OP Malhotra Class 11 Maths Solutions Chapter 2 Relations and Functions Exercise 2 (D)

S Chand Class 11 ICSE Maths Solutions Chapter 2 Relations and Functions Ex 2(d)

Question 1. A function f given as f : {(2, 7), (3, 4), (7, 9), (- 1, 6), (0, 2), (5, 3)}. In this function one-one onto ? Interchange the order of the elements in the ordered pairs and form the new relation. Is this relation a function ? If it is a function, is it one-one onto.
Answer: The given function is \( f = \{(2, 7), (3, 4), (7, 9), (-1, 6), (0, 2), (5, 3)\} \).
Since all different elements in the domain of \( f \) have different images in the codomain, the function \( f \) is one-one.
Also, every element in the codomain of \( f \) has at least one pre-image in the domain of \( f \), so \( f \) is onto.
Therefore, \( f \) is both one-one and onto.
Now, let's interchange the order of the elements to form a new relation, let's call it \( g \).
\( g = \{(7, 2), (4, 3), (9, 7), (6, -1), (2, 0), (3, 5)\} \).
To check if \( g \) is a function, we see if each first element (domain) maps to only one second element (range). In this case, each first element in \( g \) has a unique second element, so \( g \) is a function.
Since different elements in the domain of \( g \) have different images, \( g \) is one-one.
For every element in the codomain of \( g \), there is at least one element in the domain of \( g \) that maps to it, so \( g \) is onto.
Thus, the new relation \( g \) is also one-one and onto.
In simple words: The original function gives a unique output for each input, and every possible output is used. When we swap the inputs and outputs, the new relation also has a unique output for each new input, and all outputs are still used. This means both relations are "one-one" and "onto".

🎯 Exam Tip: To determine if a relation is a function, check that no two ordered pairs have the same first element but different second elements. For one-one, ensure different inputs always give different outputs. For onto, make sure every element in the codomain is an image of some element in the domain.

Question 2. Determine if each function is one-one.
(i) To each person on the earth assign the number which corresponds to his age
(ii) To each country in the world assign the latitude and longitude of its capital.
(iii) To each book written by only one author assign the author.
(iv) To each country in the world which has a prime minister assign its prime minister.
Answer:
(i) This function is not one-one. Many different people on Earth can have the same age. For example, two different people might both be 30 years old, meaning two different inputs (people) lead to the same output (age).
(ii) This function is one-one. Each country in the world has a unique latitude and longitude for its capital city. This means every different country will be assigned a different pair of latitude and longitude values.
(iii) This function is not one-one. One author can write many different books. So, different books (elements in the domain) can be assigned to the same author (element in the codomain), showing it is not one-one.
(iv) This function is one-one. Different countries in the world have different prime ministers. Therefore, each country (element in the domain) will be assigned a unique prime minister (element in the codomain).
In simple words: A function is "one-one" if every different input gives a different output. For ages, many people can be the same age. For countries and capital coordinates, each country's capital has unique coordinates. For books and authors, one author can write many books. For countries and prime ministers, each country has a unique prime minister.

🎯 Exam Tip: A function is one-one if \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\). To prove it's not one-one, just find two different inputs that give the same output.

Question 3. Let f : A→ B. Find f (A), i.e., the range of f, if f is an onto function.
Answer: If a function \( f : A \rightarrow B \) is an onto function, it means that every element in the codomain \( B \) is the image of at least one element in the domain \( A \). The range of a function, denoted \( f(A) \), is the set of all possible output values. For an onto function, the range is always equal to the entire codomain. Therefore, if \( f \) is an onto function, then \( f(A) = B \). This means every value in B is reached by the function.
In simple words: If a function is "onto," it means every item in the second set (B) gets hit by at least one item from the first set (A). So, the list of all outputs (the range) is exactly the same as the whole second set (B).

🎯 Exam Tip: Remember the definition of an onto function: the range must be equal to the codomain. This is a fundamental concept in functions.

