CBSE Class 12 Mathematics HOTs Relations and Functions Set 02

Types of Relations

Question. If \( R = \{(x, y) : x + 2y = 8\} \) is a relation on \( \mathbb{N} \), write the range of \( R \).
Answer: Here, \( R = \{(x, y) : x + 2y = 8\} \), where \( x, y \in \mathbb{N} \).
For \( x = 1, 3, 5, \dots \), \( x + 2y = 8 \) has no solution in \( \mathbb{N} \).
For \( x = 2 \), we have \( 2 + 2y = 8 \implies y = 3 \).
For \( x = 4 \), we have \( 4 + 2y = 8 \implies y = 2 \).
For \( x = 6 \), we have \( 6 + 2y = 8 \implies y = 1 \).
For \( x = 8, 10, \dots \), \( x + 2y = 8 \) has no solution in \( \mathbb{N} \).
Therefore, Range of \( R = \{y : (x, y) \in R\} = \{1, 2, 3\} \).

Question. Let \( R = \{(a, a^3) : a \text{ is a prime number less than } 5\} \) be a relation. Find the range of \( R \).
Answer: Given relation is \( R = \{(a, a^3) : a \text{ is a prime number less than } 5\} \).
The prime numbers less than 5 are 2 and 3.
Therefore, \( R = \{(2, 2^3), (3, 3^3)\} = \{(2, 8), (3, 27)\} \).
So, the range of \( R \) is \( \{8, 27\} \).

Question. Let \( R \) be the equivalence relation in the set \( A = \{0, 1, 2, 3, 4, 5\} \) given by \( R = \{(a, b) : 2 \text{ divides } (a - b)\} \). Write the equivalence class \( [0] \).
Answer: Here, \( R = \{(a, b) \in A \times A : 2 \text{ divides } (a - b)\} \) is the given equivalence relation, where \( A = \{0, 1, 2, 3, 4, 5\} \).
The equivalence class \( [0] \) is given by:
\( [0] = \{x \in A : (x, 0) \in R\} \)
\( \implies 2 \text{ divides } (x - 0) \implies x \text{ is even} \).
Therefore, \( [0] = \{0, 2, 4\} \).

Question. State the reason for the relation \( R \) in the set \( \{1, 2, 3\} \) given by \( R = \{(1, 2), (2, 1)\} \) not to be transitive.
Answer: For a relation \( R \) to be transitive, if \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \) must belong to \( R \).
We have, \( R = \{(1, 2), (2, 1)\} \).
Here, \( (1, 2) \in R \) and \( (2, 1) \in R \), but \( (1, 1) \notin R \).
Therefore, \( R \) is not transitive.

Question. Let \( A = \{1, 2, 3, \dots, 9\} \) and \( R \) be the relation in \( A \times A \) defined by \( (a, b)R(c, d) \) if \( a + d = b + c \) for \( (a, b), (c, d) \) in \( A \times A \). Prove that \( R \) is an equivalence relation. Also obtain the equivalence class \( [(2, 5)] \).
Answer: Given \( A = \{1, 2, 3, \dots, 9\} \subset \mathbb{N} \).
To show that \( R \) is an equivalence relation, we must prove that it is reflexive, symmetric, and transitive:
(i) Reflexivity: Let \( (a, b) \in A \times A \). Then \( a, b \in A \).
Since \( a + b = b + a \) (by commutativity of addition on \( \mathbb{N} \)), we have \( (a, b)R(a, b) \).
Thus, \( R \) is reflexive.
(ii) Symmetry: Let \( (a, b), (c, d) \in A \times A \) such that \( (a, b)R(c, d) \).
\( \implies a + d = b + c \implies b + c = a + d \implies c + b = d + a \) (by commutativity of addition on \( \mathbb{N} \))
\( \implies (c, d)R(a, b) \).
Thus, \( R \) is symmetric.
(iii) Transitivity: Let \( (a, b), (c, d), (e, f) \in A \times A \) such that \( (a, b)R(c, d) \) and \( (c, d)R(e, f) \).
Now, \( (a, b)R(c, d) \implies a + d = b + c \) ...(i)
and \( (c, d)R(e, f) \implies c + f = d + e \) ...(ii)
Adding (i) and (ii), we get:
\( (a + d) + (c + f) = (b + c) + (d + e) \implies a + f = b + e \implies (a, b)R(e, f) \).
Thus, \( R \) is transitive.
Since \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.
The equivalence class \( [(2, 5)] \) is given by:
\( [(2, 5)] = \{(x, y) \in A \times A : (x, y)R(2, 5)\} \)
\( \implies x + 5 = y + 2 \implies y = x + 3 \).
For \( x = 1 \implies y = 4 \)
For \( x = 2 \implies y = 5 \)
For \( x = 3 \implies y = 6 \)
For \( x = 4 \implies y = 7 \)
For \( x = 5 \implies y = 8 \)
For \( x = 6 \implies y = 9 \).
Thus, \( [(2, 5)] = \{(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)\} \).

