OP Malhotra Class 12 Maths Solutions Chapter 7 Continuity and Differentiability of Functions

Question 1. Examine the continuity of the function \( f(x) = 2x^2 - 1 \) at \( x = 3 \).
Answer:
Given the function is \( f(x) = 2x^2 - 1 \). We need to check its continuity at \( x = 3 \).
First, find the left-hand limit (LHL) at \( x = 3 \):
\( \underset{x \rightarrow 3^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 3^{-}}{\mathrm{Lt}} (2x^2 - 1) \)
Substitute \( x = 3 - h \), so as \( x \rightarrow 3^{-} \), \( h \rightarrow 0^{+} \):
\( \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} [2(3-h)^2 - 1] = 2(3-0)^2 - 1 = 2(9) - 1 = 18 - 1 = 17 \)

Next, find the right-hand limit (RHL) at \( x = 3 \):
\( \underset{x \rightarrow 3^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 3^{+}}{\mathrm{Lt}} (2x^2 - 1) \)
Substitute \( x = 3 + h \), so as \( x \rightarrow 3^{+} \), \( h \rightarrow 0^{+} \):
\( \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} [2(3+h)^2 - 1] = 2(3+0)^2 - 1 = 2(9) - 1 = 18 - 1 = 17 \)

Finally, find the value of the function at \( x = 3 \):
\( f(3) = 2(3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17 \)

Since the LHL, RHL, and the function's value at \( x = 3 \) are all equal:
\( \underset{x \rightarrow 3^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 3^{+}}{\mathrm{Lt}} f(x) = f(3) = 17 \)
Therefore, the function \( f(x) = 2x^2 - 1 \) is continuous at \( x = 3 \). This is expected because polynomial functions are always continuous everywhere.
In simple words: To check if a function is continuous at a point, we look at what happens when we get very close to that point from both sides, and also what the function's value is exactly at that point. If all three match, the function is continuous. Here, they all came out to be 17, so the function is smooth at \( x = 3 \).

🎯 Exam Tip: For polynomial functions, you can generally conclude continuity based on their nature, but for formal examination, always show the explicit calculation of LHL, RHL, and f(a) to secure full marks.

Question 2. Examine the following functions for continuity:
(a) \( f(x) = x - 5 \)
(b) \( f(x) = \frac{1}{x-5}, x \neq 5 \)
(c) \( f(x) = \frac{x^2-25}{x+5}, x \neq 5 \)
(d) \( f(x) = |x - 5| \)
Answer:
(a) Given \( f(x) = x - 5 \).
The domain of this function is all real numbers, \( D_f = \mathbb{R} \).
Let \( c \) be any arbitrary point in the domain \( D_f \).
We find the limit of \( f(x) \) as \( x \) approaches \( c \):
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} (x - 5) = c - 5 \)
The value of the function at \( c \) is \( f(c) = c - 5 \).
Since \( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = f(c) \), the function \( f(x) \) is continuous at \( x = c \).
As \( c \) represents any arbitrary point in the domain, \( f(x) \) is continuous at every point in its domain. This is true because it is a polynomial function.

(b) Given \( f(x) = \frac{1}{x-5}, x \neq 5 \).
The domain of this function is all real numbers except \( 5 \), \( D_f = \mathbb{R} - \{5\} \).
Let \( c \) be any arbitrary point in the domain \( D_f \). This means \( c \neq 5 \).
We find the limit of \( f(x) \) as \( x \) approaches \( c \):
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} \frac{1}{x-5} = \frac{1}{c-5} \)
The value of the function at \( c \) is \( f(c) = \frac{1}{c-5} \).
Since \( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = f(c) \), the function \( f(x) \) is continuous at \( x = c \).
As \( c \) represents any arbitrary point in the domain, \( f(x) \) is continuous at every point in its domain.

(c) Given \( f(x) = \frac{x^2-25}{x+5}, x \neq 5 \).
The provided solution uses \( f(x) = \frac{x^2-25}{x-5} \) for calculation. We will follow this calculation.
Here, the domain based on the calculation is \( D_f = \mathbb{R} - \{5\} \).
Let \( c \) be any arbitrary point in the domain \( D_f \), so \( c \neq 5 \).
We find the limit of \( f(x) \) as \( x \) approaches \( c \):
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} \frac{x^2-25}{x-5} = \underset{x \rightarrow c}{\mathrm{Lt}} \frac{(x-5)(x+5)}{x-5} \)
Since \( x \neq c \), we can cancel \( (x-5) \):
\( = \underset{x \rightarrow c}{\mathrm{Lt}} (x+5) = c+5 \)
The value of the function at \( c \) is \( f(c) = \frac{c^2-25}{c-5} = \frac{(c-5)(c+5)}{c-5} = c+5 \).
Since \( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = f(c) \), the function \( f(x) \) is continuous at \( x = c \).
As \( c \) represents any arbitrary point in the domain, \( f(x) \) is continuous at every point in its domain.

(d) Given \( f(x) = |x - 5| \).
This function can be written as a piecewise function:
\[ f(x) = \begin{cases} x-5 & \text{if } x \geq 5 \\ -(x-5) & \text{if } x < 5 \end{cases} \]
The domain of this function is all real numbers, \( D_f = \mathbb{R} \).
We examine the continuity in three cases:

Case I: When \( c < 5 \)
In this case, \( f(c) = -(c - 5) \).
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} -(x - 5) = -(c - 5) \)
Since \( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = f(c) \), \( f(x) \) is continuous for all \( c < 5 \).

Case II: When \( c > 5 \)
In this case, \( f(c) = c - 5 \).
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} (x - 5) = c - 5 \)
Since \( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = f(c) \), \( f(x) \) is continuous for all \( c > 5 \).

Case III: When \( c = 5 \)
First, find \( f(5) = |5 - 5| = 0 \).
Next, find the LHL at \( x = 5 \):
\( \underset{x \rightarrow 5^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 5^{-}}{\mathrm{Lt}} -(x - 5) = -(5 - 5) = 0 \)
Then, find the RHL at \( x = 5 \):
\( \underset{x \rightarrow 5^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 5^{+}}{\mathrm{Lt}} (x - 5) = 5 - 5 = 0 \)
Since \( \underset{x \rightarrow 5^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 5^{+}}{\mathrm{Lt}} f(x) = f(5) = 0 \), the function \( f(x) \) is continuous at \( x = 5 \).

By combining all three cases, the function \( f(x) = |x - 5| \) is continuous for all real numbers \( x \in \mathbb{R} \).
In simple words: A function is continuous if you can draw its graph without lifting your pen. For part (a), it's a straight line, so it's always continuous. For (b), it's a fraction that's not defined at \( x=5 \), so it's continuous everywhere else. For (c), after simplifying, it acts like a straight line too, so it's continuous where it's defined. For (d), absolute value functions make a "V" shape graph, which is smooth and doesn't have any breaks, so it's continuous everywhere.

🎯 Exam Tip: When analyzing piecewise functions or functions with absolute values, always break down the analysis into cases for each interval and especially check the points where the definition of the function changes. This ensures a thorough examination of continuity.

Question 3. A function \( f \) is defined by
\[ f(x) = \begin{cases} \frac{x^2-4x+3}{x^2-1}, & \text { for } x \neq 1 \\ 2, & \text { for } x=1 \end{cases} \]
Test the continuity of the function at \( x = 1 \).
Answer:
Given the function:
\[ f(x) = \begin{cases} \frac{x^2-4x+3}{x^2-1}, & \text { for } x \neq 1 \\ 2, & \text { for } x=1 \end{cases} \]
We need to test the continuity of \( f(x) \) at \( x = 1 \).

First, find the left-hand limit (LHL) at \( x = 1 \):
\( \text{LHL} = \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} \frac{x^2-4x+3}{x^2-1} \)
Factorize the numerator and denominator:
\( = \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} \frac{(x-1)(x-3)}{(x-1)(x+1)} \)
Since \( x \neq 1 \), we can cancel \( (x-1) \):
\( = \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} \frac{x-3}{x+1} \)
Substitute \( x = 1 - h \), so as \( x \rightarrow 1^{-} \), \( h \rightarrow 0^{+} \):
\( = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} \frac{(1-h)-3}{(1-h)+1} = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} \frac{-2-h}{2-h} = \frac{-2-0}{2-0} = \frac{-2}{2} = -1 \)

Next, find the right-hand limit (RHL) at \( x = 1 \):
\( \text{RHL} = \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} \frac{x^2-4x+3}{x^2-1} \)
Factorize the numerator and denominator:
\( = \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} \frac{(x-1)(x-3)}{(x-1)(x+1)} \)
Since \( x \neq 1 \), we can cancel \( (x-1) \):
\( = \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} \frac{x-3}{x+1} \)
Substitute \( x = 1 + h \), so as \( x \rightarrow 1^{+} \), \( h \rightarrow 0^{+} \):
\( = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} \frac{(1+h)-3}{(1+h)+1} = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} \frac{-2+h}{2+h} = \frac{-2+0}{2+0} = \frac{-2}{2} = -1 \)

Finally, find the value of the function at \( x = 1 \):
\( f(1) = 2 \) (as given in the function definition)

For continuity at \( x = 1 \), the LHL, RHL, and \( f(1) \) must all be equal.
Here, LHL = \( -1 \), RHL = \( -1 \), but \( f(1) = 2 \).
Since \( \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} f(x) \neq f(1) \), the function \( f(x) \) is discontinuous at \( x = 1 \).
In simple words: A function is continuous if its graph doesn't have any breaks or jumps. For this function, as we get closer and closer to \( x=1 \) from both sides, the value of the function approaches -1. However, the function is specifically defined as 2 at \( x=1 \). Since the approaching value (-1) is not the same as the actual value at the point (2), there's a jump in the graph, making the function discontinuous at \( x=1 \).

🎯 Exam Tip: When dealing with piecewise functions, always check the limits from both the left and right sides at the point where the definition changes, and compare them with the function's value at that exact point. If all three are equal, the function is continuous.

Question 4. Prove that the function \( f(x) = x^n \) is continuous at \( x = n \), where \( n \) is a positive integer.
Answer:
Given the function \( f(x) = x^n \), where \( n \) is a positive integer. We need to prove its continuity at \( x = n \).

First, find the left-hand limit (LHL) at \( x = n \):
\( \underset{x \rightarrow n^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow n^{-}}{\mathrm{Lt}} x^n \)
Substitute \( x = n - h \), so as \( x \rightarrow n^{-} \), \( h \rightarrow 0^{+} \):
\( = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} (n - h)^n \)
\( = (n - 0)^n = n^n \)

Next, find the right-hand limit (RHL) at \( x = n \):
\( \underset{x \rightarrow n^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow n^{+}}{\mathrm{Lt}} x^n \)
Substitute \( x = n + h \), so as \( x \rightarrow n^{+} \), \( h \rightarrow 0^{+} \):
\( = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} (n + h)^n \)
\( = (n + 0)^n = n^n \)

Finally, find the value of the function at \( x = n \):
\( f(n) = n^n \)

Since the LHL, RHL, and the function's value at \( x = n \) are all equal:
\( \underset{x \rightarrow n^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow n^{+}}{\mathrm{Lt}} f(x) = f(n) = n^n \)
Therefore, the function \( f(x) = x^n \) is continuous at \( x = n \). This applies to all integer powers, indicating smooth graphs.
In simple words: The function \( f(x) = x^n \) means "x multiplied by itself n times". For any number \( n \) that is a positive whole number, this kind of function creates a smooth curve without any gaps or breaks. When we check its value at \( x=n \), and also what its value gets close to from both sides of \( n \), they all match up perfectly, proving it's continuous.

🎯 Exam Tip: Functions of the form \( x^n \) (polynomials) are inherently continuous for all real numbers. When proving continuity, remember to show the limits from both sides and the function's value at the specific point and demonstrate their equality.

Question 5. Show that the function
\[ f(x) = \begin{cases} x^2 & \text { for } 1 \leq x<2 \\ 3x-4 & \text { for } 2 \leq x<4 \end{cases} \]
is discontinuous at \( x = 2 \) and continuous at \( x = 3 \).
Answer:
Given the function:
\[ f(x) = \begin{cases} x^2 & \text { for } 1 \leq x<2 \\ 3x-4 & \text { for } 2 \leq x<4 \end{cases} \]
We need to show it is discontinuous at \( x = 2 \) and continuous at \( x = 3 \).

**Part 1: Continuity at \( x = 2 \)**
First, find the value of the function at \( x = 2 \):
Since \( x = 2 \) falls into the second case (\( 2 \leq x<4 \)), \( f(2) = 3(2) - 4 = 6 - 4 = 2 \).

Next, find the left-hand limit (LHL) at \( x = 2 \):
Since \( x < 2 \) falls into the first case (\( 1 \leq x<2 \)),
\( \underset{x \rightarrow 2^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 2^{-}}{\mathrm{Lt}} x^2 = (2)^2 = 4 \).

Then, find the right-hand limit (RHL) at \( x = 2 \):
Since \( x > 2 \) falls into the second case (\( 2 \leq x<4 \)),
\( \underset{x \rightarrow 2^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 2^{+}}{\mathrm{Lt}} (3x - 4) = 3(2) - 4 = 6 - 4 = 2 \).

For continuity at \( x = 2 \), LHL, RHL, and \( f(2) \) must all be equal.
Here, LHL = \( 4 \), RHL = \( 2 \), and \( f(2) = 2 \).
Since \( \underset{x \rightarrow 2^{-}}{\mathrm{Lt}} f(x) \neq \underset{x \rightarrow 2^{+}}{\mathrm{Lt}} f(x) \), the function \( f(x) \) is discontinuous at \( x = 2 \). There's a clear jump in value at this point.

**Part 2: Continuity at \( x = 3 \)**
First, find the value of the function at \( x = 3 \):
Since \( x = 3 \) falls into the second case (\( 2 \leq x<4 \)), \( f(3) = 3(3) - 4 = 9 - 4 = 5 \).

Next, find the left-hand limit (LHL) at \( x = 3 \):
Since \( x < 3 \) and is close to 3, it falls into the second case, \( \underset{x \rightarrow 3^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 3^{-}}{\mathrm{Lt}} (3x - 4) = 3(3) - 4 = 9 - 4 = 5 \).

Then, find the right-hand limit (RHL) at \( x = 3 \):
Since \( x > 3 \) and is close to 3, it falls into the second case, \( \underset{x \rightarrow 3^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 3^{+}}{\mathrm{Lt}} (3x - 4) = 3(3) - 4 = 9 - 4 = 5 \).

For continuity at \( x = 3 \), LHL, RHL, and \( f(3) \) must all be equal.
Here, LHL = \( 5 \), RHL = \( 5 \), and \( f(3) = 5 \).
Since \( \underset{x \rightarrow 3^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 3^{+}}{\mathrm{Lt}} f(x) = f(3) = 5 \), the function \( f(x) \) is continuous at \( x = 3 \).
In simple words: For the function to be continuous, its graph must be smooth without any breaks. At \( x=2 \), the function changes its rule. If you check the value just before \( x=2 \) (using \( x^2 \)), it's 4. If you check it at \( x=2 \) and just after \( x=2 \) (using \( 3x-4 \)), it's 2. Since 4 is not 2, there's a break, so it's discontinuous. But at \( x=3 \), the function uses the same rule \( 3x-4 \) on both sides, and its value matches, making it continuous there.

🎯 Exam Tip: When proving discontinuity, you only need to show that one of the three conditions for continuity (LHL, RHL, or function value) is not met. For continuity, all three must be explicitly shown to be equal.

Question 6.
(a) Find all points of discontinuity of \( f \) where \( f \) is defined by
\[ f(x) = \begin{cases} 2x+3, & \text { if } x \leq 2 \\ 2x-3, & \text { if } x>2 \end{cases} \]
(b) Discuss the discontinuity of the function
\[ f(x) = \begin{cases} 2x-1, & \text { if } x<0 \\ 2x+1, & \text { if } x \geq 0 \end{cases} \]
at \( x = 0 \).
(c) Is the function defined by
\[ f(x) = \begin{cases} x+5, & \text { if } x \leq 1 \\ x-5, & \text { if } x>1 \end{cases} \]
a continuous function?
(d) Show that
\[ f(x) = \begin{cases} 5x-4, & \text { when } 0is continuous at \( x = 1 \).
Answer:
(a) Given the function:
\[ f(x) = \begin{cases} 2x+3, & \text { if } x \leq 2 \\ 2x-3, & \text { if } x>2 \end{cases} \]
The domain of \( f \) is all real numbers, \( D_f = \mathbb{R} \). We examine continuity at the point where the definition changes, \( x = 2 \), and also for arbitrary points \( c \).

Case I: When \( c < 2 \)
\( f(c) = 2c+3 \).
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} (2x+3) = 2c+3 \).
So, \( f(x) \) is continuous for all \( c < 2 \).

Case II: When \( c > 2 \)
\( f(c) = 2c-3 \).
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} (2x-3) = 2c-3 \).
So, \( f(x) \) is continuous for all \( c > 2 \).

Case III: When \( c = 2 \)
First, find the value of \( f(2) = 2(2)+3 = 4+3 = 7 \).
Next, find the left-hand limit (LHL) at \( x = 2 \):
\( \underset{x \rightarrow 2^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 2^{-}}{\mathrm{Lt}} (2x+3) = 2(2)+3 = 7 \).
Then, find the right-hand limit (RHL) at \( x = 2 \):
\( \underset{x \rightarrow 2^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 2^{+}}{\mathrm{Lt}} (2x-3) = 2(2)-3 = 1 \).
Since LHL (\( 7 \)) \( \neq \) RHL (\( 1 \)), the function \( f(x) \) is discontinuous at \( x = 2 \).
Therefore, the only point of discontinuity for \( f(x) \) is at \( x = 2 \).

