CBSE Class 12 Mathematics Relations And Functions Worksheet Set 06

Question. Let \( R \) be a relation in the set \( \mathbb{N} \) given by \( R = \{(a,b) : a = b - 2, b > 6\} \). Then,
(a) \( (8,7) \in R \)
(b) \( (6,8) \in R \)
(c) \( (3,8) \in R \)
(d) \( (2,4) \in R \)
Answer: (b) \( (6,8) \in R \)

Question. Let \( R \) be a relation from \( \mathbb{R} \) to \( \mathbb{R} \), the set of real numbers defined by \( R = \{(x,y) : x,y \in \mathbb{R} \text{ and } x - y + \sqrt{3} \text{ is an irrational number}\} \). Then, \( R \) is
(a) reflexive
(b) transitive
(c) symmetric
(d) an equivalence relation
Answer: (a) reflexive

Question. Let \( A = \{3,5\} \). Then, number of reflexive relations on \( A \) is
(a) 2
(b) 4
(c) 0
(d) 8
Answer: (b) 4

Question. The maximum number of equivalence relations on the set \( A = \{1, 2, 3\} \) are
(a) 1
(b) 2
(c) 3
(d) 5
Answer: (d) 5

Question. A relation \( R \) in set \( A = \{1, 2, 3\} \) is defined as \( R = \{(1,1), (1,2), (2,2), (3,3)\} \). Which of the following ordered pair in \( R \) shall be removed to make it an equivalence relation in \( A \)?
(a) \( (1,1) \)
(b) \( (1,2) \)
(c) \( (2,2) \)
(d) \( (3,3) \)
Answer: (b) \( (1,2) \)

Question. Let the relation \( R \) in the set \( A = \{x \in \mathbb{Z} : 0 \le x \le 12\} \), given by \( R = \{(a,b) : |a - b| \text{ is a multiple of } 4\} \). Then \( [1] \), the equivalence class containing 1, is
(a) \( \{1, 5, 9\} \)
(b) \( \{0, 1, 2, 5\} \)
(c) \( \phi \)
(d) \( A \)
Answer: (a) \( \{1, 5, 9\} \)

Question. Let \( f : \mathbb{R} \to \mathbb{R} \) be defined as \( f(x) = x^4 \). Then, the correct option is
(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

Question. A function \( f : \mathbb{R} \to \mathbb{R} \) defined as \( f(x) = x^2 - 4x + 5 \) is
(a) injective but not surjective
(b) surjective but not injective
(c) both injective and surjective
(d) neither injective nor surjective
Answer: (d) neither injective nor surjective

Question. The function \( f : \mathbb{R} \to \mathbb{R} \) defined as \( f(x) = x^3 \) is
(a) one-one but not onto
(b) not one-one but onto
(c) neither one-one nor onto
(d) both one-one and onto
Answer: (d) both one-one and onto

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 \( B \). Based on the given information, \( f \) is best defined as
(a) surjective function
(b) injective function
(c) bijective function
(d) None of the options
Answer: (b) injective function

Assertion-Reason Based Questions

Question. Assertion (A) The relation \( R \) in the set \( A = \{1, 2, 3, 4, 5, 6\} \) defined as \( R = \{(x,y) : y \text{ is divisible by } x\} \) is not an equivalence relation.
Reason (R) The relation \( R \) will be an equivalence relation, if it is reflexive, symmetric and transitive.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A) The relation \( R \) on the set \( A = \{1, 2, 3, 4, 5\} \) given by \( R = \{(a, b) : |a - b| \text{ is multiple of } 4\} \) is an equivalence relation.
Reason (R) A relation \( R \) on set \( A \) is said to be equivalence if it is reflexive, symmetric and transitive.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A) The function \( f : \mathbb{R} \to \mathbb{R} \) given by \( f(x) = x^3 \) is injective.
Reason (R) The function \( f : X \to Y \) is injective, if \( f(x) = f(y) \implies x = y \) for all \( x, y \in X \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A) The relation \( f : \{1, 2, 3, 4\} \to \{x, y, z, p\} \) defined by \( f = \{(1, x), (2, y), (3, z)\} \) is a bijective function.
Reason (R) The function \( f : \{1, 2, 3\} \to \{x, y, z, p\} \) such that \( f = \{(1, x), (2, y), (3, z)\} \) is one-one.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (d) R is correct; A is incorrect.

Question. Assertion (A) A function \( f : \mathbb{R} - \{0\} \to \mathbb{R} - \{0\} \) defined by \( f(x) = \frac{1}{x} \) is one-one and onto.
Reason (R) A function \( f : \mathbb{N} \to \mathbb{R} - \{0\} \) defined by \( f(x) = \frac{1}{x} \) is one-one and onto.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (c) A is correct; R is incorrect.

