GSEB Class 12 Maths Solutions Chapter 5 Continuity and Differentiability Exercise 5.1

Question 1. Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = -3 - 3 and at x = 5.
Answer:
(i) At \( x = 0 \):
\( \lim_{x \rightarrow 0}f(x) = \lim _{x \rightarrow 0} (5x - 3) = -3 \)
And \( f(0) = -3 \).
Since the limit of the function as \( x \) approaches 0 is equal to the function's value at 0, the function \( f \) is continuous at \( x = 0 \).
(ii) At \( x = -3 \):
\( \lim _{x \rightarrow -3}f(x) = \lim _{x \rightarrow -3} (5x - 3) = -18 \)
And \( f(-3) = -18 \).
Since the limit of the function as \( x \) approaches -3 is equal to the function's value at -3, the function \( f \) is continuous at \( x = -3 \).
(iii) At \( x = 5 \):
\( \lim _{x \rightarrow 5} f(x) = \lim _{x \rightarrow 5} (5x - 3) = 22 \)
And \( f(5) = 22 \).
Since the limit of the function as \( x \) approaches 5 is equal to the function's value at 5, the function \( f \) is continuous at \( x = 5 \).
In simple words: To check if a function is continuous at a point, we see if the limit of the function as it gets close to that point is the same as the function's value at that exact point. If they match, the function is smooth and continuous there.

Exam Tip: For polynomial functions, they are always continuous everywhere. You just need to show that the limit at the point matches the function's value at that point.

Question 2. Examine the continuity of the function f(x) = 2x² – 1 at x = 3.
Answer:
To examine continuity at \( x = 3 \):
\( \lim _{x \rightarrow 3} f(x) = \lim _{x \rightarrow 3} (2x² - 1) = 2(3)² - 1 = 2(9) - 1 = 18 - 1 = 17 \)
Now, find the function's value at \( x = 3 \):
\( f(3) = 2(3)² - 1 = 18 - 1 = 17 \)
Since \( \lim _{x \rightarrow 3} f(x) = f(3) \), the function \( f \) is continuous at \( x = 3 \).
In simple words: We check if the function's value when \( x \) is exactly 3 matches what the function gets closer to when \( x \) is almost 3. Since both are 17, the function is continuous.

Exam Tip: Always calculate both the limit of the function as x approaches the point and the exact function value at that point. If they are equal, the function is continuous.

Question 3. Examine the following functions for continuity:
(a) f(x) = x - 5
(b) f(x) = \(\frac { 1 }{ x-5 }\)
(c) f(x) = \(\frac{x^{2}-25}{x+5}\)
(d) f(x) = |x-5|
Answer:
(a) \( f(x) = x - 5 \)
This function is a polynomial. All polynomial functions are continuous everywhere in their domain.
Therefore, \( f(x) = x - 5 \) is continuous for all real numbers \( x \in R \).

(b) \( f(x) = \frac { 1 }{ x-5 } \)
This function is a rational function. A rational function is undefined when its denominator is zero. Here, the denominator is \( x - 5 \).
If \( x - 5 = 0 \), then \( x = 5 \).
At \( x = 5 \), \( f(x) \) is not defined.
Therefore, \( f \) is not continuous at \( x = 5 \).
For any \( x \neq 5 \), let \( x = c \). Then \( \lim _{x \rightarrow c} f(x) = \lim _{x \rightarrow c} \frac { 1 }{ x-5 } = \frac { 1 }{ c-5 } \).
Also, \( f(c) = \frac { 1 }{ c-5 } \).
Since \( \lim _{x \rightarrow c} f(x) = f(c) \) for \( x \neq 5 \), the function \( f \) is continuous for all \( x \in R - \{5\} \).

(c) \( f(x) = \frac{x^{2}-25}{x+5} \)
This function can be simplified: \( f(x) = \frac{(x-5)(x+5)}{x+5} \).
The function is undefined when the denominator is zero, so \( x + 5 = 0 \), which means \( x = -5 \).
At \( x = -5 \), the function is not defined.
Therefore, \( f \) is discontinuous at \( x = -5 \).
For any \( x \neq -5 \), let \( x = c \). Then \( \lim _{x \rightarrow c} f(x) = \lim _{x \rightarrow c} \frac{(x-5)(x+5)}{x+5} = \lim _{x \rightarrow c} (x-5) = c - 5 \).
Also, \( f(c) = c - 5 \) (when \( c \neq -5 \)).
Since \( \lim _{x \rightarrow c} f(x) = f(c) \) for \( x \neq -5 \), the function \( f \) is continuous for all \( x \in R - \{-5\} \).

(d) \( f(x) = |x - 5| \)
The absolute value function \( |x| \) is continuous everywhere. Since \( x - 5 \) is a polynomial (and thus continuous), the composition of these functions, \( |x - 5| \), is also continuous everywhere.
To confirm this, consider any real number \( c \):
Case 1: At \( x = c \).
\( \lim _{x \rightarrow c} |x - 5| = |c - 5| \). And \( f(c) = |c - 5| \). So \( f \) is continuous at \( x = c \).
Case 2: At \( x = 5 \).
\( f(5) = |5 - 5| = 0 \).
\( \lim _{x \rightarrow 5^{-}} |x - 5| = \lim _{x \rightarrow 5^{-}} -(x - 5) = -(5 - 5) = 0 \).
\( \lim _{x \rightarrow 5^{+}} |x - 5| = \lim _{x \rightarrow 5^{+}} (x - 5) = 5 - 5 = 0 \).
Since \( \lim _{x \rightarrow 5} |x - 5| = f(5) = 0 \), the function is continuous at \( x = 5 \).
Case 3: At \( x = c > 5 \).
\( \lim _{x \rightarrow c} |x - 5| = c - 5 \) (since \( x - 5 \) is positive). And \( f(c) = c - 5 \). So \( f \) is continuous at \( x = c > 5 \).
Case 4: At \( x = c < 5 \).
\( \lim _{x \rightarrow c} |x - 5| = -(c - 5) = 5 - c \) (since \( x - 5 \) is negative). And \( f(c) = 5 - c \). So \( f \) is continuous at \( x = c < 5 \).
Thus, \( f \) is continuous for all \( x \in R \).
In simple words: Polynomials are always continuous. For fractions, they are not continuous where the bottom part is zero. The absolute value function makes things continuous everywhere, so \( |x - 5| \) has no breaks.

Exam Tip: Remember that polynomial functions and absolute value functions are continuous everywhere. Rational functions are continuous everywhere except at points where the denominator is zero. Always simplify rational functions first to find potential discontinuities.

Question 4. Prove that the function f(x) = \( x^h \) is continuous at x = n, when n is a positive integer.
Answer:
The function given is \( f(x) = x^h \).
This is a power function, which is a type of polynomial function when \( h \) is a positive integer.
All polynomial functions are known to be continuous for all real numbers.
Since \( n \) is a positive integer, it is a real number.
Therefore, the function \( f(x) = x^h \) is continuous at \( x = n \), where \( n \) belongs to the set of natural numbers (positive integers).
In simple words: The function \( f(x) = x^h \) is a type of polynomial. All polynomial functions are smooth and unbroken everywhere. Since \( n \) is just a specific whole number, the function will also be continuous at \( x = n \).

Exam Tip: Recognize that power functions like \( x^h \) (where h is a positive integer) are essentially polynomial functions. Polynomials are fundamental continuous functions, meaning they have no breaks, jumps, or holes in their graphs.

Question 5. Is the function f defined by \( f(x) = \left\{\begin{array}{l} x, \text { if } x \leq 1 \\ 5, \text { if } x>1 \end{array}\right. \) continuous at x = 0?, At x = 1?, At x = 2?
Answer:
(i) At \( x = 0 \):
Since \( 0 \leq 1 \), we use \( f(x) = x \).
\( \lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{-}} x = 0 \)
\( \lim _{x \rightarrow 0^{+}} f(x) = \lim _{x \rightarrow 0^{+}} x = 0 \)
Also, \( f(0) = 0 \).
Since \( \lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{+}} f(x) = f(0) \), the function \( f \) is continuous at \( x = 0 \).

(ii) At \( x = 1 \):
L.H.L.: \( \lim _{x \rightarrow 1^{-}} f(x) = \lim _{x \rightarrow 1^{-}} (x) = 1 \)
R.H.L.: \( \lim _{x \rightarrow 1^{+}} f(x) = \lim _{x \rightarrow 1^{+}} (5) = 5 \)
Since \( \lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x) \), the left-hand limit is not equal to the right-hand limit.
Therefore, the function \( f \) is discontinuous at \( x = 1 \).