Question. Let \( R \) be a relation defined on the set of natural numbers \( \mathbb{N} \) as follow : \( R = \{(x, y) \mid x \in \mathbb{N}, y \in \mathbb{N} \text{ and } 2x + y = 24\} \). Find the domain and range of the relation \( R \). Also, find if \( R \) is an equivalence relation or not.
Answer: We have, \( R = \{(x, y) \mid x \in \mathbb{N}, y \in \mathbb{N} \text{ and } 2x + y = 24\} \).
Since \( 2x + y = 24 \implies y = 24 - 2x \). For \( y \in \mathbb{N} \), we must have \( 24 - 2x \ge 1 \implies 2x \le 23 \implies x \le 11.5 \).
Since \( x \in \mathbb{N} \), the possible values of \( x \) are \( \{1, 2, 3, \dots, 11\} \).
Therefore, the domain of \( R = \{1, 2, 3, 4, \dots, 11\} \).
The corresponding values of \( y \) are \( \{22, 20, 18, \dots, 2\} \).
Therefore, the range of \( R = \{2, 4, 6, 8, 10, 12, \dots, 22\} \).
To check if \( R \) is reflexive: for \( (2, 2) \in R \), we must have \( 2(2) + 2 = 6 \neq 24 \), so \( (2, 2) \notin R \). Hence, \( R \) is not reflexive, which means it is not an equivalence relation.

Question. Show that the relation \( S \) in the set \( \mathbb{R} \) of real numbers defined as \( S = \{(a, b) : a, b \in \mathbb{R} \text{ and } a \le b^3\} \) is neither reflexive, nor symmetric, nor transitive.
Answer: We have \( S = \{(a, b) : a \le b^3\} \) where \( a, b \in \mathbb{R} \).
(i) Reflexivity: For \( a = \frac{1}{2} \), we have \( \left(\frac{1}{2}\right)^3 = \frac{1}{8} \). Since \( \frac{1}{2} > \frac{1}{8} \), \( \frac{1}{2} \le \left(\frac{1}{2}\right)^3 \) is false. Thus, \( \left(\frac{1}{2}, \frac{1}{2}\right) \notin S \). So, \( S \) is not reflexive.
(ii) Symmetry: We observe that \( 1 \le 2^3 \implies (1, 2) \in S \). However, \( 2 \le 1^3 \) is false, so \( (2, 1) \notin S \). Thus, \( S \) is not symmetric.
(iii) Transitivity: We observe that \( 10 \le 3^3 = 27 \implies (10, 3) \in S \), and \( 3 \le 2^3 = 8 \implies (3, 2) \in S \). However, \( 10 \le 2^3 = 8 \) is false, so \( (10, 2) \notin S \). Thus, \( S \) is not transitive.
Hence, \( S \) is neither reflexive, nor symmetric, nor transitive.

Question. Let \( \mathbb{Z} \) be the set of all integers and \( R \) be the relation on \( \mathbb{Z} \) defined as \( R = \{(a, b) : a, b \in \mathbb{Z} \text{ and } (a - b) \text{ is divisible by } 5\} \). Prove that \( R \) is an equivalence relation.
Answer: We have \( R = \{(a, b) : (a - b) \text{ is divisible by } 5\} \).
(i) Reflexive: For any \( a \in \mathbb{Z} \), \( a - a = 0 \), which is divisible by 5. Thus, \( (a, a) \in R \). Hence, \( R \) is reflexive.
(ii) Symmetric: For any \( a, b \in \mathbb{Z} \), let \( (a, b) \in R \). Then \( a - b \) is divisible by 5 \( \implies a - b = 5m \) for some \( m \in \mathbb{Z} \).
This implies \( b - a = -5m = 5(-m) \), where \( -m \in \mathbb{Z} \). Thus, \( b - a \) is divisible by 5 \( \implies (b, a) \in R \). Hence, \( R \) is symmetric.
(iii) Transitive: For any \( a, b, c \in \mathbb{Z} \), let \( (a, b) \in R \) and \( (b, c) \in R \).
Then \( a - b = 5m \) and \( b - c = 5n \) for some \( m, n \in \mathbb{Z} \).
Adding these, we get \( a - c = (a - b) + (b - c) = 5(m + n) \). Since \( m + n \in \mathbb{Z} \), \( a - c \) is divisible by 5 \( \implies (a, c) \in R \). Hence, \( R \) is transitive.
Since \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation on \( \mathbb{Z} \).

