Samacheer Kalvi Class 10 Maths Solutions Chapter 1 Relations and Functions Exercise 1.4

 

Question 1. Determine whether the graph given below represent functions. Give reason for your answers concerning each graph.

Answer:
(i) The vertical line test helps us check if a graph represents a function. If any vertical line crosses the graph at more than one point, then it is not a function. For the first graph, a vertical line cuts the curve at two points, A and B. This means that for one x-value, there are two different y-values. So, this graph does not represent a function.
(ii) In the second graph, any vertical line drawn will intersect the curve at only one point, P. This means that for every x-value, there is only one corresponding y-value. So, this graph represents a function.
(iii) For the third graph, a vertical line can cut the curve at three different points: S, T, and U. This means one x-value has multiple y-values. So, this graph does not represent a function.
(iv) For the fourth graph, any vertical line will cross the curve at only one point, D. This shows that for each x-value, there is only one y-value. So, this graph represents a function.
In simple words: To see if a graph is a function, draw a straight line up and down. If this line crosses the graph in more than one place, it is not a function. If it only crosses once, it is a function.

🎯 Exam Tip: Remember to clearly state the vertical line test criterion and apply it specifically to each graph, mentioning the number of intersection points.

 

Question 2. Let \( f: A \rightarrow B \) be a function defined by \( f(x) = \frac {x}{ 2 } – 1 \), where \( A = \{2, 4, 6, 10, 12\} \), present f by
(i) set of ordered pairs
(ii) a table
(iii) an arrow diagram
(iv) a graph

Answer:
First, we find the output for each number in set A using the function \( f(x) = \frac {x}{ 2 } - 1 \):
Given set \( A = \{2, 4, 6, 10, 12\} \)
Let's assume set \( B = \{0, 1, 2, 4, 5, 9\} \) as given in the source for context.
For \( x = 2 \), \( f(2) = \frac{2}{2} - 1 = 1 - 1 = 0 \)
For \( x = 4 \), \( f(4) = \frac{4}{2} - 1 = 2 - 1 = 1 \)
For \( x = 6 \), \( f(6) = \frac{6}{2} - 1 = 3 - 1 = 2 \)
For \( x = 10 \), \( f(10) = \frac{10}{2} - 1 = 5 - 1 = 4 \)
For \( x = 12 \), \( f(12) = \frac{12}{2} - 1 = 6 - 1 = 5 \)

(i) Set of ordered pairs: Each input from set A is paired with its output.
\( f = \{(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)\} \)

(ii) A table: We can list the input (x) and output (f(x)) values in a table.

(iii) An arrow diagram: This shows arrows from each element in set A to its corresponding element in set B.

(iv) A graph: We plot the ordered pairs on a coordinate plane.

In simple words: We find the output for each number from set A using the function rule. Then we show these pairs as a list, a table, arrows from A to B, and dots on a graph.

🎯 Exam Tip: When presenting a function in different ways, ensure the domain, codomain, and the mapping rule are consistently applied across all representations (ordered pairs, table, arrow diagram, and graph).

 

Question 3. Represent the function \( f = \{(1,2), (2,2), (3,2), (4,3),(5,4)\} \) through (i) an arrow diagram (it) a table form (iii) a graph.

Answer:
Given the function as a set of ordered pairs: \( f = \{(1,2), (2,2), (3,2), (4,3),(5,4)\} \)
From these pairs, we can identify the domain (set A) and the range (part of set B).
Let \( A = \{1, 2, 3, 4, 5\} \) (the first numbers in each pair).
Let \( B = \{2, 3, 4\} \) (the second numbers in each pair, representing the actual outputs). The codomain might be larger, but the range is {2,3,4}.

(i) Arrow diagram: We draw arrows from each element in set A to its matching element in set B.

(ii) Table form: We list the input (X) values and their corresponding output (f(x)) values.

(iii) A graph: We plot the given ordered pairs on a coordinate plane.

In simple words: We take the given pairs and show them as arrows pointing from the first number to the second, as a list in a table, and as dots on a graph grid.