(b) Given the function:
\[ f(x) = \begin{cases} 2x-1, & \text { if } x<0 \\ 2x+1, & \text { if } x \geq 0 \end{cases} \]
We discuss discontinuity at \( x = 0 \).

First, find the value of \( f(0) = 2(0)+1 = 1 \).
Next, find the left-hand limit (LHL) at \( x = 0 \):
\( \underset{x \rightarrow 0^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 0^{-}}{\mathrm{Lt}} (2x-1) = 2(0)-1 = -1 \).
Then, find the right-hand limit (RHL) at \( x = 0 \):
\( \underset{x \rightarrow 0^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 0^{+}}{\mathrm{Lt}} (2x+1) = 2(0)+1 = 1 \).
Since LHL (\( -1 \)) \( \neq \) RHL (\( 1 \)), the function \( f(x) \) is discontinuous at \( x = 0 \).

(c) Given the function:
\[ f(x) = \begin{cases} x+5, & \text { if } x \leq 1 \\ x-5, & \text { if } x>1 \end{cases} \]
The domain of \( f \) is \( \mathbb{R} \). We examine continuity at the point where the definition changes, \( x = 1 \).

Case I: When \( c < 1 \)
\( f(c) = c+5 \).
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} (x+5) = c+5 \).
So, \( f(x) \) is continuous for all \( c < 1 \).

Case II: When \( c > 1 \)
\( f(c) = c-5 \).
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} (x-5) = c-5 \).
So, \( f(x) \) is continuous for all \( c > 1 \).

Case III: When \( c = 1 \)
First, find the value of \( f(1) = 1+5 = 6 \).
Next, find the left-hand limit (LHL) at \( x = 1 \):
\( \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} (x+5) = 1+5 = 6 \).
Then, find the right-hand limit (RHL) at \( x = 1 \):
\( \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} (x-5) = 1-5 = -4 \).
Since LHL (\( 6 \)) \( \neq \) RHL (\( -4 \)), the function \( f(x) \) is discontinuous at \( x = 1 \).
Therefore, \( f(x) \) is not a continuous function.

(d) Given the function:
\[ f(x) = \begin{cases} 5x-4, & \text { when } 0We need to show that \( f(x) \) is continuous at \( x = 1 \).

First, find the value of the function at \( x = 1 \):
Since \( x = 1 \) falls into the first case (\( 0 < x \leq 1 \)), \( f(1) = 5(1) - 4 = 5 - 4 = 1 \).

Next, find the left-hand limit (LHL) at \( x = 1 \):
Since \( x < 1 \) falls into the first case,
\( \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} (5x - 4) = 5(1) - 4 = 1 \).

Then, find the right-hand limit (RHL) at \( x = 1 \):
Since \( x > 1 \) falls into the second case (\( 1 < x < 2 \)),
\( \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} (4x^3 - 3x) = 4(1)^3 - 3(1) = 4 - 3 = 1 \).

For continuity at \( x = 1 \), LHL, RHL, and \( f(1) \) must all be equal.
Here, LHL = \( 1 \), RHL = \( 1 \), and \( f(1) = 1 \).
Since \( \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} f(x) = f(1) = 1 \), the function \( f(x) \) is continuous at \( x = 1 \).
In simple words: For each part, we check if the function's graph has any breaks. If the rule for the function changes at a specific point, like \( x=2 \) in (a) or \( x=0 \) in (b) or \( x=1 \) in (c), we compare the values it approaches from the left, from the right, and its exact value at that point. If they don't all match, the function has a break (it's discontinuous). If they match, it's smooth (continuous), as shown in (d) at \( x=1 \).

🎯 Exam Tip: Remember to clearly state the domain for each part of a piecewise function and evaluate the limits from both sides at the critical points where the function's definition changes. This structured approach helps in avoiding errors in continuity examination.

Question 7. Examine the continuity for the following functions:
(a) \( f(x) = \begin{cases} x+1, & \text { if } x \geq 1 \\ x^2+1, & \text { if } x<1 \end{cases} \)
(b) \( f(x) = \begin{cases} x^{10}-1, & \text { if } x \leq 1 \\ x^2, & \text { if } x>1 \end{cases} \)
Answer:
(a) Given the function:
\[ f(x) = \begin{cases} x+1, & \text { if } x \geq 1 \\ x^2+1, & \text { if } x<1 \end{cases} \]
The domain of \( f \) is all real numbers, \( D_f = \mathbb{R} \). We examine continuity at the point where the definition changes, \( x = 1 \), and for arbitrary points \( c \).

Case I: When \( c < 1 \)
\( f(c) = c^2+1 \).
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} (x^2+1) = c^2+1 \).
So, \( f(x) \) is continuous for all \( c < 1 \).

Case II: When \( c > 1 \)
\( f(c) = c+1 \).
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} (x+1) = c+1 \).
So, \( f(x) \) is continuous for all \( c > 1 \).

Case III: When \( c = 1 \)
First, find the value of \( f(1) = 1+1 = 2 \).
Next, find the left-hand limit (LHL) at \( x = 1 \):
\( \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} (x^2+1) = 1^2+1 = 2 \).
Then, find the right-hand limit (RHL) at \( x = 1 \):
\( \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} (x+1) = 1+1 = 2 \).
Since LHL (\( 2 \)), RHL (\( 2 \)), and \( f(1) \) (\( 2 \)) are all equal, the function \( f(x) \) is continuous at \( x = 1 \).
Therefore, by combining all three cases, the function \( f(x) \) is continuous for all \( x \in \mathbb{R} \).

(b) Given the function:
\[ f(x) = \begin{cases} x^{10}-1, & \text { if } x \leq 1 \\ x^2, & \text { if } x>1 \end{cases} \]
The domain of \( f \) is all real numbers, \( D_f = \mathbb{R} \). We examine continuity at the point where the definition changes, \( x = 1 \), and for arbitrary points \( c \).

Case I: When \( c < 1 \)
\( f(c) = c^{10}-1 \).
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} (x^{10}-1) = c^{10}-1 \).
So, \( f(x) \) is continuous for all \( c < 1 \).

Case II: When \( c > 1 \)
\( f(c) = c^2 \).
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} x^2 = c^2 \).
So, \( f(x) \) is continuous for all \( c > 1 \).

Case III: When \( c = 1 \)
First, find the value of \( f(1) = 1^{10}-1 = 1-1 = 0 \).
Next, find the left-hand limit (LHL) at \( x = 1 \):
\( \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} (x^{10}-1) = 1^{10}-1 = 0 \).
Then, find the right-hand limit (RHL) at \( x = 1 \):
\( \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} x^2 = 1^2 = 1 \).
Since LHL (\( 0 \)) \( \neq \) RHL (\( 1 \)), the function \( f(x) \) is discontinuous at \( x = 1 \).
Therefore, the function \( f(x) \) is continuous for all \( x \in \mathbb{R} \) except at \( x = 1 \).
In simple words: We check if the graph of each function has any breaks. For part (a), the function changes its rule at \( x=1 \). When we check the values from both sides of 1 and at 1 itself, they all match (they are all 2). This means the graph flows smoothly through \( x=1 \), so it's continuous everywhere. For part (b), the rule also changes at \( x=1 \). The value approaching from the left is 0, but the value approaching from the right is 1. Since these don't match, there's a break at \( x=1 \), so it's discontinuous there.

🎯 Exam Tip: Always analyze the continuity of piecewise functions by considering three cases: when \( x \) is less than the critical point, when \( x \) is greater than the critical point, and exactly at the critical point. This systematic approach ensures all conditions for continuity are checked.

Question 8. Discuss the continuity of the function \( f(x) \) at \( x = \frac{1}{2} \) when \( f(x) \) is defined as follows:
\[ f(x) = \begin{cases} \frac{1}{2}+x, & 0 \leq x<\frac{1}{2} \\ 1, & x=\frac{1}{2} \\ \frac{3}{2}+x, & \frac{1}{2}
Answer:
Given the function:
\[ f(x) = \begin{cases} \frac{1}{2}+x, & 0 \leq x<\frac{1}{2} \\ 1, & x=\frac{1}{2} \\ \frac{3}{2}+x, & \frac{1}{2}We need to discuss its continuity at \( x = \frac{1}{2} \).

First, find the value of the function at \( x = \frac{1}{2} \):
From the definition, \( f\left(\frac{1}{2}\right) = 1 \).

Next, find the left-hand limit (LHL) at \( x = \frac{1}{2} \):
Since \( x < \frac{1}{2} \) falls into the first case (\( 0 \leq x<\frac{1}{2} \)),
\( \underset{x \rightarrow \frac{1}{2}^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow \frac{1}{2}^{-}}{\mathrm{Lt}} \left(\frac{1}{2}+x\right) = \frac{1}{2} + \frac{1}{2} = 1 \).

Then, find the right-hand limit (RHL) at \( x = \frac{1}{2} \):
Since \( x > \frac{1}{2} \) falls into the third case (\( \frac{1}{2}\( \underset{x \rightarrow \frac{1}{2}^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow \frac{1}{2}^{+}}{\mathrm{Lt}} \left(\frac{3}{2}+x\right) = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2 \).

For continuity at \( x = \frac{1}{2} \), LHL, RHL, and \( f\left(\frac{1}{2}\right) \) must all be equal.
Here, LHL = \( 1 \), RHL = \( 2 \), and \( f\left(\frac{1}{2}\right) = 1 \).
Since \( \underset{x \rightarrow \frac{1}{2}^{-}}{\mathrm{Lt}} f(x) \neq \underset{x \rightarrow \frac{1}{2}^{+}}{\mathrm{Lt}} f(x) \), the function \( f(x) \) is discontinuous at \( x = \frac{1}{2} \). There's a noticeable jump in the function's value.
In simple words: To see if a function is continuous at a point, we check if the path leading to that point from the left, from the right, and the point itself all meet up. For this function, at \( x = \frac{1}{2} \), the value from the left is 1, the value from the right is 2, and the actual value at \( \frac{1}{2} \) is 1. Because the value from the right (2) doesn't match the others, the function has a break or a jump at \( x = \frac{1}{2} \), so it is not continuous.

🎯 Exam Tip: Pay close attention to the definition of piecewise functions, especially the exact values at the critical points versus the limits approaching those points. A common mistake is to confuse the function's value with one of the limits.

Question 9. Examine the continuity of the function \( f(x) = \begin{cases} -2, & \text { if } x \leq -1 \\ 2x, & \text { if } -1 < x \leq 1 \\ 2, & \text { if } x>1 \end{cases} \).
Answer:
Given the function:
\[ f(x) = \begin{cases} -2, & \text { if } x \leq -1 \\ 2x, & \text { if } -1 < x \leq 1 \\ 2, & \text { if } x>1 \end{cases} \]
The domain of \( f \) is all real numbers, \( D_f = \mathbb{R} \). We examine continuity at the points where the definition changes, \( x = -1 \) and \( x = 1 \), and for arbitrary points \( c \).

Case I: When \( c < -1 \)
\( f(c) = -2 \).
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} (-2) = -2 \).
So, \( f(x) \) is continuous for all \( c < -1 \).

Case II: When \( c = -1 \)
First, find the value of \( f(-1) = -2 \).
Next, find the left-hand limit (LHL) at \( x = -1 \):
\( \underset{x \rightarrow -1^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow -1^{-}}{\mathrm{Lt}} (-2) = -2 \).
Then, find the right-hand limit (RHL) at \( x = -1 \):
\( \underset{x \rightarrow -1^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow -1^{+}}{\mathrm{Lt}} (2x) = 2(-1) = -2 \).
Since LHL (\( -2 \)), RHL (\( -2 \)), and \( f(-1) \) (\( -2 \)) are all equal, the function \( f(x) \) is continuous at \( x = -1 \).

Case III: When \( -1 < c < 1 \)
\( f(c) = 2c \).
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} (2x) = 2c \).
So, \( f(x) \) is continuous for all \( -1 < c < 1 \).

Case IV: When \( c = 1 \)
First, find the value of \( f(1) = 2(1) = 2 \).
Next, find the left-hand limit (LHL) at \( x = 1 \):
\( \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{-}}{\mathrm{Lt}} (2x) = 2(1) = 2 \).
Then, find the right-hand limit (RHL) at \( x = 1 \):
\( \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^{+}}{\mathrm{Lt}} (2) = 2 \).
Since LHL (\( 2 \)), RHL (\( 2 \)), and \( f(1) \) (\( 2 \)) are all equal, the function \( f(x) \) is continuous at \( x = 1 \).

Case V: When \( c > 1 \)
\( f(c) = 2 \).
\( \underset{x \rightarrow c}{\mathrm{Lt}} f(x) = \underset{x \rightarrow c}{\mathrm{Lt}} (2) = 2 \).
So, \( f(x) \) is continuous for all \( c > 1 \).

By combining all five cases, the function \( f(x) \) is continuous at every point in its domain \( \mathbb{R} \). This means there are no breaks in the graph.
In simple words: This function is made of three different rules depending on the value of \( x \). To check if it's continuous everywhere, we look at the points where the rules change, which are \( x=-1 \) and \( x=1 \). At both these points, when we check the value the function approaches from the left, from the right, and its actual value, they all match perfectly. This means the graph is smooth and doesn't have any jumps or breaks anywhere.

🎯 Exam Tip: For piecewise functions with multiple critical points, it's essential to examine continuity at each point where the function's definition changes. Also, don't forget to consider the intervals where the function is defined by a single expression, as these are usually continuous by default.

Question 10. A function \( f(x) \) is defined as follows: \( f(x) = x \cos \frac{1}{x} \), when \( x \neq 0 \), \( f(0) = 0 \). Examine the continuity at \( x = 0 \).
Answer:
Given the function:
\[ f(x) = \begin{cases} x \cos \frac{1}{x}, & \text { when } x \neq 0 \\ 0, & \text { when } x=0 \end{cases} \]
We need to examine the continuity of \( f(x) \) at \( x = 0 \).

First, find the value of the function at \( x = 0 \):
\( f(0) = 0 \) (as given).

Next, find the limit of \( f(x) \) as \( x \) approaches \( 0 \):
\( \underset{x \rightarrow 0}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 0}{\mathrm{Lt}} \left(x \cos \frac{1}{x}\right) \)
We know that the cosine function is bounded, meaning its values always lie between \( -1 \) and \( 1 \), inclusive. So, \( -1 \leq \cos \frac{1}{x} \leq 1 \).
Now, consider the limit of \( x \) as \( x \rightarrow 0 \), which is \( \underset{x \rightarrow 0}{\mathrm{Lt}} x = 0 \).
By the Squeeze Theorem (or product of a function tending to zero and a bounded function), if \( \underset{x \rightarrow 0}{\mathrm{Lt}} h(x) = 0 \) and \( g(x) \) is a bounded function, then \( \underset{x \rightarrow 0}{\mathrm{Lt}} [h(x)g(x)] = 0 \).
Here, let \( h(x) = x \) and \( g(x) = \cos \frac{1}{x} \).
\( \underset{x \rightarrow 0}{\mathrm{Lt}} x = 0 \).
\( \cos \frac{1}{x} \) oscillates between \( -1 \) and \( 1 \) as \( x \rightarrow 0 \), so it is a bounded function.

Therefore,
\( \underset{x \rightarrow 0}{\mathrm{Lt}} \left(x \cos \frac{1}{x}\right) = 0 \).

Since \( \underset{x \rightarrow 0}{\mathrm{Lt}} f(x) = f(0) = 0 \), the function \( f(x) \) is continuous at \( x = 0 \). This shows how even an oscillating part can be "tamed" by multiplication with zero.
In simple words: For a function to be continuous at a point, its value at that point should match what the function is getting closer to from both sides. Here, at \( x=0 \), the function is directly set to 0. For values near \( x=0 \), the \( \cos \frac{1}{x} \) part wiggles up and down very fast. But when you multiply that wiggling part by \( x \), which itself is becoming zero, the whole thing shrinks down to zero. So, since the limit is 0 and the function's value is 0, it's continuous at \( x=0 \).

🎯 Exam Tip: For limits involving oscillating functions (like sine or cosine) multiplied by a function tending to zero, the Squeeze Theorem is often crucial. Remember that if one function approaches zero and the other is bounded, their product's limit is zero.

Question 11. The function \( f(x) \) is defined as follows:
\[ f(x) = \begin{cases} (x-a) \cos \frac{1}{x-a}, & \text { when } x \neq a \\ 0, & \text { when } x=a \end{cases} \]
Examine the continuity, when \( x = a \).
Answer:
Given the function:
\[ f(x) = \begin{cases} (x-a) \cos \frac{1}{x-a}, & \text { when } x \neq a \\ 0, & \text { when } x=a \end{cases} \]
We need to examine the continuity of \( f(x) \) at \( x = a \).

First, find the value of the function at \( x = a \):
\( f(a) = 0 \) (as given).