Question. Show that the relation \( S \) in the set \( A = \{x \in \mathbb{Z} : 0 \le x \le 12\} \) given by \( S = \{(a, b) : a, b \in A, |a - b| \text{ is divisible by } 4\} \) is an equivalence relation. Find the set of all elements related to 1.
Answer: We have \( A = \{0, 1, 2, 3, \dots, 12\} \) and \( S = \{(a, b) : |a - b| \text{ is divisible by } 4\} \).
(i) Reflexive: For any \( a \in A \), \( |a - a| = 0 \), which is divisible by 4. Thus, \( (a, a) \in S \). So, \( S \) is reflexive.
(ii) Symmetric: For any \( a, b \in A \), let \( (a, b) \in S \implies |a - b| \text{ is divisible by } 4 \implies |b - a| \text{ is divisible by } 4 \implies (b, a) \in S \). So, \( S \) is symmetric.
(iii) Transitive: Let \( (a, b) \in S \) and \( (b, c) \in S \). Then \( a - b = \pm 4k_1 \) and \( b - c = \pm 4k_2 \) for some \( k_1, k_2 \in \mathbb{N} \cup \{0\} \).
Adding these, we get \( a - c = \pm 4(k_1 + k_2) \implies |a - c| \text{ is divisible by } 4 \implies (a, c) \in S \). So, \( S \) is transitive.
Hence, \( S \) is an equivalence relation.
The set of all elements related to 1 is given by:
\( \{x \in A : (x, 1) \in S\} = \{x \in A : |x - 1| \text{ is divisible by } 4\} = \{1, 5, 9\} \).

Question. Show that the relation \( R \) defined by \( (a, b)R(c, d) \implies a + d = b + c \) on the set \( \mathbb{N} \times \mathbb{N} \) is an equivalence relation.
Answer: Refer to the proof in Types of Relations question involving \( A \times A \), substituting \( A \times A \) with \( \mathbb{N} \times \mathbb{N} \).

Question. Let \( f : X \to Y \) be a function, define a relation \( R \) on \( X \) given by \( R = \{(a, b) : f(a) = f(b)\} \). Show that \( R \) is an equivalence relation on \( X \).
Answer: We have \( R = \{(a, b) : f(a) = f(b)\} \).
(i) Reflexivity: For any \( a \in X \), we have \( f(a) = f(a) \implies (a, a) \in R \). So, \( R \) is reflexive.
(ii) Symmetry: For any \( a, b \in X \), let \( (a, b) \in R \implies f(a) = f(b) \implies f(b) = f(a) \implies (b, a) \in R \). So, \( R \) is symmetric.
(iii) Transitivity: For any \( a, b, c \in X \), let \( (a, b) \in R \) and \( (b, c) \in R \implies f(a) = f(b) \) and \( f(b) = f(c) \implies f(a) = f(c) \implies (a, c) \in R \). So, \( R \) is transitive.
Hence, \( R \) is an equivalence relation on \( X \).

Question. Prove that the relation \( R \) in the set \( A = \{1, 2, 3, 4, 5\} \) given by \( R = \{(a, b) : |a - b| \text{ is even}\} \), is an equivalence relation.
Answer: We have \( A = \{1, 2, 3, 4, 5\} \) and \( R = \{(a, b) : |a - b| \text{ is even}\} \).
(i) Reflexive: For any \( a \in A \), \( |a - a| = 0 \), which is even. Thus, \( (a, a) \in R \). So, \( R \) is reflexive.
(ii) Symmetric: Let \( (a, b) \in R \implies |a - b| \text{ is even} \implies |b - a| \text{ is even} \implies (b, a) \in R \). So, \( R \) is symmetric.
(iii) Transitive: Let \( (a, b) \in R \) and \( (b, c) \in R \implies a - b = \pm 2k_1 \) and \( b - c = \pm 2k_2 \) for some \( k_1, k_2 \in \mathbb{N} \cup \{0\} \).
Adding these, we get \( a - c = \pm 2(k_1 + k_2) \implies |a - c| \text{ is even} \implies (a, c) \in R \). So, \( R \) is transitive.
Hence, \( R \) is an equivalence relation.

