JEE Main 2020 - 6 Sep (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{2}}$ . Let $S$ denote the set of all points in $R$ , where $f$ is not differentiable. Then :

Q: Wait, why can't we just differentiate $f(x)$ directly? Why do we need to look at the intersection points?

That is a great question! The function $f(x) = \max\{x, x^2\}$ is defined piecewise. The derivative of a 'max' function is not simply the max of the derivatives. At the points where the definition switches (the intersection points), the function might have a sharp corner. We must check the limit of the difference quotient from both sides at these specific junctions to determine differentiability.

Q: Is the function continuous at $x=0$ and $x=1$? Does that matter?

Yes, the function is continuous at both points because the two curves meet at the same y-values ($0$ and $1$ respectively). However, continuity is a necessary but not sufficient condition for differentiability. A function can be continuous but still have a sharp corner, which makes it non-differentiable.

Q: What if the slopes were equal at the intersection points? Would it be differentiable then?

Exactly! If the slopes of the two functions were equal at the intersection point, the transition would be smooth, and the function would indeed be differentiable at that point. In this case, however, the slopes are different, which creates the sharp corner we are looking for.

Select Answer:

Progressive Hints

Hint 1

Visualize the function

f (x) = max {x, x^{2}}

as an 'upper envelope'. Imagine you have two graphs,

y = x

and

y = x^{2}

, drawn on the same plane. At any point

x

, the value of

f (x)

is simply the y-coordinate of the graph that is higher at that specific point.

Hint 2

Recall that a function is not differentiable at points where its graph has a 'sharp corner' or 'kink'. At these points, the slope changes abruptly, meaning the left-hand derivative (LHD) and the right-hand derivative (RHD) are not equal. You need to identify where the 'upper' graph switches from one function to the other.

Hint 3

Start by finding the intersection points of

y = x

and

y = x^{2}

by solving the equation

x^{2} = x

. Once you have the intersection points, calculate the derivative of each function at those points to see if the slopes match. If they don't, the function is not differentiable there.

Visualized Solution

Visualizing the Base Functions

We need to analyze $f (x) = max {x, x^{2}}$ .
Let's start by plotting the two base functions on the Cartesian plane.
$y = x^{2}$ is an upward-opening parabola.
$y = x$ is a straight line passing through the origin.

Understanding the $max$ Function

The function $f (x) = max {x, x^{2}}$ selects the greater value between $x$ and $x^{2}$ .
Geometrically, this means $f (x)$ will always follow the upper graph.
To trace this upper path, we first need to find where the two graphs intersect.

Setting Up the Intersection

To find the crossing points, we equate the two functions.
Set $x^{2} = x$ .
This gives the critical points where the upper function might switch.

Solving for Critical Points

Rearranging the equation: $x^{2} - x = 0$
Factoring out $x$ : $x (x - 1) = 0$
This gives two solutions: $x = 0$ and $x = 1$ .
The intersection points are $(0, 0)$ and $(1, 1)$ .

Tracing the Upper Envelope

For $x \leq 0$ , the parabola $y = x^{2}$ is above the line.
Between $x = 0$ and $x = 1$ , the line $y = x$ is above the parabola.
For $x \geq 1$ , the parabola $y = x^{2}$ is again above the line.
The thick red curve represents $f (x)$ .

Piecewise Definition of $f (x)$

Based on the visual trace, we can write $f (x)$ as a piecewise function:
$f (x) = x^{2}$ for $x \leq 0$
$f (x) = x$ for $0 < x < 1$
$f (x) = x^{2}$ for $x \geq 1$

Differentiability and Sharp Corners

A function is not differentiable at points where its graph has a sharp corner.
At these points, the left-hand derivative (LHD) and right-hand derivative (RHD) are not equal.
We suspect sharp corners at $x = 0$ and $x = 1$ .

Checking Differentiability at $x = 0$

At $x = 0$ , we check the derivatives:
LHD: $\frac{d}{d x} (x^{2}) ∣_{x = 0} = 2 (0) = 0$
RHD: $\frac{d}{d x} (x) ∣_{x = 0} = 1$
Since $0 \neq = 1$ , $f (x)$ is not differentiable at $x = 0$ .

Checking Differentiability at $x = 1$

At $x = 1$ , we check the derivatives:
LHD: $\frac{d}{d x} (x) ∣_{x = 1} = 1$
RHD: $\frac{d}{d x} (x^{2}) ∣_{x = 1} = 2 (1) = 2$
Since $1 \neq = 2$ , $f (x)$ is not differentiable at $x = 1$ .