Next, find the limit of \( f(x) \) as \( x \) approaches \( a \):
\( \underset{x \rightarrow a}{\mathrm{Lt}} f(x) = \underset{x \rightarrow a}{\mathrm{Lt}} \left((x-a) \cos \frac{1}{x-a}\right) \)
Let \( y = x-a \). As \( x \rightarrow a \), \( y \rightarrow 0 \). The expression becomes:
\( = \underset{y \rightarrow 0}{\mathrm{Lt}} \left(y \cos \frac{1}{y}\right) \)
We know that the cosine function, \( \cos \frac{1}{y} \), is bounded between \( -1 \) and \( 1 \). So, \( -1 \leq \cos \frac{1}{y} \leq 1 \).
Also, \( \underset{y \rightarrow 0}{\mathrm{Lt}} y = 0 \).
Using the property that the product of a function tending to zero and a bounded function has a limit of zero:
\( \underset{y \rightarrow 0}{\mathrm{Lt}} \left(y \cos \frac{1}{y}\right) = 0 \).

Therefore, \( \underset{x \rightarrow a}{\mathrm{Lt}} f(x) = 0 \).

Since \( \underset{x \rightarrow a}{\mathrm{Lt}} f(x) = f(a) = 0 \), the function \( f(x) \) is continuous at \( x = a \). This function demonstrates a common pattern for creating continuous functions with oscillating parts.
In simple words: To check if this function is smooth at \( x=a \), we look at its value exactly at \( a \) (which is 0) and what value it gets close to when \( x \) is just near \( a \). Even though the \( \cos \frac{1}{x-a} \) part wiggles very fast when \( x \) is close to \( a \), the \( x-a \) part becomes zero at \( a \). When you multiply a number that is going to zero by a wiggling but bounded number, the result also goes to zero. Since both the function's value at \( a \) and its limit are 0, it is continuous at \( x=a \).

🎯 Exam Tip: Functions of the form \( (x-a) \cdot g(x) \) where \( g(x) \) is bounded (like sine or cosine) and \( \underset{x \rightarrow a}{\mathrm{Lt}} (x-a) = 0 \) are frequently used to test understanding of continuity. Applying the Squeeze Theorem (or the bounded function property) is key here.

Question 12. Examine the continuity of the following functions:
(a) \( f(x) = \begin{cases} \frac{|x|}{x}, & \text{ if } x \neq 0 \\ 0, & \text{ if } x=0 \end{cases} \)
(b) \( f(x) = \begin{cases} \frac{|x|}{x}, & \text{ if } x<0 \\ -1, & \text{ if } x \geq 0 \end{cases} \)
Answer:
(a) The given function is \( f(x) = \begin{cases} \frac{|x|}{x}, & \text{ if } x \neq 0 \\ 0, & \text{ if } x=0 \end{cases} \)
We can write this function in parts:
If \( x > 0 \), then \( |x| = x \), so \( f(x) = \frac{x}{x} = 1 \).
If \( x < 0 \), then \( |x| = -x \), so \( f(x) = \frac{-x}{x} = -1 \).
If \( x = 0 \), then \( f(x) = 0 \).
So, the function can be written as: \( f(x) = \begin{cases} 1, & \text{ if } x > 0 \\ -1, & \text{ if } x < 0 \\ 0, & \text{ if } x = 0 \end{cases} \)
We need to check the continuity of \( f(x) \) at \( x=0 \).
For Left Hand Limit (LHL) at \( x=0 \):
\( \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-1) = -1 \)
For Right Hand Limit (RHL) at \( x=0 \):
\( \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (1) = 1 \)
Also, the value of the function at \( x=0 \) is \( f(0) = 0 \).
Since \( \text{LHL} \neq \text{RHL} \), the limit of the function at \( x=0 \) does not exist.
Therefore, the function \( f(x) \) is discontinuous at \( x=0 \).
For any point \( c > 0 \), \( f(c) = 1 \). The limit \( \lim_{x \to c} f(x) = \lim_{x \to c} (1) = 1 \). So it's continuous for \( c > 0 \).
For any point \( c < 0 \), \( f(c) = -1 \). The limit \( \lim_{x \to c} f(x) = \lim_{x \to c} (-1) = -1 \). So it's continuous for \( c < 0 \).
In conclusion, \( f(x) \) is continuous everywhere except at \( x=0 \).
In simple words: This function acts differently around zero. If you come from numbers smaller than zero, the answer is -1. If you come from numbers larger than zero, the answer is 1. Right at zero, the answer is 0. Since these three numbers are not the same, the function has a break at zero.

🎯 Exam Tip: When dealing with piecewise functions involving absolute values, always rewrite the function definition without the absolute value based on the intervals to clearly see its behavior.

Question 12. Examine the continuity of the following functions:
(a) \( f(x) = \begin{cases} \frac{|x|}{x}, & \text{ if } x \neq 0 \\ 0, & \text{ if } x=0 \end{cases} \)
(b) \( f(x) = \begin{cases} \frac{|x|}{x}, & \text{ if } x<0 \\ -1, & \text{ if } x \geq 0 \end{cases} \)
Answer:
(b) The given function is \( f(x) = \begin{cases} \frac{|x|}{x}, & \text{ if } x<0 \\ -1, & \text{ if } x \geq 0 \end{cases} \)
Let's simplify the first part of the function for \( x < 0 \).
If \( x < 0 \), then \( |x| = -x \). So, \( \frac{|x|}{x} = \frac{-x}{x} = -1 \).
This means the function can be rewritten as:
\( f(x) = \begin{cases} -1, & \text{ if } x<0 \\ -1, & \text{ if } x \geq 0 \end{cases} \)
So, \( f(x) = -1 \) for all real numbers \( x \).
A constant function is always continuous everywhere.
We can also check this at \( x=0 \):
For Left Hand Limit (LHL) at \( x=0 \):
\( \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-1) = -1 \)
For Right Hand Limit (RHL) at \( x=0 \):
\( \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (-1) = -1 \)
The value of the function at \( x=0 \) is \( f(0) = -1 \) (since \( x \geq 0 \) includes \( x=0 \)).
Since \( \text{LHL} = \text{RHL} = f(0) = -1 \), the function \( f(x) \) is continuous at \( x=0 \).
As it's continuous at \( x=0 \) and clearly constant for \( x<0 \) and \( x \geq 0 \), it is continuous everywhere.
In simple words: This function always gives the answer -1, no matter what x-value you pick. Functions that always give the same number are called constant functions, and they are always smooth and unbroken.

🎯 Exam Tip: Simplify piecewise functions first. Sometimes, different parts of the definition combine to form a simpler function (like a constant function), which makes checking continuity much easier.

Question 13. Examine the continuity at x = 0
(a) \( f(x) = \begin{cases} \frac{\tan 2x}{3x}, & \text{ when } x \neq 0 \\ \frac{2}{3}, & \text{ when } x=0 \end{cases} \)
(b) \( f(x) = 1 + \frac{|x|}{x} \) for \( x \neq 0 \) and \( f(0) = 1 \).
Answer:
(a) The function is \( f(x) = \begin{cases} \frac{\tan 2x}{3x}, & \text{ when } x \neq 0 \\ \frac{2}{3}, & \text{ when } x=0 \end{cases} \)
To check continuity at \( x=0 \), we need to find the limit of \( f(x) \) as \( x \to 0 \).
\( \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\tan 2x}{3x} \)
We can rewrite this limit to use the standard limit \( \lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1 \).
\( \lim_{x \to 0} \frac{\tan 2x}{3x} = \lim_{x \to 0} \frac{1}{3} \cdot \frac{\tan 2x}{x} = \lim_{x \to 0} \frac{1}{3} \cdot \frac{\tan 2x}{2x} \cdot 2 \)
\( = \frac{2}{3} \lim_{x \to 0} \frac{\tan 2x}{2x} \)
As \( x \to 0 \), \( 2x \to 0 \). So, \( \lim_{x \to 0} \frac{\tan 2x}{2x} = 1 \).
Therefore, \( \lim_{x \to 0} f(x) = \frac{2}{3} \cdot 1 = \frac{2}{3} \).
The value of the function at \( x=0 \) is given as \( f(0) = \frac{2}{3} \).
Since \( \lim_{x \to 0} f(x) = f(0) \), the function \( f(x) \) is continuous at \( x=0 \).
In simple words: For this function, as x gets super close to zero, the function's value gets closer and closer to 2/3. Since the function is also exactly 2/3 when x is zero, there are no gaps or jumps, so it is continuous.

🎯 Exam Tip: Remember standard limits like \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \) and \( \lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1 \). Manipulate expressions to fit these forms by multiplying and dividing by suitable terms.

Question 13. Examine the continuity at x = 0
(a) \( f(x) = \begin{cases} \frac{\tan 2x}{3x}, & \text{ when } x \neq 0 \\ \frac{2}{3}, & \text{ when } x=0 \end{cases} \)
(b) \( f(x) = 1 + \frac{|x|}{x} \) for \( x \neq 0 \) and \( f(0) = 1 \).
Answer:
(b) The function is \( f(x) = 1 + \frac{|x|}{x} \) for \( x \neq 0 \) and \( f(0) = 1 \).
To check continuity at \( x=0 \), we need to find the Left Hand Limit (LHL) and Right Hand Limit (RHL) at \( x=0 \).
For LHL at \( x=0 \):
\( \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left(1 + \frac{|x|}{x}\right) \)
Since \( x \to 0^- \), \( x \) is a small negative number. So, \( |x| = -x \).
\( \text{LHL} = \lim_{x \to 0^-} \left(1 + \frac{-x}{x}\right) = \lim_{x \to 0^-} (1 - 1) = \lim_{x \to 0^-} (0) = 0 \)
For RHL at \( x=0 \):
\( \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left(1 + \frac{|x|}{x}\right) \)
Since \( x \to 0^+ \), \( x \) is a small positive number. So, \( |x| = x \).
\( \text{RHL} = \lim_{x \to 0^+} \left(1 + \frac{x}{x}\right) = \lim_{x \to 0^+} (1 + 1) = \lim_{x \to 0^+} (2) = 2 \)
The value of the function at \( x=0 \) is given as \( f(0) = 1 \).
Since \( \text{LHL} \neq \text{RHL} \) (because \( 0 \neq 2 \)), the limit of the function at \( x=0 \) does not exist.
Therefore, the function \( f(x) \) is discontinuous at \( x=0 \).
In simple words: When we approach zero from the left side, the function gives zero. When we approach zero from the right side, it gives two. Since these results are different, the function has a clear break at zero.

🎯 Exam Tip: For functions involving \( |x| \), always evaluate the left and right-hand limits separately as \( |x| \) changes its definition depending on whether \( x \) is positive or negative.

Question 14. Find the value of a, if the function f(x) defined by \( f(x) = \begin{cases} 2x-1, & x<2 \\ a, & x=2 \\ x+1, & x>2 \end{cases} \) is continuous at x = 2.
Answer: The given function is \( f(x) = \begin{cases} 2x-1, & x<2 \\ a, & x=2 \\ x+1, & x>2 \end{cases} \)
For the function to be continuous at \( x=2 \), the Left Hand Limit (LHL), Right Hand Limit (RHL), and the value of the function at \( x=2 \) must all be equal.
First, calculate the LHL at \( x=2 \):
\( \text{LHL} = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x-1) \)
Substitute \( x=2 \): \( 2(2) - 1 = 4 - 1 = 3 \)
Next, calculate the RHL at \( x=2 \):
\( \text{RHL} = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x+1) \)
Substitute \( x=2 \): \( 2 + 1 = 3 \)
The value of the function at \( x=2 \) is given as \( f(2) = a \).
Since the function is continuous at \( x=2 \), we must have \( \text{LHL} = \text{RHL} = f(2) \).
So, \( 3 = 3 = a \).
Therefore, the value of \( a \) is \( 3 \).
In simple words: For a function to be smooth at a point, its value must be the same whether you approach it from the left, from the right, or are exactly at that point. Here, both sides of x=2 approach 3, so 'a' must also be 3 to keep the function connected.

🎯 Exam Tip: For continuity at a specific point in a piecewise function, always calculate the Left Hand Limit, Right Hand Limit, and the function's value at that point separately and ensure they are all equal.

Question 15. Find the relationship between a and b so that the function defined by \( f(x) = \begin{cases} ax+1, & x \leq 3 \\ bx+3, & x>3 \end{cases} \) is continuous at x = 3.
Answer: The given function is \( f(x) = \begin{cases} ax+1, & x \leq 3 \\ bx+3, & x>3 \end{cases} \)
For the function to be continuous at \( x=3 \), the Left Hand Limit (LHL), Right Hand Limit (RHL), and the value of the function at \( x=3 \) must all be equal.
First, calculate the LHL at \( x=3 \):
\( \text{LHL} = \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (ax+1) \)
Substitute \( x=3 \): \( a(3) + 1 = 3a + 1 \)
Next, calculate the RHL at \( x=3 \):
\( \text{RHL} = \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (bx+3) \)
Substitute \( x=3 \): \( b(3) + 3 = 3b + 3 \)
The value of the function at \( x=3 \) is \( f(3) = a(3) + 1 = 3a + 1 \) (since \( x \leq 3 \) includes \( x=3 \)).
Since the function is continuous at \( x=3 \), we must have \( \text{LHL} = \text{RHL} = f(3) \).
So, \( 3a + 1 = 3b + 3 \).
Now, we can find the relationship between \( a \) and \( b \):
\( 3a - 3b = 3 - 1 \)
\( 3a - 3b = 2 \)
This is the required relationship between \( a \) and \( b \).
In simple words: To make this function smooth at x=3, the two parts of the function must meet at the same height. This means their values must be equal when x is 3, leading to the equation \( 3a - 3b = 2 \).

🎯 Exam Tip: When finding a relationship between variables for continuity, equate the Left Hand Limit and Right Hand Limit at the critical point. The function's value at the point will be included in one of these limit definitions.

Question 16. For what value of \( \lambda \) is the function \( f(x) = \begin{cases} \lambda(x^2-2x), & \text{ if } x \leq 0 \\ 4x+1, & \text{ if } x>0 \end{cases} \) continuous at x = 0?
Answer: The given function is \( f(x) = \begin{cases} \lambda(x^2-2x), & \text{ if } x \leq 0 \\ 4x+1, & \text{ if } x>0 \end{cases} \)
For the function to be continuous at \( x=0 \), the Left Hand Limit (LHL), Right Hand Limit (RHL), and the value of the function at \( x=0 \) must all be equal.
First, calculate the LHL at \( x=0 \):
\( \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \lambda(x^2-2x) \)
Substitute \( x=0 \): \( \lambda(0^2 - 2(0)) = \lambda(0 - 0) = 0 \)
Next, calculate the RHL at \( x=0 \):
\( \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (4x+1) \)
Substitute \( x=0 \): \( 4(0) + 1 = 0 + 1 = 1 \)
The value of the function at \( x=0 \) is \( f(0) = \lambda(0^2 - 2(0)) = 0 \) (since \( x \leq 0 \) includes \( x=0 \)).
For continuity, we need \( \text{LHL} = \text{RHL} = f(0) \).
Here, \( \text{LHL} = 0 \) and \( \text{RHL} = 1 \).
Since \( \text{LHL} \neq \text{RHL} \) (because \( 0 \neq 1 \)), the limit of the function at \( x=0 \) does not exist.
Therefore, the function \( f(x) \) is discontinuous at \( x=0 \) for any real value of \( \lambda \). No value of \( \lambda \) can make it continuous at \( x=0 \).
In simple words: When we check the two sides of the function at x=0, they don't meet at the same height (one is 0, the other is 1). Because of this jump, the function cannot be made continuous at x=0, no matter what number we choose for lambda.

🎯 Exam Tip: If the Left Hand Limit and Right Hand Limit at a point are different constants, the function is inherently discontinuous at that point, and no parameter (like \( \lambda \)) can force continuity.

Question 17. Find the value of k, so that the function f defined by \( f(x) = \begin{cases} kx+1, & \text{ if } x \leq \pi \\ \cos x, & \text{ if } x>\pi \end{cases} \) is continuous at x = \( \pi \).
Answer: The given function is \( f(x) = \begin{cases} kx+1, & \text{ if } x \leq \pi \\ \cos x, & \text{ if } x>\pi \end{cases} \)
For the function to be continuous at \( x=\pi \), the Left Hand Limit (LHL), Right Hand Limit (RHL), and the value of the function at \( x=\pi \) must all be equal.
First, calculate the LHL at \( x=\pi \):
\( \text{LHL} = \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx+1) \)
Substitute \( x=\pi \): \( k(\pi) + 1 = k\pi + 1 \)
Next, calculate the RHL at \( x=\pi \):
\( \text{RHL} = \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} (\cos x) \)
Substitute \( x=\pi \): \( \cos(\pi) = -1 \)
The value of the function at \( x=\pi \) is \( f(\pi) = k(\pi) + 1 = k\pi + 1 \) (since \( x \leq \pi \) includes \( x=\pi \)).
Since the function is continuous at \( x=\pi \), we must have \( \text{LHL} = \text{RHL} = f(\pi) \).
So, \( k\pi + 1 = -1 \).
Now, solve for \( k \):
\( k\pi = -1 - 1 \)
\( k\pi = -2 \)
\( k = -\frac{2}{\pi} \)
Therefore, the value of \( k \) is \( -\frac{2}{\pi} \).
In simple words: For the function to be continuous at \( \pi \), the value from the left side must match the value from the right side. This means \( k\pi + 1 \) must be equal to \( -1 \). Solving this equation gives us the specific value for \( k \).

