## Relations And Functions Exercise 1.1

**Question 1. Determine whether each of the following relations is reflexive, symmetric, or transitive :**

- Relation R in the set A = {1,2, 3…13, 14} defined as R = {(x, y): 3x – y = 0}
- Relation R in the set N of natural numbers defined as R = {(x, y): y = x + 5 and x < 4}
- Relation R in the set A = (1, 2, 3, 4, 5, 6} as R = {(x, y); y is divisible by x}
- Relation R in the set Z of all integers defined as R = {(x, y): x – y is an integer}
- Relation R in the set A of human beings in a town at a particular time given by
- R = {(x, y): x and y work at the same place}
- R = {(x, y): x and y live in the same locality}
- R – {(x, y): x is exactly 7 cm taller than y}
- R = {(x, y): x is wife of y)
- R = {(x, y): x is father of y}

**Solution:**

1. A = {1,2, 3 … 13, 14}; R = {(x, y): 3x -y = 0}

**Reflexive:** R = {(1, 3), (2, 6), (3, 9), (4, 12)}

(1,1) ∉ R, So, R is not reflexive.

**Symmetric:** (1, 3) ∈ R, but (3, 1) ∉ R. So, R is not symmetric

**Transitive:** (1,3), (3, 9) ∈ R, but (1, 9) ∉ R. So, R is not transitive

Hence, R is neither reflexive, nor symmetric, nor transitive.

2. R = {(x, y): y = x + 5 and x < 4} ⇒ R ~ {(1, 6), (2, 7), (3, 8)}

**Reflexive:** (1, 1) ∉ R. So, R is not reflexive.

**Symmetric** : (1,6) ∈ R, but (6, 1) ∉ R. R is not symmetric.

**Transitive:** Since there are no three elements x, y, z ∈ N such that (x, y) ∈ R, z) ∈ R but (x, z) ∉ R

∴ R is transitive. Hence, R is neither reflexive nor symmetric but it is transitive.

3. A = {1, 2, 3,4, 5, 6}; R = {(x, y): y is divisible by x}

**Reflexive:** Let x∈ A such that if (x, x) ∈ R x is divisible by x, which is true, ∀ x∈ A

∴ R is reflexive.

**Read and Learn More Class 12 Maths Chapter Wise with Solutions**

**Symmetric:** Let x, y ∈ A such that if (x, y) ∈ R => y is divisible by x

⇒ x is not divisible by => (y, x) ∉R

For example: (2, 4) ∈ R but (4, 2) ∉R.

∴ R is not symmetric.

Transitive: Let x, y, z ∈ A such that if (x, y) ∈ R and (y, z) ∈ R

⇒ y is divisible by x and z is divisible by y.

⇒ \(\frac{y}{x}=k_1 \in I \) → Equation 1

⇒ \((y, z) \in R \Rightarrow \frac{z}{y}=k_2 \in\) I → Equation 2

Equation (1) x (2) gives

⇒ \(\frac{y}{x} \times \frac{z}{y}=k_1 \times k_2 \Rightarrow \frac{z}{x}=\left(k_1 k_2\right) \in\) I

(x, z)\( \in\)

∴ R . R is transitive.

Hence, R is reflexive and transitive but not symmetric.

(4) R = {(x, y): x – y is an integer}

**Reflexive:** If (x, x) ∈R ⇒ x – x = 0, which is an integer, ∀ x ∈ Z.

∴ R is reflexive.

**Symmetric:** If (x, y) ∈ R ⇒ (x – y) is an integer. ⇒ (y – x) is also an integer.

⇒ (v. x) ∈R

∴ R is symmetric.

**Transitive:** If (x, y) ∉ R ⇒ x -y = k_{1} ∈ (1)

and (y, z) ∈ R ⇒ y – z = k_{2} ∈ (2)

Adding equation (1) and (2), we get

(x – y) + (y – z) = (k_{1} + k_{2}) ∈

x – z = (k_{1} + k_{2}) ∈ I

⇒ (x, z) ∈ R

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

5. Given A = (x : x is a human being in a town}

(1.) R = {(x, y): x and y work at the same place}

**Reflexive:** If (x, x) e R ⇒ x and x work at the same place, which is true, V x ∈ A R is reflexive.

**Symmetric:** If (x, y) ∈ R ⇒ x and y work at the same place.

⇒ y and x also work at the same place. ⇒ (y, x)∈R.

