GSEB Class 12 Maths Solutions Chapter 1 Relations and Functions Exercise 1.2

Question 1. Show that the function f: R → R, defined by f (x) = \( \frac{1}{x} \) is one-one onto, where R is the set of all non-zero real numbers. Is the result true, if the domain R is replaced by N with co-domain being same as R ?
Answer:(i) For \( f(x) = \frac{1}{x} \), if \( f(x_1) = f(x_2) \)
\( \implies \frac{1}{x_1} = \frac{1}{x_2} \)
\( \implies x_1 = x_2 \)
Each \( x \in R \) has a unique image in the codomain. This shows \( f \) is one-one.
(ii) For each \( y \) belonging to the codomain, if \( y = \frac{1}{x} \), then \( x = \frac{1}{y} \). This means there is a unique pre-image of \( y \). This shows \( f \) is onto.
(b) When the domain \( R \) is replaced by \( N \), with the codomain \( R \) remaining the same, then \( f : N \rightarrow R \).
If \( f(x_1) = f(x_2) \)
\( \implies \frac{1}{n_1} = \frac{1}{n_2} \)
\( \implies n_1 = n_2 \), where \( n_1, n_2 \in N \). This shows \( f \) is one-one.
However, the function \( f \) is not onto because a real number like \( \frac{1}{2} \) (e.g., \( \frac{1}{2} = \frac{3}{2} \)) belonging to the codomain may not have a pre-image in \( N \).
In simple words: The function \( f(x) = \frac{1}{x} \) is both one-to-one and onto for non-zero real numbers. If we change the input numbers to only natural numbers, the function stays one-to-one, but it's no longer onto because many output values won't have a natural number as their input.

Exam Tip: To prove a function is one-one, assume \( f(x_1) = f(x_2) \) and show \( x_1 = x_2 \). To prove it's onto, for any \( y \) in the codomain, show an \( x \) in the domain exists such that \( f(x) = y \).

Question 2. Check the injectivity and surjectivity of the following functions:
(i) f : N → N is given by f(x) = x².
(ii) f : Z → Z is given by f(x) = x².
(iii) f : R → R is given by f(x) = x².
(iv) f : N → N is given by f(x) = x³.
(v) f : Z → Z is given by f(x) = x³.
Answer:
(i) \( f : N \rightarrow N \) given by \( f(x) = x^2 \).
(a) For injectivity: If \( f(x_1) = f(x_2) \)
\( \implies x_1^2 = x_2^2 \)
\( \implies x_1 = x_2 \) (since \( x_1, x_2 \in N \), negative values are not considered).
Therefore, \( f \) is one-one, which means it is injective.
(b) For surjectivity: There are some numbers in the codomain \( N \) that do not have any pre-image in the domain \( N \). For instance, 3 belongs to the codomain \( N \), but there is no integer \( x \) such that \( x^2 = 3 \) and \( x \in N \).
Thus, \( f \) is not onto, which means it is not surjective.

(ii) \( f : Z \rightarrow Z \) given by \( f(x) = x^2 \).
(a) For injectivity: Consider \( f(-1) = (-1)^2 = 1 \) and \( f(1) = (1)^2 = 1 \). Here, \( -1 \) and \( 1 \) have the same image, \( 1 \).
Therefore, \( f \) is not one-one, which means it is not injective.
(b) For surjectivity: Many elements belonging to the codomain \( Z \) do not have a pre-image in the domain \( Z \). For example, 3 belongs to \( Z \), but there is no integer \( x \) such that \( x^2 = 3 \).
Therefore, \( f \) is not onto, which means it is not surjective.

(iii) \( f : R \rightarrow R \) given by \( f(x) = x^2 \).
(a) For injectivity: Consider \( f(-1) = (-1)^2 = 1 \) and \( f(1) = (1)^2 = 1 \). Here, \( -1 \) and \( 1 \) have the same image.
Therefore, \( f \) is not one-one, which means it is not injective.
(b) For surjectivity: Negative numbers belonging to the codomain \( R \) do not have any pre-image in the domain \( R \). For example, \( -2 \in R \), but \( \sqrt{-2} \) does not belong to the domain \( R \).
Therefore, \( f \) is not onto, which means it is not surjective.

