Samacheer Kalvi Class 10 Maths Solutions Chapter 1 Relations and Functions Exercise 1.3

 

Question 1. Let \( f = \left\{ (x, y) \mid x,y \in N \text{ and } y = 2x \right\} \) be a relation on N. Find the domain, co-domain and range. Is this relation a function?
Answer: The relation is defined on natural numbers \( N \). This means both the input (domain) and potential output (co-domain) are the set of natural numbers, \( N = \{1, 2, 3, \ldots\} \). For this relation, each natural number \( x \) is linked to \( 2x \). For example, 1 is linked to 2, 2 to 4, 3 to 6, and so on. This makes each input have only one specific output.
The domain of the relation is \( \{1, 2, 3, 4, \ldots\} \), which is the set of natural numbers \( N \).
The co-domain is also \( \{1, 2, 3, 4, \ldots\} \), which is the set of natural numbers \( N \).
The range of the relation is \( \{2, 4, 6, 8, \ldots\} \), which are all the even natural numbers.
Yes, this relation is a function because every element in the domain (every natural number) is connected to exactly one element in the co-domain through the rule \( y = 2x \).
In simple words: This math rule connects each counting number (1, 2, 3...) to its double (2, 4, 6...). The numbers you start with are the domain, the numbers it could possibly go to are the co-domain, and the numbers it actually goes to are the range. Since each starting number goes to only one ending number, it is a function.

🎯 Exam Tip: Remember that for a relation to be a function, every element in the domain must map to exactly one element in the co-domain. Double-check this condition carefully.

 

Question 2. Let \( X = \{3, 4, 6, 8\} \). Determine whether the relation \( R = \{(x,f(x)) \mid x \in X, f(x) = x^2 + 1\} \) is a function from X to N?
Answer: We are given the set \( X = \{3, 4, 6, 8\} \) and a rule \( f(x) = x^2 + 1 \). We need to see if this rule makes a function from \( X \) to \( N \) (natural numbers).
Let's find the output for each number in \( X \):
For \( x = 3 \), \( f(3) = 3^2 + 1 = 9 + 1 = 10 \).
For \( x = 4 \), \( f(4) = 4^2 + 1 = 16 + 1 = 17 \).
For \( x = 6 \), \( f(6) = 6^2 + 1 = 36 + 1 = 37 \).
For \( x = 8 \), \( f(8) = 8^2 + 1 = 64 + 1 = 65 \).
All the calculated output values (10, 17, 37, 65) are natural numbers, meaning they belong to \( N \). Also, each input from \( X \) gives only one output. Therefore, the relation \( R \) is a function from \( X \) to \( N \).
In simple words: We take each number from the set X, square it, and then add 1. Since all the answers we get are simple counting numbers, and each starting number gives only one answer, it means this rule is a function.

🎯 Exam Tip: To check if a relation is a function, ensure two things: first, every element in the domain has an image; second, each element has only one image in the co-domain.

 

Question 3. Given the function \( f: x \rightarrow x^2 - 5x + 6 \), evaluate
(i) \( f(-1) \)
(ii) \( f(2a) \)
(iii) \( f(2) \)
(iv) \( f(x - 1) \)
Answer: The given function is \( f(x) = x^2 - 5x + 6 \). We need to replace \( x \) with the given values to find the answers.
(i) To find \( f(-1) \), substitute \( x = -1 \) into the function:
\( f(-1) = (-1)^2 - 5(-1) + 6 \)
\( = 1 + 5 + 6 \)
\( = 12 \)
(ii) To find \( f(2a) \), substitute \( x = 2a \) into the function:
\( f(2a) = (2a)^2 - 5(2a) + 6 \)
\( = 4a^2 - 10a + 6 \)
(iii) To find \( f(2) \), substitute \( x = 2 \) into the function:
\( f(2) = (2)^2 - 5(2) + 6 \)
\( = 4 - 10 + 6 \)
\( = 0 \)
(iv) To find \( f(x - 1) \), substitute \( x = x - 1 \) into the function:
\( f(x - 1) = (x - 1)^2 - 5(x - 1) + 6 \)
First, expand \( (x-1)^2 \): \( (x-1)(x-1) = x^2 - x - x + 1 = x^2 - 2x + 1 \).
Then, distribute \( -5 \): \( -5(x - 1) = -5x + 5 \).
So, \( f(x - 1) = x^2 - 2x + 1 - 5x + 5 + 6 \)
Combine the like terms:
\( = x^2 - 7x + 12 \)
In simple words: We replace the letter 'x' in the given math rule with new numbers or letters. Then we do the math steps like squaring and multiplying to get the final answer for each part.