🎯 Exam Tip: Pay attention to trigonometric function values at common angles like \( \pi \), \( \pi/2 \), etc. Knowing these values (e.g., \( \cos(\pi)=-1 \)) is crucial for solving continuity problems involving such functions.

Question 18. Find the value of k, for which \( f(x) = \begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x} & \text{ if } -1 \leq x<0 \\ \frac{2x+1}{x-1}, & \text{ if } 0 \leq x<1 \end{cases} \) is continuous at x = 0.
Answer: The given function is \( f(x) = \begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x} & \text{ if } -1 \leq x<0 \\ \frac{2x+1}{x-1}, & \text{ if } 0 \leq x<1 \end{cases} \)
For the function to be continuous at \( x=0 \), the Left Hand Limit (LHL), Right Hand Limit (RHL), and the value of the function at \( x=0 \) must all be equal.
First, calculate the LHL at \( x=0 \):
\( \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x} \)
To evaluate this limit, multiply the numerator and denominator by the conjugate of the numerator:
\( \text{LHL} = \lim_{x \to 0^-} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x} \cdot \frac{\sqrt{1+kx}+\sqrt{1-kx}}{\sqrt{1+kx}+\sqrt{1-kx}} \)
\( \text{LHL} = \lim_{x \to 0^-} \frac{(1+kx)-(1-kx)}{x(\sqrt{1+kx}+\sqrt{1-kx})} \)
\( \text{LHL} = \lim_{x \to 0^-} \frac{1+kx-1+kx}{x(\sqrt{1+kx}+\sqrt{1-kx})} \)
\( \text{LHL} = \lim_{x \to 0^-} \frac{2kx}{x(\sqrt{1+kx}+\sqrt{1-kx})} \)
Cancel \( x \) from the numerator and denominator (since \( x \neq 0 \) as \( x \to 0^- \)):
\( \text{LHL} = \lim_{x \to 0^-} \frac{2k}{\sqrt{1+kx}+\sqrt{1-kx}} \)
Substitute \( x=0 \): \( \frac{2k}{\sqrt{1+k(0)}+\sqrt{1-k(0)}} = \frac{2k}{\sqrt{1}+\sqrt{1}} = \frac{2k}{1+1} = \frac{2k}{2} = k \)
Next, calculate the RHL at \( x=0 \):
\( \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{2x+1}{x-1} \)
Substitute \( x=0 \): \( \frac{2(0)+1}{0-1} = \frac{1}{-1} = -1 \)
The value of the function at \( x=0 \) is defined by the second piece of the function (since \( 0 \leq x < 1 \) includes \( x=0 \)).
So, \( f(0) = \frac{2(0)+1}{0-1} = -1 \).
For the function to be continuous at \( x=0 \), we need \( \text{LHL} = \text{RHL} = f(0) \).
Therefore, \( k = -1 \).
In simple words: To make the function smooth at x=0, the left side, the right side, and the point itself must all meet at the same value. After calculating, the left side gives k, and the right side gives -1. So, k must be -1 for everything to connect.

🎯 Exam Tip: When evaluating limits involving square roots, especially when \( x \) approaches zero, rationalize the expression by multiplying by the conjugate. This helps eliminate the indeterminate form \( \frac{0}{0} \).

Question 19. If the function \( f(x) = \begin{cases} \frac{1-\cos 2x}{2x^2}, & x \neq 0 \\ k, & x=0 \end{cases} \) is continuous at x = 0, then find the value of k.
Answer: The given function is \( f(x) = \begin{cases} \frac{1-\cos 2x}{2x^2}, & x \neq 0 \\ k, & x=0 \end{cases} \)
For the function to be continuous at \( x=0 \), the limit of \( f(x) \) as \( x \to 0 \) must be equal to \( f(0) \).
First, calculate the limit of \( f(x) \) as \( x \to 0 \):
\( \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1-\cos 2x}{2x^2} \)
We use the trigonometric identity \( 1 - \cos \theta = 2\sin^2 \left(\frac{\theta}{2}\right) \). Here \( \theta = 2x \), so \( 1 - \cos 2x = 2\sin^2 \left(\frac{2x}{2}\right) = 2\sin^2 x \).
Substitute this into the limit expression:
\( \lim_{x \to 0} \frac{2\sin^2 x}{2x^2} \)
Cancel the \( 2 \) from the numerator and denominator:
\( \lim_{x \to 0} \frac{\sin^2 x}{x^2} = \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 \)
Using the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):
\( \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 = (1)^2 = 1 \)
The value of the function at \( x=0 \) is given as \( f(0) = k \).
Since the function is continuous at \( x=0 \), we must have \( \lim_{x \to 0} f(x) = f(0) \).
Therefore, \( 1 = k \).
The value of \( k \) is \( 1 \).
In simple words: For this function to be smooth at x=0, its value at x=0 (which is k) must be the same as where the function is heading when x gets very close to 0. By calculating the limit, we find it approaches 1, so k must be 1.

🎯 Exam Tip: Familiarity with trigonometric identities, especially those involving \( 1 - \cos \theta \), is essential for solving limits that appear in continuity problems. Always try to convert expressions into standard limit forms.

Question 20. Prove that the function \( f(x) = \begin{cases} \frac{x}{|x|+2x^2}, & x \neq 0 \\ k, & x=0 \end{cases} \) remains discontinuous at x = 0, regardless of the choice of k.
Answer: The given function is \( f(x) = \begin{cases} \frac{x}{|x|+2x^2}, & x \neq 0 \\ k, & x=0 \end{cases} \)
To prove that the function is discontinuous at \( x=0 \) for any value of \( k \), we need to show that the limit of \( f(x) \) as \( x \to 0 \) does not exist.
First, calculate the Left Hand Limit (LHL) at \( x=0 \):
\( \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|+2x^2} \)
For \( x < 0 \), the absolute value \( |x| = -x \).
\( \text{LHL} = \lim_{x \to 0^-} \frac{x}{-x+2x^2} \)
Factor out \( x \) from the denominator:
\( \text{LHL} = \lim_{x \to 0^-} \frac{x}{x(-1+2x)} \)
Cancel \( x \) (since \( x \neq 0 \)):
\( \text{LHL} = \lim_{x \to 0^-} \frac{1}{-1+2x} \)
Substitute \( x=0 \): \( \frac{1}{-1+2(0)} = \frac{1}{-1} = -1 \)
Next, calculate the Right Hand Limit (RHL) at \( x=0 \):
\( \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x}{|x|+2x^2} \)
For \( x > 0 \), the absolute value \( |x| = x \).
\( \text{RHL} = \lim_{x \to 0^+} \frac{x}{x+2x^2} \)
Factor out \( x \) from the denominator:
\( \text{RHL} = \lim_{x \to 0^+} \frac{x}{x(1+2x)} \)
Cancel \( x \) (since \( x \neq 0 \)):
\( \text{RHL} = \lim_{x \to 0^+} \frac{1}{1+2x} \)
Substitute \( x=0 \): \( \frac{1}{1+2(0)} = \frac{1}{1} = 1 \)
Since \( \text{LHL} = -1 \) and \( \text{RHL} = 1 \), we see that \( \text{LHL} \neq \text{RHL} \).
This means that \( \lim_{x \to 0} f(x) \) does not exist. A function cannot be continuous at a point if its limit at that point does not exist. This is true regardless of the value assigned to \( f(0) \) (which is \( k \)).
Therefore, the function \( f(x) \) is discontinuous at \( x=0 \) for any choice of \( k \).
In simple words: When we check both sides of the function as x approaches zero, they lead to different numbers (-1 from the left, 1 from the right). Because of this permanent jump, the function will always be broken at x=0, no matter what value we try to plug in for 'k' at that exact spot.

🎯 Exam Tip: To prove a function is discontinuous irrespective of a parameter, show that its Left Hand Limit and Right Hand Limit at the critical point are unequal. If they are unequal, the limit does not exist, and continuity is impossible.

Question 21. The function \( f(x) = \frac{2x^2-8}{x-2} \) is undefined at x = 2. What value should be assigned to f(2) so that f(x) is continuous at x = 2?
Answer: The given function is \( f(x) = \frac{2x^2-8}{x-2} \). It is undefined at \( x=2 \) because the denominator becomes zero.
For \( f(x) \) to be continuous at \( x=2 \), the value of \( f(2) \) must be equal to the limit of the function as \( x \to 2 \).
First, calculate the limit of \( f(x) \) as \( x \to 2 \):
\( \lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{2x^2-8}{x-2} \)
Factor out \( 2 \) from the numerator:
\( \lim_{x \to 2} \frac{2(x^2-4)}{x-2} \)
Use the difference of squares formula, \( a^2-b^2 = (a-b)(a+b) \), so \( x^2-4 = (x-2)(x+2) \):
\( \lim_{x \to 2} \frac{2(x-2)(x+2)}{x-2} \)
Cancel \( (x-2) \) from the numerator and denominator (since \( x \neq 2 \) as \( x \to 2 \)):
\( \lim_{x \to 2} 2(x+2) \)
Substitute \( x=2 \): \( 2(2+2) = 2(4) = 8 \)
For the function to be continuous at \( x=2 \), we must define \( f(2) \) to be equal to this limit.
Therefore, the value to be assigned to \( f(2) \) is \( 8 \).
In simple words: This function has a "hole" at x=2. To make it continuous, we need to fill that hole with the exact value the function is approaching. By simplifying the expression, we find that the function approaches 8 as x gets close to 2, so we should define f(2) as 8.

🎯 Exam Tip: For functions with removable discontinuities (where a factor cancels out), factorize the numerator and denominator to simplify the expression and find the limit. This limit is the value needed to make the function continuous at that point.

Question 22. The function \( f(x) = \frac{x^2-1}{x^3-1} \) at the point x = 1; what should be the value of f(1) such that f(x) may be continuous at x = 1?
Answer: The given function is \( f(x) = \frac{x^2-1}{x^3-1} \). It is undefined at \( x=1 \) because the denominator becomes zero.
For \( f(x) \) to be continuous at \( x=1 \), the value of \( f(1) \) must be equal to the limit of the function as \( x \to 1 \).
First, calculate the limit of \( f(x) \) as \( x \to 1 \):
\( \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x^2-1}{x^3-1} \)
Use the difference of squares formula \( x^2-1 = (x-1)(x+1) \) and the difference of cubes formula \( x^3-1 = (x-1)(x^2+x+1) \):
\( \lim_{x \to 1} \frac{(x-1)(x+1)}{(x-1)(x^2+x+1)} \)
Cancel \( (x-1) \) from the numerator and denominator (since \( x \neq 1 \) as \( x \to 1 \)):
\( \lim_{x \to 1} \frac{x+1}{x^2+x+1} \)
Substitute \( x=1 \): \( \frac{1+1}{1^2+1+1} = \frac{2}{1+1+1} = \frac{2}{3} \)
For the function to be continuous at \( x=1 \), we must define \( f(1) \) to be equal to this limit.
Therefore, the value to be assigned to \( f(1) \) is \( \frac{2}{3} \).
In simple words: This function has a gap at x=1. To make it continuous, we need to fill that gap with the exact value the function approaches. By factoring and simplifying, we find the function approaches 2/3 as x gets close to 1, so f(1) should be set to 2/3.

🎯 Exam Tip: Be mindful of factoring common algebraic forms like difference of squares and difference of cubes. These are frequently used to simplify rational functions and find limits at points where the function is initially undefined.

Question 23. Is the function \( f(x) = \frac{3x+2\sin x}{x} \) continuous at x = 0? If not, how may the function be defined at x = 0 to make it continuous at that point?
Answer: The given function is \( f(x) = \frac{3x+2\sin x}{x} \).
First, we check if \( f(x) \) is defined at \( x=0 \). If we substitute \( x=0 \), the denominator becomes \( 0 \), so \( f(0) \) is undefined.
Since \( f(0) \) is undefined, the function \( f(x) \) is discontinuous at \( x=0 \).
To make the function continuous at \( x=0 \), we need to define \( f(0) \) such that it equals the limit of \( f(x) \) as \( x \to 0 \).
Calculate the limit of \( f(x) \) as \( x \to 0 \):
\( \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{3x+2\sin x}{x} \)
We can split the fraction into two parts:
\( \lim_{x \to 0} \left(\frac{3x}{x} + \frac{2\sin x}{x}\right) \)
\( = \lim_{x \to 0} \left(3 + 2\frac{\sin x}{x}\right) \)
Using the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):
\( = 3 + 2(1) = 3 + 2 = 5 \)
For the function to be continuous at \( x=0 \), we must define \( f(0) = 5 \).
So, the function can be defined as: \( f(x) = \begin{cases} \frac{3x+2\sin x}{x}, & x \neq 0 \\ 5, & x=0 \end{cases} \)
In simple words: The function has a missing point at x=0 because you cannot divide by zero. To make it smooth, we find what value the function is getting close to as x approaches zero. That value is 5, so we should set f(0) to be 5.

🎯 Exam Tip: When a function is undefined at a point, check if the limit exists. If it does, you can define (or redefine) the function at that point to be equal to the limit, thereby removing the discontinuity and making the function continuous.

Question 24. Show that the function \( f(x)=|x-3|, x \in R \), is continuous at x = 3 but not differentiable at x = 3.
Answer: The given function is \( f(x) = |x-3| \).
We can write this piecewise as:
\( f(x) = \begin{cases} x-3, & \text{ if } x-3 \geq 0 \implies x \geq 3 \\ -(x-3), & \text{ if } x-3 < 0 \implies x < 3 \end{cases} \)

Part 1: Continuity at x = 3
For the function to be continuous at \( x=3 \), LHL, RHL, and \( f(3) \) must be equal.
Calculate LHL at \( x=3 \):
\( \text{LHL} = \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} -(x-3) = -(3-3) = 0 \)
Calculate RHL at \( x=3 \):
\( \text{RHL} = \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (x-3) = (3-3) = 0 \)
Calculate \( f(3) \):
\( f(3) = 3-3 = 0 \) (since \( x \geq 3 \) applies).
Since \( \text{LHL} = \text{RHL} = f(3) = 0 \), the function \( f(x) \) is continuous at \( x=3 \).

Part 2: Differentiability at x = 3
For the function to be differentiable at \( x=3 \), the Left Hand Derivative (LHD) and Right Hand Derivative (RHD) must be equal.
Calculate LHD at \( x=3 \):
\( \text{LHD} = \lim_{x \to 3^-} \frac{f(x)-f(3)}{x-3} = \lim_{x \to 3^-} \frac{-(x-3)-0}{x-3} \)
\( = \lim_{x \to 3^-} \frac{-(x-3)}{x-3} = \lim_{x \to 3^-} (-1) = -1 \)
Calculate RHD at \( x=3 \):
\( \text{RHD} = \lim_{x \to 3^+} \frac{f(x)-f(3)}{x-3} = \lim_{x \to 3^+} \frac{(x-3)-0}{x-3} \)
\( = \lim_{x \to 3^+} \frac{x-3}{x-3} = \lim_{x \to 3^+} (1) = 1 \)
Since \( \text{LHD} = -1 \) and \( \text{RHD} = 1 \), we have \( \text{LHD} \neq \text{RHD} \).
Therefore, the function \( f(x) \) is not differentiable at \( x=3 \).
In simple words: The function |x-3| is like a 'V' shape with its tip at x=3. It is continuous because you can draw it without lifting your pen. However, at the sharp tip, you cannot draw a single tangent line, meaning it's not smooth or "differentiable" there.

🎯 Exam Tip: Functions involving absolute values often have a sharp corner (or "cusp") at the point where the expression inside the absolute value changes sign. These points are typically continuous but not differentiable.

Question 25. Show that the function \( f(x)=|x-1|+|x+1| \) for all \( x \in R \), is not differentiable at the points x = -1 and x = 1.
Answer: The given function is \( f(x)=|x-1|+|x+1| \).
We need to rewrite this function piecewise based on the critical points \( x=-1 \) and \( x=1 \).
Case 1: If \( x < -1 \)
\( |x-1| = -(x-1) \) (since \( x-1 \) is negative)
\( |x+1| = -(x+1) \) (since \( x+1 \) is negative)
So, \( f(x) = -(x-1) - (x+1) = -x+1-x-1 = -2x \)
Case 2: If \( -1 \leq x < 1 \)
\( |x-1| = -(x-1) \) (since \( x-1 \) is negative)
\( |x+1| = x+1 \) (since \( x+1 \) is positive or zero)
So, \( f(x) = -(x-1) + (x+1) = -x+1+x+1 = 2 \)
Case 3: If \( x \geq 1 \)
\( |x-1| = x-1 \) (since \( x-1 \) is positive or zero)
\( |x+1| = x+1 \) (since \( x+1 \) is positive)
So, \( f(x) = (x-1) + (x+1) = x-1+x+1 = 2x \)
Combining these, the function is: \( f(x) = \begin{cases} -2x, & x < -1 \\ 2, & -1 \leq x < 1 \\ 2x, & x \geq 1 \end{cases} \)

Differentiability at x = -1
Calculate Left Hand Derivative (LHD) at \( x=-1 \):
\( \text{LHD} = \lim_{x \to -1^-} \frac{f(x)-f(-1)}{x-(-1)} \)
From the function definition, \( f(-1) = 2 \) (from the middle piece).
\( \text{LHD} = \lim_{x \to -1^-} \frac{(-2x)-2}{x+1} = \lim_{x \to -1^-} \frac{-2(x+1)}{x+1} = \lim_{x \to -1^-} (-2) = -2 \)
Calculate Right Hand Derivative (RHD) at \( x=-1 \):
\( \text{RHD} = \lim_{x \to -1^+} \frac{f(x)-f(-1)}{x-(-1)} \)
\( \text{RHD} = \lim_{x \to -1^+} \frac{2-2}{x+1} = \lim_{x \to -1^+} \frac{0}{x+1} = 0 \)
Since \( \text{LHD} = -2 \) and \( \text{RHD} = 0 \), we have \( \text{LHD} \neq \text{RHD} \).
Thus, \( f(x) \) is not differentiable at \( x=-1 \).