Question 4. Show that the function f : R → R given by f (x) = cos x for all x ∈ R, is neither one-one nor onto.
Answer: The given function is \( f : R \rightarrow R \) defined by \( f(x) = \cos x \) for all \( x \in R \).
First, let's check if it is one-one.
We know that \( \cos 0 = 1 \) and \( \cos(2\pi) = 1 \).
So, \( f(0) = 1 \) and \( f(2\pi) = 1 \).
Here, the elements \( 0 \) and \( 2\pi \) in the domain \( R \) have the same image \( 1 \) in the codomain \( R \).
Since \( 0 \neq 2\pi \) but \( f(0) = f(2\pi) \), the function \( f \) is many-one, not one-one.

Next, let's check if it is onto.
The range of the cosine function is \( [-1, 1] \). This means that for any real number \( x \), the value of \( \cos x \) will always be between \( -1 \) and \( 1 \), inclusive.
So, the range of \( f \) is \( R_f = [-1, 1] \).
The codomain of the function is \( R \) (all real numbers).
Since \( R_f = [-1, 1] \neq R \), the range is not equal to the codomain. For example, there is no real number \( x \) such that \( \cos x = 2 \).
Therefore, the function \( f \) is not onto.
Since \( f \) is neither one-one nor onto, it is neither bijective nor surjective nor injective.
In simple words: The function \( f(x) = \cos x \) is not "one-one" because different input numbers, like 0 and \( 2\pi \), can give the same answer (which is 1). It's also not "onto" because the cosine function can only give answers between -1 and 1. It can't give an answer like 5, but 5 is part of the set of all real numbers (the codomain).

🎯 Exam Tip: To prove a function is not one-one, find a counterexample (two distinct domain elements mapping to the same codomain element). To prove it's not onto, show an element in the codomain that has no pre-image in the domain.

Question 5. Let A = {- 1, 1}. Let functions f g and A of A onto A be defined by :
(i) f(x) = x, (ii) g (x) = x³, (iii) h (x) = sin x
Which function, if any, is onto ?
Answer: The given set is \( A = \{-1, 1\} \). We need to determine which of the functions \( f, g, h \) are onto when mapping from \( A \) to \( A \). A function is onto if every element in the codomain \( A \) has at least one pre-image in the domain \( A \).

(i) For \( f(x) = x \):
\( f(-1) = -1 \)
\( f(1) = 1 \)
The range is \( \{-1, 1\} \), which is equal to the codomain \( A \). So, \( f \) is onto.

(ii) For \( g(x) = x^3 \):
\( g(-1) = (-1)^3 = -1 \)
\( g(1) = (1)^3 = 1 \)
The range is \( \{-1, 1\} \), which is equal to the codomain \( A \). So, \( g \) is onto.

(iii) For \( h(x) = \sin x \):
\( h(-1) = \sin(-1) \)
\( h(1) = \sin(1) \)
The values \( \sin(-1) \) and \( \sin(1) \) are approximately \( -0.841 \) and \( 0.841 \) respectively (when using radians). Neither of these values are exactly \( -1 \) or \( 1 \).
More precisely, \( \sin x = 1 \) means \( x = \frac{\pi}{2} \) (and other values), but \( \frac{\pi}{2} \) is not in set A. Similarly, \( \sin x = -1 \) means \( x = -\frac{\pi}{2} \) (and other values), but \( -\frac{\pi}{2} \) is also not in set A.
Thus, the elements \( -1 \) and \( 1 \) in the codomain \( A \) do not have any pre-image in the domain \( A \). So, \( h \) is not onto.

Therefore, the functions \( f(x) = x \) and \( g(x) = x^3 \) are onto. They cover all possible output values in the set A.
In simple words: We have two numbers, -1 and 1. A function is "onto" if it uses up both -1 and 1 as answers when you put in -1 and 1. The functions \( f(x)=x \) and \( g(x)=x^3 \) both give -1 and 1 as answers. But \( h(x)=\sin x \) does not give exactly -1 and 1 when you put in -1 and 1, so it is not onto.