(iii) At \( x = 2 \):
Since \( 2 > 1 \), we use \( f(x) = 5 \).
L.H.L.: \( \lim _{x \rightarrow 2^{-}} f(x) = \lim _{x \rightarrow 2^{-}} (5) = 5 \)
R.H.L.: \( \lim _{x \rightarrow 2^{+}} f(x) = \lim _{x \rightarrow 2^{+}} (5) = 5 \)
Also, \( f(2) = 5 \).
Since \( \lim _{x \rightarrow 2^{-}} f(x) = \lim _{x \rightarrow 2^{+}} f(x) = f(2) \), the function \( f \) is continuous at \( x = 2 \).
In simple words: At \( x=0 \), the function is continuous because the limit and value are both 0. At \( x=1 \), it's discontinuous because the left-side limit (1) doesn't match the right-side limit (5). At \( x=2 \), it's continuous because the function is always 5 around that point, so the limit and value are both 5.

Exam Tip: For piecewise functions, always check continuity at the "junction points" (where the rule changes) by comparing L.H.L., R.H.L., and the function's value. Also check continuity within each defined interval.

Question 6. \( f(x) = \left\{\begin{array}{l} 2 x+3, x \leq 2 \\ 2 x-3, x>2 \end{array}\right. \)
Answer:
We need to examine the continuity of the function \( f(x) \) at the point where its definition changes, which is \( x = 2 \).
At \( x = 2 \):
L.H.L. (Left Hand Limit):
\( \lim _{x \rightarrow 2^{-}} f(x) = \lim _{x \rightarrow 2^{-}} (2x + 3) = 2(2) + 3 = 4 + 3 = 7 \)
R.H.L. (Right Hand Limit):
\( \lim _{x \rightarrow 2^{+}} f(x) = \lim _{x \rightarrow 2^{+}} (2x - 3) = 2(2) - 3 = 4 - 3 = 1 \)
Function value at \( x = 2 \):
\( f(2) = 2(2) + 3 = 4 + 3 = 7 \) (using the condition \( x \leq 2 \))
Since L.H.L. \( (7) \neq \) R.H.L. \( (1) \), the function \( f \) is discontinuous at \( x = 2 \).

Now, check for continuity in the intervals:
For \( x < 2 \), \( f(x) = 2x + 3 \), which is a polynomial. Hence, \( f \) is continuous for all \( x < 2 \).
For \( x > 2 \), \( f(x) = 2x - 3 \), which is also a polynomial. Hence, \( f \) is continuous for all \( x > 2 \).
Therefore, the only point of discontinuity for this function is \( x = 2 \).
In simple words: We checked the function at \( x=2 \) where its rule changes. The limit from the left was 7, but the limit from the right was 1. Since these don't match, the function has a jump at \( x=2 \) and is not continuous there. For all other points, it acts like a simple smooth line.

Exam Tip: When dealing with piecewise functions, always test the continuity at the critical points where the definition changes. If the left-hand limit, right-hand limit, and the function's value at that point are all equal, the function is continuous there.

Question 7. \( f(x) = \left\{\begin{array}{l} |x|+3, \text { if } x \leq-3 \\ -2 x, \quad \text { if }-33 \end{array}\right. \)
Answer:
We need to examine the continuity at the critical points \( x = -3 \) and \( x = 3 \).

At \( x = -3 \):
L.H.L.: \( \lim _{x \rightarrow -3^{-}} f(x) = \lim _{x \rightarrow -3^{-}} (|x|+3) = \lim _{x \rightarrow -3^{-}} (-x + 3) = -(-3) + 3 = 3 + 3 = 6 \)
R.H.L.: \( \lim _{x \rightarrow -3^{+}} f(x) = \lim _{x \rightarrow -3^{+}} (-2x) = -2(-3) = 6 \)
Function value at \( x = -3 \):
\( f(-3) = |-3| + 3 = 3 + 3 = 6 \)
Since L.H.L. \( = \) R.H.L. \( = f(-3) \), the function \( f \) is continuous at \( x = -3 \).

At \( x = 3 \):
L.H.L.: \( \lim _{x \rightarrow 3^{-}} f(x) = \lim _{x \rightarrow 3^{-}} (-2x) = -2(3) = -6 \)
R.H.L.: \( \lim _{x \rightarrow 3^{+}} f(x) = \lim _{x \rightarrow 3^{+}} (6x + 2) = 6(3) + 2 = 18 + 2 = 20 \)
Function value at \( x = 3 \):
\( f(3) \) is not defined by any of the given conditions (since \( x > 3 \) or \( x \leq -3 \) or \( -3 < x < 3 \). If \( x=3 \), none apply).
Since L.H.L. \( (-6) \neq \) R.H.L. \( (20) \), the function \( f \) is discontinuous at \( x = 3 \).

Now, check for continuity in the intervals:
For \( x < -3 \), \( f(x) = |x|+3 = -x+3 \) (since \( x < 0 \)). This is a polynomial, so continuous.
For \( -3 < x < 3 \), \( f(x) = -2x \). This is a polynomial, so continuous.
For \( x > 3 \), \( f(x) = 6x+2 \). This is a polynomial, so continuous.
Thus, the only point of discontinuity is \( x = 3 \).
In simple words: We checked where the function's rules change. At \( x=-3 \), everything matched, so it's continuous. But at \( x=3 \), the left-side limit was -6 and the right-side limit was 20, showing a clear jump, so it's not continuous there.

Exam Tip: For piecewise functions involving absolute values, make sure to correctly evaluate \( |x| \) based on the sign of \( x \) in each interval. Always check both critical points and the intervals between them.

Question 8. \( f(x) = \left\{\begin{array}{ll} \frac{|x|}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right. \)
Answer:
We need to examine continuity at \( x = 0 \), which is the critical point.
At \( x = 0 \):
L.H.L.: \( \lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{-}} \frac{|x|}{x} = \lim _{x \rightarrow 0^{-}} \frac{-x}{x} = \lim _{x \rightarrow 0^{-}} (-1) = -1 \)
R.H.L.: \( \lim _{x \rightarrow 0^{+}} f(x) = \lim _{x \rightarrow 0^{+}} \frac{|x|}{x} = \lim _{x \rightarrow 0^{+}} \frac{x}{x} = \lim _{x \rightarrow 0^{+}} (1) = 1 \)
Function value at \( x = 0 \):
\( f(0) = 0 \) (given by definition)
Since L.H.L. \( (-1) \neq \) R.H.L. \( (1) \), and neither equals \( f(0) \), the function \( f \) is discontinuous at \( x = 0 \).

Now, check for continuity in the intervals:
For \( x < 0 \), \( f(x) = \frac{|x|}{x} = \frac{-x}{x} = -1 \). This is a constant function, so it is continuous.
For \( x > 0 \), \( f(x) = \frac{|x|}{x} = \frac{x}{x} = 1 \). This is also a constant function, so it is continuous.
Thus, the only point of discontinuity is \( x = 0 \).
In simple words: At \( x=0 \), the function jumps. When \( x \) approaches 0 from the left, it's -1. When it approaches from the right, it's 1. Since these limits are different, the function is not continuous at 0. Everywhere else, it's a simple flat line.

Exam Tip: For functions involving \( \frac{|x|}{x} \), remember that \( \frac{|x|}{x} \) equals -1 for \( x<0 \) and 1 for \( x>0 \). This often creates a jump discontinuity at \( x=0 \).

Question 9. \( f(x) = \left\{\begin{array}{ll} \frac{x}{|x|}, & \text { if } x<0 \\ -1, & \text { if } x \geq 0 \end{array}\right. \)
Answer:
We need to examine continuity at \( x = 0 \), the critical point.
At \( x = 0 \):
L.H.L.: \( \lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{-}} \frac{x}{|x|} = \lim _{x \rightarrow 0^{-}} \frac{x}{-x} = \lim _{x \rightarrow 0^{-}} (-1) = -1 \)
R.H.L.: \( \lim _{x \rightarrow 0^{+}} f(x) = \lim _{x \rightarrow 0^{+}} (-1) = -1 \)
Function value at \( x = 0 \):
\( f(0) = -1 \) (using the condition \( x \geq 0 \))
Since L.H.L. \( = \) R.H.L. \( = f(0) \), the function \( f \) is continuous at \( x = 0 \).

Now, check for continuity in the intervals:
For \( x < 0 \), \( f(x) = \frac{x}{|x|} = \frac{x}{-x} = -1 \). This is a constant function, so it is continuous.
For \( x > 0 \), \( f(x) = -1 \). This is also a constant function, so it is continuous.
Since the function is continuous at \( x=0 \) and within its defined intervals, there are no points of discontinuity for this function in its domain.
In simple words: When \( x \) is less than 0, the function is -1. When \( x \) is 0 or more, it's also -1. So the function is simply -1 everywhere. This means there are no breaks or jumps, so it's continuous everywhere.

Exam Tip: Simplify piecewise functions involving absolute values before calculating limits. When a function simplifies to a constant value across its domain, it's generally continuous everywhere.