Differentiability at x = 1
Calculate Left Hand Derivative (LHD) at \( x=1 \):
\( \text{LHD} = \lim_{x \to 1^-} \frac{f(x)-f(1)}{x-1} \)
From the function definition, \( f(1) = 2(1) = 2 \) (from the third piece).
\( \text{LHD} = \lim_{x \to 1^-} \frac{2-2}{x-1} = \lim_{x \to 1^-} \frac{0}{x-1} = 0 \)
Calculate Right Hand Derivative (RHD) at \( x=1 \):
\( \text{RHD} = \lim_{x \to 1^+} \frac{f(x)-f(1)}{x-1} \)
\( \text{RHD} = \lim_{x \to 1^+} \frac{2x-2}{x-1} = \lim_{x \to 1^+} \frac{2(x-1)}{x-1} = \lim_{x \to 1^+} (2) = 2 \)
Since \( \text{LHD} = 0 \) and \( \text{RHD} = 2 \), we have \( \text{LHD} \neq \text{RHD} \).
Thus, \( f(x) \) is not differentiable at \( x=1 \).
In simple words: This function is made of two absolute value parts, creating a graph with two sharp corners at x=-1 and x=1. While the function is continuous (no breaks), these sharp corners mean it's not smooth enough to have a clear tangent line, so it's not differentiable at these points.

🎯 Exam Tip: For functions that are a sum of absolute values, identify all critical points where the expressions inside the absolute values become zero. Rewrite the function piecewise for intervals defined by these critical points, then check differentiability at each critical point using LHD and RHD.

Question 26. Show that the function \( f(x) = \begin{cases} x-1, & \text{ if } x<2 \\ 2x-3, & \text{ if } x \geq 2 \end{cases} \) is not differentiable at x = 2.
Answer: The given function is \( f(x) = \begin{cases} x-1, & \text{ if } x<2 \\ 2x-3, & \text{ if } x \geq 2 \end{cases} \)
To show that \( f(x) \) is not differentiable at \( x=2 \), we need to compare its Left Hand Derivative (LHD) and Right Hand Derivative (RHD) at this point.
First, find the value of the function at \( x=2 \):
\( f(2) = 2(2) - 3 = 4 - 3 = 1 \) (using the definition for \( x \geq 2 \)).
Calculate LHD at \( x=2 \):
\( \text{LHD} = \lim_{x \to 2^-} \frac{f(x)-f(2)}{x-2} \)
For \( x < 2 \), \( f(x) = x-1 \).
\( \text{LHD} = \lim_{x \to 2^-} \frac{(x-1)-1}{x-2} = \lim_{x \to 2^-} \frac{x-2}{x-2} \)
Cancel \( (x-2) \) (since \( x \neq 2 \) as \( x \to 2^- \)):
\( \text{LHD} = \lim_{x \to 2^-} (1) = 1 \)
Calculate RHD at \( x=2 \):
\( \text{RHD} = \lim_{x \to 2^+} \frac{f(x)-f(2)}{x-2} \)
For \( x \geq 2 \), \( f(x) = 2x-3 \).
\( \text{RHD} = \lim_{x \to 2^+} \frac{(2x-3)-1}{x-2} = \lim_{x \to 2^+} \frac{2x-4}{x-2} \)
Factor out \( 2 \) from the numerator:
\( \text{RHD} = \lim_{x \to 2^+} \frac{2(x-2)}{x-2} \)
Cancel \( (x-2) \) (since \( x \neq 2 \) as \( x \to 2^+\)):
\( \text{RHD} = \lim_{x \to 2^+} (2) = 2 \)
Since \( \text{LHD} = 1 \) and \( \text{RHD} = 2 \), we have \( \text{LHD} \neq \text{RHD} \).
Therefore, the function \( f(x) \) is not differentiable at \( x=2 \).
In simple words: This function changes its rule at x=2. If you try to find the slope of the curve coming from the left, you get 1. If you find the slope coming from the right, you get 2. Since these slopes are different, the graph has a sharp corner, meaning it's not differentiable at x=2.

🎯 Exam Tip: Differentiability implies continuity, but continuity does not imply differentiability. Always check continuity first. If a function is discontinuous, it cannot be differentiable. If it is continuous, then proceed to check LHD and RHD.

Question 27. Show that f(x) = |x - 20| is continuous at x = 20 but f'(x) does not exist at x = 20.
Answer: The given function is \( f(x) = |x-20| \).
We can write this piecewise as:
\( f(x) = \begin{cases} x-20, & \text{ if } x-20 \geq 0 \implies x \geq 20 \\ -(x-20), & \text{ if } x-20 < 0 \implies x < 20 \end{cases} \)

Part 1: Continuity at x = 20
For the function to be continuous at \( x=20 \), LHL, RHL, and \( f(20) \) must be equal.
Calculate LHL at \( x=20 \):
\( \text{LHL} = \lim_{x \to 20^-} f(x) = \lim_{x \to 20^-} -(x-20) = -(20-20) = 0 \)
Calculate RHL at \( x=20 \):
\( \text{RHL} = \lim_{x \to 20^+} f(x) = \lim_{x \to 20^+} (x-20) = (20-20) = 0 \)
Calculate \( f(20) \):
\( f(20) = 20-20 = 0 \) (since \( x \geq 20 \) applies).
Since \( \text{LHL} = \text{RHL} = f(20) = 0 \), the function \( f(x) \) is continuous at \( x=20 \).

Part 2: Differentiability at x = 20
For the function to be differentiable at \( x=20 \), the Left Hand Derivative (LHD) and Right Hand Derivative (RHD) must be equal.
Calculate LHD at \( x=20 \):
\( \text{LHD} = \lim_{x \to 20^-} \frac{f(x)-f(20)}{x-20} = \lim_{x \to 20^-} \frac{-(x-20)-0}{x-20} \)
\( = \lim_{x \to 20^-} \frac{-(x-20)}{x-20} = \lim_{x \to 20^-} (-1) = -1 \)
Calculate RHD at \( x=20 \):
\( \text{RHD} = \lim_{x \to 20^+} \frac{f(x)-f(20)}{x-20} = \lim_{x \to 20^+} \frac{(x-20)-0}{x-20} \)
\( = \lim_{x \to 20^+} \frac{x-20}{x-20} = \lim_{x \to 20^+} (1) = 1 \)
Since \( \text{LHD} = -1 \) and \( \text{RHD} = 1 \), we have \( \text{LHD} \neq \text{RHD} \).
Therefore, the function \( f(x) \) is not differentiable at \( x=20 \).
In simple words: The function \( |x-20| \) forms a 'V' shape, with its pointy end at x=20. You can draw this function without lifting your pen, so it's continuous. But because of the sharp corner, there isn't one clear slope (derivative) at x=20, which means it's not differentiable there.

🎯 Exam Tip: Absolute value functions are classic examples used to test the distinction between continuity and differentiability. They are always continuous at their 'vertex' but never differentiable at that sharp point due to differing left and right-hand derivatives.

Examples

Question 1. A real valued function f is continuous at a point x = a if it is defined at x = a and iff \( \lim _{x \rightarrow a^{-}} f(x) = \lim _{x \rightarrow a^{+}} f(x) \)
Answer: The complete definition for a real-valued function \(f(x)\) to be continuous at a point \(x=a\) is that \(f(a)\) must be defined, the limit of \(f(x)\) as \(x\) approaches \(a\) must exist, and this limit must be equal to \(f(a)\). So, \( \lim _{x \rightarrow a^{-}} f(x) = \lim _{x \rightarrow a^{+}} f(x) = f(a) \). This means the left-hand limit, the right-hand limit, and the function's value at that point are all the same.
In simple words: For a function to be continuous at a point, its value at that point, and the values it gets very close to from both sides, must all be exactly the same.

🎯 Exam Tip: Remember the three key conditions for continuity at a point: function defined, limit exists (left and right limits are equal), and limit equals the function value. All three must hold true.

Question 2. A real function f (x) is said to be differentiable at x = a, if
Answer: A real function \(f(x)\) is said to be differentiable at \(x=a\) if its left-hand derivative and right-hand derivative at \(x=a\) both exist and are equal. This is often written as \( \text{Lf}'(a) = \text{Rf}'(a) \). This ensures the function has a well-defined tangent line at that point, without sharp corners or vertical slopes.
In simple words: A function can be smoothly drawn at a point without any sharp turns if its slope from the left and its slope from the right meet at that exact point.

🎯 Exam Tip: Differentiability implies continuity, but continuity does not always imply differentiability. A function must be smooth and have no sharp points (like a corner) or breaks to be differentiable.

Question 3. If u and v are two function of x, then derivative of \( \frac { u }{ v } \) i.e., \( (\frac { u }{ v })' \) =
Answer: If \(u\) and \(v\) are two functions of \(x\), then the derivative of their quotient, \( (\frac { u }{ v })' \), is given by the quotient rule. The formula for this is: \( \left(\frac{u}{v}\right)^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^2} \), where \(v \neq 0\). This rule helps us find the rate of change of a ratio of two changing quantities.
In simple words: To find the derivative of a fraction where both top and bottom are functions, you take the bottom times the derivative of the top, minus the top times the derivative of the bottom, all divided by the bottom function squared.

🎯 Exam Tip: Memorize the quotient rule formula and be careful with the order of terms in the numerator; a common mistake is switching \(v u'\) and \(u v'\).

Question 4. If the graph of a function can be drawn around a point without lifting the pen from the paper, then the function is
Answer: If the graph of a function can be drawn around a point without lifting the pen from the paper, then the function is **Continuous**. This means there are no breaks, jumps, or holes in the graph at that point. A continuous function represents a smooth flow without sudden changes.
In simple words: If you can draw a function's line through a spot without lifting your pencil, it means the function is smooth and unbroken there.

🎯 Exam Tip: Visualizing a graph helps understand continuity. A function is continuous if its graph can be drawn without lifting your pen from the paper.

Question 5. If the graph of a function around a point cannot be drawn without lifting the pen from the paper, then function is
Answer: If the graph of a function around a point cannot be drawn without lifting the pen from the paper, then the function is **Discontinuous**. This indicates there is a break, a jump, or a hole in the graph at that specific point. Discontinuities can happen in different ways, like sudden jumps or missing points.
In simple words: If you have to lift your pencil to draw a function's line through a spot, it means the function is broken or "not continuous" there.

🎯 Exam Tip: Look for any gaps, jumps, or holes in the graph to identify points of discontinuity. These are places where the pen must be lifted to continue drawing.

Question 6. Examine the continuity of the function. \( f(x) = \left.\begin{array}{c} x^2 \text { when } x \neq 1 \\ 2 \text { when } x=1 \end{array}\right\} \) at x = 1. The function is continuous/discontinuous at x = 1.
Answer: To examine the continuity of the function \(f(x)\) at \(x=1\), we compare the limit of the function as \(x\) approaches 1 with the function's value at \(x=1\).
\( \underset{x \rightarrow 1}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1}{\mathrm{Lt}} x^2 = 1^2 = 1 \)
We are given \(f(1) = 2\).
Since \( \underset{x \rightarrow 1}{\mathrm{Lt}} f(x) \neq f(1) \) (because \(1 \neq 2\)), the function is **discontinuous** at \(x=1\).
In simple words: At the point \(x=1\), the function's value is 2, but if you look at numbers very close to 1, the function acts like \(x^2\), which gives 1. Because these two numbers are different, the function is broken at \(x=1\).

🎯 Exam Tip: For piecewise functions, always check continuity at the points where the definition changes by comparing the left limit, right limit, and the function's value at that specific point.

Question 7. The graph of a function has a jump discontinuity as shown in fig. and so it is a discontinuous function.
Answer: The graph of a function shown exhibits a **jump** discontinuity, meaning the function's value abruptly changes at a certain point. Because of this sudden change, where the left and right limits are different, it is a **discontinuous** function. The graph shows a clear gap that would require lifting a pen to draw.
In simple words: The graph of this function has a sudden step up or down, making it impossible to draw without lifting your pen. This means it's broken, or "discontinuous."

🎯 Exam Tip: A jump discontinuity occurs when the left-hand limit and the right-hand limit both exist but are not equal at a point. This is a common type of discontinuity.

Question 8. The graph of a function has a hole discontinuity as shown in fig. and so it is a discontinuous function.
Answer: The graph of a function shown has a **hole** discontinuity, which means there is a single point missing from the graph. Because of this missing point, where the limit exists but does not equal the function's value (or the function is undefined), it is a **discontinuous** function. This type of discontinuity is often called a removable discontinuity.
In simple words: This graph has a tiny empty circle, like a missing spot, which means it's broken and "not continuous" at that exact place.

🎯 Exam Tip: A hole discontinuity (removable discontinuity) happens when the limit of the function exists at a point, but either the function is not defined there or its value at that point is different from the limit.

Question 9. The function \( f(x) = x³ – 7x² + 5 \) being a polynomial function is continuous.
Answer: The function \( f(x) = x³ – 7x² + 5 \) is a **polynomial** function, and all polynomial functions are inherently **continuous**. This means their graphs can be drawn without any breaks, jumps, or holes across their entire domain, which is all real numbers. These functions are very smooth and predictable.
In simple words: Because this function is a polynomial (made of powers of \(x\) added and subtracted), it's always smooth and has no breaks anywhere.

🎯 Exam Tip: A fundamental property of polynomial functions is that they are continuous everywhere. This is a crucial concept to remember for analyzing function behavior.

Question 10. The composition gof of two continuous functions is continuous/discontinuous.
Answer: The composition gof of two continuous functions is **continuous**. If you have two functions, \(g\) and \(f\), and both are continuous, then putting one inside the other (like \(g(f(x))\)) will also result in a continuous function. This property is very useful in advanced calculus.
In simple words: If you combine two functions that are both smooth and unbroken, the new function you create will also be smooth and unbroken.

🎯 Exam Tip: The composition of continuous functions is a continuous function. This is a powerful theorem in calculus for determining the continuity of complex functions.

Question 11. If \( f(x) = \left\{\begin{array}{c} 3 x,-8, \text { if } x \leq 5 \\ 2 k, \text { if } x>5 \end{array}\right. \) is continuous, find k
(a) \( \frac { 3 }{ 7 } \)
(b) \( \frac { 7 }{ 2 } \)
(c) Missing Option (d) \( \frac { 4 }{ 7 } \)
Answer: (b) \( \frac { 7 }{ 2 } \)
For the function \(f(x)\) to be continuous at \(x=5\), the left-hand limit, the right-hand limit, and the function's value at \(x=5\) must all be equal.
L.H.L. \( = \underset{x \rightarrow 5^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 5^{-}}{\mathrm{Lt}} (3x - 8) = 3(5) - 8 = 15 - 8 = 7 \)
R.H.L. \( = \underset{x \rightarrow 5^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 5^{+}}{\mathrm{Lt}} (2k) = 2k \)
Also, \(f(5) = 3(5) - 8 = 7 \).
Since \(f(x)\) is continuous at \(x=5\), we must have L.H.L. = R.H.L. = \(f(5)\).
So, \( 7 = 2k \)
\( \implies k = \frac{7}{2} \)
In simple words: For the function to be unbroken at \(x=5\), the value it approaches from the left side (which is 7) must be the same as the value it approaches from the right side (which is \(2k\)). Setting these equal gives us \(k = \frac{7}{2}\).

🎯 Exam Tip: When a piecewise function is defined to be continuous at a boundary point, set the left-hand limit, right-hand limit, and the function's value at that point equal to each other to solve for any unknown constants.

Question 12. If \(f(x) = 2x\) and \(g (x) = \frac { x² }{ 2 } + 1\), then which of the following can be a discontinuous function
(a) \(f(x) + g(x)\)
(b) \(f(x) – g(x)\)
(c) \(f(x).g(x)\)
(d) \( \frac{g(x)}{f(x)} \)
Answer: (d) \( \frac{g(x)}{f(x)} \)
Both \(f(x) = 2x\) and \(g(x) = \frac{x^2}{2} + 1\) are polynomial functions, which means they are continuous everywhere.
(a) \(f(x) + g(x) = 2x + \frac{x^2}{2} + 1\). This is a polynomial, so it is continuous.
(b) \(f(x) - g(x) = 2x - (\frac{x^2}{2} + 1)\). This is also a polynomial, so it is continuous.
(c) \(f(x) \cdot g(x) = 2x (\frac{x^2}{2} + 1) = x^3 + 2x\). This is a polynomial, so it is continuous.
(d) \( \frac{g(x)}{f(x)} = \frac{\frac{x^2}{2} + 1}{2x} = \frac{x^2 + 2}{4x} \). This function becomes undefined when the denominator is zero, which happens at \(4x = 0 \implies x=0\). Therefore, this function is discontinuous at \(x=0\).
In simple words: When you add, subtract, or multiply two smooth functions, the result is still a smooth function. But if you divide them, the new function can break if the bottom part becomes zero. In this case, \( \frac{g(x)}{f(x)} \) breaks when \(f(x)\) is zero, which is at \(x=0\).