🎯 Exam Tip: When dealing with small finite sets, list out all the images of the domain elements. If this set of images (the range) matches the codomain exactly, the function is onto.

Question 6. Given A = {2, 3, 4}, B = {2, 5, 6, 7}, construct an example of each of the following.
(i) A one-one mapping from A to B
(ii) A mapping from A to B which is not one-one
(iii) A mapping from B to A.
Answer: Given sets are \( A = \{2, 3, 4\} \) and \( B = \{2, 5, 6, 7\} \).

(i) A one-one mapping from A to B:
Let \( f = \{(2, 5), (3, 6), (4, 7)\} \).
In this mapping, each different element in the domain \( A \) (2, 3, 4) is mapped to a different image in the codomain \( B \) (5, 6, 7). No two inputs share an output. Thus, \( f \) is a one-one mapping from A to B.

(ii) A mapping from A to B which is not one-one:
Let \( g = \{(2, 5), (3, 5), (4, 7)\} \).
Here, the elements \( 2 \) and \( 3 \) in the domain \( A \) both map to the same image, \( 5 \), in the codomain \( B \). Since different inputs (2 and 3) give the same output (5), this function is not one-one; it is a many-one function.

(iii) A mapping from B to A:
Let \( h = \{(2, 2), (5, 3), (6, 4), (7, 4)\} \).
For this to be a mapping from B to A, every element in B must be assigned to exactly one element in A. Here, \( 2 \in B \) maps to \( 2 \in A \), \( 5 \in B \) maps to \( 3 \in A \), \( 6 \in B \) maps to \( 4 \in A \), and \( 7 \in B \) maps to \( 4 \in A \). Each element in B has a unique image in A, ensuring it's a valid function. Since \( 6 \) and \( 7 \) both map to \( 4 \), this function is many-one.
In simple words: We need to show different ways to connect numbers between two sets, A and B.
(i) For a "one-one" connection, each number from A goes to a unique number in B. Like 2 to 5, 3 to 6, 4 to 7.
(ii) For a connection that is "not one-one," some numbers from A go to the same number in B. Like both 2 and 3 go to 5, and 4 goes to 7.
(iii) For a connection from B to A, each number from B must go to a number in A. Like 2 from B goes to 2 in A, 5 from B goes to 3 in A, and 6 and 7 from B both go to 4 in A.

🎯 Exam Tip: When constructing examples for mappings, ensure that for a function, each element in the domain maps to *exactly one* element in the codomain. To make it "not one-one," ensure at least two distinct domain elements map to the same codomain element.

Question 7. Are these sets of ordered pairs functions? If so, examine whether the mapping is onto or one-one.
(i) \({(x, y) : x \text{ is a person, y is the mother of x}}\)
(ii) \({(a, b) : a \text{ is a person, b is an ancestor of a)}}\)
Answer:
(i) Let \( f = \{(x, y): x \text{ is a person, y is the mother of x}\} \).
Is it a function? Yes, because every person has one and only one biological mother. So, each \( x \) (person) corresponds to a unique \( y \) (mother).
Is it one-one? No. Different people (e.g., siblings) can have the same mother. So, different \( x \) values can map to the same \( y \) value.
Is it onto? Yes. Every mother \( y \) must be the mother of some person \( x \) in the set of all people. So, all possible 'mothers' are used as output values.

(ii) Let \( g = \{(a, b): a \text{ is a person, b is an ancestor of a}\} \).
Is it a function? No. A person \( a \) can have many ancestors (e.g., a father, a mother, grandfathers, grandmothers, etc.). This means one element \( a \) in the domain would correspond to multiple elements \( b \) in the codomain. For a relation to be a function, each input must have only one output. Therefore, this set of ordered pairs does not form a function.
In simple words:
(i) The relation "is the mother of" is a function because everyone has just one mother. It is not "one-one" because siblings share a mother. It is "onto" because every mother is someone's mother.
(ii) The relation "is an ancestor of" is not a function. This is because one person has many ancestors (like parents, grandparents, etc.), so one input would have many outputs.

