JEE Advanced 2016

LEVELJEE Advanced

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : [- 1/2, 2] \to R$ and $g : [- 1/2, 2] \to R$ be functions defined by $f (x) = [x^{2} - 3]$ and $g (x) = ∣ x ∣ f (x) + ∣4 x - 7∣ f (x)$ , where $[y]$ denotes the greatest integer less than or equal to $y$ . Then

Q: Wait, why is $f(x)$ continuous at $x=0$ even though $x^2-3$ is an integer there?

Great observation! A function $[u(x)]$ is discontinuous at $x_0$ only if the limit from the left and right are different. At $x=0$, $x^2-3 = -3$. For $x$ slightly less than $0$ or slightly greater than $0$, $x^2$ is slightly positive, so $x^2-3$ is slightly greater than $-3$. Thus, the greatest integer remains $-3$ from both sides. Continuity is about the limit, not just the value!

Q: Why can't we just say $g(x)$ is non-differentiable everywhere $f(x)$ is discontinuous?

You are almost right, but there is a catch! If $f(x)$ is discontinuous at $x_0$ but the other factor $h(x)$ is zero at that exact point, the product $h(x)f(x)$ might actually be differentiable. Always check if the 'multiplier' vanishes at the point of discontinuity.

Q: How do I know if $x=7/4$ is a point of non-differentiability for $g(x)$?

We look at the neighborhood of $7/4$. Since $7/4 = 1.75$ and $\sqrt{3} \approx 1.732$, $7/4$ falls in the interval $(\sqrt{3}, 2)$. In this specific interval, $f(x) = 0$. Since $g(x) = h(x) \cdot 0 = 0$, the function is constant zero in that neighborhood, making it perfectly differentiable!

Q: Why do we only check points in the open interval $(-1/2, 2)$ for differentiability?

Differentiability is a local property defined by the limit of the difference quotient. By definition, we require a two-sided neighborhood to exist, which is why we focus on the open interval $(a, b)$ rather than the closed endpoints.

Q: Is it possible for $f(x)$ to be discontinuous at $x=2$?

Yes! The domain is $[-1/2, 2]$. At $x=2$, $x^2-3 = 1$. As $x$ approaches $2$ from the left, $x^2-3$ is slightly less than $1$, so $[x^2-3] = 0$. But at $x=2$, the value is $1$. Since the left-hand limit ($0$) does not equal the function value ($1$), it is a jump discontinuity.

Select Answer:

* Multiple Correct

Visualized Solution

Defining the Functions and Domain

We are given two functions: $f (x) = [x^{2} - 3]$ and $g (x) = (∣ x ∣ + ∣4 x - 7∣) f (x)$ .
The domain for both functions is restricted to the closed interval $[- \frac{1}{2}, 2]$ .
Here, $[y]$ represents the Greatest Integer Function (GIF), which outputs the largest integer less than or equal to $y$ .

Finding the Range of $x^{2} - 3$

To analyze the Greatest Integer Function $f (x) = [x^{2} - 3]$ , we must first find the range of the inner expression $x^{2} - 3$ .
Since $x \in [- \frac{1}{2}, 2]$ , the term $x^{2}$ lies in the interval $[0, 4]$ .
Subtracting $3$ from all parts: $- 3 \leq x^{2} - 3 \leq 1$ .
Thus, the range of the inner function is $[- 3, 1]$ .

Identifying Critical Integer Points

The Greatest Integer Function $[y]$ is discontinuous at integer values of $y$ .
Therefore, $f (x) = [x^{2} - 3]$ can only be discontinuous where $x^{2} - 3 = k$ , for integers $k \in {- 3, - 2, - 1, 0, 1}$ .
Let's solve for $x$ in each case within our domain $[- \frac{1}{2}, 2]$ :
1. $x^{2} - 3 = - 3 ⟹ x = 0$
2. $x^{2} - 3 = - 2 ⟹ x = 1$
3. $x^{2} - 3 = - 1 ⟹ x = 2$
4. $x^{2} - 3 = 0 ⟹ x = 3$
5. $x^{2} - 3 = 1 ⟹ x = 2$

Checking Continuity at $x = 0$

Let's check the behavior of $f (x)$ around $x = 0$ .
At $x = 0$ : $f (0) = [0^{2} - 3] = [- 3] = - 3$ .
For any $x$ near $0$ (either positive or negative), $x^{2} \geq 0 ⟹ x^{2} - 3 \geq - 3$ .
Since $x^{2} - 3$ is slightly greater than or equal to $- 3$ for small $x$ , its greatest integer value remains exactly $- 3$ .
Therefore, $lim_{x \to 0} f (x) = - 3 = f (0)$ .
Conclusion: $f (x)$ is continuous at $x = 0$ .