(iv) \( f : N \rightarrow N \) given by \( f(x) = x^3 \).
(a) For injectivity: If \( f(x_1) = f(x_2) \)
\( \implies x_1^3 = x_2^3 \)
\( \implies x_1 = x_2 \)
This means every \( x \in N \) has a unique image in its codomain.
Therefore, \( f \) is one-one, which means it is injective.
(b) For surjectivity: There are many members of the codomain \( N \) that do not have a pre-image in its domain \( N \). For instance, 2, 3, etc. have no pre-image in \( N \).
Therefore, \( f \) is not onto, which means it is not surjective.

(v) \( f : Z \rightarrow Z \) given by \( f(x) = x^3 \).
(a) For injectivity: Here, if \( f(x_1) = f(x_2) \)
\( \implies x_1^3 = x_2^3 \)
\( \implies x_1 = x_2 \)
Therefore, \( f \) is one-one, which means it is injective.
(b) For surjectivity: Many members of the codomain \( Z \) do not have any pre-image in its domain \( Z \). For example, 2 belonging to the codomain \( Z \) has no integer pre-image in \( Z \) (as \( \sqrt[3]{2} \) is not an integer).
Therefore, \( f \) is not surjective.
In simple words: We checked if different functions like \( x^2 \) and \( x^3 \) are "injective" (each input gives a unique output) and "surjective" (all possible outputs are hit). We found that \( x^2 \) is often not injective or surjective when dealing with integers or real numbers, but it can be injective for natural numbers. \( x^3 \) is generally more injective, but still not always surjective, depending on the number set it operates on.

Exam Tip: Remember to test both injectivity (one-one) and surjectivity (onto) independently. For injectivity, look for cases where different inputs give the same output. For surjectivity, look for elements in the codomain that have no pre-image in the domain.

Question 3. Prove that greatest integer function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Answer: The function is \( f : R \rightarrow R \) defined by \( f(x) = [x] \).
(a) To check if it is one-one: Consider \( f(1.2) = 1 \) and \( f(1.5) = 1 \). Here, different inputs (1.2 and 1.5) give the same output (1).
Therefore, \( f \) is not one-one.
(b) To check if it is onto: All the images of \( x \in R \) belonging to its domain have integers as the values in the codomain. However, no proper or improper fraction belonging to the codomain of \( f \) has any pre-image in its domain. For example, there is no \( x \) such that \( [x] = 0.5 \).
Therefore, \( f \) is not onto.
Hence, \( f \) is neither one-one nor onto.
In simple words: The greatest integer function, which rounds numbers down to the nearest whole number, isn't unique (e.g., 1.2 and 1.5 both round to 1). Also, it can't create all numbers, like decimals (e.g., no number rounds down to 0.5). So, it's neither one-to-one nor onto.

Exam Tip: For greatest integer functions, remember that many real numbers map to the same integer, making them not one-one. Also, their range only includes integers, so they are not onto for a codomain of all real numbers.

Question 4. Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither one-one nor onto, where |x| is x if x is positive and |x| is –x if x is negative.
Answer: The function is \( f : R \rightarrow R \) given by \( f(x) = |x| \).
(a) To check if it is one-one: Consider \( f(-1) = |-1| = 1 \) and \( f(1) = |1| = 1 \). Here, \( -1 \) and \( 1 \) have the same image.
Therefore, \( f \) is not one-one.
(b) To check if it is onto: No negative value belonging to the codomain of \( f \) has any pre-image in its domain. For example, there is no real number \( x \) such that \( |x| = -2 \). The modulus function always produces non-negative output values.
Therefore, \( f \) is not onto.
Hence, \( f \) is neither one-one nor onto.
In simple words: The modulus function, which gives the positive value of any number, isn't one-to-one because both a positive number and its negative counterpart give the same output (e.g., |1| and |-1| both equal 1). It's also not onto because it can never produce negative numbers as an output, even though negative numbers are part of the set it's supposed to map to.

Exam Tip: The modulus function is a classic example of a function that is not one-one due to symmetry (\( |x| = |-x| \)) and not onto for a real codomain because its range is limited to non-negative values.