🎯 Exam Tip: To check if a relation is a function, ensure that each element in the domain maps to exactly one element in the codomain. If an element maps to more than one, it's not a function. Then, apply the definitions of one-one and onto if it is a function.

Question 8. Is the function f : N → N (N is the set of natural numbers) defined by f (n) = 2n + 3 for all n ∈ N onto ?
Answer: The given function is \( f : N \rightarrow N \) defined by \( f(n) = 2n + 3 \) for all \( n \in N \). Here, \( N \) represents the set of natural numbers \( \{1, 2, 3, ...\} \).
To check if the function is onto, we need to see if every element in the codomain \( N \) has a pre-image in the domain \( N \).
Let \( y \) be an arbitrary element in the codomain \( N \). We want to find an \( n \in N \) such that \( f(n) = y \).
So, \( 2n + 3 = y \).
This implies \( 2n = y - 3 \).
This implies \( n = \frac{y - 3}{2} \).
Now, let's consider an element in the codomain, for example, \( y = 1 \).
If \( y = 1 \), then \( n = \frac{1 - 3}{2} = \frac{-2}{2} = -1 \).
However, \( -1 \) is not a natural number (it is not in the set \( N = \{1, 2, 3, ...\} \)).
This means that the element \( 1 \) in the codomain \( N \) does not have a pre-image in the domain \( N \). Therefore, not all elements in the codomain are "hit" by the function.
So, the function \( f \) is not onto.
In simple words: The function takes a natural number, multiplies it by 2, and adds 3. For the function to be "onto," every natural number must be a possible answer. But if we try to get an answer like 1, the starting number would have to be -1, which is not a natural number. So, some natural numbers can't be made by this function, meaning it's not onto.

🎯 Exam Tip: To prove a function \( f: A \to B \) is not onto, find an element \( y \in B \) for which there is no \( x \in A \) such that \( f(x) = y \). This shows that the range is a proper subset of the codomain.

Question 9. Let A = {x : 0 ≤ x ≤ 2} and B = {1}. Give an example of a function from A to B. Can you define a function from B and A which is onto ? Give reasons for your answer.
Answer: Given sets are \( A = \{x : 0 \le x \le 2\} \) and \( B = \{1\} \). Set A contains all real numbers from 0 to 2, inclusive, which is an infinite set. Set B contains only one element, the number 1.

Example of a function from A to B:
Let's define a function \( f : A \rightarrow B \) as \( f(x) = 1 \) for all \( x \in A \).
This is a valid function because every element in \( A \) (e.g., 0, 0.5, 1, 1.5, 2) maps to the single element \( 1 \) in \( B \). Each element in A has a unique image (which is 1). This is a constant function.

Can we define a function from B to A which is onto?
No, we cannot define a function \( g : B \rightarrow A \) which is onto.
**Reason:** For a function to be onto, every element in the codomain \( A \) must have at least one pre-image in the domain \( B \).
The domain \( B \) only has one element, which is \( 1 \). So, any function \( g \) from \( B \) to \( A \) can only map this single element \( 1 \) to exactly one value in \( A \). For example, \( g(1) = c \), where \( c \) is some value in \( A \).
However, the codomain \( A = \{x : 0 \le x \le 2\} \) contains an infinite number of elements. A single element from \( B \) can only map to one element in \( A \), so it's impossible for a function from \( B \) to \( A \) to "cover" all the infinite elements in \( A \). The range of \( g \) would only contain one element, which cannot be equal to the entire set A.
In simple words: Set A has many numbers between 0 and 2. Set B just has the number 1.
A function from A to B could just be: no matter what number you pick from A, the answer is always 1. This works.
But a function from B to A cannot be "onto." This is because B only has one number (1), so it can only point to one number in A. Since A has many, many numbers, that single number from B can't possibly "hit" all of them.