Final Conclusion

The function fails to be differentiable exactly at the sharp corners.
The set $S$ of all points where $f$ is not differentiable is ${0, 1}$ .
Correct Option: ${0, 1}$

00:00 / 00:00

The Sigma Insight: Differentiability of a Function

Solution Diagram

Analyzing the Setup

The function

f (x) = max {x, x^{2}}

represents the "upper envelope" of the line

y = x

and the parabola

y = x^{2}

. As we traverse the

x

-axis, the function selects the greater value between these two paths at every point.

Finding the Junctions

To identify where the function might fail to be smooth, we must locate the intersection points of the two curves. We set the functions equal to each other:

x^{2} = x

Rearranging this yields the quadratic equation

x^{2} - x = 0

, which factors as

x (x - 1) = 0

. This reveals our critical junctions at

x = 0

and

x = 1

The Anatomy of the Function

By analyzing the relative positions of the curves, we define

f (x)

as a piecewise function:

f (x) = ⎩ ⎨ ⎧ x^{2}, x, x^{2}, x \leq 0 0 < x < 1 x \geq 1

A function is differentiable only if it is smooth, meaning the left-hand derivative (LHD) must equal the right-hand derivative (RHD) at every point.

The Verdict at $x = 0$

At the junction

x = 0

, we calculate the derivatives from both sides:

The LHD is the derivative of

x^{2}

x = 0

, which is

\frac{d}{d x} (x^{2})_{x = 0} = 2 (0) = 0

The RHD is the derivative of

x

x = 0

, which is

\frac{d}{d x} (x)_{x = 0} = 1

Since $0 eq 1$, the function possesses a sharp corner at

x = 0

and is not differentiable there.

The Verdict at $x = 1$

At the junction

x = 1

, we repeat the process:

The LHD is the derivative of

x

x = 1

, which is

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

The RHD is the derivative of

x^{2}

x = 1

, which is

\frac{d}{d x} (x^{2})_{x = 1} = 2 (1) = 2

Since $1 eq 2$, the function possesses a sharp corner at

x = 1

and is not differentiable there.

Final Conclusion

The set

S

of points where

f (x)

is not differentiable is ${0, 1}$ . We have successfully navigated the geometry of the function and confirmed its behavior at these critical junctions.

Common Traps & FAQs

Wait, why can't we just differentiate

f (x)

directly? Why do we need to look at the intersection points?

That is a great question! The function

f (x) = max {x, x^{2}}

is defined piecewise. The derivative of a 'max' function is not simply the max of the derivatives. At the points where the definition switches (the intersection points), the function might have a sharp corner. We must check the limit of the difference quotient from both sides at these specific junctions to determine differentiability.

Is the function continuous at

x = 0

and

x = 1

? Does that matter?

Yes, the function is continuous at both points because the two curves meet at the same y-values (

0

and

1

respectively). However, continuity is a necessary but not sufficient condition for differentiability. A function can be continuous but still have a sharp corner, which makes it non-differentiable.

What if the slopes were equal at the intersection points? Would it be differentiable then?

Exactly! If the slopes of the two functions were equal at the intersection point, the transition would be smooth, and the function would indeed be differentiable at that point. In this case, however, the slopes are different, which creates the sharp corner we are looking for.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{2}}$ . Let $S$ denote the set of all points in $R$ , where $f$ is not differentiable. Then :

Select Answer:

Visualized Solution

Visualizing the Base Functions

Understanding the $max$ Function

Setting Up the Intersection

Solving for Critical Points

Tracing the Upper Envelope

Piecewise Definition of $f (x)$

Differentiability and Sharp Corners

Checking Differentiability at $x = 0$

Checking Differentiability at $x = 1$

Final Conclusion

The Sigma Insight: Differentiability of a Function

Analyzing the Setup

Finding the Junctions

The Anatomy of the Function

The Verdict at $x = 0$

The Verdict at $x = 1$

Final Conclusion

Similar Questions

Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{3}}$ . The set of all points where $f (x)$ is NOT differentiable is

Let $S = {t \in R : f (x) = ∣ x - π ∣ \cdot (e^{∣ x ∣} - 1) sin ∣ x ∣ is not differentiable at t}$ . Then the set $S$ is equal to :

Let $f (x) = {max {∣ x ∣, x^{2}}, 8 - 2∣ x ∣, ∣ x ∣ \leq 2 2 < ∣ x ∣ \leq 4$ . Let S be the set of points in the interval (-4, 4) at which f is not differentiable. Then S:

Let $f : (- 1, 1) \to R$ be a function defined by $f (x) = max {- ∣ x ∣, - 1 - x^{2}}$ . If K be the set of all points at which f is not differentiable, then K has exactly :

Let $K$ be the set of all real values of $x$ where the function $f (x) = sin ∣ x ∣ - ∣ x ∣ + 2 (x - π) cos ∣ x ∣$ is not differentiable. Then the set $K$ is equal to:

Let $S$ be the set of all points in $(- π, π)$ at which the function, $f (x) = min {sin x, cos x}$ is not differentiable. Then $S$ is a subset of which of the following?