🎯 Exam Tip: Remember that sums, differences, and products of continuous functions are always continuous. However, quotients of continuous functions are continuous only where the denominator is non-zero. Always check for points where the denominator might be zero.

Question 13. If \( f(x) = \left\{\begin{array}{c} 3 x,-8, \text { if } x \leq 5 \\ 2 k, \text { if } x>5 \end{array}\right. \) is continuous, find k
(a) \( \frac { 3 }{ 7 } \)
(b) \( \frac { 7 }{ 2 } \)
(c) Missing Option (d) \( \frac { 4 }{ 7 } \)
Answer: (b) \( \frac { 7 }{ 2 } \)
For \(f(x)\) to be continuous at \(x=5\), the left-hand limit, right-hand limit, and \(f(5)\) must be equal.
L.H.L. \( = \underset{x \rightarrow 5^{-}}{\mathrm{Lt}} (3x - 8) = 3(5) - 8 = 7 \)
R.H.L. \( = \underset{x \rightarrow 5^{+}}{\mathrm{Lt}} (2k) = 2k \)
\(f(5) = 3(5) - 8 = 7 \).
For continuity, \(7 = 2k \implies k = \frac{7}{2}\).
In simple words: For the function to be unbroken at \(x=5\), the value it approaches from the left side (which is 7) must be the same as the value it approaches from the right side (which is \(2k\)). Setting these equal gives us \(k = \frac{7}{2}\).

🎯 Exam Tip: This type of problem is very common. Always ensure you are evaluating the correct function definition for the left and right limits, and at the point itself.

Question 14. The function \( f (x) = \frac{4-x^2}{4 x-x^3} \)
(a) discontinuous at only one point
(b) discontinuous exactly at two points
(c) discontinuous exactly at three points
(d) none of these.
Answer: (c) discontinuous exactly at three points
The function \(f(x) = \frac{4-x^2}{4x-x^3}\) is a rational function. Rational functions are discontinuous where their denominator is zero. Let's find the values of \(x\) for which the denominator is zero:
\( 4x - x^3 = 0 \)
Factor out \(x\):
\( x(4 - x^2) = 0 \)
This gives \(x=0\) or \(4 - x^2 = 0\).
\( 4 - x^2 = 0 \implies x^2 = 4 \implies x = \pm 2 \).
So, the denominator is zero at \(x = 0, x = 2,\) and \(x = -2\). These are the points of discontinuity. Thus, there are exactly three points of discontinuity.
In simple words: This function is a fraction. It breaks down and is "not continuous" whenever its bottom part becomes zero. By solving for when the bottom part is zero, we find three such points.

🎯 Exam Tip: For rational functions, always factorize the denominator to find all values of \(x\) where it becomes zero. These values are the points of discontinuity for the function.

Question 15. Let f and g be two real functions at a real number c. Then, which of the following are true?
I. f+ g is a continuous at x = c.
II. f- g is continuous at x = c.
III. f.g is continuous at x = c.
IV. \( \frac { f }{ g } \) is continuous at x = c provided g (x) \(\neq\) 0
(a) I and II
(b) II and III
(c) I, II and III
(d) All
Answer: (d) All
These are standard results from the algebra of continuous functions. If two functions, \(f\) and \(g\), are continuous at a point \(x=c\), then their sum (\(f+g\)), difference (\(f-g\)), and product (\(f \cdot g\)) are also continuous at \(x=c\). Additionally, their quotient (\(\frac{f}{g}\)) is continuous at \(x=c\), provided that the denominator \(g(c)\) is not equal to zero. All four statements are correct based on these mathematical rules.
In simple words: If two functions are smooth at a point, then adding, subtracting, multiplying, or even dividing them (as long as you don't divide by zero) will also result in a smooth function at that same point.

🎯 Exam Tip: Understanding the algebra of continuous functions is fundamental. Remember that continuity is preserved under addition, subtraction, multiplication, and division (provided the denominator is not zero).

Question 16. A real function f is said to be continuous, if it is continuous at every point in the
(a) domain of f
(b) co-domain of f
(c) range off
(d) none of these
Answer: (a) domain of f
A real function \(f\) is defined as continuous if it is continuous at every single point within its **domain**. This means there are no breaks, jumps, or holes anywhere the function is supposed to exist. The domain defines the set of all possible input values for which the function is defined.
In simple words: A function is called "continuous" overall if it's smooth and unbroken everywhere that it's allowed to have an input number.

🎯 Exam Tip: The definition of a continuous function refers to its behavior throughout its entire domain, not just at specific points. Pay attention to the distinction between continuity at a point and continuity over an interval or domain.

Question 17. If \( f(x) = \left\{\begin{array}{rll} k x^2, & \text { if } & x \leq 2 \\ 3, & \text { if } & x>2 \end{array}\right. \) is continuous at x = 2, then the value of k is
Answer: For \(f(x)\) to be continuous at \(x=2\), the left-hand limit, the right-hand limit, and the function's value at \(x=2\) must all be equal.
L.H.L. \( = \underset{x \rightarrow 2^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 2^{-}}{\mathrm{Lt}} (kx^2) = k(2^2) = 4k \)
R.H.L. \( = \underset{x \rightarrow 2^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 2^{+}}{\mathrm{Lt}} (3) = 3 \)
Also, \(f(2) = k(2^2) = 4k \).
Since \(f(x)\) is continuous at \(x=2\), we must have L.H.L. = R.H.L. = \(f(2)\).
So, \( 4k = 3 \)
\( \implies k = \frac{3}{4} \)
In simple words: For the function to be unbroken at \(x=2\), the value it approaches from the left (\(4k\)) must match the value it approaches from the right (3). Solving this tells us that \(k\) must be \(\frac{3}{4}\).

🎯 Exam Tip: Always evaluate all three conditions (LHL, RHL, and function value) at the point of interest for piecewise functions to ensure full continuity.

Question 18. Which of the following functions is/are continuous?
I. constant function
II. polynomial function
III. modulus function
IV. sine function
(a) only I
(b) only II
(c) II, III and IV
(d) All of these
Answer: (d) All of these
All the listed types of functions are continuous everywhere within their domains.
I. **Constant functions** (e.g., \(f(x)=c\)) are always continuous, as their graph is a straight horizontal line.
II. **Polynomial functions** (e.g., \(f(x)=ax^n + ...\)) are continuous everywhere, having smooth, unbroken graphs.
III. **Modulus functions** (e.g., \(f(x)=|x|\)) are continuous everywhere, though they may have sharp corners where they are not differentiable.
IV. **Sine functions** (e.g., \(f(x)=\sin x\)) are continuous everywhere, displaying a smooth, oscillating wave.
Therefore, all of these function types are continuous.
In simple words: Constant lines, polynomials (like \(x^2\)), absolute value functions (like \(|x|\)), and sine waves are all smooth and unbroken everywhere they are defined. So, all of them are continuous.

🎯 Exam Tip: It is crucial to remember the basic continuous function types: polynomials, trigonometric functions (sine, cosine), exponential functions, logarithmic functions (in their domain), and modulus functions. Their continuity is a fundamental property.

Question 19. The function \( f (x) = 2x² + \cos x + e^x – 2 \) is
(a) discontinuous at x = 0
(b) discontinuous at x = \( \pi \)
(c) discontinuous at x = \( \frac { \pi }{ 2 } \)
(d) continuous at all points
Answer: (d) continuous at all points
The given function \(f(x) = 2x^2 + \cos x + e^x - 2\) is a combination of several basic functions known to be continuous everywhere:
- \(2x^2\) is a polynomial function, so it's continuous everywhere.
- \(\cos x\) is a trigonometric function, continuous everywhere.
- \(e^x\) is an exponential function, continuous everywhere.
- \(-2\) is a constant function, continuous everywhere.
Since the sum and difference of continuous functions are also continuous, \(f(x)\) is continuous at all points. This means its graph has no breaks or jumps anywhere.
In simple words: This function is made by adding and subtracting parts that are all smooth by themselves (like \(x^2\), cosine, and \(e^x\)). When you combine smooth parts, the whole function stays smooth everywhere.

🎯 Exam Tip: Remember that sums, differences, and products of continuous functions are continuous. This property is very useful for quickly determining the continuity of composite functions.

Question 20. If the function \( f(x) = \left\{\begin{array}{ccc} \frac{x^2-1}{x-1}, & \text { when } & x \neq 1 \\ k, & \text { when } & x=1 \end{array}\right. \) is given to be continuous at x = 1, then the value of k is
(a) - 2
(b) 3
(c) \( \frac { 1 }{ 2 } \)
(d) 2
Answer: (d) 2
For the function \(f(x)\) to be continuous at \(x=1\), the limit of \(f(x)\) as \(x\) approaches 1 must be equal to \(f(1)\).
First, let's find the limit of \(f(x)\) as \(x \rightarrow 1\):
\( \underset{x \rightarrow 1}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1}{\mathrm{Lt}} \frac{x^2-1}{x-1} \)
We can factor the numerator using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\):
\( \underset{x \rightarrow 1}{\mathrm{Lt}} \frac{(x-1)(x+1)}{x-1} \)
Since \(x \neq 1\), we can cancel out the \((x-1)\) terms:
\( \underset{x \rightarrow 1}{\mathrm{Lt}} (x+1) \)
Now, substitute \(x=1\):
\( = 1 + 1 = 2 \)
We are given \(f(1) = k\).
For continuity, \( \underset{x \rightarrow 1}{\mathrm{Lt}} f(x) = f(1) \).
Therefore, \( 2 = k \).
In simple words: For this function to be smooth at \(x=1\), its value at \(x=1\) (which is \(k\)) must be the same as the value it gets very close to from both sides. When we simplify the function for values near 1, we find it approaches 2. So, \(k\) must be 2.

🎯 Exam Tip: When dealing with removable discontinuities, factorize and simplify the function to find the limit. The value of \(k\) for continuity will be equal to this limit.

Question 21. The number of points of discontinuity of \(f(x) = |x| - |x+1|\) is
(a) 1
(b) 2
(c) 0
(d) None of these
Answer: (c) 0
Let \(f(x) = |x| - |x+1|\). We need to analyze this piecewise function at critical points \(x=0\) and \(x=-1\).
1. When \(x < -1\): \(|x| = -x\), \(|x+1| = -(x+1)\).
\(f(x) = -x - (-(x+1)) = -x + x + 1 = 1\)
2. When \(-1 \leq x < 0\): \(|x| = -x\), \(|x+1| = x+1\).
\(f(x) = -x - (x+1) = -2x - 1\)
3. When \(x \geq 0\): \(|x| = x\), \(|x+1| = x+1\).
\(f(x) = x - (x+1) = -1\)
So, \( f(x) = \left\{\begin{array}{ll} 1, & \text { if } x<-1 \\ -2x-1, & \text { if } -1 \leq x<0 \\ -1, & \text { if } x \geq 0 \end{array}\right. \)
Now, let's check continuity at \(x=-1\) and \(x=0\).
At \(x=-1\):
LHL \( = \underset{x \rightarrow -1^{-}}{\mathrm{Lt}} (1) = 1 \)
RHL \( = \underset{x \rightarrow -1^{+}}{\mathrm{Lt}} (-2x-1) = -2(-1)-1 = 2-1 = 1 \)
\(f(-1) = -2(-1)-1 = 1 \)
Since LHL = RHL = \(f(-1)\), \(f(x)\) is continuous at \(x=-1\).
At \(x=0\):
LHL \( = \underset{x \rightarrow 0^{-}}{\mathrm{Lt}} (-2x-1) = -2(0)-1 = -1 \)
RHL \( = \underset{x \rightarrow 0^{+}}{\mathrm{Lt}} (-1) = -1 \)
\(f(0) = -1 \)
Since LHL = RHL = \(f(0)\), \(f(x)\) is continuous at \(x=0\).
Modulus functions are continuous everywhere. Therefore, the function \(f(x) = |x| - |x+1|\) is continuous for all real numbers, meaning it has 0 points of discontinuity.
In simple words: This function is made by subtracting two absolute value functions. Absolute value functions are always smooth. When you analyze where the definition changes, you find that the pieces connect perfectly, so the whole function is smooth everywhere and has no breaks.

🎯 Exam Tip: When dealing with sums or differences of modulus functions, break the function into piecewise definitions based on the points where the expressions inside the modulus signs become zero. Then, check continuity at these boundary points.

Question 22. If a function f is differentiable at c, then it is
(a) discontinuous at c
(b) continuous at c
(c) not defined at c
(d) None of these
Answer: (b) continuous at c
A fundamental theorem in calculus states that if a function \(f\) is differentiable at a point \(c\), then it must also be continuous at that point \(c\). Differentiability implies that the function is smooth and has no sharp corners, breaks, or holes at \(c\), which are all conditions for continuity. The reverse is not always true; a function can be continuous but not differentiable (e.g., \(f(x)=|x|\) at \(x=0\)).
In simple words: If a function can have a clear slope at a certain point, it must also be a smooth, unbroken line at that point. You can't have a slope if there's a gap or a jump.

🎯 Exam Tip: Always remember the implication: differentiability \( \implies \) continuity. This means if a function is not continuous at a point, it cannot be differentiable at that point either.

Question 23. Which of the following is not a correct statement? A function is not differentiable at x = a, if
(a) either or both Rf' (a) and Lf' (a) do not exist
(b) both Rf' (a) and Lf' exist but are not equal
(c) Rf' (a) and Lf' (a) both exist and are equal
(d) either or both Rf' (a) and Lf' (a) are not finite
Answer: (c) Rf' (a) and Lf' (a) both exist and are equal
A function is not differentiable at \(x=a\) if:
(a) The left-hand derivative (\(\text{Lf}'(a)\)) or the right-hand derivative (\(\text{Rf}'(a)\)) (or both) do not exist.
(b) Both \(\text{Lf}'(a)\) and \(\text{Rf}'(a)\) exist but are not equal (indicating a sharp point or cusp).
(d) Either \(\text{Lf}'(a)\) or \(\text{Rf}'(a)\) (or both) are not finite (meaning the tangent is vertical).
Statement (c), "Rf' (a) and Lf' (a) both exist and are equal", describes the condition where a function *is* differentiable at \(x=a\). Therefore, this is **not** a correct statement for a function being *not* differentiable at \(x=a\).
In simple words: A function is "not smooth" at a point if its slope from the left and right are different, or if the slope doesn't exist, or if the slope is straight up/down. The statement that says both slopes are equal and exist means the function IS smooth, so it's the wrong answer for what makes a function NOT smooth.

🎯 Exam Tip: Differentiability requires the left-hand and right-hand derivatives to be equal and finite. Any scenario where these conditions are not met results in non-differentiability.

Question 24. Let \( f(x) = \left\{\begin{array}{cll} c x^2+2 x, & \text { if } & x<2 \\ 2 x+4, & \text { if } & x \geq 2 \end{array}\right.\). If the function is continuous on \( (-\infty, \infty) \), then the value of c is equal to
(a) 4
(b) 2
(c) 3
(d) 1
Answer: (d) 1
For the function \(f(x)\) to be continuous on \( (-\infty, \infty) \), it must be continuous at the point where its definition changes, which is \(x=2\).
For continuity at \(x=2\), the left-hand limit, the right-hand limit, and the function's value at \(x=2\) must all be equal.
L.H.L. \( = \underset{x \rightarrow 2^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 2^{-}}{\mathrm{Lt}} (cx^2+2x) = c(2^2) + 2(2) = 4c + 4 \)
R.H.L. \( = \underset{x \rightarrow 2^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 2^{+}}{\mathrm{Lt}} (2x+4) = 2(2)+4 = 4+4 = 8 \)
Also, \(f(2) = 2(2)+4 = 8 \).
For \(f(x)\) to be continuous at \(x=2\), we set L.H.L. = R.H.L. = \(f(2)\):
\( 4c + 4 = 8 \)
Subtract 4 from both sides:
\( 4c = 4 \)
Divide by 4:
\( c = 1 \)
In simple words: For this function to be smooth everywhere, the two parts must meet perfectly at \(x=2\). The value from the left side (\(4c+4\)) must match the value from the right side (8). Solving this equation tells us that \(c\) must be 1.

🎯 Exam Tip: When a piecewise function is continuous over its entire domain, always focus on the boundary points. The limits from both sides and the function's value at these points must be equal.

Question 25. The function \( f(x) = \left\{\begin{array}{cl} x-1, & x<2 \\ 2 x-3, & x \geq 2 \end{array}\right.\) is
(a) continuous at x = 2 only
(b) for all integral values of x only
(c) for all real values of x
(d) for all real values of x such that x \(\neq\) 2
Answer: (c) for all real values of x
Let's examine the continuity of \(f(x)\):
1. For \(x < 2\), \(f(x) = x-1\). This is a polynomial function, so it is continuous for all \(x < 2\).
2. For \(x > 2\), \(f(x) = 2x-3\). This is also a polynomial function, so it is continuous for all \(x > 2\).
3. Now, we need to check continuity at the boundary point \(x=2\):
L.H.L. \( = \underset{x \rightarrow 2^{-}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 2^{-}}{\mathrm{Lt}} (x-1) = 2-1 = 1 \)
R.H.L. \( = \underset{x \rightarrow 2^{+}}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 2^{+}}{\mathrm{Lt}} (2x-3) = 2(2)-3 = 4-3 = 1 \)
Also, \(f(2) = 2(2)-3 = 1 \).
Since L.H.L. = R.H.L. = \(f(2)\), the function is continuous at \(x=2\).
Combining these three observations, the function \(f(x)\) is continuous for all real values of \(x\).
In simple words: The function is made of two simple, smooth lines. When we check where they meet at \(x=2\), they join up perfectly without any breaks. So, the whole function is smooth and unbroken everywhere.