🎯 Exam Tip: Be careful with signs, especially when substituting negative numbers or expressions like \( (x-1) \) where squaring and multiplication steps can introduce common errors.

 

Question 4. A graph representing the function \( f(x) \) is given. It is clear that \( f(9) = 2 \).
(i) Find the following values of the function
(a) \( f(0) \)
(b) \( f(7) \)
(c) \( f(2) \)
(d) \( f(10) \)
(ii) For what value of \( x \) is \( f(x) = 1 \)?
(iii) Describe the following
(i) Domain
(ii) Range.
(iv) What is the image of 6 under \( f \)?
Answer: We will read the values directly from the provided graph.
(i) To find the values of the function:
(a) Looking at the graph, when \( x = 0 \), the graph is at \( y = 9 \). So, \( f(0) = 9 \).
(b) When \( x = 7 \), the graph is at \( y = 6 \). So, \( f(7) = 6 \).
(c) When \( x = 2 \), the graph is at \( y = 6 \). So, \( f(2) = 6 \).
(d) When \( x = 10 \), the graph is at \( y = 0 \). So, \( f(10) = 0 \).
(ii) We need to find the value of \( x \) for which \( f(x) = 1 \). On the graph, find where the y-value is 1. The line \( y = 1 \) intersects the graph at \( x = 9.5 \). So, when \( f(x) = 1 \), the value of \( x \) is \( 9.5 \).
(iii) To describe the domain and range from the graph:
(i) The domain refers to all possible x-values the function uses. From the graph, the x-values start from 0 and go up to 10. So, the domain is \( \{x \mid 0 \le x \le 10, x \in R\} \). This can also be represented as a set of discrete integer points based on the graph gridlines, so \( \{0, 1, 2, 3, \ldots, 10\} \).
(ii) The range refers to all possible y-values the function produces. From the graph, the y-values start from 0 and go up to 9. So, the range is \( \{y \mid 0 \le y \le 9, y \in R\} \). This can also be represented as a set of discrete integer points based on the graph gridlines, so \( \{0, 1, 2, 3, \ldots, 9\} \).
(iv) The image of 6 under \( f \) means \( f(6) \). Looking at the graph, when \( x = 6 \), the corresponding y-value is \( 5 \). So, the image of 6 under \( f \) is \( 5 \).
In simple words: We look at the given graph. For part (i), we find the height (y-value) at specific points (x-values). For part (ii), we find the x-value when the height is 1. For part (iii), the domain is all the x-values the graph covers, and the range is all the y-values the graph covers. For part (iv), we find the y-value when x is 6.

🎯 Exam Tip: When reading values from a graph, always pay close attention to the scales on both the x-axis and y-axis to ensure accuracy, especially for points that fall between grid lines.

 