🎯 Exam Tip: When given a function as a set of ordered pairs, make sure to correctly identify the domain (all x-values) and range (all y-values) for accurate representation in diagrams and tables.

 

Question 4. Show that the function \( f: N \rightarrow N \) defined by \( f(x) = 2x - 1 \) is one-one but not onto.

Answer:
We are given the function \( f: N \rightarrow N \), where \( N = \{1, 2, 3, 4, 5, ...\} \) is the set of natural numbers, and the rule is \( f(x) = 2x - 1 \).

Let's find the values of \( f(x) \) for some natural numbers:
For \( x = 1 \), \( f(1) = 2(1) - 1 = 2 - 1 = 1 \)
For \( x = 2 \), \( f(2) = 2(2) - 1 = 4 - 1 = 3 \)
For \( x = 3 \), \( f(3) = 2(3) - 1 = 6 - 1 = 5 \)
For \( x = 4 \), \( f(4) = 2(4) - 1 = 8 - 1 = 7 \)
For \( x = 5 \), \( f(5) = 2(5) - 1 = 10 - 1 = 9 \)
The set of ordered pairs is \( f = \{(1,1), (2,3), (3,5), (4,7), (5,9), ...\} \)

(i) To check if it is a one-one function:
A function is one-one if every different element in the domain (N) has a different image in the codomain (N). From our calculations, we see that for every unique input \( x \), we get a unique output \( f(x) \). For example, 1 maps to 1, 2 maps to 3, 3 maps to 5, and so on. No two different inputs map to the same output. This means that \( f \) is a one-one function.
(ii) To check if it is an onto function:
A function is onto if every element in the codomain (N) has at least one pre-image in the domain (N). The outputs (range) we got are \( \{1, 3, 5, 7, 9, ...\} \), which are all odd natural numbers. The codomain is the set of all natural numbers \( N = \{1, 2, 3, 4, 5, ...\} \). Since the even natural numbers (like 2, 4, 6, 8) in the codomain do not have any pre-image in the domain, the range is not equal to the codomain. Therefore, the function \( f \) is not an onto function.
The function \( f: N \rightarrow N \) defined by \( f(x) = 2x - 1 \) is one-one but not onto.
In simple words: Each starting number gives a different ending number, so it's "one-one". But not all numbers in the ending set are hit by an arrow (like even numbers), so it's not "onto".

🎯 Exam Tip: To prove a function is one-one, show that \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). To prove it's not onto, find an element in the codomain that has no pre-image in the domain.

 

Question 5. Show that the function \( f: N \Rightarrow N \) defined by \( f(m) = m^2 + m + 3 \) is one-one function.

Answer:
We are given the function \( f: N \rightarrow N \), where \( N = \{1, 2, 3, 4, 5, ...\} \) and the rule is \( f(m) = m^2 + m + 3 \).

Let's calculate the values of \( f(m) \) for some natural numbers:
For \( m = 1 \), \( f(1) = 1^2 + 1 + 3 = 1 + 1 + 3 = 5 \)
For \( m = 2 \), \( f(2) = 2^2 + 2 + 3 = 4 + 2 + 3 = 9 \)
For \( m = 3 \), \( f(3) = 3^2 + 3 + 3 = 9 + 3 + 3 = 15 \)
For \( m = 4 \), \( f(4) = 4^2 + 4 + 3 = 16 + 4 + 3 = 23 \)
The set of ordered pairs is \( f = \{(1, 5), (2, 9), (3, 15), (4, 23), ...\} \)