∴R is symmetric.

**Transitive:** If (x, y) ∈ R and (y, z) ∈ R

⇒ x and y work at the same place and y and z work at the same place.

⇒ x and z work at the same place. ⇒ (x, z) ∈ R

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

(2) R = {(x, y): x and y live in the same locality}

**Reflexive:** If (x, x) e R => x and x live in the same locality, which is true, ∀ x ∈ A

∴ R is reflexive.

**Symmetric:** If (x, y) ∈ R ⇒ x and y live in the same locality.

⇒ y and x also live in the same locality ⇒ (y, x) ∈ R

∴ R is symmetric.

**Transitive:** Let (x, y) ∈ R and (y, z) ∈ R.

x and y live in the same locality and y and z live in the same locality.

⇒ x and z live in the same locality. ⇒ (x, z) ∈ R

R is transitive. Hence, R is reflexive, symmetric, and transitive.

(3.) R = {(x, y): x is exactly 7 cm taller than y}

**Reflexive:** If (x, x) ∈ R ⇒ x is exactly 7 cm taller than x, which is not true, for any x ∈ A. Since human beings x cannot be taller than themselves.

∴ So, R is not reflexive.

**Symmetric:** If (x, y) ∈ R ⇒ x is exactly 7 cm taller than y ⇒ y is exactly 7 cm smaller

⇒ (y, x)∉ R

∴ R is not symmetric.

**Transitive:** If (x, y), (y, z) ∈ R

⇒ x is exactly 7 cm taller than y and y is exactly 7 cm taller than z.

⇒ x is exactly 14 cm taller than z.

∴ (x, z) ∉ R

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(4) R = ((x, y): x is wife of y)

**Reflexive:** If (x, x) ∈ R ⇒ x is wife of x, which is not true, for any x ∈ A

∴ R is not reflexive.

**Symmetric;** If (x, y) ∈ R ⇒ x is the wife of y ⇒ y is the husband of x.

(y, x) ∉ R

R is not symmetric.

**Transitive:** There are no three elements x, y, z ∈ A such that (x, y) R and (y, z) ∈ R but (x, z) ∉ R

So, R is transitive. Hence, R is neither reflexive, nor symmetric, but R is transitive.

(5) R = {(x, y): x is father of y}

**Reflexive:** If (x, x) ∈ R ⇒ x is the father of x, which, is not true, for any x ∈ A

∴ R is not reflexive.

**Symmetric:** If (x, y) ∈ R ⇒ x is the father of y ⇒ his son or daughter of x.

(y,x) ∉ R R is not symmetric.

**Transitive:** If (x, y) ∈ R and (y, z) ∈ R,

x is the father of y, and y is the father of z ⇒ x is the grandfather of z.

(x, z) ∉ R

R is not transitive. Hence, R is neither reflexive, symmetric, nor transitive.

**Question 2. Show that the relation R in the set R of real numbers, defined as R = {(a, b): a < b} is neither reflexive nor symmetric nor transitive.**

**Solution:**

**Reflexive:** R = {(a, b): a \(\leq b^2\)}

Let a ∈ R such that if (a, a) ∈ R ⇒ a ≤ a² is not true for all a ∈ R

i.e. (a, a) \(\notin R, \forall a \in\) R

Let a=\(\frac{1}{2} \in\) R

⇒ \(\frac{1}{2}\) ≰ \(\frac{1}{2^2}\) ⇒ \(\frac{1}{2}\) ≰ \(\frac{1}{4}\) ∈ R

∴ R is Not Reflexive

**Symmetric:** Let a, b ∈ R such that if (a, b) ∈ R ⇒ a ≤ b² ⇒ b≰ a²

**For example. **2≰ 5² ⇒(2,5) ∈ R but 5≰ 2²

⇒ (5,2) ∉ R

∴ R is not symmetric

**Transitive:** Let a,b,csR such that if (a, b) ∈ R an nd (b. c) ∈ R

⇒ a ≤ b² and b ≤ c² ⇒ a ≰ c²

**For Example: **5 ≤ 3² and 3 ≤ 2² but 5 ≰ 2²

(5, 3) e R and (3, 2) ∈ R but (5, 2) ∉R

∴ R is not transitive

∴ Hence, R is neither reflexive nor symmetric nor transitive.