Question. Assertion (A) If \( f : X \to X \) is onto, then \( f \) is one-one and if \( f \) is one-one, then \( f \) is onto, where \( X \) is a finite set.
Reason (R) Every one-one function is always onto and every onto function is always one-one.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (c) A is correct; R is incorrect.

Case Study Based Questions - I

In a school, madam Sunita take lecture in Class XII. In this class, students divided into two groups \( A \) and \( B \). Group \( A \) has 20 students and group \( B \) has 25 students. The relation between students of these groups are defined as \( R : \text{group } A \to \text{group } B \) or \( R : A \to B \), \( R = \{(x,y) : x \in A \text{ and } y \in B\} \).

Question. Find the number of elements in codomain of \( R : A \to B \).
Answer: Group \( B \) has 25 students, so the codomain of \( R : A \to B \) is \( B \). Thus, the number of elements in the codomain is 25.

Question. Find the type of relation of the set of all identity elements.
Answer: The set of all identity elements is an equivalence relation.

Question. Find the total number of relations that can be formed from set \( A \) to \( B \).
Answer: Total number of relations \( = 2^{mn} = 2^{20 \times 25} = 2^{500} \).

Question. Suppose the relation defined from set \( A \) to set \( B \) is \( R_1 = \{(a_1, b_1), (a_2, b_3), (a_3, b_4)\} \), then find the inverse relation.
Answer: The inverse relation is \( R_1^{-1} = \{(b_1, a_1), (b_3, a_2), (b_4, a_3)\} \).

Case Study Based Questions - II

Students of a school are taken to a railway museum to learn about railways heritage and its history. An exhibit in the museum depicted many rail lines on the track near the railway station. Let \( L \) be the set of all rail lines on the railway track and \( R \) be the relation on \( L \) defined by \( R = \{(l_1, l_2) : l_1 \text{ is parallel to } l_2\} \).

Question. Find whether the relation \( R \) is symmetric or not.
Answer: Yes, \( R \) is symmetric because if line \( l_1 \) is parallel to \( l_2 \), then \( l_2 \) is parallel to \( l_1 \). Thus, \( (l_1, l_2) \in R \implies (l_2, l_1) \in R \).

Question. Find whether the relation \( R \) is transitive or not.
Answer: Yes, \( R \) is transitive. If \( l_1 \) is parallel to \( l_2 \) and \( l_2 \) is parallel to \( l_3 \), then \( l_1 \) is parallel to \( l_3 \). Thus, \( (l_1, l_2) \in R \) and \( (l_2, l_3) \in R \implies (l_1, l_3) \in R \).

Question. If one of the rail lines on the railway track is represented by the equation \( y = 3x + 2 \), then find the set of rail lines in \( R \) related to it.
Answer: The set of rail lines related to \( y = 3x + 2 \) is the set of all lines parallel to it. Since parallel lines have the same slope, the required set of lines is \( \{y = 3x + c : c \in \mathbb{R}\} \).

Question. Let \( S \) be the relation defined by \( S = \{(l_1, l_2) : l_1 \text{ is perpendicular to } l_2\} \). Check whether the relation \( S \) is symmetric or transitive.
Answer: \( S \) is symmetric but not transitive.
- **Symmetric:** If \( l_1 \perp l_2 \), then \( l_2 \perp l_1 \). Thus, \( (l_1, l_2) \in S \implies (l_2, l_1) \in S \).
- **Transitive:** If \( l_1 \perp l_2 \) and \( l_2 \perp l_3 \), then \( l_1 \) is parallel to (or coincident with) \( l_3 \), not perpendicular. Thus, \( S \) is not transitive.

Subjective Questions

Very Short Answer Type Questions

Question. If the relation \( R \) is defined on the set \( A = \{1, 2, 3, 4, 5\} \) by \( R = \{(a, b) : |a^2 - b^2| < 8\} \). Then, find the relation \( R \).
Answer: We need to find pairs \( (a,b) \in A \times A \) such that \( |a^2 - b^2| < 8 \).
- For \( a = 1 \): \( |1 - b^2| < 8 \implies b = 1, 2 \) (since \( |1-1|=0 < 8 \), \( |1-4|=3 < 8 \), \( |1-9|=8 \not< 8 \)). Pairs: \( (1, 1), (1, 2) \).
- For \( a = 2 \): \( |4 - b^2| < 8 \implies b = 1, 2, 3 \) (since \( |4-1|=3 \), \( |4-4|=0 \), \( |4-9|=5 \)). Pairs: \( (2, 1), (2, 2), (2, 3) \).
- For \( a = 3 \): \( |9 - b^2| < 8 \implies b = 2, 3, 4 \) (since \( |9-4|=5 \), \( |9-9|=0 \), \( |9-16|=7 \)). Pairs: \( (3, 2), (3, 3), (3, 4) \).
- For \( a = 4 \): \( |16 - b^2| < 8 \implies b = 3, 4 \) (since \( |16-9|=7 \), \( |16-16|=0 \)). Pairs: \( (4, 3), (4, 4) \).
- For \( a = 5 \): \( |25 - b^2| < 8 \implies b = 5 \) (since \( |25-25|=0 \)). Pairs: \( (5, 5) \).
Thus, the relation is: \( R = \{(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (3, 4), (4, 3), (4, 4), (5, 5)\} \)