To show that \( f \) is a one-one function, we need to prove that if \( f(m_1) = f(m_2) \), then \( m_1 = m_2 \).
Let \( m_1, m_2 \in N \).
Assume \( f(m_1) = f(m_2) \)
\( m_1^2 + m_1 + 3 = m_2^2 + m_2 + 3 \)
\( m_1^2 + m_1 = m_2^2 + m_2 \)
\( m_1^2 - m_2^2 + m_1 - m_2 = 0 \)
\( (m_1 - m_2)(m_1 + m_2) + (m_1 - m_2) = 0 \)
\( (m_1 - m_2)(m_1 + m_2 + 1) = 0 \)
Since \( m_1, m_2 \in N \), both \( m_1 \) and \( m_2 \) are positive integers. So, \( m_1 + m_2 + 1 \) will always be a positive integer and cannot be zero.
Therefore, for the product to be zero, \( (m_1 - m_2) \) must be zero.
\( m_1 - m_2 = 0 \)
\( m_1 = m_2 \)
This proves that the function \( f(m) = m^2 + m + 3 \) is a one-one function. Each natural number input gives a unique natural number output, so no two different inputs will ever map to the same output.
In simple words: If you start with two different natural numbers and put them into this function, you will always get two different answers. This means the function pairs each input with a unique output, making it a "one-one" function.

🎯 Exam Tip: To demonstrate a function is one-one algebraically, set \( f(m_1) = f(m_2) \) and show through logical steps that this forces \( m_1 = m_2 \).

 

Question 6. Let \( A = \{1, 2, 3, 4\} \) and \( B = N \). Let \( f: A \rightarrow B \) be defined by \( f(x) = x^3 \).
(i) find the range of \( f \)
(ii) identify the type of function

Answer:
Given set \( A = \{1, 2, 3, 4\} \) and codomain \( B = N \) (set of natural numbers).
The function is defined as \( f(x) = x^3 \).

Let's calculate the function values for each element in set A:
For \( x = 1 \), \( f(1) = 1^3 = 1 \)
For \( x = 2 \), \( f(2) = 2^3 = 8 \)
For \( x = 3 \), \( f(3) = 3^3 = 27 \)
For \( x = 4 \), \( f(4) = 4^3 = 64 \)

(i) The range of \( f \): The range is the set of all output values.
The range of \( f = \{1, 8, 27, 64\} \).

(ii) Identify the type of function:
To determine the type of function, we check if it is one-one and onto.
**One-one:** All different elements in set A (1, 2, 3, 4) have distinct images (1, 8, 27, 64) in set B (N). No two elements from A map to the same element in B. Thus, the function is one-one.
**Onto:** The codomain is \( N = \{1, 2, 3, 4, ...\} \). The range is \( \{1, 8, 27, 64\} \). Since the range is not equal to the codomain (many numbers in N like 2, 3, 4, 5, 6, 7 etc., are not images of any element in A), the function is not onto.
Therefore, the function is a one-one function but not an onto function.

In simple words: The function takes numbers from {1,2,3,4} and cubes them. The outputs are {1,8,27,64}. Since each input gives a unique output, it is a one-one function. However, many natural numbers (like 2, 3, 4, etc.) are not outputs, so it is not an onto function.

🎯 Exam Tip: Clearly define the domain and codomain when identifying function types. For "onto", always compare the calculated range with the given codomain.

 

Question 7. In each of the following cases state whether the function is bijective or not. Justify your answer.
(i) \( f: R \rightarrow R \) defined by \( f(x) = 2x + 1 \)
(ii) \( f: R \rightarrow R \) defined by \( f(x) = 3 - 4x^2 \)

Answer:
A function is bijective if it is both one-one (injective) and onto (surjective).

(i) Given function: \( f: R \rightarrow R \) defined by \( f(x) = 2x + 1 \)
**One-one (Injective):**
Let \( x_1, x_2 \in R \). Assume \( f(x_1) = f(x_2) \).
\( 2x_1 + 1 = 2x_2 + 1 \)
\( 2x_1 = 2x_2 \)
\( x_1 = x_2 \)
Since \( f(x_1) = f(x_2) \implies x_1 = x_2 \), the function \( f(x) = 2x + 1 \) is one-one.

Let's test with some values:
\( f(0) = 2(0) + 1 = 1 \)
\( f(1) = 2(1) + 1 = 3 \)
\( f(2) = 2(2) + 1 = 5 \)
\( f(3) = 2(3) + 1 = 7 \)
Each different element has a different image, confirming it is one-one.