**Question 3. 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.**

**Solution:**

**Reflexive:** Given A = {1, 2, 3, 4, 5, 6}.

R = {(a, b): b = a + 1} ⇒ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} v (1,1) ∉ R, where 1 ∈ A

∴ R is not reflexive.

**Symmetric:** (1, 2) 6 R but (2, 1) ∉ R, where 1, 2 ∉ A.

∴ So R is not symmetric

**Transitive:** (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R, where 1,2,3 ∈ A.

R is not transitive Hence R is neither reflexive, symmetric, nor transitive.

**Question 4. Show that the relation R in R defined as R = {(a, b): a < b}, is reflexive and transitive but not symmetric.**

**Solution:**

**Reflexive:** Given R = {(a, b): a < b}

Let a ∈ R such that if (a, a) ∈ R ⇒ a < a, which is true, ∀a∈R

∴ R is reflexive

**Symmetric:** Let a, b e R such that if (a, b) e R ⇒ a < b ⇒ b > a ⇒ (b, a) ∉ R

For example: 1 < 2 ⇒ (1, 2) e R But 2 ∉1 ⇒ (2,1)

∴ R is not symmetric

**Transitive:** Let a, b, c ∈ R such that if (a, b) ∈ R and (b, c) ∈ R ⇒ a < b and b<c ⇒ a<c ⇒ (a, c) ∈ R

∴ R is transitive

Flence R is reflexive and transitive but not symmetric.

**Question 5. Check whether the relation R in R defined as R = {(a, b): a < b³} is reflexive, symmetric, or transitive.**

**Solution:**

**Reflexive:** Given R = {(a, b): a < b³}

Let a∈ R such that if (a, a) ∈ R ⇒ a < a’

Not true for all a ∈ R ⇒ (a, ) R, ∀ a ∈ R 1

Let a = \(\frac{1}{2} \in \mathrm{R}\)

⇒ \(\frac{1}{2} \notin \frac{1}{2^3} \Rightarrow \frac{1}{2} \notin \frac{1}{8} \Rightarrow\left(\frac{1}{2}, \frac{1}{2}\right) \notin \mathrm{R}\)

∴ R is not Reflexive.

**Symmetric:** Let a, b ∈ R such that if (a, b) ∈ R ⇒ a < b³ ⇒ b∉ a³

For example: 2 < 9J but ∈ ∉ 21 => (2, 9)∈R but (9, 2) ∉ R

∴ R is not symmetric

**Transitiv** e: Let a, b, c ∈ R such that if (a, b) ∈ R and (b, c) ∈ R

⇒ a < b³ and b < c³ ⇒ a ∉ c³

For example: 9 < 33 and 3 < 2 but 9

⇒ (9,3) ∈ R and (3,2) ∈ R but (9,2) ∉ R

∴ R is not transitive

∴ Hence R is neither reflexive nor symmetric nor transitive.

**Question 6. Show that the relation R in the set {1,2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.**

**Solution:**

**Reflexive:** Let A = {1, 2, 3}.

A relation R on A is defined as R = {(1, 2), (2,1)}.

v (1,1) ∉ R, where 1 ∈ A R is not reflexive

**Symmetric:** (1,2) ∈ R and (2,1) ∈ R

∴ R is symmetric

**Transitive:** (1,2) ∈ R and (2,1) ∈ R but (1,1) ∉ R

∴ R is not transitive

Hence, R is symmetric but neither reflexive nor transitive.

**Question 7. Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y): x and y have same number of pages} is an equivalence relation.**

**Solution:**

**Reflexive:** Set A is the set of all books in the library of a college. R = {(x, y): x and y have the same number of pages}

Let x ∉ A such that if (x, x) ∈ R ⇒ x and x have the same number of pages.

Which is the true, V x ∈ A

∴R is reflexive

**Symmetric:** Let x, y ∈ A such that if (x, y) ∈ R ⇒ x and y have the same number of pages.

⇒ y and x have the same number of pages. ⇒ (y, x)∈ R

∴ R is symmetric.