Question 10. \( f(x) = \left\{\begin{array}{ll} x+1, & \text { if } x \geq 1 \\ x^{2}+1, & \text { if } x<1 \end{array}\right. \)
Answer:
We need to examine the continuity at the critical point \( x = 1 \).
At \( x = 1 \):
L.H.L.: \( \lim _{x \rightarrow 1^{-}} f(x) = \lim _{x \rightarrow 1^{-}} (x² + 1) = (1)² + 1 = 1 + 1 = 2 \)
R.H.L.: \( \lim _{x \rightarrow 1^{+}} f(x) = \lim _{x \rightarrow 1^{+}} (x + 1) = 1 + 1 = 2 \)
Function value at \( x = 1 \):
\( f(1) = 1 + 1 = 2 \) (using the condition \( x \geq 1 \))
Since L.H.L. \( = \) R.H.L. \( = f(1) \), the function \( f \) is continuous at \( x = 1 \).

Now, check for continuity in the intervals:
For \( x < 1 \), \( f(x) = x² + 1 \). This is a polynomial function, so it is continuous for all \( x < 1 \).
For \( x > 1 \), \( f(x) = x + 1 \). This is also a polynomial function, so it is continuous for all \( x > 1 \).
Since the function is continuous at \( x=1 \) and within its defined intervals, there are no points of discontinuity for this function for any real number \( x \in R \).
In simple words: The function changes its rule at \( x=1 \). We found that the limits from both sides and the function's actual value at \( x=1 \) are all 2. This means the pieces join together smoothly, so the function is continuous everywhere.

Exam Tip: Polynomial parts of piecewise functions are always continuous within their open intervals. The key is to test for continuity at the join points where the function's definition switches.

Question 11. \( f(x) = \left\{\begin{array}{l} x^{3}-3, \text { if } x \leq 2 \\ x^{2}+1, \text { if } x>2 \end{array}\right. \)
Answer:
We need to examine the continuity at the critical point \( x = 2 \).
At \( x = 2 \):
L.H.L.: \( \lim _{x \rightarrow 2^{-}} f(x) = \lim _{x \rightarrow 2^{-}} (x³ - 3) = (2)³ - 3 = 8 - 3 = 5 \)
R.H.L.: \( \lim _{x \rightarrow 2^{+}} f(x) = \lim _{x \rightarrow 2^{+}} (x² + 1) = (2)² + 1 = 4 + 1 = 5 \)
Function value at \( x = 2 \):
\( f(2) = (2)³ - 3 = 8 - 3 = 5 \) (using the condition \( x \leq 2 \))
Since L.H.L. \( = \) R.H.L. \( = f(2) \), the function \( f \) is continuous at \( x = 2 \).

Now, check for continuity in the intervals:
For \( x < 2 \), \( f(x) = x³ - 3 \). This is a polynomial function, so it is continuous for all \( x < 2 \).
For \( x > 2 \), \( f(x) = x² + 1 \). This is also a polynomial function, so it is continuous for all \( x > 2 \).
Since the function is continuous at \( x=2 \) and within its defined intervals, there are no points of discontinuity for this function for any real number \( x \in R \).
In simple words: The function changes its definition at \( x=2 \). We checked the limits from both sides and the exact value at \( x=2 \), and they all came out to be 5. This shows that the two pieces of the function connect smoothly, so there are no breaks in its graph anywhere.

Exam Tip: When both parts of a piecewise function are polynomials, the primary check for continuity is at the point where the definition changes. If the limits and value match there, it's continuous everywhere.

Question 12. \( f(x) = \left\{\begin{array}{l} x^{10}-1, \text { if } x \leq 1 \\ x^{2}, \quad \text { if } x>1 \end{array}\right. \)
Answer:
We need to examine the continuity at the critical point \( x = 1 \).
At \( x = 1 \):
L.H.L.: \( \lim _{x \rightarrow 1^{-}} f(x) = \lim _{x \rightarrow 1^{-}} (x^{10} - 1) = (1)^{10} - 1 = 1 - 1 = 0 \)
R.H.L.: \( \lim _{x \rightarrow 1^{+}} f(x) = \lim _{x \rightarrow 1^{+}} (x²) = (1)² = 1 \)
Function value at \( x = 1 \):
\( f(1) = (1)^{10} - 1 = 1 - 1 = 0 \) (using the condition \( x \leq 1 \))
Since L.H.L. \( (0) \neq \) R.H.L. \( (1) \), the left-hand limit is not equal to the right-hand limit.
Therefore, the function \( f \) is discontinuous at \( x = 1 \).

Now, check for continuity in the intervals:
For \( x < 1 \), \( f(x) = x^{10} - 1 \). This is a polynomial function, so it is continuous for all \( x < 1 \).
For \( x > 1 \), \( f(x) = x² \). This is also a polynomial function, so it is continuous for all \( x > 1 \).
Thus, the only point of discontinuity is \( x = 1 \).
In simple words: We checked the function at \( x=1 \), where its rule changes. The limit as \( x \) approaches 1 from the left was 0, but from the right, it was 1. Because these limits are different, the function has a jump at \( x=1 \) and is not continuous there. Elsewhere, it behaves like smooth polynomial functions.

Exam Tip: A key aspect of continuity is that the left-hand limit and right-hand limit must be equal at a point for the overall limit to exist. If they differ, the function is discontinuous at that point.

Question 13. Is the function defined by \( f(x) = \left\{\begin{array}{l} x+5, \text { if } x \leq 1 \\ x-5, \text { if } x> 1 \end{array}\right. \) a continuous function?
Answer:
We need to examine the continuity at the critical point \( x = 1 \).
At \( x = 1 \):
L.H.L.: \( \lim _{x \rightarrow 1^{-}} f(x) = \lim _{x \rightarrow 1^{-}} (x + 5) = 1 + 5 = 6 \)
R.H.L.: \( \lim _{x \rightarrow 1^{+}} f(x) = \lim _{x \rightarrow 1^{+}} (x - 5) = 1 - 5 = -4 \)
Function value at \( x = 1 \):
\( f(1) = 1 + 5 = 6 \) (using the condition \( x \leq 1 \))
Since L.H.L. \( (6) \neq \) R.H.L. \( (-4) \), the function \( f \) is discontinuous at \( x = 1 \).

Now, check for continuity in the intervals:
For \( x < 1 \), \( f(x) = x+5 \). This is a polynomial function, so it is continuous for all \( x < 1 \).
For \( x > 1 \), \( f(x) = x-5 \). This is also a polynomial function, so it is continuous for all \( x > 1 \).
Therefore, the function \( f \) is continuous for all real numbers \( x \in R \), except at \( x = 1 \).
In simple words: The function changes its rule at \( x=1 \). The limit from the left side gives 6, while the limit from the right side gives -4. Since these two values are not the same, the function has a break or jump at \( x=1 \), meaning it is not continuous there. It is smooth everywhere else.

Exam Tip: A function is discontinuous at a point if the left-hand limit, right-hand limit, or the function's value at that point are not all equal. A common error is to incorrectly conclude continuity if any of these conditions are violated.

Question 14. \( f(x) = \left\{\begin{array}{l} 3, \text { if } 0 \leq x \leq 1 \\ 4, \text { {if } 1
Answer:
We need to examine continuity at the critical points \( x = 1 \) and \( x = 3 \).

Continuity in intervals:
In the interval \( 0 \leq x < 1 \), \( f(x) = 3 \). This is a constant function, so it is continuous in this interval.
In the interval \( 1 < x < 3 \), \( f(x) = 4 \). This is a constant function, so it is continuous in this interval.
In the interval \( 3 \leq x \leq 10 \), \( f(x) = 5 \). This is a constant function, so it is continuous in this interval.

At \( x = 1 \):
L.H.L.: \( \lim _{x \rightarrow 1^{-}} f(x) = \lim _{x \rightarrow 1^{-}} (3) = 3 \)
R.H.L.: \( \lim _{x \rightarrow 1^{+}} f(x) = \lim _{x \rightarrow 1^{+}} (4) = 4 \)
Function value at \( x = 1 \):
\( f(1) = 3 \) (using the condition \( 0 \leq x \leq 1 \))
Since L.H.L. \( (3) \neq \) R.H.L. \( (4) \), the function \( f \) is discontinuous at \( x = 1 \).

At \( x = 3 \):
L.H.L.: \( \lim _{x \rightarrow 3^{-}} f(x) = \lim _{x \rightarrow 3^{-}} (4) = 4 \)
R.H.L.: \( \lim _{x \rightarrow 3^{+}} f(x) = \lim _{x \rightarrow 3^{+}} (5) = 5 \)
Function value at \( x = 3 \):
\( f(3) = 5 \) (using the condition \( 3 \leq x \leq 10 \))
Since L.H.L. \( (4) \neq \) R.H.L. \( (5) \), the function \( f \) is discontinuous at \( x = 3 \).