**Onto (Surjective):**
Let \( y \) be an arbitrary element in the codomain \( R \). We need to find an \( x \) in the domain \( R \) such that \( f(x) = y \).
\( 2x + 1 = y \)
\( 2x = y - 1 \)
\( x = \frac{y - 1}{2} \)
For any real number \( y \), \( \frac{y - 1}{2} \) is also a real number. This means that for every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f(x) = y \). Thus, the function is onto.
Since \( f \) is both one-one and onto, it is a bijective function.

(ii) Given function: \( f: R \rightarrow R \) defined by \( f(x) = 3 - 4x^2 \)
**One-one (Injective):**
Let's test with some values:
\( f(1) = 3 - 4(1)^2 = 3 - 4 = -1 \)
\( f(-1) = 3 - 4(-1)^2 = 3 - 4 = -1 \)
Here, \( f(1) = f(-1) = -1 \), but \( 1 \neq -1 \). Since two different elements in the domain map to the same image, the function is not one-one.

**Onto (Surjective):**
Let's consider the range of the function. For any real number \( x \), \( x^2 \ge 0 \).
So, \( 4x^2 \ge 0 \).
Then, \( -4x^2 \le 0 \).
Therefore, \( 3 - 4x^2 \le 3 \).
This means that \( f(x) \le 3 \). The range of the function is \( (-\infty, 3] \).
The codomain is \( R \), which includes numbers greater than 3. For example, if we take \( y = 5 \) (which is in the codomain), there is no real \( x \) such that \( f(x) = 5 \), because \( 3 - 4x^2 = 5 \implies -4x^2 = 2 \implies x^2 = -\frac{1}{2} \), which has no real solutions for \( x \).
So, the positive numbers in R (the codomain) that are greater than 3 do not have any pre-image in the domain. Thus, the function is not onto.
Since \( f \) is neither one-one nor onto, it is not a bijective function.
In simple words: For the first function, each starting number gives a unique ending number, and every possible ending number can be reached. So it's "bijective". For the second function, different starting numbers can give the same ending number (like 1 and -1 both give -1), and you can't reach all possible ending numbers (like 5). So it's not "bijective".

🎯 Exam Tip: To prove a function is not one-one, find a counterexample (two different inputs that give the same output). To prove it is not onto, find an element in the codomain that has no pre-image.

 

Question 9. If the function \( f: A \rightarrow B \) defined by \( f(x) = ax + b \) is an onto function? Find a and b.

Answer:
Given sets \( A = \{-1, 1\} \) and \( B = \{0, 2\} \).
The function is \( f(x) = ax + b \).
Since \( f \) is an onto function from A to B, this means every element in B must be an image of some element in A. Also, since A and B have the same number of elements (2 each), if it's onto, it must also be one-one, making it bijective.

We need to map the elements of A to the elements of B.
For \( x = -1 \), \( f(-1) = a(-1) + b = -a + b \)
For \( x = 1 \), \( f(1) = a(1) + b = a + b \)

Since B has elements {0, 2}, there are two possible mappings for an onto function:
Case 1: \( f(-1) = 0 \) and \( f(1) = 2 \)
This gives us two equations:
1. \( -a + b = 0 \)
2. \( a + b = 2 \)
Add equation (1) and (2):
\( (-a + b) + (a + b) = 0 + 2 \)
\( 2b = 2 \)
\( b = 1 \)
Substitute \( b = 1 \) into equation (1):
\( -a + 1 = 0 \)
\( a = 1 \)
So, \( a = 1, b = 1 \). In this case, \( f(x) = x + 1 \).
Let's check: \( f(-1) = -1+1=0 \) and \( f(1) = 1+1=2 \). This mapping works.