Evaluating $f (x)$ at $x = 1, 2, 3, 2$

Let's write down the piecewise definition of $f (x)$ across the intervals:
1. For $x \in [- \frac{1}{2}, 1)$ : $x^{2} - 3 \in [- 3, - 2) ⟹ f (x) = - 3$
2. For $x \in [1, 2)$ : $x^{2} - 3 \in [- 2, - 1) ⟹ f (x) = - 2$
3. For $x \in [2, 3)$ : $x^{2} - 3 \in [- 1, 0) ⟹ f (x) = - 1$
4. For $x \in [3, 2)$ : $x^{2} - 3 \in [0, 1) ⟹ f (x) = 0$
5. For $x = 2$ : $x^{2} - 3 = 1 ⟹ f (2) = 1$

Discontinuity Points of $f (x)$

From the piecewise definition, we see clear jump discontinuities at:
1. $x = 1$ (jumps from $- 3$ to $- 2$ )
2. $x = 2$ (jumps from $- 2$ to $- 1$ )
3. $x = 3$ (jumps from $- 1$ to $0$ )
4. $x = 2$ (jumps from $0$ to $1$ )
Thus, $f (x)$ is discontinuous at exactly four points in $[- \frac{1}{2}, 2]$ .
This confirms that Option 2 is correct!

Analyzing the Structure of $g (x)$

The function $g (x)$ is defined as: $g (x) = (∣ x ∣ + ∣4 x - 7∣) f (x)$ .
Let $h (x) = ∣ x ∣ + ∣4 x - 7∣$ . Then $g (x) = h (x) \cdot f (x)$ .
The potential points of non-differentiability for $g (x)$ in the open interval $(- \frac{1}{2}, 2)$ are:
1. Points where $h (x)$ is non-differentiable: $x = 0$ and $x = \frac{7}{4}$ .
2. Points where $f (x)$ is discontinuous: $x = 1, 2, 3$ .

Checking Differentiability at $x = 0$ and $x = \frac{7}{4}$

At $x = 0$ :
In a small neighborhood of $0$ , $f (x) = - 3$ (constant).
Thus, $g (x) = - 3 (∣ x ∣ + ∣4 x - 7∣)$ .
Since $∣ x ∣$ has a sharp corner at $x = 0$ , $g (x)$ is non-differentiable at $x = 0$ .
At $x = \frac{7}{4}$ :
Since $3 \approx 1.732 < 1.75 < 2$ , $x = \frac{7}{4}$ lies in $(3, 2)$ .
In this interval, $f (x) = 0$ (constant).
Thus, $g (x) = 0$ in a neighborhood of $\frac{7}{4} ⟹$ differentiable.

Checking Differentiability at $x = 1, 2, 3$

At $x_{0} \in {1, 2, 3}$ , $f (x)$ has a jump discontinuity.
The multiplier $h (x) = ∣ x ∣ + ∣4 x - 7∣$ is non-zero at these points:
1. $h (1) = 1 + 3 = 4 \neq = 0$
2. $h (2) = 2 + ∣4 2 - 7∣ \neq = 0$
3. $h (3) = 3 + ∣4 3 - 7∣ \neq = 0$
Since $g (x) = h (x) \cdot f (x)$ , and $h (x_{0}) \neq = 0$ while $f (x)$ has a jump discontinuity, $g (x)$ must also be discontinuous at these points.
Discontinuity directly implies non-differentiability at $x = 1, 2, 3$ .

Final Conclusion and Correct Options

Let's summarize our findings:
1. $f (x)$ is discontinuous at exactly four points: ${1, 2, 3, 2}$ .
2. $g (x)$ is non-differentiable at exactly four points in $(- \frac{1}{2}, 2)$ : ${0, 1, 2, 3}$ .
Therefore, both Option 2 and Option 3 are correct!

00:00 / 00:00

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

Welcome, fellow traveler on the JEE journey! Today, we are going to dissect a problem that seems like a simple exercise in function analysis but is actually a beautiful study of how functions behave under pressure.

We are looking at

f (x) = [x^{2} - 3]

and its partner

g (x) = (∣ x ∣ + ∣4 x - 7∣) f (x)

over the interval

x \in [- 1/2, 2]

. Let's peel back the layers.

The Geometry of the Greatest Integer

First, let us focus on

f (x) = [x^{2} - 3]

. The Greatest Integer Function (GIF) is notorious for its 'jumpy' nature; it stays flat, then suddenly leaps to the next integer.