Question 5. Let \( f(x) = 2x + 5 \). If \( x \neq 0 \) then find \( \frac{f(x+2)-f(2)}{x} \).
Answer: We are given the function \( f(x) = 2x + 5 \). We need to find an expression involving \( f(x+2) \) and \( f(2) \).
First, let's find \( f(x+2) \) by replacing \( x \) with \( (x+2) \) in the function:
\( f(x+2) = 2(x+2) + 5 \)
\( = 2x + 4 + 5 \)
\( = 2x + 9 \)
Next, let's find \( f(2) \) by replacing \( x \) with \( 2 \) in the function:
\( f(2) = 2(2) + 5 \)
\( = 4 + 5 \)
\( = 9 \)
Now, we can substitute these into the given expression:
\( \frac{f(x+2)-f(2)}{x} = \frac{(2x+9) - (9)}{x} \)
\( = \frac{2x+9-9}{x} \)
\( = \frac{2x}{x} \)
Since it is given that \( x \neq 0 \), we can cancel \( x \) from the numerator and denominator:
\( = 2 \)
The value of the expression is 2. This is a common form used in calculus to find the slope of a line or derivative.
In simple words: We are given a rule for \( f(x) \). We first find what \( f \) gives us for \( (x+2) \) and for \( 2 \). Then we put these results into the fraction and simplify it. In the end, after all the steps, the answer is 2.

🎯 Exam Tip: When simplifying algebraic fractions, always check for conditions like \( x \neq 0 \) that allow you to cancel terms. This problem is similar to finding the difference quotient.

 

Question 6. A function \( f \) is defined by \( f(x) = 2x - 3 \).
(i) find \( \frac{f(0)+f(1)}{2} \)
(ii) find \( x \) such that \( f(x) = 0 \).
(iii) find \( x \) such that \( f(x) = x \).
(iv) find \( x \) such that \( f(x) = f(1 - x) \).
Answer: The given function is \( f(x) = 2x - 3 \). We will solve each part by substituting or setting up equations.
(i) First, find \( f(0) \) and \( f(1) \):
\( f(0) = 2(0) - 3 = 0 - 3 = -3 \)
\( f(1) = 2(1) - 3 = 2 - 3 = -1 \)
Now substitute these values into the expression \( \frac{f(0)+f(1)}{2} \):
\( \frac{f(0)+f(1)}{2} = \frac{-3 + (-1)}{2} \)
\( = \frac{-3 - 1}{2} \)
\( = \frac{-4}{2} \)
\( = -2 \)
(ii) To find \( x \) such that \( f(x) = 0 \), set the function equal to 0:
\( 2x - 3 = 0 \)
Add 3 to both sides:
\( 2x = 3 \)
Divide by 2:
\( x = \frac{3}{2} \)
(iii) To find \( x \) such that \( f(x) = x \), set the function equal to \( x \):
\( 2x - 3 = x \)
Subtract \( x \) from both sides:
\( 2x - x - 3 = 0 \)
\( x - 3 = 0 \)
Add 3 to both sides:
\( x = 3 \)
(iv) To find \( x \) such that \( f(x) = f(1 - x) \), first find \( f(1 - x) \):
\( f(1 - x) = 2(1 - x) - 3 \)
\( = 2 - 2x - 3 \)
\( = -2x - 1 \)
Now set \( f(x) = f(1 - x) \):
\( 2x - 3 = -2x - 1 \)
Add \( 2x \) to both sides:
\( 2x + 2x - 3 = -1 \)
\( 4x - 3 = -1 \)
Add 3 to both sides:
\( 4x = -1 + 3 \)
\( 4x = 2 \)
Divide by 4:
\( x = \frac{2}{4} \)
\( x = \frac{1}{2} \)
In simple words: We use the given rule \( f(x) = 2x - 3 \) to solve different problems. For part (i), we put 0 and 1 into the rule, then add those results and divide by 2. For parts (ii), (iii), and (iv), we set up equations where \( f(x) \) is equal to something else, then we solve for \( x \).

🎯 Exam Tip: Always perform operations like distribution and combining like terms carefully to avoid algebraic errors, especially when dealing with expressions involving \( (1-x) \).