🎯 Exam Tip: When the codomain has more elements than the domain, particularly if the domain is finite and the codomain is infinite (as in this case), it's generally impossible for a function to be onto. The number of possible outputs cannot exceed the number of inputs if it's to cover all outputs.

Question 10. Prove that the function f : R → R, f (x) = x² + x is a many-one into function.
Answer: The given function is \( f : R \rightarrow R \) defined by \( f(x) = x^2 + x \) for all \( x \in R \).
First, let's show it is many-one.
To prove it is many-one, we need to find two different elements in the domain that map to the same image.
Consider \( x = 1 \): \( f(1) = 1^2 + 1 = 1 + 1 = 2 \).
Consider \( x = -2 \): \( f(-2) = (-2)^2 + (-2) = 4 - 2 = 2 \).
Since \( f(1) = 2 \) and \( f(-2) = 2 \), but \( 1 \neq -2 \), the function \( f \) is many-one.
Alternatively, we can set \( f(x) = f(y) \) and show that \( x \neq y \).
\( x^2 + x = y^2 + y \)

\( \implies \) \( x^2 - y^2 + x - y = 0 \)

\( \implies \) \( (x - y)(x + y) + (x - y) = 0 \)

\( \implies \) \( (x - y)(x + y + 1) = 0 \)
This means either \( x - y = 0 \) or \( x + y + 1 = 0 \).
So, \( x = y \) or \( x = -y - 1 \).
Since \( x \) can be equal to \( -y - 1 \) (for example, if \( y=1 \), \( x=-2 \)), there exist cases where \( x \neq y \) but \( f(x) = f(y) \). Therefore, \( f \) is many-one.

Next, let's show it is into (i.e., not onto).
For \( f \) to be onto, every element \( y \) in the codomain \( R \) must have a pre-image \( x \) in the domain \( R \).
Let \( f(x) = y \). So, \( x^2 + x = y \).
Rearranging this, we get \( x^2 + x - y = 0 \).
This is a quadratic equation for \( x \). For \( x \) to be a real number, the discriminant \( \Delta = b^2 - 4ac \) must be greater than or equal to zero.
Here, \( a=1, b=1, c=-y \). So, \( \Delta = (1)^2 - 4(1)(-y) = 1 + 4y \).
For \( x \) to be a real number, \( 1 + 4y \ge 0 \).
This implies \( 4y \ge -1 \).
This implies \( y \ge -\frac{1}{4} \).
This means that the function \( f(x) \) can only produce output values \( y \) that are greater than or equal to \( -\frac{1}{4} \).
However, the codomain is \( R \) (all real numbers), which includes numbers less than \( -\frac{1}{4} \), such as \( -1 \).
If we take \( y = -1 \), then \( 1 + 4(-1) = 1 - 4 = -3 < 0 \), meaning there is no real \( x \) for which \( f(x) = -1 \).
Since there are elements in the codomain \( R \) (e.g., \( -1 \)) that have no pre-image in the domain \( R \), the function \( f \) is not onto. When a function is not onto, it is called an into function.
Therefore, \( f \) is a many-one into function.
In simple words: The function \( f(x) = x^2 + x \) is "many-one" because different starting numbers can give the same answer. For example, both 1 and -2 give the answer 2. It is "into" (meaning not "onto") because not all real numbers can be answers. For instance, you can never get an answer like -1, no matter what real number you plug in for \( x \).

🎯 Exam Tip: When proving a function is many-one, you can either find specific counterexamples or show that \(f(x) = f(y)\) does not always imply \(x = y\). To prove it's into, determine the actual range of the function and show it's a proper subset of the codomain.