To find where

f (x)

breaks, we identify when the input

x^{2} - 3

hits an integer. Given

x \in [- 1/2, 2]

, the range of

x^{2}

[0, 4]

, which implies the input

x^{2} - 3

spans the interval

[- 3, 1]

The function

f (x)

will attempt to jump whenever

x^{2} - 3

equals the integers

- 3, - 2, - 1, 0,

1

. Solving the equation

x^{2} - 3 = k

for these integers yields the following candidate points:

x \in {0, 1, 2, 3, 2}

The Illusion of Discontinuity

Here is where many students stumble. They see

x = 0

and immediately mark it as a discontinuity, but we must look closer.

x = 0

, the function value is

f (0) = [0^{2} - 3] = - 3

. As we approach

0

from either side,

x^{2}

is a tiny positive value, meaning

x^{2} - 3

is slightly larger than

- 3

Since the greatest integer of a number slightly larger than

- 3

is still

- 3

, the limit matches the function value. Therefore,

f (x)

is actually continuous at

x = 0

. The true jumps occur at

x = 1, 2, 3,

and

2

. Thus,

f (x)

has exactly four points of discontinuity.

The Complexity of $g (x)$

Now, let's analyze

g (x) = (∣ x ∣ + ∣4 x - 7∣) f (x)

. This function is a product of

h (x) = ∣ x ∣ + ∣4 x - 7∣

and

f (x)

For

g (x)

to be non-differentiable, we check two conditions: where

h (x)

has sharp corners (at

x = 0

and

x = 7/4

) and where

f (x)

is discontinuous.

We know

f (x)

is discontinuous at

x = 1, 2, 3

. At these points, $h(x) eq 0$, so the product

g (x)

inherits the discontinuity, making it non-differentiable. This provides three points.

x = 0

h (x)

has a sharp corner due to the

∣ x ∣

term. Since

f (0) = - 3

(a non-zero constant), the sharp corner is preserved in the product

g (x)

. This gives us our fourth point of non-differentiability.

The Grand Finale

Finally, we check

x = 7/4

. This point lies in the interval

(3, 2)

, where

f (x) = 0

In a small neighborhood around

x = 7/4

f (x)

is identically zero, which forces

g (x) = h (x) \cdot 0 = 0

. Since a constant function is perfectly differentiable,

x = 7/4

is not a point of non-differentiability.

By carefully analyzing the behavior of these functions, we conclude that

f (x)

is discontinuous at four points, and

g (x)

is non-differentiable at four points. The math is elegant, the logic is sound, and you have successfully navigated the traps.

Common Traps & FAQs

Wait, why is

f (x)

continuous at

x = 0

even though

x^{2} - 3

is an integer there?

Great observation! A function

[u (x)]

is discontinuous at

x_{0}

only if the limit from the left and right are different. At

x = 0

x^{2} - 3 = - 3

. For

x

slightly less than

0

or slightly greater than

0

x^{2}

is slightly positive, so

x^{2} - 3

is slightly greater than

- 3

. Thus, the greatest integer remains

- 3

from both sides. Continuity is about the limit, not just the value!

Why can't we just say

g (x)

is non-differentiable everywhere

f (x)

is discontinuous?

You are almost right, but there is a catch! If

f (x)

is discontinuous at

x_{0}

but the other factor

h (x)

is zero at that exact point, the product

h (x) f (x)

might actually be differentiable. Always check if the 'multiplier' vanishes at the point of discontinuity.

How do I know if

x = 7/4

is a point of non-differentiability for

g (x)

We look at the neighborhood of

7/4

. Since

7/4 = 1.75

and

3 \approx 1.732

7/4

falls in the interval

(3, 2)

. In this specific interval,

f (x) = 0

. Since

g (x) = h (x) \cdot 0 = 0

, the function is constant zero in that neighborhood, making it perfectly differentiable!

Why do we only check points in the open interval

(- 1/2, 2)

for differentiability?

Differentiability is a local property defined by the limit of the difference quotient. By definition, we require a two-sided neighborhood to exist, which is why we focus on the open interval

(a, b)

rather than the closed endpoints.

Is it possible for

f (x)

to be discontinuous at

x = 2

Yes! The domain is

[- 1/2, 2]

. At

x = 2

x^{2} - 3 = 1

. As

x

approaches

2

from the left,

x^{2} - 3

is slightly less than

1

, so

[x^{2} - 3] = 0

. But at

x = 2

, the value is

1

. Since the left-hand limit (

0

) does not equal the function value (

1

), it is a jump discontinuity.