Question 5. Show that the Signum function f : R → R, given by \( f(x) = \begin{cases} 1, & \text{if } x>0 \\ 0, & \text{if } x=0 \\ -1, & \text{if } x<0 \end{cases} \) is neither one-one nor onto.
Answer: The function is \( f : R \rightarrow R \), given by \( f(x) = \begin{cases} 1, & \text{if } x>0 \\ 0, & \text{if } x=0 \\ -1, & \text{if } x<0 \end{cases} \).
(a) To check if it is one-one: Consider \( f(1) = 1 \) and \( f(2) = 1 \). Here, \( 1 \) and \( 2 \) have the same image, \( 1 \). This means if \( f(x_1) = f(x_2) = 1 \) for \( x > 0 \), then \( x_1 \neq x_2 \) (e.g., \( f(3) = 1 \) and \( f(5) = 1 \), but \( 3 \neq 5 \)).
Similarly, if \( f(x_1) = f(x_2) = -1 \) for \( x < 0 \), then \( x_1 \neq x_2 \).
Therefore, \( f \) is not one-one.
(b) To check if it is onto: Except for \( -1, 0 \), and \( 1 \), no other members of the codomain of \( f \) have any pre-image in its domain. For example, there is no \( x \) such that \( f(x) = 0.5 \).
Therefore, \( f \) is not onto.
Thus, \( f \) is neither one-one nor onto.
In simple words: The Signum function, which outputs only -1, 0, or 1, is not one-to-one because many different positive numbers all give '1', and many different negative numbers all give '-1'. It's also not onto because its outputs are limited to just those three numbers, meaning it can't produce any other real number in its codomain.

Exam Tip: For step functions like the Signum function, multiple inputs often map to the same output (not one-one), and the range is usually a very small subset of the codomain (not onto).

Question 6. 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.
Answer: The given sets are \( A = \{1, 2, 3\} \) and \( B = \{4, 5, 6, 7\} \). The function is \( f = \{(1, 4), (2, 5), (3, 6)\} \).
In this function, every member of set \( A \) has a unique image in set \( B \), as shown in the figure.
Element 1 from A maps to 4 in B.
Element 2 from A maps to 5 in B.
Element 3 from A maps to 6 in B.
Since no two distinct elements in A map to the same element in B, the function \( f \) is one-one.
In simple words: For the given function, each number in set A (1, 2, 3) goes to a different number in set B (1 to 4, 2 to 5, 3 to 6). Because no two numbers from A share the same number in B, the function is one-to-one.

Exam Tip: A function is one-one (injective) if distinct elements in the domain always map to distinct elements in the codomain. Visually, this means no two arrows point to the same element.

Question 7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x.
(ii) f : R → R defined by f(x) = 1 + x².
Answer:
(i) \( f : R \rightarrow R \) defined by \( f(x) = 3 - 4x \).
(a) To check if it is one-one: Let \( f(x_1) = f(x_2) \).
\( \implies 3 - 4x_1 = 3 - 4x_2 \)
\( \implies -4x_1 = -4x_2 \)
\( \implies x_1 = x_2 \)
This shows that \( f \) is one-one.
(b) To check if it is onto: Let \( y = f(x) \). Then \( y = 3 - 4x \).
We want to find \( x \) in terms of \( y \):
\( 4x = 3 - y \)
\( x = \frac{3-y}{4} \)
For every value of \( y \) belonging to its codomain \( R \), there is a pre-image \( x \) in its domain \( R \).
Therefore, \( f \) is onto.
Since \( f \) is both one-one and onto, it is a bijective function.

(ii) \( f : R \rightarrow R \) defined by \( f(x) = 1 + x^2 \).
(a) To check if it is one-one: Consider \( f(1) = 1 + (1)^2 = 2 \) and \( f(-1) = 1 + (-1)^2 = 2 \).
Here, \( -1 \) and \( 1 \) have the same image, \( 2 \).
Therefore, \( f \) is not one-one.
(b) To check if it is onto: The minimum value of \( 1 + x^2 \) is \( 1 \) (when \( x = 0 \)). This means the range of \( f \) is \( [1, \infty) \).
No negative number belonging to its codomain \( R \) has a pre-image in its domain. For example, there is no real number \( x \) such that \( 1 + x^2 = 0 \).
Therefore, \( f \) is not onto.
Thus, \( f \) is neither one-one nor onto.
In simple words: For \( f(x) = 3 - 4x \), it's both one-to-one (different inputs always give different outputs) and onto (every possible output can be reached), making it bijective. For \( f(x) = 1 + x^2 \), it's neither one-to-one (e.g., \( f(1) \) and \( f(-1) \) both give 2) nor onto (it can't produce values less than 1).