 

Question 7. A square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown. Express the volume V of the box as a function of \( x \).
Answer: We start with a square piece of material that has sides of 24 cm. When we cut equal squares of side length \( x \) from each corner and fold up the sides, we create an open box.
The original length of each side was 24 cm. When we cut \( x \) cm from both ends of each side, the new length and width of the base of the box become \( (24 - 2x) \) cm. The height of the box will be the side length of the square cut from the corners, which is \( x \) cm.
So, for the cuboid (box) formed:
Length \( (l) = (24 - 2x) \) cm
Width \( (b) = (24 - 2x) \) cm
Height \( (h) = x \) cm
The volume \( V \) of a cuboid is given by the formula \( V = l \times b \times h \).
Substitute the dimensions into the formula:
\( V(x) = (24 - 2x) \times (24 - 2x) \times x \)
\( V(x) = (24 - 2x)^2 \times x \)
First, expand \( (24 - 2x)^2 \):
\( (24 - 2x)^2 = (24)^2 - 2(24)(2x) + (2x)^2 \)
\( = 576 - 96x + 4x^2 \)
Now, multiply this by \( x \):
\( V(x) = (576 - 96x + 4x^2)x \)
\( V(x) = 576x - 96x^2 + 4x^3 \)
It is standard practice to write polynomials in descending order of power:
\( V(x) = 4x^3 - 96x^2 + 576x \)
This expression gives the volume of the box as a function of the side length \( x \) of the cut squares.
In simple words: Imagine a square piece of paper. If you cut out small squares from its corners and then fold up the sides, you get an open box. The size of the cuts (let's call it \( x \)) changes the length, width, and height of the box. We use these new dimensions to write a math rule (a function) that tells us the volume of the box based on \( x \).

🎯 Exam Tip: Remember to express the length and width of the base in terms of \( x \) by subtracting \( 2x \) from the original side length, as \( x \) is removed from both ends. The height is simply \( x \).

 

Question 8. A function \( f \) is defined by \( f(x) = 3 - 2x \). Find \( x \) such that \( f(x^2) = (f(x))^2 \).
Answer: We are given the function \( f(x) = 3 - 2x \). We need to find the value of \( x \) that satisfies the equation \( f(x^2) = (f(x))^2 \).
First, let's find the expression for \( f(x^2) \) by replacing \( x \) with \( x^2 \) in the function:
\( f(x^2) = 3 - 2(x^2) \)
\( = 3 - 2x^2 \)
Next, let's find the expression for \( (f(x))^2 \) by squaring the entire function \( f(x) \):
\( (f(x))^2 = (3 - 2x)^2 \)
Expand the square: \( (a-b)^2 = a^2 - 2ab + b^2 \)
\( (3 - 2x)^2 = 3^2 - 2(3)(2x) + (2x)^2 \)
\( = 9 - 12x + 4x^2 \)
Now, set \( f(x^2) \) equal to \( (f(x))^2 \), as per the question:
\( 3 - 2x^2 = 9 - 12x + 4x^2 \)
To solve for \( x \), move all terms to one side of the equation to form a quadratic equation. Let's move everything to the right side to keep the \( x^2 \) term positive:
\( 0 = 9 - 12x + 4x^2 - 3 + 2x^2 \)
Combine like terms:
\( 0 = 6x^2 - 12x + 6 \)
We can simplify this equation by dividing all terms by 6:
\( \frac{0}{6} = \frac{6x^2}{6} - \frac{12x}{6} + \frac{6}{6} \)
\( 0 = x^2 - 2x + 1 \)
This is a perfect square trinomial, which can be factored as \( (x - 1)^2 \):
\( 0 = (x - 1)(x - 1) \)
For the product to be zero, at least one of the factors must be zero:
\( x - 1 = 0 \)
\( \implies x = 1 \)
So, the value of \( x \) that satisfies the given condition is 1. This shows how algebraic functions can be solved using quadratic principles.
In simple words: We have a rule for \( f(x) \). We need to find a number \( x \) where plugging \( x^2 \) into the rule gives the same answer as plugging \( x \) into the rule and then squaring the whole result. We set up an equation, do the algebra, and find that \( x \) must be 1.

🎯 Exam Tip: When solving equations involving functions, always calculate \( f(g(x)) \) and \( (f(x))^n \) separately and carefully before setting up the final equation to avoid errors.