Question. Find the domain and range of the following relation \( R = \{(c, d) : d = |c - 1|, c \in \mathbb{Z} \text{ and } |c| < 4\} \).
Answer: Given \( c \in \mathbb{Z} \) and \( |c| < 4 \), we have \( c \in \{-3, -2, -1, 0, 1, 2, 3\} \). Calculating \( d = |c - 1| \): - For \( c = -3 \), \( d = |-3 - 1| = 4 \) - For \( c = -2 \), \( d = |-2 - 1| = 3 \) - For \( c = -1 \), \( d = |-1 - 1| = 2 \) - For \( c = 0 \), \( d = |0 - 1| = 1 \) - For \( c = 1 \), \( d = |1 - 1| = 0 \) - For \( c = 2 \), \( d = |2 - 1| = 1 \) - For \( c = 3 \), \( d = |3 - 1| = 2 \)
Domain \( = \{-3, -2, -1, 0, 1, 2, 3\} \) Range \( = \{0, 1, 2, 3, 4\} \)

Question. Let \( R = \{(a, a^3) : a \text{ is a prime number less than } 5\} \) be a relation. Then, find the range of \( R \).
Answer: Prime numbers less than 5 are 2 and 3. Thus, \( a \) can take values 2 and 3. The relation \( R = \{(2, 2^3), (3, 3^3)\} = \{(2, 8), (3, 27)\} \). Hence, the range of \( R \) is \( \{8, 27\} \).

Question. Check whether the relation \( R \) defined in set \( A = \{4, 5, 6\} \) as \( R = \{(4, 4), (4, 5), (5, 4), (4, 6), (6, 4)\} \) is reflexive, symmetric or transitive.
Answer: - **Reflexive:** Here, \( (5, 5) \notin R \) and \( (6, 6) \notin R \), so \( R \) is not reflexive. - **Symmetric:** Since \( (4, 5) \in R \) and \( (5, 4) \in R \), and \( (4, 6) \in R \) and \( (6, 4) \in R \), \( R \) is symmetric. - **Transitive:** Here, \( (5, 4) \in R \) and \( (4, 6) \in R \) but \( (5, 6) \notin R \), so \( R \) is not transitive.

Question. An equivalence relation \( R \) in \( A \) divides it into equivalence classes \( A_1, A_2, A_3 \). What is the value of \( A_1 \cup A_2 \cup A_3 \) and \( A_1 \cap A_2 \cap A_3 \)?
Answer: Since equivalence classes partition the set \( A \): - \( A_1 \cup A_2 \cup A_3 = A \) - \( A_1 \cap A_2 \cap A_3 = \phi \)

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: The equivalence class \( [0] = \{b \in A : (0, b) \in R\} = \{b \in A : 2 \text{ divides } (0 - b)\} = \{b \in A : 2 \text{ divides } -b\} \). Within \( A = \{0, 1, 2, 3, 4, 5\} \), the numbers divisible by 2 are 0, 2, and 4. Hence, \( [0] = \{0, 2, 4\} \).

Question. A relation \( R \) in the set of real numbers \( \mathbb{R} \) defined as \( R = \{(a, b) : \sqrt{a} = b\} \) is a function or not. Justify.
Answer: Since \( \sqrt{a} \) is only defined for \( a \ge 0 \), elements in the domain \( (-\infty, 0) \) have no image under this relation. Therefore, \( R \) is not a function.

Question. Let \( A = \{1, 2, 3\} \), \( B = \{4, 5, 6, 7\} \) and \( f \) be a function from set \( A \) to set \( B \) is \( f = \{(1, 4), (2, 5), (3, 6)\} \). Then, check whether \( f(x) \) is one-to-one or not.
Answer: Since every distinct element of set \( A \) has a distinct image in set \( B \), \( f \) is one-to-one.