Thus, the function is continuous in the specified intervals, but it has discontinuities at \( x = 1 \) and \( x = 3 \).
In simple words: This function is like a staircase with flat steps. It's smooth and continuous on each step. But at \( x=1 \), it jumps from 3 to 4. And at \( x=3 \), it jumps from 4 to 5. So, it's not continuous at those two points.

Exam Tip: For step functions (piecewise functions with constant values), discontinuities occur at the points where the function's definition changes, unless the values on either side of the change point happen to be identical.

Question 15. \( f(x) = \left\{\begin{array}{l} 2 x, \text { if } x<0 \\ 0, \text { if } 0 \leq x \leq 1 \\ 4 x, \text { if } x>1 \end{array}\right. \)
Answer:
We need to examine continuity at the critical points \( x = 0 \) and \( x = 1 \).

At \( x = 0 \):
L.H.L.: \( \lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{-}} (2x) = 2(0) = 0 \)
R.H.L.: \( \lim _{x \rightarrow 0^{+}} f(x) = \lim _{x \rightarrow 0^{+}} (0) = 0 \)
Function value at \( x = 0 \):
\( f(0) = 0 \) (using the condition \( 0 \leq x \leq 1 \))
Since L.H.L. \( = \) R.H.L. \( = f(0) \), the function \( f \) is continuous at \( x = 0 \).

At \( x = 1 \):
L.H.L.: \( \lim _{x \rightarrow 1^{-}} f(x) = \lim _{x \rightarrow 1^{-}} (0) = 0 \)
R.H.L.: \( \lim _{x \rightarrow 1^{+}} f(x) = \lim _{x \rightarrow 1^{+}} (4x) = 4(1) = 4 \)
Function value at \( x = 1 \):
\( f(1) = 0 \) (using the condition \( 0 \leq x \leq 1 \))
Since L.H.L. \( (0) \neq \) R.H.L. \( (4) \), the function \( f \) is discontinuous at \( x = 1 \).

Continuity in intervals:
For \( x < 0 \), \( f(x) = 2x \). This is a polynomial, so it is continuous.
For \( 0 < x < 1 \), \( f(x) = 0 \). This is a constant function, so it is continuous.
For \( x > 1 \), \( f(x) = 4x \). This is a polynomial, so it is continuous.
Thus, the only point of discontinuity is \( x = 1 \).
In simple words: We checked the points where the function's rule changes. At \( x=0 \), everything matched, so it's continuous. But at \( x=1 \), the limit from the left was 0, and from the right was 4, showing a jump. So, it's not continuous at \( x=1 \).

Exam Tip: Pay close attention to the strict inequalities (e.g., \( x<0 \)) versus inequalities that include the boundary (e.g., \( 0 \leq x \leq 1 \)) when determining which part of the piecewise function to use for limits and function values.

Question 16. \( f(x) = \left\{\begin{array}{ll} -2, & \text { if } x<-1 \\ 2 x, & \text { if }-11 \end{array}\right. \)
Answer:
We need to examine continuity at the critical points \( x = -1 \) and \( x = 1 \).

At \( x = -1 \):
L.H.L.: \( \lim _{x \rightarrow -1^{-}} f(x) = \lim _{x \rightarrow -1^{-}} (-2) = -2 \)
R.H.L.: \( \lim _{x \rightarrow -1^{+}} f(x) = \lim _{x \rightarrow -1^{+}} (2x) = 2(-1) = -2 \)
Function value at \( x = -1 \):
The function is not defined at \( x = -1 \) by any of the given conditions (since it specifies \( x<-1 \), \( -11 \)). For continuity, \( f(-1) \) must exist and be equal to the limit.
Since \( f(-1) \) is undefined, the function \( f \) is discontinuous at \( x = -1 \).

At \( x = 1 \):
L.H.L.: \( \lim _{x \rightarrow 1^{-}} f(x) = \lim _{x \rightarrow 1^{-}} (2x) = 2(1) = 2 \)
R.H.L.: \( \lim _{x \rightarrow 1^{+}} f(x) = \lim _{x \rightarrow 1^{+}} (2) = 2 \)
Function value at \( x = 1 \):
The function is not defined at \( x = 1 \) by any of the given conditions (since it specifies \( x<-1 \), \( -11 \)).
Since \( f(1) \) is undefined, the function \( f \) is discontinuous at \( x = 1 \).

Continuity in intervals:
For \( x < -1 \), \( f(x) = -2 \). This is a constant function, so it is continuous.
For \( -1 < x < 1 \), \( f(x) = 2x \). This is a polynomial, so it is continuous.
For \( x > 1 \), \( f(x) = 2 \). This is a constant function, so it is continuous.
Thus, the points of discontinuity are \( x = -1 \) and \( x = 1 \).
In simple words: The function has breaks at both \( x=-1 \) and \( x=1 \). Even though the limits from both sides match at these points, the function itself is not defined at these exact points, which makes it discontinuous. Everywhere else, it's a smooth line.

Exam Tip: For continuity, three conditions must be met: the function must be defined at the point, the limit must exist at the point, and the limit must equal the function's value. If the function is not defined at the point, it is automatically discontinuous there.

Question 17. Find the values of a and b so that the function defined by \( f(x) = \left\{\begin{array}{ll} a x+1, & \text { if } x \leq 3 \\ b x+3, & \text { if } x>3 \end{array}\right. \) is continuous at x = 3.
Answer:
For the function to be continuous at \( x = 3 \), the left-hand limit, the right-hand limit, and the function's value at \( x = 3 \) must all be equal.

At \( x = 3 \):
L.H.L.: \( \lim _{x \rightarrow 3^{-}} f(x) = \lim _{x \rightarrow 3^{-}} (ax + 1) = a(3) + 1 = 3a + 1 \)
R.H.L.: \( \lim _{x \rightarrow 3^{+}} f(x) = \lim _{x \rightarrow 3^{+}} (bx + 3) = b(3) + 3 = 3b + 3 \)
Function value at \( x = 3 \):
\( f(3) = a(3) + 1 = 3a + 1 \) (using the condition \( x \leq 3 \))

For continuity, L.H.L. \( = \) R.H.L. \( = f(3) \).
So, \( 3a + 1 = 3b + 3 \)
Subtract 1 from both sides:
\( 3a = 3b + 2 \)
Divide by 3:
\( a = b + \frac{2}{3} \)
This means that for any arbitrary value of \( b \), we can find a corresponding value of \( a \) that will make the function continuous at \( x = 3 \). Thus, there are infinitely many such pairs of values for \( a \) and \( b \).
In simple words: For the function to be smooth at \( x=3 \), the left part and the right part must meet at the same point. This gives us a relationship between 'a' and 'b'. We found that 'a' must always be 'b' plus 2/3. Many pairs of 'a' and 'b' can satisfy this, so there are countless ways for the function to be continuous.

Exam Tip: When a function is continuous at a point, it means the pieces "meet" at that point. Set the left-hand limit, right-hand limit, and the function's value equal to each other to form equations and solve for the unknown variables.

Question 18. For what value of \( \lambda \), is the function \( f(x) = \left\{\begin{array}{l} \lambda\left(x^{2}-2 x\right), \text { if } x \leq 0 \\ 4 x+1, \quad \text { if } x>0 \end{array}\right. \) continuous at x = 0? What about continuity at x = 1?
Answer:
**Continuity at \( x = 0 \):**
For continuity at \( x = 0 \), L.H.L., R.H.L., and \( f(0) \) must be equal.
L.H.L.: \( \lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{-}} \lambda(x² - 2x) = \lambda((0)² - 2(0)) = \lambda(0 - 0) = 0 \)
R.H.L.: \( \lim _{x \rightarrow 0^{+}} f(x) = \lim _{x \rightarrow 0^{+}} (4x + 1) = 4(0) + 1 = 1 \)
Function value at \( x = 0 \):
\( f(0) = \lambda((0)² - 2(0)) = 0 \) (using the condition \( x \leq 0 \))

Here, L.H.L. \( (0) \neq \) R.H.L. \( (1) \).
Since the left-hand limit and right-hand limit are not equal, the function is not continuous at \( x = 0 \) for any value of \( \lambda \).

**Continuity at \( x = 1 \):**
For \( x = 1 \), we use the condition \( x > 0 \), so \( f(x) = 4x + 1 \).
\( \lim _{x \rightarrow 1} f(x) = \lim _{x \rightarrow 1} (4x + 1) = 4(1) + 1 = 4 + 1 = 5 \)
Function value at \( x = 1 \):
\( f(1) = 4(1) + 1 = 5 \)
Since \( \lim _{x \rightarrow 1} f(x) = f(1) \), the function \( f \) is continuous at \( x = 1 \) for all values of \( \lambda \) (because \( \lambda \) does not affect the function's definition for \( x > 0 \)).
In simple words: At \( x=0 \), the limits from the left and right don't match (0 vs. 1), no matter what value \( \lambda \) has. So, the function is never continuous at \( x=0 \). At \( x=1 \), the function follows a simple rule (4x+1) which is continuous, so it's always continuous at \( x=1 \), regardless of \( \lambda \).