Case 2: \( f(-1) = 2 \) and \( f(1) = 0 \)
This gives us two equations:
1. \( -a + b = 2 \)
2. \( a + b = 0 \)
Add equation (1) and (2):
\( (-a + b) + (a + b) = 2 + 0 \)
\( 2b = 2 \)
\( b = 1 \)
Substitute \( b = 1 \) into equation (2):
\( a + 1 = 0 \)
\( a = -1 \)
So, \( a = -1, b = 1 \). In this case, \( f(x) = -x + 1 \).
Let's check: \( f(-1) = -(-1)+1 = 1+1=2 \) and \( f(1) = -(1)+1=0 \). This mapping also works.

Therefore, there are two possible pairs of (a, b) for the function to be onto:
1. \( a = 1 \) and \( b = 1 \)
2. \( a = -1 \) and \( b = 1 \)
Both pairs ensure that the function \( f(x) = ax+b \) maps A onto B.
In simple words: For the function to be "onto", all numbers in the second set (B) must be the answer for some number from the first set (A). We found two ways to pick 'a' and 'b' so that this happens. Either \( a=1, b=1 \) or \( a=-1, b=1 \).

🎯 Exam Tip: When given an onto function with finite domain and codomain, remember that every element in the codomain must be an image of at least one element from the domain. If the sizes are equal, it implies a one-to-one correspondence.

 

Question 10. A function \( f: [-5, 9] \rightarrow R \) is defined as follows:
\[ f(x) = \begin{cases} 6x+1 & \text{if } -5 \le x < 2 \\ 5x^2-1 & \text{if } 2 \le x < 6 \\ 3x-4 & \text{if } 6 \le x \le 9 \end{cases} \]
Find (i) \( f(-3) + f(2) \) (ii) \( f(7) - f(1) \) (iii) \( 2f(4) + f(8) \) (iv) \( \frac{2f(-2) - f(6)}{f(4)+f(-2)} \)

Answer:
We use the correct part of the function definition based on the value of \( x \).

(i) Find \( f(-3) + f(2) \)
For \( f(-3) \): Since \( -5 \le -3 < 2 \), use \( f(x) = 6x + 1 \).
\( f(-3) = 6(-3) + 1 = -18 + 1 = -17 \)
For \( f(2) \): Since \( 2 \le 2 < 6 \), use \( f(x) = 5x^2 - 1 \).
\( f(2) = 5(2)^2 - 1 = 5(4) - 1 = 20 - 1 = 19 \)
Now, calculate the sum:
\( f(-3) + f(2) = -17 + 19 = 2 \)

(ii) Find \( f(7) - f(1) \)
For \( f(7) \): Since \( 6 \le 7 \le 9 \), use \( f(x) = 3x - 4 \).
\( f(7) = 3(7) - 4 = 21 - 4 = 17 \)
For \( f(1) \): Since \( -5 \le 1 < 2 \), use \( f(x) = 6x + 1 \).
\( f(1) = 6(1) + 1 = 6 + 1 = 7 \)
Now, calculate the difference:
\( f(7) - f(1) = 17 - 7 = 10 \)

(iii) Find \( 2f(4) + f(8) \)
For \( f(4) \): Since \( 2 \le 4 < 6 \), use \( f(x) = 5x^2 - 1 \).
\( f(4) = 5(4)^2 - 1 = 5(16) - 1 = 80 - 1 = 79 \)
For \( f(8) \): Since \( 6 \le 8 \le 9 \), use \( f(x) = 3x - 4 \).
\( f(8) = 3(8) - 4 = 24 - 4 = 20 \)
Now, calculate the expression:
\( 2f(4) + f(8) = 2(79) + 20 = 158 + 20 = 178 \)