**Transitive:** Let x, y, z ∈ A such that if (x, y) ⇒ R and (y, z)∈ R.

⇒ x and y have the same number of pages and y and z have the same number of pages.

⇒ x and z have the same number of pages ⇒ (x, z) ∈ R

∴ R is transitive.

Since, R is reflexive, symmetric, and transitive on A

∴ Hence, R is an equivalence relation to A.

**Question 8. Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b): |a – b| is even}, is an equivalence relation. Show that all the elements of {1,3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2,4}.**

**Solution:**

**Reflexive:** A = {1,2,3,4, 5}, R = [{a, b);| a — b| is even}

Let a ∈ A such that if (a, a) ∈ R ⇒ |a – a| = 0, is even

which is true, ∀ a ∈ A

∴ R is reflexive.

**Symmetric:** Let a, b ∈ A such that if (a, b) ∈ R ⇒ |a- b| is even |-(b – a)| = |b – a| is also even

⇒ (b, a) ∈ R

∴ R is symmetric.

**Transitive:** Let a, b, c ∈ R such that if (a, b) ∉ R and (b, c) ∈ It

⇒ |a — b| and |b – c| both are even

⇒ |a-b| = 2k, and | b — c| = 2k_{2}, (k_{1},k_{2} ∈ Z)

⇒ (a-b) = ±2k,(1) and (b-c) = ± 2 k_{2} (2)

On adding equations (1) & (2), we get:

(a-c) = ± 2(k_{1}+ k_{2}) ⇒ |a-c| = 2(k_{1} + k_{2})

⇒ |a – c| is even ⇒ (a, c) ∈ R (v k_{1} + k_{2} ∈ Z)

∴ R is transitive

Since R is reflexive, symmetric, and transitive on A.

Hence, R is an equivalence relation to A.

Since the difference of two even (or odd) is always even.

So, every element of set {1, 3, 5} is related to each other and also every element of set {2,4} is related to each other but no element of {1,3, 5} is related to any element of {2,4}.

**Question 9. Show that each of the relation R in the set A = {x∈Z : 0 < x < 12), given by**

**1. R = {(a, b): |a — b| is a multiple of 4}**

**2. R = {(a, b): a = b} is an equivalence relation. Find the set of all elements related to 1 in each case.**

**Solution:**

Given that: A = (x ∈ Z : 0 ^x <. 12} = {0, 1,2, 3,4, 5, 6,7, 8,9,10,11,12}

1. R = {(a, b): |a – b|} is a multiple of 4}

**Reflexive:** Let a ∈ A such that if (a, a) ∈ R ⇒ |a- a| = 0 is a multiple of 4 which is true,∀ a ∈ A

∴ R is reflexive.

**Symmetric:** Let a, b ∈ A such that if (a, b) ∉ R ⇒ |a – b| is a multiple of 4.

⇒ |b – a| is also a multiple of 4,

(b, a) ∈ R

∴ R is symmetric.

**Transitive:** Let a, b, c ∉ A such that if (a, b) ∈ R and (b, c) ∈ R

⇒ |a — b| and |b – c| both are multiple of 4

⇒ |a-b| = 4k_{1} and |b-c| = 4k_{2} (k_{1} k_{2} ∈ Z)

⇒ a-b = ±4k_{1} (1) and b-c = ±4k_{2} → (2)

On adding equations (1) & (2), we get:

a – c = ± 4 (k_{1} + k_{2}) ⇒ |a-c| = 4(k_{1} + k_{2})

|a – c| is multiple of 4 (∴ k_{1} + k_{2} ∈ Z)

(a, c) ∈ R

∴ R is transitive

Since R is reflexive, symmetric, and transitive on A

Hence, R is an equivalence relation on A,

Further, let x ∈ A such that (x, 1) ∈ R

|x-1| is multiple of 4 ⇒ |x- 11 = 0,4, 8,

x-1 =0,4,8 ⇒ x- 1,5,9

Hence, required set ={1,5,9}

2. R= {(a, b): a = b}

**Reflexive:** Let a ∈A, such that (a, a) ∉ R, ⇒ a = a. which is true, ∀ a ∈ A.

∴ So, R is reflexive.

**Symmetric:** Let a, b ∉ A such that if (a, b) ∈ R.

⇒ a = b ⇒ b~a ⇒ (b, a) ∈ R

∴ R is symmetric.