Question. Let \( A = \{-3, -2, 1, 2\} \) and \( B = \{1, 4, 9\} \), then function \( f \) is defined from set \( A \) to set \( B \) such that \( f(x) = x^2 \). Show that \( f \) is many-one function.
Answer: We have \( f(-2) = (-2)^2 = 4 \) and \( f(2) = (2)^2 = 4 \). Since two different inputs \( -2 \) and \( 2 \) map to the same output \( 4 \), the function \( f \) is many-one.

Question. Show that the function \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = 3x - 4 \) is bijective.
Answer: - **One-one (Injective):** Let \( x_1, x_2 \in \mathbb{R} \) such that \( f(x_1) = f(x_2) \implies 3x_1 - 4 = 3x_2 - 4 \implies 3x_1 = 3x_2 \implies x_1 = x_2 \). Hence, \( f \) is one-one. - **Onto (Surjective):** Let \( y \in \mathbb{R} \) (codomain). Set \( y = 3x - 4 \implies x = \frac{y + 4}{3} \). Since \( y \in \mathbb{R} \), \( x = \frac{y+4}{3} \in \mathbb{R} \) (domain) exists such that \( f(x) = 3\left(\frac{y+4}{3}\right) - 4 = y \). Hence, \( f \) is onto.
Since the function is both one-one and onto, it is bijective.

Short Answer Type Questions

Question. Show that relation \( R \) in the set of real numbers, defined as \( R = \{(a, b) : a \le b^2\} \) is neither reflexive nor symmetric nor transitive.
Answer: - **Reflexive:** Let \( a = \frac{1}{2} \). Since \( \frac{1}{2} > \left(\frac{1}{2}\right)^2 = \frac{1}{4} \), \( \left(\frac{1}{2}, \frac{1}{2}\right) \notin R \). Thus, \( R \) is not reflexive. - **Symmetric:** Let \( a = -1, b = 3 \). We have \( -1 \le 3^2 = 9 \), so \( (-1, 3) \in R \). However, \( 3 > (-1)^2 = 1 \), so \( (3, -1) \notin R \). Thus, \( R \) is not symmetric. - **Transitive:** Let \( a = 2, b = -3, c = 1 \). We have \( 2 \le (-3)^2 = 9 \), so \( (2, -3) \in R \), and \( -3 \le 1^2 = 1 \), so \( (-3, 1) \in R \). But \( 2 > 1^2 = 1 \), so \( (2, 1) \notin R \). Thus, \( R \) is not transitive.

Question. Check whether the relation \( R \) in the set \( \mathbb{Z} \) of integers defined as \( R = \{(a,b) : a + b \text{ is divisible by 2}\} \) is reflexive, symmetric or transitive. Write the equivalence class containing 0 i.e. \( [0] \).
Answer: - **Reflexive:** For any \( a \in \mathbb{Z} \), \( a + a = 2a \), which is divisible by 2. Thus, \( (a, a) \in R \) for all \( a \in \mathbb{Z} \). Hence, \( R \) is reflexive. - **Symmetric:** Let \( (a, b) \in R \implies a + b \) is divisible by 2. Since \( b + a = a + b \), \( b + a \) is also divisible by 2, meaning \( (b, a) \in R \). Hence, \( R \) is symmetric. - **Transitive:** Let \( (a, b) \in R \) and \( (b, c) \in R \). Then \( a + b = 2\lambda \) and \( b + c = 2\mu \) for some integers \( \lambda, \mu \). Adding these gives \( a + 2b + c = 2(\lambda + \mu) \implies a + c = 2(\lambda + \mu - b) = 2k \). So, \( a + c \) is divisible by 2, meaning \( (a, c) \in R \). Hence, \( R \) is transitive.
Thus, \( R \) is an equivalence relation. The equivalence class \( [0] = \{b \in \mathbb{Z} : 0 + b \text{ is divisible by 2}\} = \{b \in \mathbb{Z} : b \text{ is even}\} = \{\dots, -6, -4, -2, 0, 2, 4, 6, \dots\} \).

Question. Show that the relation \( R \) on the set \( A = \{1, 2, 3, 4, 5\} \) given by \( R = \{(a, b) : |a - b| \text{ is even}\} \) is an equivalence relation.
Answer: - **Reflexive:** For any \( a \in A \), \( |a - a| = 0 \), which is even. Thus, \( (a, a) \in R \) for all \( a \in A \). Hence, \( R \) is reflexive. - **Symmetric:** If \( (a, b) \in R \), then \( |a - b| \) is even. Since \( |b - a| = |a - b| \), \( |b - a| \) is also even, meaning \( (b, a) \in R \). Hence, \( R \) is symmetric. - **Transitive:** If \( (a, b) \in R \) and \( (b, c) \in R \), then \( |a - b| = 2k \) and \( |b - c| = 2m \) for some integers \( k, m \). Thus, \( a - b = \pm 2k \) and \( b - c = \pm 2m \). Adding these, \( a - c = (a - b) + (b - c) = \pm 2k \pm 2m = 2(\pm k \pm m) \). Hence, \( |a - c| \) is even, meaning \( (a, c) \in R \). Hence, \( R \) is transitive.
Since \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.