Exam Tip: Linear functions of the form \( f(x) = ax + b \) (where \( a \neq 0 \)) are almost always bijective. Quadratic functions like \( f(x) = x^2 \) or \( 1+x^2 \) are generally neither one-one nor onto over real numbers due to symmetry and restricted range.

Question 8. Let A and B be two sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.
Answer: We have \( f : A \times B \rightarrow B \times A \) such that \( f(a, b) = (b, a) \).
(a) To check if it is one-one: Let \( f(a_1, b_1) = f(a_2, b_2) \).
\( \implies (b_1, a_1) = (b_2, a_2) \)
By equality of ordered pairs, we must have \( b_1 = b_2 \) and \( a_1 = a_2 \).
This means \( (a_1, b_1) = (a_2, b_2) \).
Therefore, \( f \) is one-one.
(b) To check if it is onto: Let \( (x, y) \) be an arbitrary element in the codomain \( B \times A \).
We need to find an element \( (a, b) \) in the domain \( A \times B \) such that \( f(a, b) = (x, y) \).
From the definition of \( f \), we have \( f(a, b) = (b, a) \).
So, we need \( (b, a) = (x, y) \).
This implies \( b = x \) and \( a = y \).
Since \( x \in B \) and \( y \in A \), the element \( (y, x) \) is in the domain \( A \times B \).
And \( f(y, x) = (x, y) \).
Therefore, for every element \( (x, y) \) in the codomain, there exists a pre-image \( (y, x) \) in the domain.
Thus, \( f \) is onto.
Since \( f \) is both one-one and onto, it is a bijective function.
In simple words: This function just swaps the order of elements in a pair, like turning (a, b) into (b, a). It's one-to-one because if two pairs map to the same swapped pair, their original pairs must also be identical. It's onto because any swapped pair you get can always be traced back to an original pair by just swapping them back. So, it's a bijective function.

Exam Tip: A function that swaps elements in ordered pairs is often bijective. Focus on demonstrating that \( (a_1, b_1) = (a_2, b_2) \) if \( f(a_1, b_1) = f(a_2, b_2) \) for one-one, and finding a pre-image \( (a, b) \) for any \( (y, x) \) in the codomain for onto.

Question 9. Let f: N → N be defined by \( f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases} \) State whether the function f is bijective. Justify your answer.
Answer: The function \( f : N \rightarrow N \) is defined by \( f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases} \).
(a) To check if it is one-one: Consider \( f(1) \). Since 1 is odd, \( f(1) = \frac{1+1}{2} = \frac{2}{2} = 1 \).
Consider \( f(2) \). Since 2 is even, \( f(2) = \frac{2}{2} = 1 \).
Here, the elements 1 and 2, which belong to the domain \( N \), have the same image, \( 1 \), in the codomain.
Therefore, \( f \) is not one-one (not injective).
(b) To check if it is onto: Let \( y \) be any natural number in the codomain \( N \). If \( y \in N \), we want to find \( n \in N \) such that \( f(n) = y \). Case 1: If \( y \) is the image of an odd number \( n \), then \( \frac{n+1}{2} = y \implies n+1 = 2y \implies n = 2y-1 \). Since \( y \in N \), \( 2y-1 \) will always be an odd natural number (and thus in \( N \)). For example, if \( y=1 \), \( n = 2(1)-1 = 1 \). Case 2: If \( y \) is the image of an even number \( n \), then \( \frac{n}{2} = y \implies n = 2y \). Since \( y \in N \), \( 2y \) will always be an even natural number (and thus in \( N \)). For example, if \( y=1 \), \( n = 2(1) = 2 \). So, every member \( y \) in the codomain has at least one pre-image in its domain. For instance, for \( y=1 \), both \( n=1 \) (odd) and \( n=2 \) (even) map to 1. For \( y=2 \), \( n=3 \) (odd, \( f(3)=2 \)) and \( n=4 \) (even, \( f(4)=2 \)).
Therefore, \( f \) is onto.
Since \( f \) is onto but not one-one, it is not a bijective function.
In simple words: This function takes a natural number, adds one and halves it if odd, or just halves it if even. It's not one-to-one because both 1 and 2 map to 1. However, it is onto because every natural number in the output can be reached from at least one number in the input. Since it's not both, it's not bijective.