🎯 Exam Tip: When dealing with piecewise functions, examine the continuity of each piece within its defined interval and then critically check for continuity at the boundary points between the pieces.

Question 26. Let \(f (x)\) be a function differentiable at \(x = c\), then \( \lim _{x \rightarrow c} f(x) \) equals
(a) \(f(c)\)
(b) \(f '(x)\)
(c) \( \frac { 1 }{ f(c) } \)
(d) None of these
Answer: (a) \(f(c)\)
If a function \(f(x)\) is differentiable at \(x=c\), it implies that \(f(x)\) is also continuous at \(x=c\). A property of continuous functions is that the limit of the function as \(x\) approaches \(c\) is equal to the function's value at \(c\). That is, \( \lim _{x \rightarrow c} f(x) = f(c) \). This relationship is a direct consequence of differentiability implying continuity.
In simple words: If a function has a clear slope at a point, it means the function is also unbroken there. For an unbroken function, the value it gets close to is exactly its value at that point.

🎯 Exam Tip: Always remember that differentiability at a point guarantees continuity at that same point. This means if a limit exists and the function is defined, they must be equal for a continuous function.

Question 27. If \( f(x) = \left\{\begin{array}{cl} \frac{1-\cos p x}{x \sin x}, & x \neq 0 \\ \frac{1}{2}, & x=0 \end{array}\right.\) is continuous at x = 0, then p is equal to
(a) 2
(b) – 2
(c) 1, – 1
(d) none of these
Answer: (c) 1, – 1
For \(f(x)\) to be continuous at \(x=0\), the limit of \(f(x)\) as \(x\) approaches 0 must be equal to \(f(0)\).
We are given \(f(0) = \frac{1}{2}\).
Now, let's find the limit:
\( \underset{x \rightarrow 0}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 0}{\mathrm{Lt}} \frac{1-\cos px}{x \sin x} \)
We use the identity \(1-\cos \theta = 2 \sin^2 (\frac{\theta}{2})\) and standard limits \( \underset{\theta \rightarrow 0}{\mathrm{Lt}} \frac{\sin \theta}{\theta} = 1 \).
\( = \underset{x \rightarrow 0}{\mathrm{Lt}} \frac{2 \sin^2 (\frac{px}{2})}{x \sin x} \)
To use the standard limit, we need \( (\frac{px}{2})^2 \) in the denominator for \( \sin^2 (\frac{px}{2}) \) and \( x \) in the denominator for \( \sin x \):
\( = \underset{x \rightarrow 0}{\mathrm{Lt}} \frac{2 \sin^2 (\frac{px}{2})}{(\frac{px}{2})^2} \cdot \frac{(\frac{px}{2})^2}{1} \cdot \frac{1}{x \sin x} \)
\( = \underset{x \rightarrow 0}{\mathrm{Lt}} 2 \left( \frac{\sin (\frac{px}{2})}{\frac{px}{2}} \right)^2 \cdot \frac{p^2 x^2}{4} \cdot \frac{1}{x \sin x} \)
\( = \underset{x \rightarrow 0}{\mathrm{Lt}} 2 (1)^2 \cdot \frac{p^2 x^2}{4} \cdot \frac{1}{x \cdot \frac{\sin x}{x} \cdot x} \)
\( = \underset{x \rightarrow 0}{\mathrm{Lt}} 2 \cdot \frac{p^2 x^2}{4} \cdot \frac{1}{x^2 \cdot 1} \)
\( = \underset{x \rightarrow 0}{\mathrm{Lt}} \frac{2 p^2 x^2}{4 x^2} \)
\( = \frac{2 p^2}{4} = \frac{p^2}{2} \)
For continuity, \( \frac{p^2}{2} = f(0) \).
\( \frac{p^2}{2} = \frac{1}{2} \)
\( p^2 = 1 \)
\( p = \pm 1 \)
In simple words: For the function to be smooth at \(x=0\), its value at \(x=0\) (which is \(\frac{1}{2}\)) must be the same as what the function gets close to as \(x\) approaches 0. By using special limit rules for sine and cosine, we find this limit is \(\frac{p^2}{2}\). Setting these equal helps us find that \(p\) can be either 1 or -1.

🎯 Exam Tip: When evaluating limits involving trigonometric functions, remember to use standard limits like \( \underset{\theta \rightarrow 0}{\mathrm{Lt}} \frac{\sin \theta}{\theta} = 1 \) and the identity \(1-\cos \theta = 2 \sin^2 (\frac{\theta}{2})\). Always ensure the argument of sine matches the denominator for the limit property to apply.

Question 28. The function \( f (x) = | \sin x | \)
(a) f is differentiable everywhere
(b) f is continuous everywhere but not differentiable at \(x = n\pi, n \in Z\)
(c) f is continuous everywhere but not differentiable at \(x = (2n + 1) \frac { \pi }{ 2 }, n \in Z\)
(d) none of these
Answer: (b) f is continuous everywhere but not differentiable at \(x = n\pi, n \in Z\)
The function \(f(x) = |\sin x|\) can be expressed as:
\( f(x) = \left\{\begin{array}{ll} \sin x, & \text { if } \sin x \geq 0 \\ -\sin x, & \text { if } \sin x < 0 \end{array}\right. \)
Since \(\sin x\) is continuous everywhere, and \(|x|\) is continuous everywhere, their composition \(|\sin x|\) is also continuous everywhere.
Now, let's check for differentiability. A function of the form \(|g(x)|\) is generally not differentiable where \(g(x)=0\) and \(g'(x) \neq 0\). Here, \(g(x) = \sin x\), and \(g'(x) = \cos x\).
\(g(x) = \sin x = 0\) when \(x = n\pi\) for any integer \(n\).
At these points, \(g'(n\pi) = \cos(n\pi) = (-1)^n\), which is \(\pm 1\) and not zero.
Thus, \(f(x) = |\sin x|\) is not differentiable at \(x = n\pi\), for any integer \(n\). This is because at these points, the graph forms a sharp corner (cusp).
Let's formally check the left-hand derivative (LHD) and right-hand derivative (RHD) at \(x = n\pi\).
**Case 1: \(n\) is an even integer (e.g., \(x=0, 2\pi, ...\))**
Around \(x=n\pi\) (where \(\sin x\) changes from negative to positive if \(n\) is even and for \(x < n\pi\), \(\sin x\) is negative):
\( \text{LHD} = \underset{x \rightarrow n\pi^{-}}{\mathrm{Lt}} \frac{f(x) - f(n\pi)}{x - n\pi} = \underset{x \rightarrow n\pi^{-}}{\mathrm{Lt}} \frac{-\sin x - 0}{x - n\pi} \)
Let \(x = n\pi - h\), where \(h \rightarrow 0^+\).
\( = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} \frac{-\sin(n\pi - h)}{n\pi - h - n\pi} = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} \frac{-(-\sin h)}{-h} = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} \frac{\sin h}{-h} = -1 \)
\( \text{RHD} = \underset{x \rightarrow n\pi^{+}}{\mathrm{Lt}} \frac{f(x) - f(n\pi)}{x - n\pi} = \underset{x \rightarrow n\pi^{+}}{\mathrm{Lt}} \frac{\sin x - 0}{x - n\pi} \)
Let \(x = n\pi + h\), where \(h \rightarrow 0^+\).
\( = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} \frac{\sin(n\pi + h)}{n\pi + h - n\pi} = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} \frac{\sin h}{h} = 1 \)
Since \(\text{LHD} \neq \text{RHD}\) at \(x=n\pi\) for even \(n\), \(f(x)\) is not differentiable.
**Case 2: \(n\) is an odd integer (e.g., \(x=\pi, 3\pi, ...\))**
Around \(x=n\pi\) (where \(\sin x\) changes from positive to negative if \(n\) is odd):
\( \text{LHD} = \underset{x \rightarrow n\pi^{-}}{\mathrm{Lt}} \frac{f(x) - f(n\pi)}{x - n\pi} = \underset{x \rightarrow n\pi^{-}}{\mathrm{Lt}} \frac{\sin x - 0}{x - n\pi} \)
Let \(x = n\pi - h\), where \(h \rightarrow 0^+\).
\( = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} \frac{\sin(n\pi - h)}{n\pi - h - n\pi} = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} \frac{\sin h}{-h} = -1 \)
\( \text{RHD} = \underset{x \rightarrow n\pi^{+}}{\mathrm{Lt}} \frac{f(x) - f(n\pi)}{x - n\pi} = \underset{x \rightarrow n\pi^{+}}{\mathrm{Lt}} \frac{-\sin x - 0}{x - n\pi} \)
Let \(x = n\pi + h\), where \(h \rightarrow 0^+\).
\( = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} \frac{-\sin(n\pi + h)}{n\pi + h - n\pi} = \underset{h \rightarrow 0^{+}}{\mathrm{Lt}} \frac{-(-\sin h)}{h} = 1 \)
Since \(\text{LHD} \neq \text{RHD}\) at \(x=n\pi\) for odd \(n\), \(f(x)\) is not differentiable.
Therefore, \(f(x) = |\sin x|\) is continuous everywhere but not differentiable at \(x = n\pi\) for all integers \(n\).
In simple words: The function \(|\sin x|\) is smooth everywhere because the sine wave itself is smooth, and taking the absolute value just flips the negative parts up. However, wherever \(\sin x\) crosses the x-axis (at \(n\pi\)), the graph creates a sharp corner because it reflects the wave upwards. At these sharp corners, you can't find a single, clear slope, so the function is not differentiable there.

🎯 Exam Tip: Functions involving absolute values, like \(|g(x)|\), are always continuous. However, they are typically not differentiable at points where \(g(x)=0\) and \(g'(x) \neq 0\), as these points correspond to sharp corners or cusps in the graph.

Question 29. If \( f(x) = \left\{\begin{array}{cc} m x+1, & x \geq \frac{\pi}{2} \\ \sin x+n, & x<\frac{\pi}{2} \end{array}\right. \) is continuous at \( x = \frac { \pi }{ 2 }\), then
(a) m = 1, n = 0
(b) m = \( \frac { n\pi }{ 2 } + 1 \)
(c) n = \( \frac { m\pi }{ 2 } \)
(d) m = n = \( \frac { \pi }{ 2 } \)
Answer: Solution not provided in the given page range.

Question 29. If \( f(x) = \left\{\begin{array}{cc} m x+1, & x \geq \frac{\pi}{2} \\ \sin x+n, & x<\frac{\pi}{2} \end{array}\right. \) is continuous at \( x = \frac{\pi}{2} \), then
(a) \( m = 1, n = 0 \)
(b) \( m = \frac{\eta\pi}{2} + 1 \)
(c) \( n = \frac{m\pi}{2} \)
(d) \( m = n = \frac{\pi}{2} \)
Answer: (c) \( n = \frac{m\pi}{2} \)
L.H.L. at \( x = \frac{\pi}{2} \):
\( \underset{x \rightarrow \frac{\pi}{2}^-}{\mathrm{Lt}} f(x) = \underset{x \rightarrow \frac{\pi}{2}^-}{\mathrm{Lt}} (\sin x+n) \)
\( = \sin \frac{\pi}{2} + n = 1 + n \)
R.H.L. at \( x = \frac{\pi}{2} \):
\( \underset{x \rightarrow \frac{\pi}{2}^+}{\mathrm{Lt}} f(x) = \underset{x \rightarrow \frac{\pi}{2}^+}{\mathrm{Lt}} (m x+1) \)
\( = m \frac{\pi}{2} + 1 \)
Value of the function at \( x = \frac{\pi}{2} \):
\( f(\frac{\pi}{2}) = m \frac{\pi}{2} + 1 \)
For \( f(x) \) to be continuous at \( x = \frac{\pi}{2} \), the L.H.L., R.H.L., and \( f(\frac{\pi}{2}) \) must all be equal.
So, \( 1+n = m \frac{\pi}{2} + 1 \)
\( \implies n = m \frac{\pi}{2} \)
This means the relationship between m and n is \( n = \frac{m\pi}{2} \).
In simple words: For a function to be smooth and continuous, its value from the left side, right side, and exactly at that point must all be the same. By setting these equal, we find the condition that links m and n for continuity.

🎯 Exam Tip: When a function is defined piecewise, always check the limits from both sides and the function's value at the critical point where the definition changes. For continuity, all three must match.

Question 30. Determine the value of constant 'k' so that the function \( f(x) = \left\{\begin{array}{r} \frac{k x}{|x|}, x<0 \\ 3, x \geq 0 \end{array}\right. \) is continuous at \( x = 0 \).
Answer: Given the function \( f(x) = \left\{\begin{array}{r} \frac{k x}{|x|}, x<0 \\ 3, x \geq 0 \end{array}\right. \).
For the function to be continuous at \( x=0 \), the left-hand limit, the right-hand limit, and the value of the function at \( x=0 \) must all be equal.
First, let's find the left-hand limit (L.H.L.) as \( x \) approaches 0 from the left side (where \( x<0 \)):
L.H.L. \( = \underset{x \rightarrow 0^-}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 0^-}{\mathrm{Lt}} \frac{k x}{|x|} \)
When \( x<0 \), \( |x| = -x \).
\( = \underset{x \rightarrow 0^-}{\mathrm{Lt}} \frac{k x}{-x} = \underset{x \rightarrow 0^-}{\mathrm{Lt}} (-k) = -k \)
Next, let's find the right-hand limit (R.H.L.) as \( x \) approaches 0 from the right side (where \( x \geq 0 \)):
R.H.L. \( = \underset{x \rightarrow 0^+}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 0^+}{\mathrm{Lt}} (3) = 3 \)
Now, find the value of the function at \( x=0 \):
\( f(0) = 3 \)
For continuity, L.H.L. = R.H.L. = \( f(0) \).
So, we must have \( -k = 3 \).
\( \implies k = -3 \).
In simple words: For the function to be continuous at zero, the values it approaches from the left, from the right, and its actual value at zero must all be the same. By making them equal, we find that 'k' must be -3.

🎯 Exam Tip: When dealing with absolute values in limits, remember to define the absolute value function piecewise according to whether the expression inside is positive or negative. This is crucial for correctly evaluating left and right-hand limits.

Question 31. If the function f defined as \( f(x) = \left\{\begin{array}{cc} \frac{x^2-9}{x-3}, & x \neq 3 \\ k & x=3 \end{array}\right. \) is continuous at \( x = 3 \), find the value of A.
Answer: Given the function \( f(x) = \left\{\begin{array}{cc} \frac{x^2-9}{x-3}, & x \neq 3 \\ k & x=3 \end{array}\right. \). (Note: The question asks for 'A', but 'k' is used in the function, so we will find 'k').
For a function to be continuous at a point \( x=c \), the limit of the function as \( x \) approaches \( c \) must be equal to the value of the function at \( c \). So, \( \underset{x \rightarrow 3}{\mathrm{Lt}} f(x) = f(3) \).
Let's find the limit of \( f(x) \) as \( x \) approaches 3:
\( \underset{x \rightarrow 3}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 3}{\mathrm{Lt}} \frac{x^2-9}{x-3} \)
We can factor the numerator \( x^2-9 \) as \( (x-3)(x+3) \).
\( = \underset{x \rightarrow 3}{\mathrm{Lt}} \frac{(x-3)(x+3)}{x-3} \)
Since \( x \neq 3 \), we can cancel out the \( (x-3) \) term.
\( = \underset{x \rightarrow 3}{\mathrm{Lt}} (x+3) \)
Now, substitute \( x=3 \) into the expression:
\( = 3+3 = 6 \)
The value of the function at \( x=3 \) is given as \( f(3) = k \).
For continuity, we set the limit equal to the function's value:
\( 6 = k \)
So, the value of k (or A, as intended by the question) is 6.
In simple words: If a function is continuous at a certain point, its value at that point is the same as what it approaches from both sides. We simplified the function to find what it approaches, and that value must be 'k'.

🎯 Exam Tip: For functions with removable discontinuities (like this one), factor the numerator and denominator to simplify the expression and find the limit. The value of k will be this limit if the function is continuous.

Question 32. If \( f(x) = \left\{\begin{array}{cc} x^2, & \text { when } x \neq 1 \\ 0 & \text { when } x=1 \end{array}\right. \) find whether it is continuous or discontinuous at \( x = 1 \).
Answer: Given the function \( f(x) = \left\{\begin{array}{cc} x^2, & x \neq 1 \\ 0, & x=1 \end{array}\right. \).
To check the continuity of \( f(x) \) at \( x=1 \), we need to compare the limit of \( f(x) \) as \( x \) approaches 1 with the value of \( f(1) \).
First, let's find the limit of \( f(x) \) as \( x \) approaches 1:
\( \underset{x \rightarrow 1}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1}{\mathrm{Lt}} x^2 \)
Since \( x \) is approaching 1 (but not equal to 1), we use the first part of the function definition.
\( = 1^2 = 1 \)
Next, let's find the value of the function at \( x=1 \):
\( f(1) = 0 \) (from the second part of the function definition)
For continuity, \( \underset{x \rightarrow 1}{\mathrm{Lt}} f(x) \) must be equal to \( f(1) \).
Here, \( \underset{x \rightarrow 1}{\mathrm{Lt}} f(x) = 1 \) and \( f(1) = 0 \).
Since \( 1 \neq 0 \), we have \( \underset{x \rightarrow 1}{\mathrm{Lt}} f(x) \neq f(1) \).
Therefore, the function \( f(x) \) is discontinuous at \( x=1 \). A function is considered continuous if you can draw its graph without lifting your pen.
In simple words: The function is not continuous at x=1 because the value the function is heading towards (which is 1) is not the same as its actual value at that point (which is 0). There is a break in the graph at x=1.

