- 1

, while the Right-Hand Derivative (RHD) of

0

0

. Since $-1 eq 0$,

x = 0

is a sharp corner.

x = 1

: The LHD of

0

0

, while the RHD of

x - 1

1

. Since $0 eq 1$,

x = 1

is a sharp corner.

The Final Tally

We have identified one point of discontinuity (

m = 1

x = - 1

) and three points of non-differentiability (

n = 3

x = - 1, 0, 1

Summing these values, we obtain:

m + n = 1 + 3 = 4

The total number of points where the function is non-differentiable is 4.

Common Traps & FAQs

Wait, why is

x = - 1

a point of discontinuity? The function is defined there.

Great question! Continuity at a point

a

requires that the limit as

x \to a

exists and equals the function value

f (a)

. At

x = - 1

, the left-hand limit (approaching from

(- 2, - 1)

) is

2

, while the right-hand limit (approaching from

[- 1, 0)

) is

1

. Since

2 \neq = 1

, the limit does not exist, creating a jump discontinuity.

Why is

x = 0

non-differentiable even though the function is continuous there?

Continuity is necessary but not sufficient for differentiability. At

x = 0

, the function is continuous because the LHL, RHL, and

f (0)

all equal

0

. However, for differentiability, the slope must be consistent. The left-hand derivative is the slope of

- x

, which is

- 1

. The right-hand derivative is the slope of

0

, which is

0

. Since

- 1 \neq = 0

, there is a sharp corner, making it non-differentiable.

Do I need to worry about the absolute value sign changing the function?

In this specific case, no. If you analyze the function

f (x)

in each interval, you will find that

f (x) \geq 0

for all

x \in (- 2, 2)

. Since

f (x)

is always non-negative,

∣ f (x) ∣

is identical to

f (x)

. This simplifies our task significantly, as we only need to analyze the original piecewise function.

Enter Numerical Value:

Visualized Solution

Defining the Piecewise Function $f (x)$

Interval Analysis: $(- 2, - 1)$

Interval Analysis: $[- 1, 0)$

Interval Analysis: $[0, 1)$

Interval Analysis: $[1, 2)$

Consolidating $∣ f (x) ∣$

Checking Continuity at $x = - 1$

Checking Continuity at $x = 0$ and $x = 1$

Differentiability at $x = 0$

Differentiability at $x = 1$

Final Calculation of $m + n$

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

Deconstructing the Staircase

The Positive Territory

The Modulus and Sharp Corners

The Final Tally

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

Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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

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

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

Final Calculation of .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

Deconstructing the Staircase

The Positive Territory

The Modulus and Sharp Corners

The Final Tally

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 :

Defining the Piecewise Function $f (x)$

Interval Analysis: $(- 2, - 1)$

Interval Analysis: $[- 1, 0)$

Interval Analysis: $[0, 1)$

Interval Analysis: $[1, 2)$

Consolidating $∣ f (x) ∣$

Checking Continuity at $x = - 1$

Checking Continuity at $x = 0$ and $x = 1$

Differentiability at $x = 0$

Differentiability at $x = 1$

Final Calculation of $m + n$

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 :