Exam Tip: For functions defined piecewise, always check both parts of the definition for injectivity and surjectivity. Pay special attention to boundary cases or values that might map differently from what seems obvious.

Question 10. Let A = R-{3} and B = R-{1}. consider the function f: A → B defined by f (x) = \( \frac{x-2}{x-3} \). Is f one-one and onto? Justify your answer.
Answer: The sets are \( A = R-\{3\} \) and \( B = R-\{1\} \). The function is \( f : A \rightarrow B \) defined by \( f(x) = \frac{x-2}{x-3} \).
(a) To check if it is one-one: Let \( f(x_1) = f(x_2) \).
\( \implies \frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3} \)
\( \implies (x_1-2)(x_2-3) = (x_2-2)(x_1-3) \)
\( \implies x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 2x_1 - 3x_2 + 6 \)
\( \implies -3x_1 - 2x_2 = -2x_1 - 3x_2 \)
\( \implies -x_1 = -x_2 \)
\( \implies x_1 = x_2 \)
Therefore, \( f \) is one-one.
(b) To check if it is onto: Let \( y = f(x) \). Then \( y = \frac{x-2}{x-3} \).
We want to find \( x \) in terms of \( y \):
\( y(x-3) = x-2 \)
\( xy - 3y = x - 2 \)
\( xy - x = 3y - 2 \)
\( x(y-1) = 3y - 2 \)
\( x = \frac{3y-2}{y-1} \)
For every value of \( y \) in the codomain \( B = R-\{1\} \), \( y \neq 1 \), so the denominator \( y-1 \) is never zero. This means \( x \) is always a defined real number. We also need to ensure \( x \neq 3 \).
If \( x = 3 \), then \( 3 = \frac{3y-2}{y-1} \)
\( \implies 3(y-1) = 3y-2 \)
\( \implies 3y-3 = 3y-2 \)
\( \implies -3 = -2 \), which is a contradiction.
So, \( x \) will never be equal to 3. Therefore, for every value of \( y \) in the codomain \( B \), there is a pre-image \( x \) in the domain \( A \).
Thus, \( f \) is onto.
Since \( f \) is both one-one and onto, it is a bijective function.
In simple words: This function maps from all real numbers except 3 to all real numbers except 1. It is one-to-one because if two different inputs gave the same output, a contradiction would arise. It is onto because for any possible output, we can always find an input that produces it, and that input will never be 3. Thus, the function is bijective.

Exam Tip: When dealing with rational functions like \( f(x) = \frac{ax+b}{cx+d} \), to check for onto, first solve for \( x \) in terms of \( y \). Then, ensure that for every \( y \) in the codomain, a valid \( x \) in the domain exists. Also, confirm that the \( x \) value obtained never equals any excluded values from the domain.

Question 11. Let f: R → R be defined as f(x) = x⁴. Choose the correct answer:
(a) f is one-one onto.
(b) f is many-one onto.
(c) f is one-one but not onto.
(d) f is neither one-one nor onto.
Answer: (d) f is neither one-one nor onto.
In simple words: The function \( f(x) = x^4 \) is not one-to-one because different inputs like 1 and -1 both give the same output (1). It's also not onto because it can never produce negative numbers as outputs, even though the output set includes all real numbers.

Exam Tip: Functions with even powers (like \( x^2, x^4, x^6 \)) are typically not one-one because \( f(x) = f(-x) \). Their range is restricted to non-negative values, so they are not onto for a codomain of all real numbers.

Question 12. Let f: R → R be defined as f(x) = 3x. Choose the correct answer:
(a) f is one-one onto.
(b) f is many-one onto.
(c) f is one-one but not onto.
(d) f is neither one-one nor onto.
Answer: (a) f is one-one onto.
In simple words: The function \( f(x) = 3x \) is one-to-one because each input gives a unique output. It is also onto because any real number can be an output, meaning you can always find an input that maps to it. Thus, it's both one-to-one and onto.

Exam Tip: Linear functions of the form \( f(x) = ax + b \), where \( a \neq 0 \), are always bijective (both one-one and onto) when the domain and codomain are real numbers.