(iv) Find \( \frac{2f(-2) - f(6)}{f(4)+f(-2)} \)
For \( f(-2) \): Since \( -5 \le -2 < 2 \), use \( f(x) = 6x + 1 \).
\( f(-2) = 6(-2) + 1 = -12 + 1 = -11 \)
For \( f(6) \): Since \( 6 \le 6 \le 9 \), use \( f(x) = 3x - 4 \).
\( f(6) = 3(6) - 4 = 18 - 4 = 14 \)
We already found \( f(4) = 79 \) and \( f(-2) = -11 \).
Now, calculate the numerator:
\( 2f(-2) - f(6) = 2(-11) - 14 = -22 - 14 = -36 \)
Now, calculate the denominator:
\( f(4) + f(-2) = 79 + (-11) = 79 - 11 = 68 \)
Finally, calculate the fraction:
\( \frac{2f(-2) - f(6)}{f(4)+f(-2)} = \frac{-36}{68} = \frac{-9}{17} \)
In simple words: This question asks us to use a function that has different rules for different ranges of numbers. For each part, we first check which rule applies based on the input number. Then we calculate the value using that rule and combine them as asked. Make sure to choose the correct rule for each number.

🎯 Exam Tip: Pay close attention to the inequality signs (e.g., \( < \), \( \le \)) in piecewise functions to correctly identify which rule applies for each input value, especially at the boundary points.

 

Question 11. The distance S an object travels under the influence of gravity in time t seconds is given by \( S(t) = \frac{1}{2} gt^2 + at + b \) where, (g is the acceleration due to gravity), a, b are constants. Check if the function S (t)is one-one.

Answer:
Given the function for distance \( S(t) = \frac{1}{2} gt^2 + at + b \). Here, \( t \) represents time, which is usually a positive value or zero, so \( t \ge 0 \).
We need to check if \( S(t) \) is a one-one function. A function is one-one if \( S(t_1) = S(t_2) \) implies \( t_1 = t_2 \).

Let \( t_1 \) and \( t_2 \) be two different values of time.
Assume \( S(t_1) = S(t_2) \)
\( \frac{1}{2} gt_1^2 + at_1 + b = \frac{1}{2} gt_2^2 + at_2 + b \)
Subtract \( b \) from both sides:
\( \frac{1}{2} gt_1^2 + at_1 = \frac{1}{2} gt_2^2 + at_2 \)
Multiply by 2 to clear the fraction:
\( gt_1^2 + 2at_1 = gt_2^2 + 2at_2 \)
Rearrange the terms to one side:
\( gt_1^2 - gt_2^2 + 2at_1 - 2at_2 = 0 \)
Factor out \( g \) and \( 2a \):
\( g(t_1^2 - t_2^2) + 2a(t_1 - t_2) = 0 \)
Use the difference of squares formula \( t_1^2 - t_2^2 = (t_1 - t_2)(t_1 + t_2) \):
\( g(t_1 - t_2)(t_1 + t_2) + 2a(t_1 - t_2) = 0 \)
Factor out \( (t_1 - t_2) \):
\( (t_1 - t_2) [g(t_1 + t_2) + 2a] = 0 \)

For this product to be zero, either \( (t_1 - t_2) = 0 \) or \( [g(t_1 + t_2) + 2a] = 0 \).
If \( t_1 - t_2 = 0 \), then \( t_1 = t_2 \), which implies the function is one-one.
However, if \( g(t_1 + t_2) + 2a = 0 \), then we could have \( t_1 \neq t_2 \) but still \( S(t_1) = S(t_2) \).
Since \( g \) is acceleration due to gravity (usually a positive constant like 9.8 or 32) and \( t_1, t_2 \) are positive values for time, \( t_1 + t_2 \) is also positive.
If \( g(t_1 + t_2) + 2a = 0 \), this would mean \( t_1 + t_2 = -\frac{2a}{g} \). If \( a \) is negative and \( 2a/g \) is a positive value, then there could be different \( t_1, t_2 \) such that their sum equals \( -\frac{2a}{g} \), leading to \( S(t_1) = S(t_2) \) even when \( t_1 \neq t_2 \). For example, if \( a=-g \), then \( t_1+t_2=2 \). We could have \( t_1=0.5, t_2=1.5 \) or \( t_1=0.1, t_2=1.9 \).