Question. Check whether the relation \( R \) defined in the set \( \{1, 2, 3, 4, 5, 6\} \) as \( R = \{(a, b) : b = a + 1\} \) is reflexive, symmetric or transitive.
Answer: The set is \( A = \{1, 2, 3, 4, 5, 6\} \) and \( R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\} \).
(i) Reflexivity: Since \( (a, a) \notin R \) for all \( a \in A \) (e.g., \( (1, 1) \notin R \)), \( R \) is not reflexive.
(ii) Symmetry: Since \( (1, 2) \in R \) but \( (2, 1) \notin R \), \( R \) is not symmetric.
(iii) Transitivity: Since \( (1, 2) \in R \) and \( (2, 3) \in R \), but \( (1, 3) \notin R \), \( R \) is not transitive.

LA (6 marks)

Question. Let \( \mathbb{N} \) denote the set of all natural numbers and \( R \) be the relation on \( \mathbb{N} \times \mathbb{N} \) defined by \( (a, b)R(c, d) \) if \( ad(b + c) = bc(a + d) \). Show that \( R \) is an equivalence relation.
Answer: (i) Reflexivity: Let \( (a, b) \in \mathbb{N} \times \mathbb{N} \). Since \( ab(b + a) = ba(a + b) \) (by commutativity of addition and multiplication on \( \mathbb{N} \")), we have \( (a, b)R(a, b) \). So, \( R \) is reflexive.
(ii) Symmetry: Let \( (a, b), (c, d) \in \mathbb{N} \times \mathbb{N} \) such that \( (a, b)R(c, d) \).
\( \implies ad(b + c) = bc(a + d) \implies cb(d + a) = da(c + b) \) (by commutativity of addition and multiplication on \( \mathbb{N} \))
\( \implies (c, d)R(a, b) \). So, \( R \) is symmetric.
(iii) Transitivity: Let \( (a, b), (c, d), (e, f) \in \mathbb{N} \times \mathbb{N} \) such that \( (a, b)R(c, d) \) and \( (c, d)R(e, f) \).
Now, \( (a, b)R(c, d) \implies ad(b + c) = bc(a + d) \implies \frac{b+c}{bc} = \frac{a+d}{ad} \implies \frac{1}{c} + \frac{1}{b} = \frac{1}{d} + \frac{1}{a} \) ...(i)
and \( (c, d)R(e, f) \implies cf(d + e) = de(c + f) \implies \frac{d+e}{de} = \frac{c+f}{cf} \implies \frac{1}{e} + \frac{1}{d} = \frac{1}{f} + \frac{1}{c} \) ...(ii)
Adding (i) and (ii), we get:
\( \frac{1}{b} + \frac{1}{e} = \frac{1}{a} + \frac{1}{f} \implies \frac{b+e}{be} = \frac{a+f}{af} \implies af(b + e) = be(a + f) \implies (a, b)R(e, f) \).
So, \( R \) is transitive.
Hence, \( R \) is an equivalence relation on \( \mathbb{N} \times \mathbb{N} \).

Question. Show that the relation \( R \) in the set \( A = \{1, 2, 3, 4, 5\} \) given by \( R = \{(a, b) : |a - b| \text{ is divisible by } 2\} \) is an equivalence relation. Write all the equivalence classes of \( R \).
Answer: Refer to Question 12 for the proof that \( R \) is an equivalence relation.
The equivalence classes of \( R \) are:
\( [1] = \{x \in A : |x - 1| \text{ is divisible by } 2\} = \{1, 3, 5\} \)
\( [2] = \{x \in A : |x - 2| \text{ is divisible by } 2\} = \{2, 4\} \).
Since any element of \( A \) is either even or odd, there are only these two equivalence classes: \( [1] = \{1, 3, 5\} \) and \( [2] = \{2, 4\} \).

Types of Functions

Question. 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 \( B \), state whether \( f \) is one-one or not.
Answer: Here, \( f(1) = 4 \), \( f(2) = 5 \), and \( f(3) = 6 \). Since distinct elements in the domain \( A \) have distinct images in the co-domain \( B \), \( f \) is a one-one function.

Question. What is the range of the function \( f(x) = \frac{|x - 1|}{x - 1} \)?
Answer: We have, \( |x - 1| = \begin{cases} x - 1, & x \ge 1 \\ -(x - 1), & x < 1 \end{cases} \).
Therefore, for \( x \neq 1 \):
\[ f(x) = \frac{|x - 1|}{x - 1} = \begin{cases} 1, & x > 1 \\ -1, & x < 1 \end{cases} \]
Thus, the range of \( f \) is \( \{-1, 1\} \).