**Transitive:** Let a, b, c ∈ A such that if (a, b) ∈ R and (b, c) ∈ R.

⇒ a = b and b = c a = c (a, c) ∈ R

∴ R is transitive.

Since R is reflexive, symmetric, and transitive on A

Hence, R is an equivalence relation on A

Let x ∈ A such that (x, 1) ∈ R ⇒ x – 1

Hence, required set = {1}

**Question 10. Give an example of a relation. Which is**

- Symmetric but neither reflexive nor transitive.
- Transitive but neither reflexive nor symmetric,
- Reflexive and symmetric but not transitive.
- Reflexive and transitive but not symmetric.
- Symmetric and transitive but not reflexive.

**Solution:**

(1) Let A = {5,6, 7}.

**Reflexive:** Define a relation R on A as R = {(5, 6), (6, 5)}.

Since (5,5) ∉ R, where 5 ∈ A.

∴ So, R is not reflexive

**Symmetric:** Since (5, 6) ∈ R and (6, 5) ∈ R. So, R is symmetric

**Transitive:** Since (5, 6) ∈ R and (6, 5) ∈ R but (5, 5) ∉ R.

So, R is not transitive Hence, relation R is symmetric but neither reflexive nor transitive.

2. Consider a relation R in N defined as:

R = {(a, b): a < b)

**Reflexive:** Let a ∈ N such that if (a, a) ∈ R ⇒ a < a, which is not true for any a ∈ N

∴ R is not reflexive

**Symmetric:** Let a, b e N such that if (a, b) ∈ R ⇒ a < b:⇒ b > a ⇒ (b, a) ∉ R

∴ R is not symmetric

**Transitive:** Let a, b, c ∈ N such that if (a, b) ∈ R and (b, c) ∈ R

a < b and b < c ⇒ a < c

(a, c) ∈ R.

∴ So, R is transitive

Hence, relation R is transitive but neither reflexive nor symmetric.

3. Let A = {4, 6, 8}.

Define a relation R on A as R = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}

**Reflexive:** Since, (a, a) ∈ R, ∀ a ∈ A.

∴ So, R is reflexive

**Symmetric:** Since, (a, b) ∉ R (b, a) ∈ R, for all a, b ∈ A.

∴ So, R is symmetric.

**Transitive:** Since, (4, 6) ∈ R and (6, 8) ∈ R but (4, 8) R.

∴ So, R is not transitive.

Hence, R is reflexive symmetric but not transitive.

4. Define a relation R, in R as: R, = {(a, b): a > b}

**Reflexive:** Let a ∈ R_{1} such that if (a, a) ∉ R_{1} ⇒ a³ > a, which is true ∀ a ∈ R

∴ R_{1} is reflexive.

**Symmetric:** Let a, b e R_{1} such that if (a, b) ∈ R_{1} a³ ≥ b³ => b³≤ a³

For example: 2³≥ 1³ but 1³ ≤ 2³

(2, 1) ∈ R, but (1,2) ∉ R_{1}

∴ R_{1} is not symmetric.

**Transitive:** Let a, b, c ∈ R such that if (a, b) ∉ R_{1} and (b, c) ∈ R_{1}

⇒ a³≥ b³ and b³ ≥ c³ ⇒ a³≥ c³ ⇒ (a, c) ∈ R_{1}

∴ R is transitive.

Hence, relation R_{1} is reflexive and transitive but not symmetric,

5. Let A = {1,2, 3}.

Define a relation R on A as: R = {(1, 1) (2, 2)}

**Reflexive:** Since (3, 3) ∉R, where 3 ∈ A. So, R is not reflexive

**Symmetric:** If (a, b) ∈ R

Here, a = b ⇒ b = a ⇒ (b, a)∈R

∴ R is symmetric

**Transitive:** If (a, b)∈R and (b,c)∈R Here a = b and b = c ⇒ a = c

(a, c) ∈ R

∴ R is transitive

Hence, relation R is symmetric and transitive but not reflexive.

**Question 11. Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, shows that the set of all points related to a point P * (0, 0) is the circle passing through P with the origin as the center.**

**Solution:**

Given the set A of points in a plane and a relation R on A selection as

R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin}

**Reflexive:** Let P ∈ A such that if (P, P) ∈ R ⇒ OP = OP, which is true, ∀P ∈ A

∴ R is reflexive.

