JEE Main 2022 (26 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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

Q: Can I just differentiate $1 + x \sin x$ and check where it's zero?

Not quite! That would only tell you where the function $1+x\sin x$ has a local maximum or minimum. It doesn't tell you anything about the non-differentiability of the *minimum* function. You must always check the points where the two functions intersect, as that is where the 'minimum' logic forces a change in the function's behavior.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Imagine two paths: a flat road at

y = 1

and a winding, oscillating path

y = 1 + x sin x

. The function

f (x)

is like a traveler who always chooses the lower path. Where do these paths cross? Visualize the intersection point in the interval

[0, 2 π]

Hint 2

Hint 2 (The Arsenal): To determine if the traveler stumbles (non-differentiability) or falls off a cliff (discontinuity), you must analyze the behavior exactly at the intersection point

x = π

. Remember the definitions of continuity and differentiability: continuity requires the limit to exist and equal the function value, while differentiability requires the left-hand and right-hand derivatives to be equal.

Hint 3

Hint 3 (The Execution): Start by calculating the Left Hand Derivative (LHD) and Right Hand Derivative (RHD) at

x = π

. For

x < π

f (x) = 1

. For

x > π

f (x) = 1 + x sin x

. Compare

f^{'} (π^{-})

and

f^{'} (π^{+})

to see if the slopes match.

Visualized Solution

Understanding the Function

Given: $f (x) = min {1, 1 + x sin x}$
Interval: $x \in [0, 2 π]$
Goal: Find $m$ (non-differentiable points) and $n$ (discontinuous points)

Visualizing the Components

We are comparing two functions:
$y_{1} = 1$ (Horizontal line)
$y_{2} = 1 + x sin x$ (Oscillating curve)

Analyzing the First Half $(0, π)$

For $x \in (0, π)$ , $sin x > 0$
$⟹ x sin x > 0$
$⟹ 1 + x sin x > 1$

Analyzing the Second Half $(π, 2 π)$

For $x \in (π, 2 π)$ , $sin x < 0$
$⟹ x sin x < 0$
$⟹ 1 + x sin x < 1$

Applying the Minimum Logic

$f (x)$ takes the minimum of the two values.
In $[0, π]$ , $min {1, 1 + x sin x} = 1$
In $(π, 2 π]$ , $min {1, 1 + x sin x} = 1 + x sin x$

Defining the Piecewise Function

$f (x) = {1 1 + x sin x 0 \leq x \leq π π < x \leq 2 π$
Critical point to check is $x = π$

Checking Continuity at $x = π$

Left Hand Limit (LHL): $lim_{x \to π^{-}} 1 = 1$
Right Hand Limit (RHL): $lim_{x \to π^{+}} (1 + x sin x) = 1 + π (0) = 1$
$f (π) = 1$

Continuity Conclusion

Since $L H L = R H L = f (π)$ , $f (x)$ is continuous at $x = π$ .
It is continuous everywhere in $[0, 2 π]$ .
Number of discontinuous points, $n = 0$ .

Checking Differentiability (LHD)

Left Hand Derivative (LHD) at $x = π$ :
For $x < π$ , $f (x) = 1$
$⟹ f^{'} (x) = 0$
$L H D = 0$

Checking Differentiability (RHD)

Right Hand Derivative (RHD) at $x = π$ :
For $x > π$ , $f (x) = 1 + x sin x$
Using product rule: $f^{'} (x) = sin x + x cos x$

Evaluating RHD at $x = π$

$R H D = lim_{x \to π^{+}} (sin x + x cos x)$
$R H D = sin π + π cos π$
$R H D = 0 + π (- 1) = - π$

Differentiability Conclusion

$L H D = 0$ and $R H D = - π$
Since $L H D \neq = R H D$ , $f (x)$ is not differentiable at $x = π$ .
Number of non-differentiable points, $m = 1$ .

Final Answer

We found $m = 1$ and $n = 0$ .
The ordered pair $(m, n)$ is $(1, 0)$ .

00:00 / 00:00

The Sigma Insight: Relationship Between Continuity and Differentiability

Solution Diagram

Analyzing the Setup

We are tasked with analyzing the function defined by:

f (x) = min {1, 1 + x sin x}

on the interval

x \in [0, 2 π]

. This function represents the lower of the two paths defined by

y_{1} = 1

and

y_{2} = 1 + x sin x

The Visual Landscape

In the interval

[0, π]

, the term

sin x \geq 0

. Consequently,

x sin x \geq 0

, which implies that

1 + x sin x \geq 1

. Since

f (x)

is the minimum of the two paths, we have:

f (x) = 1, for x \in [0, π]

In the interval

(π, 2 π]

, the term

sin x < 0

. Consequently,

x sin x < 0

, which implies that

1 + x sin x < 1

. In this region, the function follows the oscillating path:

f (x) = 1 + x sin x, for x \in (π, 2 π]