Exam Tip: If the left and right-hand limits at a critical point are different and do not involve the unknown variable, then the function cannot be made continuous at that point, regardless of the variable's value.

Question 19. Show that function defined by g(x) = x − [x] is discontinuous at all integral points. Here, [x] denotes the greatest integer less than or equal to x.
Answer:
Let \( c \) be any integer. We need to check the continuity of \( g(x) = x - [x] \) at \( x = c \).

At \( x = c \):
L.H.L. (Left Hand Limit):
\( \lim _{x \rightarrow c^{-}} g(x) = \lim _{x \rightarrow c^{-}} (x - [x]) \)
Let \( x = c - h \), where \( h \rightarrow 0^{+} \).
\( = \lim _{h \rightarrow 0^{+}} ((c - h) - [c - h]) \)
Since \( h \) is a small positive number, \( c - h \) is slightly less than \( c \). The greatest integer less than or equal to \( c - h \) is \( c - 1 \).
\( = \lim _{h \rightarrow 0^{+}} (c - h - (c - 1)) \)
\( = \lim _{h \rightarrow 0^{+}} (c - h - c + 1) \)
\( = \lim _{h \rightarrow 0^{+}} (1 - h) = 1 - 0 = 1 \)

R.H.L. (Right Hand Limit):
\( \lim _{x \rightarrow c^{+}} g(x) = \lim _{x \rightarrow c^{+}} (x - [x]) \)
Let \( x = c + h \), where \( h \rightarrow 0^{+} \).
\( = \lim _{h \rightarrow 0^{+}} ((c + h) - [c + h]) \)
Since \( h \) is a small positive number, \( c + h \) is slightly greater than \( c \). The greatest integer less than or equal to \( c + h \) is \( c \).
\( = \lim _{h \rightarrow 0^{+}} (c + h - c) \)
\( = \lim _{h \rightarrow 0^{+}} (h) = 0 \)

Function value at \( x = c \):
\( g(c) = c - [c] \)
Since \( c \) is an integer, \( [c] = c \).
\( g(c) = c - c = 0 \)

Since L.H.L. \( (1) \neq \) R.H.L. \( (0) \), and neither equals \( g(c) \), the function \( g(x) \) is discontinuous at all integral points \( x = c \).
In simple words: We checked the function at any whole number, let's call it 'c'. When we looked at numbers just below 'c', the function's value was 1. When we looked at numbers just above 'c', it was 0. Since these limits are different, the function takes a jump at every whole number, making it discontinuous at all integers.

Exam Tip: For greatest integer functions like \( [x] \), discontinuities occur at every integer. When calculating limits for \( [x] \) at an integer \( c \), remember that \( [c-h] = c-1 \) and \( [c+h] = c \) for a small positive \( h \).

Question 20. Is the function defined by x² – sin x + 5 continuous at x = π?
Answer:
Let \( f(x) = x² - \sin x + 5 \).
To check continuity at \( x = \pi \), we need to evaluate the left-hand limit, right-hand limit, and the function's value at \( x = \pi \).

L.H.L.:
\( \lim _{x \rightarrow \pi^{-}} f(x) = \lim _{x \rightarrow \pi^{-}} (x² - \sin x + 5) \)
Let \( x = \pi - h \), where \( h \rightarrow 0^{+} \).
\( = \lim _{h \rightarrow 0^{+}} ((\pi - h)² - \sin (\pi - h) + 5) \)
\( = \lim _{h \rightarrow 0^{+}} (\pi² - 2\pi h + h² - \sin h + 5) \)
As \( h \rightarrow 0 \), \( \sin h \rightarrow 0 \).
\( = \pi² - 2\pi(0) + (0)² - 0 + 5 = \pi² + 5 \)

R.H.L.:
\( \lim _{x \rightarrow \pi^{+}} f(x) = \lim _{x \rightarrow \pi^{+}} (x² - \sin x + 5) \)
Let \( x = \pi + h \), where \( h \rightarrow 0^{+} \).
\( = \lim _{h \rightarrow 0^{+}} ((\pi + h)² - \sin (\pi + h) + 5) \)
\( = \lim _{h \rightarrow 0^{+}} (\pi² + 2\pi h + h² + \sin h + 5) \)
As \( h \rightarrow 0 \), \( \sin h \rightarrow 0 \).
\( = \pi² + 2\pi(0) + (0)² + 0 + 5 = \pi² + 5 \)

Function value at \( x = \pi \):
\( f(\pi) = \pi² - \sin \pi + 5 = \pi² - 0 + 5 = \pi² + 5 \)

Since L.H.L. \( = \) R.H.L. \( = f(\pi) \), the function \( f(x) = x² - \sin x + 5 \) is continuous at \( x = \pi \).

Alternatively:
Let \( g(x) = x² + 5 \) and \( h(x) = \sin x \).
We know that \( x² + 5 \) is a polynomial function, so \( g(x) \) is continuous for all real numbers \( x \in R \).
We also know that \( \sin x \) is a trigonometric function that is continuous for all real numbers \( x \in R \).
The difference of two continuous functions is also continuous.
Therefore, \( f(x) = g(x) - h(x) = x² + 5 - \sin x \) is continuous for all real numbers \( x \in R \), and thus continuous at \( x = \pi \).
In simple words: We checked the limits from both sides and the function's value at \( x=\pi \). All these came out to be \( \pi^2 + 5 \). Since they match, the function is continuous at \( x=\pi \). Also, because it's made up of simpler functions (like \( x^2 \) and \( \sin x \)) that are always continuous, the whole function is continuous everywhere.

Exam Tip: You can either use the limit definition directly or apply properties of continuous functions. If a function is a sum, difference, product, or quotient of known continuous functions, it will also be continuous (provided the denominator is not zero for quotients).

Question 21. Discuss the continuity of the following functions:
(a) f(x) = sin x + cos x
(b) f(x) = sin x – cos x
(c) f(x) = sin x.cos x
Answer:
We know that the sine function, \( \sin x \), and the cosine function, \( \cos x \), are continuous for all real numbers \( x \in R \).

(a) \( f(x) = \sin x + \cos x \)
The sum of two continuous functions is always continuous. Since \( \sin x \) and \( \cos x \) are continuous everywhere, their sum \( f(x) = \sin x + \cos x \) is also continuous for all \( x \in R \).
Alternatively, we can write \( \sin x + \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right) = \sqrt{2} \left( \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} \right) = \sqrt{2} \sin(x+\frac{\pi}{4}) \). The sine function is continuous everywhere, so \( f(x) \) is continuous.

(b) \( f(x) = \sin x - \cos x \)
The difference of two continuous functions is always continuous. Since \( \sin x \) and \( \cos x \) are continuous everywhere, their difference \( f(x) = \sin x - \cos x \) is also continuous for all \( x \in R \).
Alternatively, we can write \( \sin x - \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x \right) = \sqrt{2} \left( \sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} \right) = \sqrt{2} \sin(x-\frac{\pi}{4}) \). The sine function is continuous everywhere, so \( f(x) \) is continuous.

(c) \( f(x) = \sin x \cdot \cos x \)
The product of two continuous functions is always continuous. Since \( \sin x \) and \( \cos x \) are continuous everywhere, their product \( f(x) = \sin x \cdot \cos x \) is also continuous for all \( x \in R \).
Alternatively, we can use the identity \( \sin x \cos x = \frac{1}{2} \sin(2x) \). Since \( \sin(2x) \) is a composition of continuous functions (\( 2x \) is polynomial, \( \sin u \) is trigonometric), it is continuous. Multiplying by a constant \( \frac{1}{2} \) maintains continuity, so \( f(x) \) is continuous.
In simple words: Since \( \sin x \) and \( \cos x \) are always smooth functions, their sum, difference, and product will also be smooth and continuous everywhere. You can also rewrite them to see they are just continuous sine functions.

Exam Tip: Remember the basic properties of continuous functions: sums, differences, products, and compositions of continuous functions are continuous in their domains. This can save time by avoiding detailed limit calculations.

Question 22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Answer:
**1. Cosine function: \( f(x) = \cos x \)**
The cosine function is continuous for all real numbers \( x \in R \).
To prove this, let \( c \) be any real number.
We know that \( \lim _{x \rightarrow c} \cos x = \cos c \).
Also, \( f(c) = \cos c \).
Since \( \lim _{x \rightarrow c} f(x) = f(c) \), the cosine function is continuous for all real values of \( x \).