Question 11. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one but not onto.
Answer: Given sets are \( A = \{1, 2, 3\} \) and \( B = \{4, 5, 6, 7\} \).
The function \( f \) is defined as \( f = \{(1, 4), (2, 5), (3, 6)\} \).

To show that \( f \) is one-one:
The domain of \( f \) is \( D_f = \{1, 2, 3\} \).
The images of the elements in the domain are:
\( f(1) = 4 \)
\( f(2) = 5 \)
\( f(3) = 6 \)
Clearly, different elements in the domain \( D_f \) (1, 2, 3) have different images (4, 5, 6). No two inputs map to the same output. Therefore, \( f \) is a one-one function.

To show that \( f \) is not onto:
The range of \( f \) is \( R_f = \{4, 5, 6\} \). This is the set of all actual output values.
The codomain of \( f \) is the set \( B = \{4, 5, 6, 7\} \). This is the set of all possible output values.
For \( f \) to be onto, the range must be equal to the codomain (\( R_f = B \)).
However, we can see that \( R_f = \{4, 5, 6\} \) is not equal to \( B = \{4, 5, 6, 7\} \), because the element \( 7 \) is in \( B \) but it is not an image of any element in \( A \). There is no \( x \in A \) such that \( f(x) = 7 \).
Since there is an element in the codomain \( B \) (which is 7) that does not have a pre-image in the domain \( A \), the function \( f \) is not onto.
Thus, \( f \) is one-one but not onto.
In simple words: We have two groups of numbers, A and B, and a way to connect them called function \( f \).
It is "one-one" because each number in group A (1, 2, 3) connects to a different, unique number in group B (4, 5, 6).
It is not "onto" because the number 7 is in group B but no number from group A connects to it. So, 7 is left out of the answers.

🎯 Exam Tip: Always clearly state the domain, range, and codomain of the function. For one-one, match each distinct input to a distinct output. For onto, explicitly check if the range covers every element in the codomain.

Question 12. Show that the function f : R → R by f(x) = 3 – 4x is one-one onto and hence bijective.
Answer: The given function is \( f : R \rightarrow R \) defined by \( f(x) = 3 - 4x \) for all \( x \in R \).

First, let's show it is one-one.
Let \( x, y \in R \) such that \( f(x) = f(y) \).
\( 3 - 4x = 3 - 4y \)

\( \implies \) \( -4x = -4y \)

\( \implies \) \( x = y \)
Since \( f(x) = f(y) \) implies \( x = y \), the function \( f \) is one-one.

Next, let's show it is onto.
Let \( y \) be any arbitrary element in the codomain \( R \). We need to find an \( x \in R \) such that \( f(x) = y \).
\( 3 - 4x = y \)

\( \implies \) \( -4x = y - 3 \)

\( \implies \) \( 4x = 3 - y \)

\( \implies \) \( x = \frac{3 - y}{4} \)
Since \( y \) is a real number, \( (3 - y)/4 \) will also always be a real number. This means for every \( y \) in the codomain \( R \), there exists an \( x = \frac{3 - y}{4} \) in the domain \( R \) such that \( f(x) = y \).
Therefore, the function \( f \) is onto.

Since the function \( f \) is both one-one and onto, it is a bijective function.
In simple words: We need to show that the function \( f(x) = 3 - 4x \) is "one-one" and "onto". It's "one-one" because if two inputs give the same answer, the inputs must have been the same number. It's "onto" because for any answer \( y \) you want, you can always find a starting number \( x \) that will give you that \( y \). Because it's both one-one and onto, it is called a "bijective" function.

🎯 Exam Tip: For linear functions like \( f(x) = ax + b \) (where \( a \neq 0 \)), they are almost always both one-one and onto on real numbers. For one-one, assume \( f(x_1) = f(x_2) \) and show \( x_1 = x_2 \). For onto, set \( f(x) = y \) and solve for \( x \) in terms of \( y \), ensuring \( x \) is always in the domain for any \( y \) in the codomain.