The Critical Intersection

To check for continuity at

x = π

, we evaluate the limits from both sides:

x \to π^{-} lim f (x) = 1

x \to π^{+} lim f (x) = 1 + π sin (π) = 1 + 0 = 1

Since the left-hand limit, the right-hand limit, and the function value

f (π) = 1

are all equal, the function is continuous at

x = π

. Thus, the number of points of discontinuity is

n = 0

The Sharp Corner

To determine differentiability, we examine the derivatives at

x = π

: For

x < π

f (x) = 1

, so the left-hand derivative (LHD) is:

f^{'} (π^{-}) = 0

For

x > π

f (x) = 1 + x sin x

. Using the product rule, the derivative is

f^{'} (x) = sin x + x cos x

. Evaluating this at

x = π

f^{'} (π^{+}) = sin (π) + π cos (π) = 0 + π (- 1) = - π

Since the LHD (

0

) is not equal to the RHD (

- π

), the function is not differentiable at

x = π

. This indicates a sharp corner, meaning the number of points of non-differentiability is

m = 1

The Final Revelation

We have successfully navigated the landscape and identified the properties of the function. We found

m = 1

and

n = 0

The final ordered pair

(m, n)

is $(1, 0)$ .

Common Traps & FAQs

Wait, if the derivatives are different, doesn't that mean the function is discontinuous?

Great question! A function can be continuous (the graph is connected) but still have a 'sharp corner' where the slope changes abruptly. Think of a 'V' shape—it's connected, but you can't define a single tangent at the vertex. Here, the function is continuous at

x = π

because the limits match, but the slopes don't, creating that sharp corner.

Why do we only check

x = π

Because that is the only point where the 'minimum' rule switches from one function to the other. Everywhere else, the function is either just

1

or just

1 + x sin x

, both of which are smooth and continuous. We only need to investigate the 'seams' where the definition changes.

Can I just differentiate

1 + x sin x

and check where it's zero?

Not quite! That would only tell you where the function

1 + x sin x

has a local maximum or minimum. It doesn't tell you anything about the non-differentiability of the *minimum* function. You must always check the points where the two functions intersect, as that is where the 'minimum' logic forces a change in the function's behavior.

Select Answer:

Visualized Solution

Understanding the Function

Visualizing the Components

Analyzing the First Half $(0, π)$

Analyzing the Second Half $(π, 2 π)$

Applying the Minimum Logic

Defining the Piecewise Function

Checking Continuity at $x = π$

Continuity Conclusion

Checking Differentiability (LHD)

Checking Differentiability (RHD)

Evaluating RHD at $x = π$

Differentiability Conclusion

Final Answer

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Visual Landscape

The Critical Intersection

The Sharp Corner

The Final Revelation

Similar Questions

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

If the function $f (x) = {k_{1} (x - π)^{2} - 1, k_{2} cos x, x \leq π x > π$ is twice differentiable, then the ordered pair $(k_{1}, k_{2})$ is equal to:

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 $f (x) = {(x - 1)^{2} sin \frac{1}{( x - 1 )} - ∣ x ∣ - 1 if x \neq = 1 if x = 1$ be a real-valued function. Then the set of points where $f (x)$ is not differentiable is .........

The function $f (x) = 1 + ∣ sin x ∣$ is

Select Answer:

Visualized Solution

Understanding the Function

Visualizing the Components

Analyzing the First Half .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (0,π)

Analyzing the Second Half .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (π,2π)

Applying the Minimum Logic

Defining the Piecewise Function

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

Continuity Conclusion

Checking Differentiability (LHD)

Checking Differentiability (RHD)

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

Differentiability Conclusion

Final Answer

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Visual Landscape

The Critical Intersection

The Sharp Corner

The Final Revelation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f(x)={k1​(x−π)2−1,k2​cosx,​x≤πx>π​ is twice differentiable, then the ordered pair (k1​,k2​) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={(x−1)2sin(x−1)1​−∣x∣−1​if x=1if x=1​ be a real-valued function. Then the set of points where f(x) is not differentiable is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=1+∣sinx∣ is

Analyzing the First Half $(0, π)$

Analyzing the Second Half $(π, 2 π)$

Checking Continuity at $x = π$

Evaluating RHD at $x = π$

If the function $f (x) = {k_{1} (x - π)^{2} - 1, k_{2} cos x, x \leq π x > π$ is twice differentiable, then the ordered pair $(k_{1}, k_{2})$ is equal to:

Let $f (x) = {(x - 1)^{2} sin \frac{1}{( x - 1 )} - ∣ x ∣ - 1 if x \neq = 1 if x = 1$ be a real-valued function. Then the set of points where $f (x)$ is not differentiable is .........

The function $f (x) = 1 + ∣ sin x ∣$ is