🎯 Exam Tip: For functions defined piecewise at a point, always explicitly calculate the limit at that point and the function's value at that point, then compare them to determine continuity. A mismatch indicates discontinuity.

Question 33. If the function \( f(x) = \left\{\begin{array}{cc} \frac{\sin \frac{3 x}{2}}{x}, & x \neq 0 \\ k, & x=0 \end{array}\right. \) is continuous at \( x = 0 \), then write the value of A.
Answer: Given the function \( f(x) = \left\{\begin{array}{cc} \frac{\sin \frac{3 x}{2}}{x}, & x \neq 0 \\ k, & x=0 \end{array}\right. \). (Note: The question asks for 'A', but 'k' is used in the function definition, so we will find 'k').
For the function \( f(x) \) to be continuous at \( x=0 \), the limit of \( f(x) \) as \( x \) approaches 0 must be equal to the value of \( f(0) \). So, \( \underset{x \rightarrow 0}{\mathrm{Lt}} f(x) = f(0) \).
Let's find the limit of \( f(x) \) as \( x \) approaches 0:
\( \underset{x \rightarrow 0}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 0}{\mathrm{Lt}} \frac{\sin \frac{3 x}{2}}{x} \)
To evaluate this limit, we can use the standard limit identity \( \underset{\theta \rightarrow 0}{\mathrm{Lt}} \frac{\sin \theta}{\theta} = 1 \). We need to make the denominator match the argument of the sine function.
Multiply and divide the expression by \( \frac{3}{2} \):
\( = \underset{x \rightarrow 0}{\mathrm{Lt}} \frac{\sin \frac{3 x}{2}}{x} \times \frac{\frac{3}{2}}{\frac{3}{2}} \)
\( = \underset{x \rightarrow 0}{\mathrm{Lt}} \frac{\sin \frac{3 x}{2}}{\frac{3 x}{2}} \times \frac{3}{2} \)
As \( x \rightarrow 0 \), \( \frac{3x}{2} \rightarrow 0 \). So, \( \underset{x \rightarrow 0}{\mathrm{Lt}} \frac{\sin \frac{3 x}{2}}{\frac{3 x}{2}} = 1 \).
\( = 1 \times \frac{3}{2} = \frac{3}{2} \)
Now, let's find the value of the function at \( x=0 \):
\( f(0) = k \)
For continuity, we set the limit equal to the function's value:
\( k = \frac{3}{2} \)
So, the value of k is \( \frac{3}{2} \). Trigonometric limits are often used to define values for continuity.
In simple words: For the function to be smooth at x=0, the value it approaches must be equal to its actual value at x=0. We used a special rule for sine limits to find this approaching value, which tells us what 'k' must be.

🎯 Exam Tip: Remember the fundamental trigonometric limit \( \underset{\theta \rightarrow 0}{\mathrm{Lt}} \frac{\sin \theta}{\theta} = 1 \). When solving continuity problems involving trigonometric functions, manipulate the expression to fit this form by multiplying and dividing by the necessary term in the denominator.

Question 34. Examine the continuity of the function \( f(x) = x^3 + 2x^2 - 1 \).
Answer: Given the function \( f(x) = x^3 + 2x^2 - 1 \).
This function is a polynomial function because it is made up of terms involving \( x \) raised to non-negative integer powers (like \( x^3 \), \( x^2 \), and a constant term) combined with addition and subtraction. Polynomials are among the most basic continuous functions.
A key property of polynomial functions is that they are continuous everywhere in their domain, which for any polynomial is all real numbers (R). This means there are no breaks, jumps, or holes in their graphs.
Therefore, the function \( f(x) = x^3 + 2x^2 - 1 \) is continuous at every real point \( x \in R \).
In simple words: This function is a polynomial, which means it's a smooth curve without any gaps or breaks. Because of this, it is continuous everywhere, meaning you can draw its graph without ever lifting your pen.

🎯 Exam Tip: Recognize common continuous functions like polynomials, exponential functions, sine, and cosine functions. Their continuity over their entire domain (or specific intervals) is a standard result that simplifies continuity examination.

Question 35. Examine the continuity of the function \( f(x) = \left\{\begin{array}{rll} 3 x+5 & \text { if } & x \geq 2 \\ x^2 & \text { if } & x<2 \end{array}\right. \)
Answer: Given the function \( f(x) = \left\{\begin{array}{rll} 3 x+5 & \text { if } & x \geq 2 \\ x^2 & \text { if } & x<2 \end{array}\right. \).
To examine the continuity of this piecewise function, we need to check three cases:
**Case I: When \( x < 2 \)**
In this interval, \( f(x) = x^2 \). This is a polynomial function, and all polynomial functions are continuous everywhere. So, \( f(x) \) is continuous for all \( x < 2 \).
**Case II: When \( x > 2 \)**
In this interval, \( f(x) = 3x+5 \). This is also a polynomial function, and thus it is continuous for all \( x > 2 \).
**Case III: At \( x = 2 \)**
This is the critical point where the function's definition changes. For continuity at \( x=2 \), the left-hand limit, the right-hand limit, and the value of the function at \( x=2 \) must all be equal.
1. Left-Hand Limit (L.H.L.) as \( x \rightarrow 2^- \):
\( \underset{x \rightarrow 2^-}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 2^-}{\mathrm{Lt}} x^2 \)
\( = 2^2 = 4 \)
2. Right-Hand Limit (R.H.L.) as \( x \rightarrow 2^+ \):
\( \underset{x \rightarrow 2^+}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 2^+}{\mathrm{Lt}} (3x+5) \)
\( = 3(2)+5 = 6+5 = 11 \)
3. Value of the function at \( x=2 \):
\( f(2) = 3(2)+5 = 11 \) (since \( x \geq 2 \) applies)
Comparing the limits: L.H.L. \( = 4 \) and R.H.L. \( = 11 \).
Since L.H.L. \( \neq \) R.H.L. (4 is not equal to 11), the limit of \( f(x) \) as \( x \) approaches 2 does not exist.
Therefore, the function \( f(x) \) is discontinuous at \( x=2 \). A function must have matching limits from both sides to be continuous. Combining all cases, \( f(x) \) is continuous everywhere except at \( x=2 \).
In simple words: This function is made of two smooth parts, but they don't meet up at x=2. The left side goes to 4, but the right side goes to 11. Because these don't match, the function has a jump at x=2 and is not continuous there.

🎯 Exam Tip: Always analyze piecewise functions by checking the continuity of each "piece" in its defined interval, and then critically examine the points where the function's definition changes. These boundary points are where discontinuities most often occur.

Question 36. Examine the continuity of \( f(x) = \left\{\begin{array}{cl} \frac{|x-4|}{2(x-4)}, & \text { if } x \neq 4 \\ 0, & \text { if } x=4 \end{array}\right. \) at \( x=4 \).
Answer: Given the function \( f(x) = \left\{\begin{array}{cl} \frac{|x-4|}{2(x-4)}, & x \neq 4 \\ 0, & x=4 \end{array}\right. \).
To examine the continuity of \( f(x) \) at \( x=4 \), we need to check if the limit of \( f(x) \) as \( x \) approaches 4 is equal to \( f(4) \).
First, let's find the value of the function at \( x=4 \):
\( f(4) = 0 \) (from the second part of the definition).
Next, we need to find the left-hand limit (L.H.L.) as \( x \) approaches 4 from the left side (where \( x<4 \)):
L.H.L. \( = \underset{x \rightarrow 4^-}{\mathrm{Lt}} \frac{|x-4|}{2(x-4)} \)
When \( x<4 \), the expression \( (x-4) \) is negative. Therefore, \( |x-4| = -(x-4) \).
\( = \underset{x \rightarrow 4^-}{\mathrm{Lt}} \frac{-(x-4)}{2(x-4)} \)
We can cancel \( (x-4) \) from the numerator and denominator since \( x \neq 4 \).
\( = \underset{x \rightarrow 4^-}{\mathrm{Lt}} (-\frac{1}{2}) = -\frac{1}{2} \)
Now, let's find the right-hand limit (R.H.L.) as \( x \) approaches 4 from the right side (where \( x>4 \)):
R.H.L. \( = \underset{x \rightarrow 4^+}{\mathrm{Lt}} \frac{|x-4|}{2(x-4)} \)
When \( x>4 \), the expression \( (x-4) \) is positive. Therefore, \( |x-4| = (x-4) \).
\( = \underset{x \rightarrow 4^+}{\mathrm{Lt}} \frac{(x-4)}{2(x-4)} \)
Again, we can cancel \( (x-4) \).
\( = \underset{x \rightarrow 4^+}{\mathrm{Lt}} (\frac{1}{2}) = \frac{1}{2} \)
For the limit to exist, the L.H.L. and R.H.L. must be equal. Here, L.H.L. \( = -\frac{1}{2} \) and R.H.L. \( = \frac{1}{2} \).
Since L.H.L. \( \neq \) R.H.L., the limit of \( f(x) \) as \( x \) approaches 4 does not exist. Even if it existed, it would also need to equal \( f(4) \).
Therefore, the function \( f(x) \) is discontinuous at \( x=4 \).
In simple words: This function acts differently when you approach x=4 from the left versus from the right. Because these approaches lead to different values, the function has a jump at x=4 and is not continuous.

🎯 Exam Tip: Absolute value functions often lead to different left and right-hand limits at the point where the expression inside the absolute value changes sign. Always evaluate these limits separately to check for continuity or discontinuity.

Question 37. Examine the continuity of \( f(x) = |x| + |x - 1| \), at \( x = 1 \).
Answer: Given the function \( f(x) = |x| + |x - 1| \).
To examine the continuity of \( f(x) \) at \( x=1 \), we first define the function piecewise by considering the signs of \( x \) and \( (x-1) \).
1. If \( x < 0 \): Then \( |x| = -x \) and \( |x-1| = -(x-1) = -x+1 \).
So, \( f(x) = -x + (-x+1) = -2x+1 \).
2. If \( 0 \leq x < 1 \): Then \( |x| = x \) and \( |x-1| = -(x-1) = -x+1 \).
So, \( f(x) = x + (-x+1) = 1 \).
3. If \( x \geq 1 \): Then \( |x| = x \) and \( |x-1| = (x-1) \).
So, \( f(x) = x + (x-1) = 2x-1 \).
The piecewise definition of the function is:
\[ f(x) = \begin{cases} -2x+1, & x < 0 \\ 1, & 0 \leq x < 1 \\ 2x-1, & x \geq 1 \end{cases} \]
Now, let's check for continuity at \( x=1 \):
1. Left-Hand Limit (L.H.L.) as \( x \rightarrow 1^- \):
As \( x \) approaches 1 from the left, \( 0 \leq x < 1 \), so \( f(x) = 1 \).
\( \underset{x \rightarrow 1^-}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^-}{\mathrm{Lt}} (1) = 1 \)
2. Right-Hand Limit (R.H.L.) as \( x \rightarrow 1^+ \):
As \( x \) approaches 1 from the right, \( x \geq 1 \), so \( f(x) = 2x-1 \).
\( \underset{x \rightarrow 1^+}{\mathrm{Lt}} f(x) = \underset{x \rightarrow 1^+}{\mathrm{Lt}} (2x-1) = 2(1)-1 = 1 \)
3. Value of the function at \( x=1 \):
Since \( x \geq 1 \) for \( x=1 \), \( f(1) = 2(1)-1 = 1 \).
Since L.H.L. = R.H.L. = \( f(1) = 1 \), the function \( f(x) \) is continuous at \( x=1 \). Absolute value functions are inherently smooth, which helps ensure continuity.
In simple words: This function combines two absolute values. When we check its behavior near x=1, we see that the value it approaches from the left, from the right, and its actual value at x=1 are all the same (which is 1). This means the function is smooth and continuous at x=1.

🎯 Exam Tip: When a function involves multiple absolute values, defining it piecewise is the most effective strategy. Carefully identify the critical points where the expressions inside the absolute values change sign to correctly define each piece of the function.

Question 38. If \( f(x) = 2|x| + 3|\sin x| + 6 \), then the right hand derivative of \( f(x) \) at \( x=0 \) is
Answer: Given the function \( f(x) = 2|x| + 3|\sin x| + 6 \).
We need to find the right-hand derivative of \( f(x) \) at \( x=0 \), denoted as \( R f'(0) \). The formula for the right-hand derivative is:
\( R f'(0) = \underset{h \rightarrow 0^+}{\mathrm{Lt}} \frac{f(0+h) - f(0)}{h} \)
First, let's find \( f(0) \):
\( f(0) = 2|0| + 3|\sin 0| + 6 = 2(0) + 3(0) + 6 = 6 \)
Next, let's find \( f(0+h) \) for \( h > 0 \). Since \( h \) is a small positive number, \( |h| = h \) and \( |\sin h| = \sin h \) (as \( \sin h > 0 \) for small positive \( h \)).
So, \( f(0+h) = f(h) = 2h + 3\sin h + 6 \).
Now, substitute these into the right-hand derivative formula:
\( R f'(0) = \underset{h \rightarrow 0^+}{\mathrm{Lt}} \frac{(2h + 3\sin h + 6) - 6}{h} \)
\( = \underset{h \rightarrow 0^+}{\mathrm{Lt}} \frac{2h + 3\sin h}{h} \)
We can split the fraction:
\( = \underset{h \rightarrow 0^+}{\mathrm{Lt}} (\frac{2h}{h} + \frac{3\sin h}{h}) \)
\( = \underset{h \rightarrow 0^+}{\mathrm{Lt}} (2 + 3 \frac{\sin h}{h}) \)
Using the standard limit \( \underset{h \rightarrow 0}{\mathrm{Lt}} \frac{\sin h}{h} = 1 \):
\( = 2 + 3(1) = 5 \)
Thus, the right-hand derivative of \( f(x) \) at \( x=0 \) is 5.
In simple words: The right-hand derivative tells us the slope of the function's graph when we approach x=0 from the positive side. By calculating the limit, we find that this slope is 5.

🎯 Exam Tip: When calculating derivatives from first principles, especially right-hand or left-hand derivatives, ensure you correctly interpret absolute value functions and trigonometric functions like sine for small positive (or negative) h, as this affects their simplified form.

Question 39. Let \( f(x) = x|x| \) for all \( x \in R \), check its differentiability at \( x = 0 \).
Answer: Given the function \( f(x) = x|x| \) for all \( x \in R \).
To check its differentiability at \( x=0 \), we need to determine if the left-hand derivative (LHD) and the right-hand derivative (RHD) at \( x=0 \) are equal.
First, let's define \( f(x) \) piecewise:
1. If \( x < 0 \): Then \( |x| = -x \). So, \( f(x) = x(-x) = -x^2 \).
2. If \( x \geq 0 \): Then \( |x| = x \). So, \( f(x) = x(x) = x^2 \).
So, the piecewise function is:
\[ f(x) = \begin{cases} -x^2, & x < 0 \\ x^2, & x \geq 0 \end{cases} \]
Also, the value of the function at \( x=0 \) is \( f(0) = 0|0| = 0 \).
Now, let's calculate the Left-Hand Derivative (LHD) at \( x=0 \):
\( L f'(0) = \underset{h \rightarrow 0^-}{\mathrm{Lt}} \frac{f(0+h) - f(0)}{h} = \underset{h \rightarrow 0^-}{\mathrm{Lt}} \frac{f(h) - 0}{h} \)
For \( h < 0 \), \( f(h) = -h^2 \).
\( = \underset{h \rightarrow 0^-}{\mathrm{Lt}} \frac{-h^2}{h} = \underset{h \rightarrow 0^-}{\mathrm{Lt}} (-h) = 0 \)
Next, let's calculate the Right-Hand Derivative (RHD) at \( x=0 \):
\( R f'(0) = \underset{h \rightarrow 0^+}{\mathrm{Lt}} \frac{f(0+h) - f(0)}{h} = \underset{h \rightarrow 0^+}{\mathrm{Lt}} \frac{f(h) - 0}{h} \)
For \( h > 0 \), \( f(h) = h^2 \).
\( = \underset{h \rightarrow 0^+}{\mathrm{Lt}} \frac{h^2}{h} = \underset{h \rightarrow 0^+}{\mathrm{Lt}} (h) = 0 \)
Since the LHD at \( x=0 \) is 0 and the RHD at \( x=0 \) is also 0, they are equal.
Therefore, the function \( f(x) = x|x| \) is differentiable at \( x=0 \). It has a smooth transition at the origin.
In simple words: To check if a function is differentiable (smooth, no sharp corners) at a point, we look at the slope from the left and the slope from the right. For this function, both slopes at x=0 are 0, which means they match, and so the function is differentiable there.

🎯 Exam Tip: For functions involving absolute values, always define the function piecewise first. Then, use the definitions of left-hand and right-hand derivatives to check if they are equal at the point in question. If they are equal, the function is differentiable at that point.