**2. Secant function: \( f(x) = \sec x \)**
We know that \( \sec x = \frac{1}{\cos x} \).
The secant function is undefined when \( \cos x = 0 \).
\( \cos x = 0 \) when \( x = (2n + 1)\frac{\pi}{2} \), where \( n \) is an integer.
Therefore, \( f(x) = \sec x \) is discontinuous at \( x = (2n + 1)\frac{\pi}{2} \), \( n \in Z \).
For any other point \( x = c \neq (2n + 1)\frac{\pi}{2} \), \( \cos c \neq 0 \).
Since \( \cos x \) is continuous, \( \frac{1}{\cos x} \) is also continuous where \( \cos x \neq 0 \).
Thus, \( f(x) = \sec x \) is continuous for all \( x \in R \) except at points \( x = (2n + 1)\frac{\pi}{2} \), \( n \in Z \).

**3. Cosecant function: \( f(x) = \csc x \)**
We know that \( \csc x = \frac{1}{\sin x} \).
The cosecant function is undefined when \( \sin x = 0 \).
\( \sin x = 0 \) when \( x = n\pi \), where \( n \) is an integer.
Therefore, \( f(x) = \csc x \) is discontinuous at \( x = n\pi \), \( n \in Z \).
For any other point \( x = c \neq n\pi \), \( \sin c \neq 0 \).
Since \( \sin x \) is continuous, \( \frac{1}{\sin x} \) is also continuous where \( \sin x \neq 0 \).
Thus, \( f(x) = \csc x \) is continuous for all \( x \in R \) except at points \( x = n\pi \), \( n \in Z \).

**4. Cotangent function: \( f(x) = \cot x \)**
We know that \( \cot x = \frac{\cos x}{\sin x} \).
The cotangent function is undefined when \( \sin x = 0 \).
\( \sin x = 0 \) when \( x = n\pi \), where \( n \) is an integer.
Therefore, \( f(x) = \cot x \) is discontinuous at \( x = n\pi \), \( n \in Z \).
For any other point \( x = c \neq n\pi \), \( \sin c \neq 0 \).
Since \( \sin x \) and \( \cos x \) are continuous, their quotient \( \frac{\cos x}{\sin x} \) is also continuous where \( \sin x \neq 0 \).
Thus, \( f(x) = \cot x \) is continuous for all \( x \in R \) except at points \( x = n\pi \), \( n \in Z \).
In simple words: The cosine function is smooth everywhere. Secant, cosecant, and cotangent functions are continuous only where they are defined. They have breaks (discontinuities) at points where their denominators (\( \cos x \) or \( \sin x \)) become zero.

Exam Tip: Remember that rational functions (like reciprocal trigonometric functions) are discontinuous at points where their denominators are zero. Identify these points by setting the denominator to zero and solving for \( x \).

Question 23. Find all the points of discontinuity of f, where \( f(x) = \left\{\begin{array}{l} \frac{\sin x}{x}, \text { if } x<0 \\ x+1, \text { if } x \geq 0 \end{array}\right. \)
Answer:
We need to examine continuity at the critical point \( x = 0 \).

At \( x = 0 \):
L.H.L.: \( \lim _{x \rightarrow 0^{-}} f(x) = \lim _{x \rightarrow 0^{-}} \frac{\sin x}{x} \)
We know the standard limit \( \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 \).
So, \( \lim _{x \rightarrow 0^{-}} \frac{\sin x}{x} = 1 \).
R.H.L.: \( \lim _{x \rightarrow 0^{+}} f(x) = \lim _{x \rightarrow 0^{+}} (x+1) = 0 + 1 = 1 \)
Function value at \( x = 0 \):
\( f(0) = 0 + 1 = 1 \) (using the condition \( x \geq 0 \))
Since L.H.L. \( = \) R.H.L. \( = f(0) \), the function \( f \) is continuous at \( x = 0 \).

Continuity in intervals:
For \( x < 0 \), \( f(x) = \frac{\sin x}{x} \). Both \( \sin x \) and \( x \) are continuous. The quotient is continuous as long as the denominator is not zero. For \( x < 0 \), \( x \) is never zero, so \( f(x) \) is continuous for all \( x < 0 \).
For \( x > 0 \), \( f(x) = x+1 \). This is a polynomial function, so it is continuous for all \( x > 0 \).
Since the function is continuous at \( x=0 \) and in its defined intervals, there are no points of discontinuity for this function.
In simple words: We checked the function where its rule changes, at \( x=0 \). The limit from the left (using \( \frac{\sin x}{x} \)) was 1, and the limit from the right (using \( x+1 \)) was also 1. The function's value at \( x=0 \) was also 1. Since all these match, the function is smooth and has no breaks anywhere.

Exam Tip: Remember the fundamental trigonometric limit \( \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 \). This is crucial when dealing with continuity of functions defined with \( \frac{\sin x}{x} \) at \( x=0 \).

Question 24. Determine if the function defined by \( f(x) = \left\{\begin{array}{l} x^{2} \sin \frac{1}{x}, \text { if } x \neq 0 \\ 0, \quad \text { if } x=0 \end{array}\right. \) is a continuous function?
Answer: We need to check the continuity of \( f(x) \) at \( x=0 \).
L.H.L. \( = \lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} x^{2} \sin \frac{1}{x} \)
Let \( x = -h \). As \( x \rightarrow 0^{-} \), \( h \rightarrow 0^{+} \).
\( = \lim_{h \rightarrow 0^{+}} (-h)^{2} \sin \left(\frac{1}{-h}\right) \)
\( = \lim_{h \rightarrow 0^{+}} h^{2} \left(-\sin \frac{1}{h}\right) \)
\( = \lim_{h \rightarrow 0^{+}} -h^{2} \sin \frac{1}{h} \)
Since \( -1 \leq \sin \frac{1}{h} \leq 1 \) (a finite quantity) and \( \lim_{h \rightarrow 0^{+}} h^{2} = 0 \),
L.H.L. \( = 0 \times (\text{finite quantity}) = 0 \).
R.H.L. \( = \lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} x^{2} \sin \frac{1}{x} \)
Let \( x = h \). As \( x \rightarrow 0^{+} \), \( h \rightarrow 0^{+} \).
\( = \lim_{h \rightarrow 0^{+}} h^{2} \sin \frac{1}{h} \)
Since \( -1 \leq \sin \frac{1}{h} \leq 1 \) and \( \lim_{h \rightarrow 0^{+}} h^{2} = 0 \),
R.H.L. \( = 0 \times (\text{finite quantity}) = 0 \).
Also, \( f(0) = 0 \) (as given).
Since L.H.L. \( = \) R.H.L. \( = f(0) = 0 \), the function \( f(x) \) is continuous at \( x=0 \).
For \( x \neq 0 \), \( x^2 \) is a continuous function and \( \sin \frac{1}{x} \) is also continuous.
The product of two continuous functions is continuous. Therefore, \( f(x) \) is continuous for all \( x \in R \).
In simple words: We check if the function is smooth at \( x=0 \) by looking at values very close to it from both sides. Because the values from the left and right, and the function's actual value at \( x=0 \), are all the same, the function is continuous everywhere.

Exam Tip: For functions involving \( \sin(\frac{1}{x}) \) or \( \cos(\frac{1}{x}) \) multiplied by powers of x, remember that \( \sin(\frac{1}{x}) \) is bounded between -1 and 1, which helps in evaluating limits at \( x=0 \).

Question 25. Examine the continuity of f, where f is defined by \( f(x) = \left\{\begin{array}{r} \sin x-\cos x, \text { if } x \neq 0 \\ -1, \quad \text { if } x=0 \end{array}\right. \)
Answer: We need to check the continuity of \( f(x) \) at \( x=0 \).
L.H.L. \( = \lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} (\sin x - \cos x) \)
Let \( x = -h \). As \( x \rightarrow 0^{-} \), \( h \rightarrow 0^{+} \).
\( = \lim_{h \rightarrow 0^{+}} (\sin(-h) - \cos(-h)) \)
\( = \lim_{h \rightarrow 0^{+}} (-\sin h - \cos h) \)
\( = -\sin(0) - \cos(0) = 0 - 1 = -1 \).
R.H.L. \( = \lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} (\sin x - \cos x) \)
Let \( x = h \). As \( x \rightarrow 0^{+} \), \( h \rightarrow 0^{+} \).
\( = \lim_{h \rightarrow 0^{+}} (\sin h - \cos h) \)
\( = \sin(0) - \cos(0) = 0 - 1 = -1 \).
Also, \( f(0) = -1 \) (as given).
Since L.H.L. \( = \) R.H.L. \( = f(0) = -1 \), the function \( f(x) \) is continuous at \( x=0 \).
Alternatively, we know that \( \sin x \) and \( \cos x \) are both continuous functions for all \( x \in R \).
The difference of two continuous functions is also continuous. Therefore, \( f(x) = \sin x - \cos x \) is continuous for all \( x \in R \).
In simple words: We check if the function is smooth at \( x=0 \). By looking at values from both sides (left and right) and the function's value right at \( x=0 \), we see they all match to -1. This means the function is connected and smooth at \( x=0 \), making it continuous everywhere.