Question. Let \( T \) be the set of all triangles in a plane with \( R \), a relation in \( T \) given by \( R = \{(T_1, T_2) : T_1 \cong T_2\} \). Show that \( R \) is an equivalence relation.
Answer: - **Reflexive:** Every triangle \( T_1 \) is congruent to itself, i.e., \( T_1 \cong T_1 \implies (T_1, T_1) \in R \). Thus, \( R \) is reflexive. - **Symmetric:** If \( (T_1, T_2) \in R \implies T_1 \cong T_2 \implies T_2 \cong T_1 \implies (T_2, T_1) \in R \). Thus, \( R \) is symmetric. - **Transitive:** If \( (T_1, T_2) \in R \) and \( (T_2, T_3) \in R \), then \( T_1 \cong T_2 \) and \( T_2 \cong T_3 \implies T_1 \cong T_3 \implies (T_1, T_3) \in R \). Thus, \( R \) is transitive.
Since \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.

Question. Show that the function \( f : \mathbb{N} \to \mathbb{N} \), given by \( f(1) = f(2) = 1 \) and \( f(x) = x - 1 \) for every \( x > 2 \), is onto but not one-one.
Answer: - **Not one-one:** Since \( f(1) = 1 \) and \( f(2) = 1 \), two distinct elements (1 and 2) have the same image. Hence, \( f \) is not one-one. - **Onto:** Let \( y \in \mathbb{N} \). If \( y = 1 \), its pre-images are 1 and 2. For \( y > 1 \), we can choose \( x = y + 1 \in \mathbb{N} \) (since \( x > 2 \)), such that \( f(x) = f(y + 1) = (y + 1) - 1 = y \). Thus, every element in the codomain has a pre-image. Hence, \( f \) is onto.

Question. Let \( A = \mathbb{R} - \{3\} \), \( B = \mathbb{R} - \{1\} \). If \( f : A \to B \) is the function defined by \( f(x) = \frac{x-2}{x-3} \), \( \forall x \in A \). Then, show that \( f \) is bijective.
Answer: - **Injective:** Let \( f(x_1) = f(x_2) \implies \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} \) \( \implies (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) \) \( \implies x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6 \) \( \implies -3x_1 - 2x_2 = -3x_2 - 2x_1 \) \( \implies -x_1 = -x_2 \implies x_1 = x_2 \). Hence, \( f \) is injective. - **Surjective:** Let \( y \in B = \mathbb{R} - \{1\} \). We set \( y = \frac{x - 2}{x - 3} \) \( \implies y(x - 3) = x - 2 \implies xy - 3y = x - 2 \) \( \implies x(y - 1) = 3y - 2 \implies x = \frac{3y - 2}{y - 1} \). Since \( y \ne 1 \), \( x \) is well-defined. Also, if \( x = 3 \implies 3y - 2 = 3y - 3 \implies -2 = -3 \), which is impossible, so \( x \in A = \mathbb{R} - \{3\} \). We verify \( f(x) = f\left(\frac{3y - 2}{y - 1}\right) = \frac{\frac{3y - 2}{y - 1} - 2}{\frac{3y - 2}{y - 1} - 3} = \frac{3y - 2 - 2y + 2}{3y - 2 - 3y + 3} = \frac{y}{1} = y \). Thus, \( f \) is surjective.
Since \( f \) is both injective and surjective, it is bijective.

Question. Let \( A = \mathbb{R} - \{2\} \) and \( B = \mathbb{R} - \{1\} \). If \( f : A \to B \) is a function defined by \( f(x) = \frac{x - 1}{x - 2} \), show that \( f \) is one-one and onto.
Answer: - **One-one:** Let \( x_1, x_2 \in A \) such that \( f(x_1) = f(x_2) \implies \frac{x_1 - 1}{x_1 - 2} = \frac{x_2 - 1}{x_2 - 2} \) \( \implies (x_1 - 1)(x_2 - 2) = (x_2 - 1)(x_1 - 2) \) \( \implies x_1 x_2 - 2x_1 - x_2 + 2 = x_1 x_2 - 2x_2 - x_1 + 2 \) \( \implies -2x_1 - x_2 = -2x_2 - x_1 \implies -x_1 = -x_2 \implies x_1 = x_2 \). Hence, \( f \) is one-one. - **Onto:** Let \( y \in B = \mathbb{R} - \{1\} \). Set \( y = \frac{x - 1}{x - 2} \implies y(x - 2) = x - 1 \implies xy - 2y = x - 1 \implies x(y - 1) = 2y - 1 \implies x = \frac{2y - 1}{y - 1} \). Since \( y \ne 1 \), \( x \) is well-defined. Also, \( x \ne 2 \) because if \( \frac{2y-1}{y-1} = 2 \implies 2y-1 = 2y-2 \implies -1 = -2 \), which is impossible. Hence, \( f \) is onto.