Therefore, the function \( S(t) = \frac{1}{2} gt^2 + at + b \) is generally not a one-one function for all possible constants \( a \) and \( b \) over the domain of \( t \ge 0 \). A simple real-world example is projectile motion, where an object reaches the same height (distance from ground) at two different times (one going up, one coming down).
In simple words: This function describes how distance changes over time, like when you throw a ball up. The ball can be at the same height (distance) at two different times – once going up and once coming down. Because two different times can give the same distance, this function is generally not "one-one".

🎯 Exam Tip: To show a quadratic function is not one-one, look for symmetry or multiple inputs yielding the same output. If the coefficient of \( t^2 \) is non-zero, it usually indicates a parabolic path, which is not one-one over its entire domain.

 

Question 12. The function 't' which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by t(C) = F where F = \( \frac { 9 }{ 5 } \) C + 32. Find,
(i) t(0)
(ii) t(28)
(iii) t(-10)
(iv) the value of C when t(C) = 212
(v) the temperature when the Celsius value is equal to the Fahrenheit value.
Answer:
Given the formula for temperature conversion: \( t(C) = \frac{ 9C }{ 5 } + 32 \)
(i) To find t(0), we substitute C = 0 into the formula:
\( t(0) = \frac{ 9(0) }{ 5 } + 32 \)
\( \implies t(0) = 0 + 32 \)
\( \implies t(0) = 32^\circ F \)
So, 0 degrees Celsius is equal to 32 degrees Fahrenheit, which is the freezing point of water.
(ii) To find t(28), we substitute C = 28 into the formula:
\( t(28) = \frac{ 9(28) }{ 5 } + 32 \)
\( \implies t(28) = \frac{ 252 }{ 5 } + 32 \)
\( \implies t(28) = 50.4 + 32 \)
\( \implies t(28) = 82.4^\circ F \)
This means 28 degrees Celsius converts to 82.4 degrees Fahrenheit.
(iii) To find t(-10), we substitute C = -10 into the formula:
\( t(-10) = \frac{ 9(-10) }{ 5 } + 32 \)
\( \implies t(-10) = \frac{ -90 }{ 5 } + 32 \)
\( \implies t(-10) = -18 + 32 \)
\( \implies t(-10) = 14^\circ F \)
This shows that -10 degrees Celsius is 14 degrees Fahrenheit.
(iv) To find the value of C when t(C) = 212, we set the formula equal to 212 and solve for C:
\( \frac{ 9C }{ 5 } + 32 = 212 \)
\( \implies \frac{ 9C }{ 5 } = 212 - 32 \)
\( \implies \frac{ 9C }{ 5 } = 180 \)
\( \implies 9C = 180 \times 5 \)
\( \implies 9C = 900 \)
\( \implies C = \frac{ 900 }{ 9 } \)
\( \implies C = 100^\circ C \)
This indicates that 212 degrees Fahrenheit is equal to 100 degrees Celsius, which is the boiling point of water.
(v) To find the temperature when the Celsius value is equal to the Fahrenheit value, let C = F = x. We substitute x into the formula:
\( x = \frac{ 9x }{ 5 } + 32 \)
Multiply all terms by 5 to remove the fraction:
\( 5x = 9x + (32 \times 5) \)
\( \implies 5x = 9x + 160 \)
Now, move the 'x' terms to one side:
\( 5x - 9x = 160 \)
\( \implies -4x = 160 \)
Divide by -4 to find x:
\( x = \frac{ 160 }{ -4 } \)
\( \implies x = -40 \)
Therefore, the temperature at which Celsius and Fahrenheit values are equal is -40 degrees. This specific temperature is unique where both scales align.
In simple words: First, use the given formula to convert Celsius to Fahrenheit for specific temperatures. Then, for part (iv), set the formula equal to 212 and solve to find the Celsius temperature. For part (v), set both Celsius and Fahrenheit to the same variable, 'x', in the formula and solve for 'x' to find the temperature where both scales are identical.

🎯 Exam Tip: Always remember the temperature conversion formula and practice rearranging it to solve for different variables. Pay close attention to negative signs and the order of operations to avoid errors.