Exam Tip: Remember that \( \sin(-h) = -\sin h \) and \( \cos(-h) = \cos h \). These identities are crucial when evaluating limits for left-hand and right-hand continuity checks.

Question 26. Find the value of k such that the function \( f(x) = \left\{\begin{array}{r} \frac{k \cos x}{\pi-2 x}, \text { if } x \neq \frac{\pi}{2} \\ 3, \text { if } x=\frac{\pi}{2} \end{array}\right. \) is continuous at \( x = \frac{\pi}{2} \).
Answer: For \( f(x) \) to be continuous at \( x = \frac{\pi}{2} \), we must have \( \lim_{x \rightarrow \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) \).
We are given \( f\left(\frac{\pi}{2}\right) = 3 \).
Now, let's find the limit:
\( \lim_{x \rightarrow \frac{\pi}{2}} f(x) = \lim_{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x} \)
Let \( x = \frac{\pi}{2} - h \). As \( x \rightarrow \frac{\pi}{2} \), \( h \rightarrow 0 \).
\( = \lim_{h \rightarrow 0} \frac{k \cos\left(\frac{\pi}{2} - h\right)}{\pi - 2\left(\frac{\pi}{2} - h\right)} \)
We know \( \cos\left(\frac{\pi}{2} - h\right) = \sin h \).
\( = \lim_{h \rightarrow 0} \frac{k \sin h}{\pi - \pi + 2h} \)
\( = \lim_{h \rightarrow 0} \frac{k \sin h}{2h} \)
\( = \frac{k}{2} \lim_{h \rightarrow 0} \frac{\sin h}{h} \)
Since \( \lim_{h \rightarrow 0} \frac{\sin h}{h} = 1 \),
\( = \frac{k}{2} \times 1 = \frac{k}{2} \).
For continuity, \( \frac{k}{2} = 3 \).
Solving for \( k \), we get \( k = 6 \).
In simple words: To make the function smooth at \( x = \frac{\pi}{2} \), the function's value at that point must equal its limit as we get closer to it. By using a special limit rule for \( \sin h / h \), we can figure out that k must be 6 for this to happen.

Exam Tip: When evaluating limits of trigonometric functions involving \( \frac{0}{0} \) indeterminate form, substitution with \( h \) (where \( x=a \pm h \)) and applying standard limits like \( \lim_{h \rightarrow 0} \frac{\sin h}{h} = 1 \) is a common and effective technique.

Question 27. Find the value of k such that the function \( f(x) = \left\{\begin{array}{l} k x^{2}, \text { if } x \leq 2 \\ 3, \text { if } x>2 \end{array}\right. \) is continuous at \( x=2 \).
Answer: For \( f(x) \) to be continuous at \( x=2 \), we must have L.H.L. \( = \) R.H.L. \( = f(2) \).
First, find \( f(2) \):
When \( x \leq 2 \), \( f(x) = kx^2 \). So, \( f(2) = k(2)^2 = 4k \).
Next, find the Left-Hand Limit (L.H.L.):
L.H.L. \( = \lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2^{-}} kx^2 \)
Substitute \( x=2 \): \( = k(2)^2 = 4k \).
Next, find the Right-Hand Limit (R.H.L.):
R.H.L. \( = \lim_{x \rightarrow 2^{+}} f(x) = \lim_{x \rightarrow 2^{+}} 3 \)
\( = 3 \).
For continuity at \( x=2 \), L.H.L. \( = \) R.H.L. \( = f(2) \).
So, \( 4k = 3 \).
Solving for \( k \), we get \( k = \frac{3}{4} \).
In simple words: For the function to be smooth and connected at \( x=2 \), the value it approaches from the left, the value it approaches from the right, and its actual value at \( x=2 \) must all be the same. Setting these equal, we find that k needs to be \( \frac{3}{4} \).

Exam Tip: When dealing with piecewise functions, always check the three conditions for continuity at the point where the definition changes: the limit from the left, the limit from the right, and the function's value at that specific point. All three must be equal.

Question 28. Find the value of k such that the function \( f(x) = \left\{\begin{array}{ll} k x+1, & \text { if } x \leq \pi \\ \cos x, & \text { if } x>\pi \end{array}\right. \) is continuous at \( x=\pi \).
Answer: For \( f(x) \) to be continuous at \( x=\pi \), we must have L.H.L. \( = \) R.H.L. \( = f(\pi) \).
First, find \( f(\pi) \):
When \( x \leq \pi \), \( f(x) = kx+1 \). So, \( f(\pi) = k\pi+1 \).
Next, find the Left-Hand Limit (L.H.L.):
L.H.L. \( = \lim_{x \rightarrow \pi^{-}} f(x) = \lim_{x \rightarrow \pi^{-}} (kx+1) \)
Substitute \( x=\pi \): \( = k\pi+1 \).
Next, find the Right-Hand Limit (R.H.L.):
R.H.L. \( = \lim_{x \rightarrow \pi^{+}} f(x) = \lim_{x \rightarrow \pi^{+}} \cos x \)
Substitute \( x=\pi \): \( = \cos(\pi) = -1 \).
For continuity at \( x=\pi \), L.H.L. \( = \) R.H.L. \( = f(\pi) \).
So, \( k\pi+1 = -1 \).
Subtract 1 from both sides: \( k\pi = -2 \).
Solve for \( k \): \( k = -\frac{2}{\pi} \).
In simple words: To ensure the function flows smoothly at \( x=\pi \), the part of the function for \( x \leq \pi \) and the part for \( x > \pi \) must meet perfectly at that point. By making the limits from both sides, and the function's exact value at \( \pi \), equal, we find that k must be \( -\frac{2}{\pi} \).

Exam Tip: Pay careful attention to the definitions of the piecewise function for \( x \leq \pi \) and \( x > \pi \) to correctly choose which expression to use for \( f(\pi) \), L.H.L., and R.H.L.

Question 29. Find the value of k such that the function \( f(x) = \left\{\begin{array}{l} k x+1, \text { if } x \leq 5 \\ 3 x-5, \text { if } x>5 \end{array}\right. \) is continuous at \( x=5 \).
Answer: For \( f(x) \) to be continuous at \( x=5 \), we must have L.H.L. \( = \) R.H.L. \( = f(5) \).
First, find \( f(5) \):
When \( x \leq 5 \), \( f(x) = kx+1 \). So, \( f(5) = k(5)+1 = 5k+1 \).
Next, find the Left-Hand Limit (L.H.L.):
L.H.L. \( = \lim_{x \rightarrow 5^{-}} f(x) = \lim_{x \rightarrow 5^{-}} (kx+1) \)
Substitute \( x=5 \): \( = k(5)+1 = 5k+1 \).
Next, find the Right-Hand Limit (R.H.L.):
R.H.L. \( = \lim_{x \rightarrow 5^{+}} f(x) = \lim_{x \rightarrow 5^{+}} (3x-5) \)
Substitute \( x=5 \): \( = 3(5)-5 = 15-5 = 10 \).
For continuity at \( x=5 \), L.H.L. \( = \) R.H.L. \( = f(5) \).
So, \( 5k+1 = 10 \).
Subtract 1 from both sides: \( 5k = 9 \).
Solve for \( k \): \( k = \frac{9}{5} \).
In simple words: To make the function connected at \( x=5 \), the values it approaches from the left, from the right, and its actual value at \( x=5 \) must all be identical. By setting these values equal, we find that k must be \( \frac{9}{5} \).

Exam Tip: Always make sure to use the correct part of the piecewise function for evaluating \( f(a) \), \( \lim_{x \rightarrow a^{-}} f(x) \), and \( \lim_{x \rightarrow a^{+}} f(x) \). A common error is mixing up the conditions for \( \leq \) and \( > \).