Question. State whether the function \( f : \mathbb{N} \to \mathbb{N} \) given by \( f(x) = 5x \) is injective, surjective or both.
Answer: (i) Injectivity: Let \( f(x_1) = f(x_2) \implies 5x_1 = 5x_2 \implies x_1 = x_2 \) for any \( x_1, x_2 \in \mathbb{N} \). Thus, \( f \) is injective.
(ii) Surjectivity: For \( 2 \in \mathbb{N} \) (co-domain), there does not exist any \( x \in \mathbb{N} \) (domain) such that \( f(x) = 5x = 2 \) because \( x = 2/5 \notin \mathbb{N} \). Thus, \( f \) is not surjective.
Therefore, the function is injective but not surjective.

SA (4 marks)

Question. Show that \( f : \mathbb{N} \to \mathbb{N} \), given by \( f(x) = \begin{cases} x + 1, & \text{if } x \text{ is odd} \\ x - 1, & \text{if } x \text{ is even} \end{cases} \) is both one-one and onto.
Answer: Let \( x, y \in \mathbb{N} \) such that \( f(x) = f(y) \).
(i) If \( x \) and \( y \) are both even: \( f(x) = f(y) \implies x - 1 = y - 1 \implies x = y \).
(ii) If \( x \) and \( y \) are both odd: \( f(x) = f(y) \implies x + 1 = y + 1 \implies x = y \).
(iii) If \( x \) is odd and \( y \) is even: \( f(x) = f(y) \implies x + 1 = y - 1 \implies y - x = 2 \), which is impossible because the difference between an even and an odd number must be odd. Hence, this case does not arise.
In any case, \( f(x) = f(y) \implies x = y \), so \( f \) is one-one.

For any \( y \in \mathbb{N} \) (co-domain):
If \( y \) is odd, then \( y + 1 \) is even, and \( f(y + 1) = (y + 1) - 1 = y \).
If \( y \) is even, then \( y - 1 \) is odd, and \( f(y - 1) = (y - 1) + 1 = y \).
Thus, every element of the co-domain has a pre-image in the domain, so \( f \) is onto.
Hence, \( f \) is both one-one and onto.

Question. Let \( f : \mathbb{N} \to \mathbb{N} \) be defined by \( f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases} \) for all \( n \in \mathbb{N} \). Find whether the function \( f \) is bijective.
Answer: Let's check injectivity. We have \( f(1) = \frac{1+1}{2} = 1 \) and \( f(2) = \frac{2}{2} = 1 \).
Since \( f(1) = f(2) = 1 \) but \( 1 \neq 2 \), the function \( f \) is not one-one.
Therefore, \( f \) is not bijective.

Question. If \( f : \mathbb{R} \to \mathbb{R} \) be the function defined by \( f(x) = 4x^3 + 7 \), show that \( f \) is a bijection.
Answer: (i) One-one: Let \( x_1, x_2 \in \mathbb{R} \) such that \( f(x_1) = f(x_2) \implies 4x_1^3 + 7 = 4x_2^3 + 7 \implies x_1^3 = x_2^3 \implies x_1 = x_2 \) (since the cube root of a real number is unique). Thus, \( f \) is one-one.
(ii) Onto: Let \( y \in \mathbb{R} \). We find \( x \in \mathbb{R} \) such that \( f(x) = y \implies 4x^3 + 7 = y \implies x = \left(\frac{y-7}{4}\right)^{1/3} \in \mathbb{R} \). Since this \( x \) is a real number for any real \( y \), \( f \) is onto.
Hence, \( f \) is bijective.

Question. Show that the function \( f : W \to W \) defined by \( f(n) = \begin{cases} n + 1, & \text{if } n \text{ is even} \\ n - 1, & \text{if } n \text{ is odd} \end{cases} \) is a bijective function.
Answer: Refer to Question 19. The proof is identical with the domain \( W \) of whole numbers (including 0, which is even).