 

Question 9. A plane is flying at a speed of 500 km per hour. Express the distance \( d \) travelled by the plane as function of time \( t \) in hours.
Answer: We are given the speed of the plane as 500 km per hour. We need to express the distance \( d \) travelled as a function of time \( t \) (in hours).
The basic formula relating speed, distance, and time is:
\( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
We can rearrange this formula to solve for distance:
\( \text{Distance} = \text{Speed} \times \text{Time} \)
In this problem:
Speed \( = 500 \) km/hour
Time \( = t \) hours
Distance \( = d \)
Substitute these values into the distance formula:
\( d = 500 \times t \)
So, the distance \( d \) travelled by the plane as a function of time \( t \) is \( d(t) = 500t \). This is a linear function, meaning the distance increases steadily with time.
In simple words: If a plane flies at 500 km every hour, to find out how far it goes, you just multiply its speed by the number of hours it flies. So, distance is 500 times the time.

🎯 Exam Tip: Remember the fundamental formula "Distance = Speed × Time" (D=ST) and make sure units are consistent (km, hours, km/h) before substituting values.

 

Question 10. The data in the adjacent table depicts the length of a woman's forehand and her corresponding height. Based on this data, a student finds a relationship between the height (\( y \)) and the forehand length (\( x \)) as \( y = ax + b \), where \( a, b \) are constants.
(i) Check if this relation is a function.
(ii) Find \( a \) and \( b \).
(iii) Find the height of a woman whose forehand length is 40 cm.
(iv) Find the length of forehand of a woman if her height is 53.3 inches.
Answer: We will use the given table and the linear relationship \( y = ax + b \) to answer the questions.
Here is the data table:

(i) To check if the relation is a function, we look at the table. For each unique forehand length (\( x \)), there is only one corresponding height (\( y \)). For example, a forehand length of 35 cm always gives a height of 56 inches. Since each input \( x \) has exactly one output \( y \), this relation is a function.
(ii) We are given the relation \( y = ax + b \). We can use any two data points from the table to find \( a \) and \( b \). Let's use \( (x_1, y_1) = (35, 56) \) and \( (x_2, y_2) = (45, 65) \).
First, find the slope \( a \):
\( a = \frac{y_2 - y_1}{x_2 - x_1} = \frac{65 - 56}{45 - 35} = \frac{9}{10} = 0.9 \)
Now use one point and the value of \( a \) to find \( b \). Let's use \( (35, 56) \):
\( y = ax + b \)
\( 56 = 0.9(35) + b \)
\( 56 = 31.5 + b \)
\( b = 56 - 31.5 \)
\( b = 24.5 \)
So, the constants are \( a = 0.9 \) and \( b = 24.5 \). The relationship is \( y = 0.9x + 24.5 \).
(iii) To find the height of a woman whose forehand length is 40 cm, substitute \( x = 40 \) into the equation \( y = 0.9x + 24.5 \):
\( y = 0.9(40) + 24.5 \)
\( y = 36 + 24.5 \)
\( y = 60.5 \) inches
So, the height of a woman with a 40 cm forehand length is 60.5 inches. (The source specified "feet" here, but since all other units for height are inches, and the question uses "inches", we will assume it's a typo and use "inches").
(iv) To find the length of forehand of a woman if her height is 53.3 inches, substitute \( y = 53.3 \) into the equation \( y = 0.9x + 24.5 \):
\( 53.3 = 0.9x + 24.5 \)
Subtract 24.5 from both sides:
\( 53.3 - 24.5 = 0.9x \)
\( 28.8 = 0.9x \)
Divide by 0.9:
\( x = \frac{28.8}{0.9} \)
\( x = 32 \) cm
So, a woman with a height of 53.3 inches has a forehand length of 32 cm.
In simple words: First, we check the table to see if it's a function (each forehand length has only one height). Then, we use the numbers in the table to find the missing parts 'a' and 'b' for the height rule \( y = ax + b \). After finding the rule, we use it to calculate height if we know forehand length, or forehand length if we know height.

🎯 Exam Tip: When given a table of values and asked to find a linear relationship \( y = ax + b \), always calculate the slope \( a \) first using two points, then use one of the points to find the y-intercept \( b \). Always verify the consistency of units in the problem.