Question 30. Find the values of a and b such that the function defined by \( f(x) = \left\{\begin{array}{cl} 5, & \text { if } x \leq 2 \\ a x+b, & \text { if } 2
Answer: For \( f(x) \) to be continuous, it must be continuous at the transition points \( x=2 \) and \( x=10 \).
**Continuity at \( x=2 \):**
L.H.L. \( = \lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2^{-}} 5 = 5 \).
R.H.L. \( = \lim_{x \rightarrow 2^{+}} f(x) = \lim_{x \rightarrow 2^{+}} (ax+b) = a(2)+b = 2a+b \).
\( f(2) = 5 \) (from the definition \( x \leq 2 \)).
For continuity at \( x=2 \), L.H.L. \( = \) R.H.L. \( = f(2) \).
So, \( 2a+b = 5 \) ... (1)
**Continuity at \( x=10 \):**
L.H.L. \( = \lim_{x \rightarrow 10^{-}} f(x) = \lim_{x \rightarrow 10^{-}} (ax+b) = a(10)+b = 10a+b \).
R.H.L. \( = \lim_{x \rightarrow 10^{+}} f(x) = \lim_{x \rightarrow 10^{+}} 21 = 21 \).
\( f(10) = 21 \) (from the definition \( x \geq 10 \)).
For continuity at \( x=10 \), L.H.L. \( = \) R.H.L. \( = f(10) \).
So, \( 10a+b = 21 \) ... (2)
Now we have a system of two linear equations:
1) \( 2a+b = 5 \)
2) \( 10a+b = 21 \)
Subtract equation (1) from equation (2):
\( (10a+b) - (2a+b) = 21 - 5 \)
\( 8a = 16 \)
\( a = \frac{16}{8} = 2 \).
Substitute \( a=2 \) into equation (1):
\( 2(2)+b = 5 \)
\( 4+b = 5 \)
\( b = 5-4 = 1 \).
Thus, the values are \( a=2 \) and \( b=1 \).
In simple words: For the function to be smooth and continuous everywhere, it must be continuous at the points where its definition changes, which are \( x=2 \) and \( x=10 \). By setting the left and right limits equal at both these points, we get two equations. Solving these equations helps us find the exact values for 'a' and 'b' that make the function connected.

Exam Tip: For piecewise functions with multiple conditions, ensure continuity is checked at all boundary points where the function definition changes. This typically leads to a system of linear equations that you need to solve to find the unknown constants.

Question 31. Show that the function defined by \( f(x) = \cos x^2 \) is a continuous function.
Answer: We can show this by using the property that the composition of continuous functions is continuous.
Let \( g(x) = \cos x \) and \( h(x) = x^2 \).
We know that \( g(x) = \cos x \) is a continuous function for all real numbers \( x \in R \).
We also know that \( h(x) = x^2 \) is a polynomial function, which is continuous for all real numbers \( x \in R \).
Now, the given function \( f(x) = \cos x^2 \) can be written as a composite function \( f(x) = g(h(x)) \).
This means \( f(x) = g(x^2) = \cos(x^2) \).
Since both \( g(x) \) and \( h(x) \) are continuous functions, their composition \( f(x) = g(h(x)) \) is also continuous.
Therefore, \( f(x) = \cos x^2 \) is a continuous function.
In simple words: We can think of \( \cos x^2 \) as two simpler functions put together: first squaring 'x' (which is always smooth), and then taking the cosine of that result (which is also always smooth). When you combine two smooth functions this way, the new function is also smooth and continuous.

Exam Tip: A powerful technique to prove continuity of complex functions is to break them down into simpler, known continuous functions (like polynomials, trigonometric functions, exponentials) and use properties of continuous functions (sum, difference, product, quotient, and composition).

Question 32. Show that the function defined by \( f(x) = | \cos x | \) is a continuous function.
Answer: We can demonstrate this by using the property that the composition of continuous functions is continuous.
Let \( g(x) = |x| \) and \( h(x) = \cos x \).
We know that \( h(x) = \cos x \) is a continuous function for all real numbers \( x \in R \).
We also know that \( g(x) = |x| \) (the absolute value function) is continuous for all real numbers \( x \in R \).
The given function \( f(x) = | \cos x | \) can be written as a composite function \( f(x) = g(h(x)) \).
This means \( f(x) = g(\cos x) = |\cos x| \).
Since both \( g(x) \) and \( h(x) \) are continuous functions, their composition \( f(x) = g(h(x)) \) is also continuous.
Therefore, \( f(x) = |\cos x| \) is a continuous function.
In simple words: Imagine \( |\cos x| \) as combining two functions: first calculating \( \cos x \) (which is always a smooth curve) and then taking the absolute value of that result (which is also a smooth operation, just reflecting negative parts above the x-axis). Since both steps are continuous, the final combined function is also continuous.

Exam Tip: Remember that the absolute value function \( |x| \) is continuous everywhere. This is a common component in demonstrating the continuity of functions like \( |f(x)| \) where \( f(x) \) is continuous.

Question 33. Examine if \( \sin |x| \) is a continuous function.
Answer: We can examine the continuity of \( f(x) = \sin |x| \) by considering it as a composition of two functions.
Let \( g(x) = \sin x \) and \( h(x) = |x| \).
We know that \( g(x) = \sin x \) is a continuous function for all real numbers \( x \in R \).
We also know that \( h(x) = |x| \) (the absolute value function) is continuous for all real numbers \( x \in R \).
The given function \( f(x) = \sin |x| \) can be written as a composite function \( f(x) = g(h(x)) \).
This means \( f(x) = g(|x|) = \sin |x| \).
Since both \( g(x) \) and \( h(x) \) are continuous functions, their composition \( f(x) = g(h(x)) \) is also continuous.
Therefore, \( f(x) = \sin |x| \) is a continuous function.
In simple words: This function is made by taking the absolute value of 'x' first (which is a smooth process) and then finding the sine of that result (which is also smooth). Since both steps are continuous, the whole function \( \sin |x| \) is also continuous and has no breaks or jumps.

Exam Tip: The composition property for continuous functions is highly useful: if \( g \) is continuous at \( h(c) \) and \( h \) is continuous at \( c \), then the composite function \( f(x) = g(h(x)) \) is continuous at \( c \).

Question 34. Find all the points of discontinuity of f defined by \( f(x) = |x| - |x+1| \).
Answer: We need to find points where \( f(x) \) might not be continuous. The absolute value function \( |x| \) is continuous everywhere. The sum or difference of continuous functions is also continuous. Therefore, \( f(x) = |x| - |x+1| \) is continuous everywhere because both \( |x| \) and \( |x+1| \) are continuous.
Let's analyze the function by considering different intervals based on where the expressions inside the absolute values change sign:
The critical points are \( x=0 \) (for \( |x| \)) and \( x=-1 \) (for \( |x+1| \)).
**Case 1: \( x < -1 \)**
In this interval, \( x < 0 \) and \( x+1 < 0 \).
So, \( |x| = -x \) and \( |x+1| = -(x+1) \).
\( f(x) = -x - (-(x+1)) = -x + x + 1 = 1 \).
**Case 2: \( -1 \leq x < 0 \)**
In this interval, \( x < 0 \) but \( x+1 \geq 0 \).
So, \( |x| = -x \) and \( |x+1| = x+1 \).
\( f(x) = -x - (x+1) = -x - x - 1 = -2x - 1 \).
**Case 3: \( x \geq 0 \)**
In this interval, \( x \geq 0 \) and \( x+1 > 0 \).
So, \( |x| = x \) and \( |x+1| = x+1 \).
\( f(x) = x - (x+1) = x - x - 1 = -1 \).
So, the piecewise definition of \( f(x) \) is:
\( f(x) = \left\{\begin{array}{ll} 1, & \text { if } x < -1 \\ -2x - 1, & \text { if } -1 \leq x < 0 \\ -1, & \text { if } x \geq 0 \end{array}\right. \)
Now, let's check continuity at the transition points \( x=-1 \) and \( x=0 \).
**At \( x=-1 \):**
L.H.L. \( = \lim_{x \rightarrow -1^{-}} f(x) = \lim_{x \rightarrow -1^{-}} 1 = 1 \).
R.H.L. \( = \lim_{x \rightarrow -1^{+}} f(x) = \lim_{x \rightarrow -1^{+}} (-2x-1) = -2(-1)-1 = 2-1 = 1 \).
\( f(-1) = -2(-1)-1 = 2-1 = 1 \) (from the definition for \( -1 \leq x < 0 \)).
Since L.H.L. \( = \) R.H.L. \( = f(-1) = 1 \), \( f(x) \) is continuous at \( x=-1 \).
**At \( x=0 \):**
L.H.L. \( = \lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} (-2x-1) = -2(0)-1 = -1 \).
R.H.L. \( = \lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} -1 = -1 \).
\( f(0) = -1 \) (from the definition for \( x \geq 0 \)).
Since L.H.L. \( = \) R.H.L. \( = f(0) = -1 \), \( f(x) \) is continuous at \( x=0 \).
Since \( f(x) \) is continuous at all critical points and is a polynomial in other intervals, there are no points of discontinuity.
Hence, \( f(x) \) is continuous for all \( x \in R \).
In simple words: To understand this function, we split it into three sections based on where the absolute value signs might change. After rewriting the function for each section, we checked the points where these sections meet. We found that the function remains connected and smooth at all these meeting points. This shows that the function is continuous everywhere and has no breaks.

Exam Tip: When a function involves multiple absolute value expressions, define the function piecewise by considering intervals determined by the roots of the expressions inside the absolute values. Then, check continuity at each boundary point of these intervals. Remember that a sum/difference of continuous functions is always continuous.