Select Answer:

Visualized Solution

Defining the Functions and Domain

Finding the Range of $x^{2} - 3$

Identifying Critical Integer Points

Checking Continuity at $x = 0$

Evaluating $f (x)$ at $x = 1, 2, 3, 2$

Discontinuity Points of $f (x)$

Analyzing the Structure of $g (x)$

Checking Differentiability at $x = 0$ and $x = \frac{7}{4}$

Checking Differentiability at $x = 1, 2, 3$

Final Conclusion and Correct Options

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Geometry of the Greatest Integer

The Illusion of Discontinuity

The Complexity of $g (x)$

The Grand Finale

Similar Questions

Consider the function $f : (0, 2) \to R$ defined by $f (x) = \frac{x}{2} + \frac{2}{x}$ and the function $g (x)$ defined by $g (x) = {min {f (t)}, \frac{3}{2} + x, 0 < t \leq x and 0 < x \leq 1 1 < x < 2$ . Then

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 the functions $f : R \to R$ and $g : R \to R$ be defined as : $f (x) = {x + 2, x^{2}, x < 0 x \geq 0$ and $g (x) = {x^{3}, 3 x - 2, x < 1 x \geq 1$ . Then, the number of points in $R$ where $(f \circ g) (x)$ is NOT differentiable is equal to :

Let $f : [a, b] \to [1, \infty)$ be a continuous function and let $g : R \to R$ be defined as $g (x) = ⎩ ⎨ ⎧ 0, \int_{a}^{x} f (t) d t, \int_{a}^{b} f (t) d t, if x < a, if a \leq x \leq b; if x > b .$ then

The function $f (x) = {∣ x - 3∣, \frac{x ^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}, x \geq 1 x < 1$ is

Let $[t]$ denote the greatest integer less than or equal to $t$ . Let $f (x) = x - [x]$ , $g (x) = 1 - x + [x]$ , and $h (x) = min {f (x), g (x)}$ , $x \in [- 2, 2]$ . Then $h$ is :

Select Answer:

Visualized Solution

Defining the Functions and Domain

Finding the Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2−3

Identifying Critical Integer Points

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

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1,2​,3​,2

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

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

Checking Differentiability 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=47​

Checking Differentiability at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1,2​,3​

Final Conclusion and Correct Options

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Geometry of the Greatest Integer

The Illusion of Discontinuity

The Complexity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

The Grand Finale

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the function f:(0,2)→R defined by f(x)=2x​+x2​ and the function g(x) defined by g(x)={min{f(t)},23​+x,​0<t≤x and 0<x≤11<x<2​. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:[a,b]→[1,∞) be a continuous function and let g:R→R be defined as g(x)=⎩⎨⎧​0,∫ax​f(t)dt,∫ab​f(t)dt,​if x<a,if a≤x≤b;if x>b.​ then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)={∣x−3∣,4x2​−23x​+413​,​x≥1x<1​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let [t] denote the greatest integer less than or equal to t. Let f(x)=x−[x], g(x)=1−x+[x], and h(x)=min{f(x),g(x)}, x∈[−2,2]. Then h is :

Finding the Range of $x^{2} - 3$

Checking Continuity at $x = 0$

Evaluating $f (x)$ at $x = 1, 2, 3, 2$

Discontinuity Points of $f (x)$

Analyzing the Structure of $g (x)$

Checking Differentiability at $x = 0$ and $x = \frac{7}{4}$

Checking Differentiability at $x = 1, 2, 3$

The Complexity of $g (x)$

Consider the function $f : (0, 2) \to R$ defined by $f (x) = \frac{x}{2} + \frac{2}{x}$ and the function $g (x)$ defined by $g (x) = {min {f (t)}, \frac{3}{2} + x, 0 < t \leq x and 0 < x \leq 1 1 < x < 2$ . Then

Let $f : [a, b] \to [1, \infty)$ be a continuous function and let $g : R \to R$ be defined as $g (x) = ⎩ ⎨ ⎧ 0, \int_{a}^{x} f (t) d t, \int_{a}^{b} f (t) d t, if x < a, if a \leq x \leq b; if x > b .$ then

The function $f (x) = {∣ x - 3∣, \frac{x ^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}, x \geq 1 x < 1$ is

Let $[t]$ denote the greatest integer less than or equal to $t$ . Let $f (x) = x - [x]$ , $g (x) = 1 - x + [x]$ , and $h (x) = min {f (x), g (x)}$ , $x \in [- 2, 2]$ . Then $h$ is :