Question. Show that the modulus function \( f : \mathbb{R} \to \mathbb{R} \) given by \( f(x) = |x| \), is neither one-one nor onto, where \( |x| \) is \( x \), if \( x \) is positive or 0 and \( |x| \) is \( -x \), if \( x \) is negative.
Answer: - **Not one-one:** Let \( x_1 = 1 \) and \( x_2 = -1 \). We have \( f(1) = |1| = 1 \) and \( f(-1) = |-1| = 1 \). Since \( f(1) = f(-1) \) but \( 1 \ne -1 \), \( f \) is not one-one. - **Not onto:** Since \( f(x) = |x| \ge 0 \) for all \( x \in \mathbb{R} \), the range of \( f \) is \( [0, \infty) \). This is not equal to the codomain \( \mathbb{R} \) (e.g., negative numbers like \( -1 \) have no pre-images). Hence, \( f \) is not onto.

Question. Let \( A = \{1, 3, 5, 7, \dots\} \) and \( B = \{2, 4, 6, 8, \dots\} \). Define a function from \( A \) to \( B \) that is neither one-one nor onto.
Answer: We can define \( f : A \to B \) by \( f(x) = 2 \) for all \( x \in A \). Here, all elements of \( A \) map to 2, so the function is not one-one. Furthermore, any element in \( B \) other than 2 has no pre-image in \( A \), so the function is not onto.

Question. Show that the function \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = 4x^3 - 5 \), \( \forall x \in \mathbb{R} \) is one-one and onto.
Answer: - **One-one:** Let \( f(x_1) = f(x_2) \implies 4x_1^3 - 5 = 4x_2^3 - 5 \implies 4x_1^3 = 4x_2^3 \implies x_1^3 = x_2^3 \implies (x_1 - x_2)(x_1^2 + x_1 x_2 + x_2^2) = 0 \). Since \( x_1^2 + x_1 x_2 + x_2^2 > 0 \) for real non-zero \( x_1, x_2 \), we must have \( x_1 - x_2 = 0 \implies x_1 = x_2 \). Hence, \( f \) is one-one. - **Onto:** Let \( y \in \mathbb{R} \). Set \( y = 4x^3 - 5 \implies 4x^3 = y + 5 \implies x = \left(\frac{y + 5}{4}\right)^{1/3} \), which is a real number for any real \( y \). Thus, \( f(x) = y \), showing that \( f \) is onto.

Question. Suppose \( A = B = \{x \in \mathbb{R} : -1 \le x \le 1\} \). Show that \( f : A \to B \) given by \( f(x) = x|x| \) is a bijection.
Answer: We can write \( f(x) = \begin{cases} x^2, & \text{if } x \ge 0 \\ -x^2, & \text{if } x < 0 \end{cases} \) - **One-one:** Let \( x, y \in A \) with \( x \ne y \). If both are non-negative, \( x^2 \ne y^2 \). If both are negative, \( -x^2 \ne -y^2 \). If one is negative and the other is positive, one output is negative and the other is positive, so they cannot be equal. Thus, \( f(x) \ne f(y) \), meaning \( f \) is one-one. - **Onto:** For \( 0 \le x \le 1 \), \( f(x) = x^2 \) takes all values in \( [0, 1] \). For \( -1 \le x < 0 \), \( f(x) = -x^2 \) takes all values in \( [-1, 0) \). Thus, the range of \( f \) is \( [-1, 1] \), which is equal to \( B \). Hence, \( f \) is onto.
Since \( f \) is both one-one and onto, it is a bijection.