Let $S = {(λ, μ) \in R \times R : f (t) = (∣ λ ∣ e^{∣ t ∣} - μ) sin (2∣ t ∣), t \in R, is a differentiable function}$ . Then S is a subset of :

Let $f : R \to R$ be a function such that $f (x + y) = f (x) + f (y), \forall x, y \in R$ . If $f (x)$ is differentiable at $x = 0$ , then

Let a function $g : [0, 4] \to R$ be defined as $g (x) = {max_{0 \leq t \leq x} {t^{3} - 6 t^{2} + 9 t - 3}, 4 - x, 0 \leq x \leq 3 3 < x \leq 4$ then the number of points in the interval $(0, 4)$ where $g (x)$ is NOT differentiable, is

Let $f (x) = {x^{2} cos \frac{π}{x} 0 x \neq = 0 x = 0, x \in R$ then $f$ is

Select Answer:

Visualized Solution

Visualizing the Base Functions

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

Setting Up the Intersection

Solving for Critical Points

Tracing the Upper Envelope

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

Differentiability and Sharp Corners

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

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

Final Conclusion

The Sigma Insight: Differentiability of a Function

Analyzing the Setup

Finding the Junctions

The Anatomy of the Function

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

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

Final Conclusion

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={t∈R:f(x)=∣x−π∣⋅(e∣x∣−1)sin∣x∣ is not differentiable at t}. Then the set S is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={max{∣x∣,x2},8−2∣x∣,​∣x∣≤22<∣x∣≤4​. Let S be the set of points in the interval (-4, 4) at which f is not differentiable. Then S:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(−1,1)→R be a function defined by f(x)=max{−∣x∣,−1−x2​}. If K be the set of all points at which f is not differentiable, then K has exactly :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let K be the set of all real values of x where the function f(x)=sin∣x∣−∣x∣+2(x−π)cos∣x∣ is not differentiable. Then the set K is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S be the set of all points in (−π,π) at which the function, f(x)=min{sinx,cosx} is not differentiable. Then S is a subset of which of the following?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={(λ,μ)∈R×R:f(t)=(∣λ∣e∣t∣−μ)sin(2∣t∣),t∈R, is a differentiable function}. Then S is a subset of :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function such that f(x+y)=f(x)+f(y),∀x,y∈R. If f(x) is differentiable at x=0, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={x2cosxπ​0​x=0x=0​,x∈R then f is

Understanding the $max$ Function

Piecewise Definition of $f (x)$

Checking Differentiability at $x = 0$

Checking Differentiability at $x = 1$

The Verdict at $x = 0$

The Verdict at $x = 1$

Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{3}}$ . The set of all points where $f (x)$ is NOT differentiable is

Let $S = {t \in R : f (x) = ∣ x - π ∣ \cdot (e^{∣ x ∣} - 1) sin ∣ x ∣ is not differentiable at t}$ . Then the set $S$ is equal to :

Let $f (x) = {max {∣ x ∣, x^{2}}, 8 - 2∣ x ∣, ∣ x ∣ \leq 2 2 < ∣ x ∣ \leq 4$ . Let S be the set of points in the interval (-4, 4) at which f is not differentiable. Then S:

Let $f : (- 1, 1) \to R$ be a function defined by $f (x) = max {- ∣ x ∣, - 1 - x^{2}}$ . If K be the set of all points at which f is not differentiable, then K has exactly :

Let $K$ be the set of all real values of $x$ where the function $f (x) = sin ∣ x ∣ - ∣ x ∣ + 2 (x - π) cos ∣ x ∣$ is not differentiable. Then the set $K$ is equal to:

Let $S$ be the set of all points in $(- π, π)$ at which the function, $f (x) = min {sin x, cos x}$ is not differentiable. Then $S$ is a subset of which of the following?

Let $S = {(λ, μ) \in R \times R : f (t) = (∣ λ ∣ e^{∣ t ∣} - μ) sin (2∣ t ∣), t \in R, is a differentiable function}$ . Then S is a subset of :

Let $f : R \to R$ be a function such that $f (x + y) = f (x) + f (y), \forall x, y \in R$ . If $f (x)$ is differentiable at $x = 0$ , then

Let $f (x) = {x^{2} cos \frac{π}{x} 0 x \neq = 0 x = 0, x \in R$ then $f$ is