Question. Show that the function \( f : \mathbb{R} \to \mathbb{R} \) given by \( f(x) = ax + b \), where \( a, b \in \mathbb{R}, a \neq 0 \) is a bijective function.
Answer: (i) Injectivity: Let \( x_1, x_2 \in \mathbb{R} \) such that \( f(x_1) = f(x_2) \implies ax_1 + b = ax_2 + b \implies ax_1 = ax_2 \implies x_1 = x_2 \) (as \( a \neq 0 \)). Thus, \( f \) is one-one.
(ii) Surjectivity: Let \( y \in \mathbb{R} \). We set \( f(x) = y \implies ax + b = y \implies x = \frac{y-b}{a} \). Since \( a \neq 0 \), \( x \in \mathbb{R} \) exists for all \( y \in \mathbb{R} \). Thus, \( f \) is onto.
Hence, \( f \) is a bijective function.

Composition of Functions and Invertible Function

Question. Let \( f : \{1, 3, 4\} \to \{1, 2, 5\} \) and \( g : \{1, 2, 5\} \to \{1, 3\} \) given by \( f = \{(1, 2), (3, 5), (4, 1)\} \) and \( g = \{(1, 3), (2, 3), (5, 1)\} \). Write down \( g \circ f \).
Answer: We have:
\( (g \circ f)(1) = g(f(1)) = g(2) = 3 \)
\( (g \circ f)(3) = g(f(3)) = g(5) = 1 \)
\( (g \circ f)(4) = g(f(4)) = g(1) = 3 \)
Therefore, \( g \circ f = \{(1, 3), (3, 1), (4, 3)\} \).

Question. If \( f : \mathbb{R} \to \mathbb{R} \) defined as \( f(x) = \frac{2x - 7}{4} \) is an invertible function, write \( f^{-1}(x) \).
Answer: Let \( y = f(x) = \frac{2x - 7}{4} \implies 4y = 2x - 7 \implies x = \frac{4y + 7}{2} \).
Since \( f \) is invertible, \( f^{-1}(y) = \frac{4y + 7}{2} \).
Hence, \( f^{-1}(x) = \frac{4x + 7}{2} \).

Question. If \( f : \mathbb{R} \to \mathbb{R} \) is defined by \( f(x) = (3 - x^3)^{1/3} \), then find \( f \circ f(x) \).
Answer: We have, \( f(x) = (3 - x^3)^{1/3} \).
\[ (f \circ f)(x) = f(f(x)) = f\left((3 - x^3)^{1/3}\right) = \left[3 - \left((3 - x^3)^{1/3}\right)^3\right]^{1/3} \]
\[ = \left[3 - (3 - x^3)\right]^{1/3} = (x^3)^{1/3} = x \].

Question. If \( f : \mathbb{R} \to \mathbb{R} \) is defined by \( f(x) = 3x + 2 \), find \( f(f(x)) \.
Answer: \( f(f(x)) = f(3x + 2) = 3(3x + 2) + 2 = 9x + 6 + 2 = 9x + 8 \).

Question. If the function \( f : \mathbb{R} \to \mathbb{R} \), defined by \( f(x) = 3x - 4 \), is invertible, find \( f^{-1} \).
Answer: Let \( y = f(x) = 3x - 4 \implies y + 4 = 3x \implies x = \frac{y + 4}{3} \).
Since \( f \) is invertible, \( f^{-1}(y) = \frac{y + 4}{3} \).
Hence, \( f^{-1}(x) = \frac{x + 4}{3} \).

Question. If \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = \frac{3x + 5}{2} \) is an invertible function, find \( f^{-1} \).
Answer: Let \( y = f(x) = \frac{3x + 5}{2} \implies 2y = 3x + 5 \implies x = \frac{2y - 5}{3} \).
Since \( f \) is invertible, \( f^{-1}(y) = \frac{2y - 5}{3} \).
Hence, \( f^{-1}(x) = \frac{2x - 5}{3} \).

Question. If \( f(x) = x + 7 \) and \( g(x) = x - 7 \), \( x \in \mathbb{R} \), find \( (f \circ g)(7) \).
Answer: We have, \( (f \circ g)(x) = f(g(x)) = f(x - 7) = (x - 7) + 7 = x \).
Therefore, \( (f \circ g)(7) = 7 \).

Question. If \( f : \mathbb{R} \to \mathbb{R} \) is defined by \( f(x) = x^2 - 3x + 2 \), find \( f(f(x)) \.
Answer: \( f(f(x)) = f(x^2 - 3x + 2) = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2 \).
Expanding: \( (x^2 - 3x + 2)^2 = x^4 + 9x^2 + 4 - 6x^3 + 4x^2 - 12x \)
Adding terms:
\( f(f(x)) = (x^4 - 6x^3 + 13x^2 - 12x + 4) - 3x^2 + 9x - 6 + 2 = x^4 - 6x^3 + 10x^2 - 3x \).