Long Answer Type Questions

Question. Check whether the relation \( R \) in the set \( \mathbb{N} \) of natural numbers given by \( R = \{(a, b) : a \text{ is divisor of } b\} \) is reflexive, symmetric or transitive. Also, determine whether \( R \) is an equivalence relation.
Answer: - **Reflexive:** Every natural number \( a \) divides itself. Thus, \( aRa \implies (a, a) \in R \) for all \( a \in \mathbb{N} \). It is reflexive. - **Symmetric:** Let \( (a, b) \in R \implies a \) is a divisor of \( b \). This does not imply that \( b \) is a divisor of \( a \). For example, \( (2, 4) \in R \) (2 divides 4) but \( (4, 2) \notin R \) (4 does not divide 2). Hence, it is not symmetric. - **Transitive:** Let \( (a, b) \in R \) and \( (b, c) \in R \). Then \( b = k_1 a \) and \( c = k_2 b \) for some positive integers \( k_1, k_2 \). Substituting \( b \), we get \( c = k_2(k_1 a) = (k_1 k_2) a \). Since \( k_1 k_2 \) is a positive integer, \( a \) is a divisor of \( c \), so \( (a, c) \in R \). Hence, it is transitive.
Since \( R \) is reflexive and transitive but not symmetric, it is not an equivalence relation.

Question. Given, a non-empty set \( X \), define the relation \( R \) in \( P(X) \) as follows: For \( A, B \in P(X) \), \( (A, B) \in R \iff A \subset B \). Prove that \( R \) is reflexive, transitive but not symmetric.
Answer: - **Reflexive:** For any set \( A \in P(X) \), \( A \subset A \). Hence, \( (A, A) \in R \). Thus, \( R \) is reflexive. - **Symmetric:** Let \( (A, B) \in R \implies A \subset B \). This does not necessarily imply \( B \subset A \). For example, if \( A = \phi \) and \( B \ne \phi \), then \( \phi \subset B \) is true so \( (\phi, B) \in R \), but \( B \subset \phi \) is false, so \( (B, \phi) \notin R \). Thus, \( R \) is not symmetric. - **Transitive:** Let \( (A, B) \in R \) and \( (B, C) \in R \implies A \subset B \) and \( B \subset C \). By set theory, \( A \subset C \), which means \( (A, C) \in R \). Thus, \( R \) is transitive.

Question. Show that the relation \( R \) on the set \( \mathbb{Z} \) of all integers defined by \( (x,y) \in R \iff (x-y) \text{ is divisible by 3} \) is an equivalence relation.
Answer: - **Reflexive:** For any \( x \in \mathbb{Z} \), \( x - x = 0 \), which is divisible by 3. Thus, \( (x, x) \in R \). - **Symmetric:** Let \( (x, y) \in R \implies x - y \) is divisible by 3. Thus, \( x - y = 3\lambda \) for some integer \( \lambda \). Then \( y - x = -3\lambda = 3(-\lambda) \), which is also divisible by 3. Thus, \( (y, x) \in R \). - **Transitive:** Let \( (x, y) \in R \) and \( (y, z) \in R \). Then \( x - y = 3\lambda \) and \( y - z = 3\mu \) for some integers \( \lambda, \mu \). Adding these equations gives \( (x - y) + (y - z) = 3\lambda + 3\mu \implies x - z = 3(\lambda + \mu) \), which is divisible by 3. Thus, \( (x, z) \in R \).
Since \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.

Question. Prove that function \( f : [0, \infty) \to [-5, \infty) \) defined as \( f(x) = 4x^2 + 4x - 5 \) is both one-one and onto.
Answer: - **One-one:** Let \( x_1, x_2 \in [0, \infty) \) such that \( f(x_1) = f(x_2) \implies 4x_1^2 + 4x_1 - 5 = 4x_2^2 + 4x_2 - 5 \) \( \implies 4(x_1^2 - x_2^2) + 4(x_1 - x_2) = 0 \) \( \implies 4(x_1 - x_2)(x_1 + x_2 + 1) = 0 \). Since \( x_1, x_2 \ge 0 \), we have \( x_1 + x_2 + 1 \ge 1 > 0 \). Thus, we must have \( x_1 - x_2 = 0 \implies x_1 = x_2 \). Thus, \( f \) is one-one. - **Onto:** Let \( y \in [-5, \infty) \). Set \( y = 4x^2 + 4x - 5 \implies 4x^2 + 4x - (5 + y) = 0 \). Using the quadratic formula: \( x = \frac{-4 \pm \sqrt{16 + 16(5 + y)}}{8} = \frac{-4 \pm 4\sqrt{1 + 5 + y}}{8} = \frac{-1 \pm \sqrt{6 + y}}{2} \). Since \( y \ge -5 \), \( \sqrt{6 + y} \ge 1 \). Since \( x \ge 0 \), we take the positive sign: \( x = \frac{-1 + \sqrt{6+y}}{2} \ge 0 \), which is a valid element of the domain. Hence, \( f \) is onto.

Question. A function \( f : [-4, 4] \to [0, 4] \) is given by \( f(x) = \sqrt{16 - x^2} \). Show that \( f \) is an onto function but not a one-one function. Further, find all possible values of \( a \) for which \( f(a) = \sqrt{7} \).
Answer: - **Not one-one:** Let \( x_1 = 2 \) and \( x_2 = -2 \). Both lie in \( [-4, 4] \). We have \( f(2) = \sqrt{16 - 4} = \sqrt{12} \) and \( f(-2) = \sqrt{16 - 4} = \sqrt{12} \). Since \( f(2) = f(-2) \) but \( 2 \ne -2 \), the function is not one-one. - **Onto:** Let \( y \in [0, 4] \). Set \( y = \sqrt{16 - x^2} \implies y^2 = 16 - x^2 \implies x^2 = 16 - y^2 \implies x = \pm \sqrt{16 - y^2} \). Since \( y \in [0, 4] \), \( 16 - y^2 \in [0, 16] \), so \( x = \pm \sqrt{16 - y^2} \) is a real number in \( [-4, 4] \). Hence, every \( y \) has a pre-image, so \( f \) is onto. - **Find \( a \) for which \( f(a) = \sqrt{7} \):** \( \sqrt{16 - a^2} = \sqrt{7} \implies 16 - a^2 = 7 \implies a^2 = 9 \implies a = \pm 3 \). Both \( 3 \) and \( -3 \) lie in \( [-4, 4] \). Thus, the possible values of \( a \) are \( \pm 3 \).

Question. Consider \( f : \mathbb{R}_+ \to [-9, \infty) \) given by \( f(x) = 5x^2 + 6x - 9 \). Show that \( f \) is bijective.
Answer: - **One-one:** Let \( x_1, x_2 \in \mathbb{R}_+ \) such that \( f(x_1) = f(x_2) \implies 5x_1^2 + 6x_1 - 9 = 5x_2^2 + 6x_2 - 9 \) \( \implies 5(x_1^2 - x_2^2) + 6(x_1 - x_2) = 0 \implies (x_1 - x_2)(5x_1 + 5x_2 + 6) = 0 \). Since \( x_1, x_2 > 0 \), \( 5x_1 + 5x_2 + 6 > 0 \). Hence, \( x_1 - x_2 = 0 \implies x_1 = x_2 \). So \( f \) is one-one. - **Onto:** Let \( y \in [-9, \infty) \). Set \( y = 5x^2 + 6x - 9 \implies 5x^2 + 6x - (9 + y) = 0 \). Using the quadratic formula: \( x = \frac{-6 \pm \sqrt{36 + 20(9 + y)}}{10} \). Since \( y \ge -9 \), the term under the square root is non-negative. To ensure \( x \in \mathbb{R}_+ \) (i.e., \( x > 0 \)), we choose the positive sign: \( x = \frac{-6 + \sqrt{36 + 20(9 + y)}}{10} \). Since \( y \ge -9 \), \( \sqrt{36 + 20(9+y)} \ge \sqrt{36} = 6 \). For \( y > -9 \), \( x > 0 \). Thus, \( f \) is onto.
Since \( f \) is both one-one and onto, it is bijective.

Question. Show that the function \( f : \mathbb{N} \to \mathbb{N} \) given by \( f(n) = n - (-1)^n \), \( \forall n \in \mathbb{N} \) is a bijection.
Answer: We can write \( f(n) = \begin{cases} n + 1, & \text{if } n \text{ is odd} \\ n - 1, & \text{if } n \text{ is even} \end{cases} \) - **One-one:** Let \( f(n_1) = f(n_2) \). - Case I: If \( n_1 \) and \( n_2 \) are both odd, \( n_1 + 1 = n_2 + 1 \implies n_1 = n_2 \). - Case II: If \( n_1 \) and \( n_2 \) are both even, \( n_1 - 1 = n_2 - 1 \implies n_1 = n_2 \). - Case III: If \( n_1 \) is odd and \( n_2 \) is even, then \( n_1 + 1 \) is even and \( n_2 - 1 \) is odd, so \( f(n_1) = f(n_2) \) is impossible because an even number cannot equal an odd number. Thus, \( f(n_1) = f(n_2) \implies n_1 = n_2 \). Hence, \( f \) is one-one. - **Onto:** Let \( y \in \mathbb{N} \). - If \( y \) is odd, then \( y + 1 \) is even. Let \( n = y + 1 \). Since \( n \) is even, \( f(n) = n - 1 = (y + 1) - 1 = y \). - If \( y \) is even, then \( y - 1 \) is odd. Let \( n = y - 1 \). Since \( n \) is odd, \( f(n) = n + 1 = (y - 1) + 1 = y \). In both cases, every \( y \in \mathbb{N} \) has a pre-image in \( \mathbb{N} \). Hence, \( f \) is onto.
Since \( f \) is both one-one and onto, it is a bijection.