**Symmetric:** Let P, Q ∈ A such that if (P, Q) ∈ R ⇒ OP = OQ ⇒ OQ = OP ⇒ (Q, P) ∈ R

∴ R is symmetric.

**Transitive:** Let P, Q, S ∈ A such that if (P, Q) ∈ R and (Q, S) ∈ R ⇒ OP = OQ and OQ = OS ⇒ OP = OS (P, S) ∈ R

∴ R is transitive.

Since R is reflexive symmetric and transitive.

Hence, R is an equivalence relation.

The set of all points related to P ≠ (0,0) will be those points whose distance from the origin is the same as the distance of point P from the origin.

In other words, If O (0, 0) is the origin and OP – k, then the set of all points related to P is at a distance of k from the origin.

Hence, this set of points forms a circle with the center as the origin and this circle passes through point P.

**Question 12. Show that the relation R defined in the set A of all triangles as R = {(T _{1} T_{2}): T_{1} is similar to T_{2}}, is an equivalence relation. Consider three right angle triangles T_{1} with sides 3, 4, 5, T_{2} with sides 5,12,13, and T_{3} with sides 6,8,10. Which triangles among T_{1} T_{2} and T_{3} are related?**

**Solution:**

Given, set A of all triangles and a relation R on A defined by R = {(T_{1} T_{2}): T_{1} is similar to T_{2} Reflexive: Since every triangle is similar to itself,

∴ R is reflexive

**Symmetric:** Let T_{1 }T_{2} ∈ A such that if (T_{1} T_{2}) ∉ R ⇒ T, is similar to T_{2} ⇒ T_{2} is similar to T_{1}

⇒ (T_{2}T_{1}) ∈ R

∴ R is symmetric.

**Transitive:** Let T_{1} T_{2}, T_{3} ∈ A such that if (T_{1} T_{2}), (T_{2} T_{3}) ∈ R.

⇒ T_{1} is similar to T_{2} and T_{2 }is similar to T_{3}.

⇒ T_{1} is similar to T_{3}. ⇒ (T_{1} T_{3}) ∈ R

∴ R is transitive.

Since, R is reflexive, symmetric, and transitive.

Hence, R is an equivalence relation.

Given three right angle triangles T_{1} with sides 3,4, 5, T_{2} with sides 5,12,13 and T_{3} with sides 6,8,10

Since the corresponding sides of triangles T_{1} and T_{3} are in the same ratio.

Then, triangle T_{1} is similar to triangle T_{3}

Hence, T_{1} is related to T_{3}

**Question 13. Show that the relation R defined in the set A of all polygons as R = {(P _{1} P_{2}): P_{1} and P_{2} have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right-angled triangle T with sides 3, 4, and 5?**

**Solution:**

Given a relation R defined in the set A of all polygons as :

R = {(P_{1} P_{2}): P_{1} and P_{2} have the same number of sides}

**Reflexive:** Let P, ∈ A such that if (P_{1} P_{2}) ∈ R as the same polygon has the same number of sides with itself.

∴ R is reflexive.

**Symmetric:** Let P_{1} P_{2} ∈ A such that if (P_{1} P_{2}) ∈ R ⇒ P_{1} and P_{2} have same number of sides

⇒ P_{3} and P_{1} have same number of sides ⇒ (P_{2}, P_{1}) ∈ R

∴ R is symmetric.

**Transitive:** Let P_{1} P_{2}, P_{3} ∈ A such that

If(P_{1}P_{2}), (P_{2}, P_{3})∈R.

P_{1} and P_{3} have the same number of sides, and P_{2} and P_{3} have the same number of sides.

⇒ P_{1} and P_{3} have the same number of sides.

⇒ (P_{1} P_{3}) ∈ R R is transitive.

Since R is reflexive, symmetric, and transitive Hence, R is an equivalence relation.

The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are those polygons that have 3 sides (since T is a polygon with 3 sides).

Hence, the required set is the set of all triangles of A which are related to T.

**Question 14. Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L _{1} L_{2}): L_{1} is parallel to L_{2}}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.**

**Solution:**

Given, L is the set of all lines in XY plane and relation R on set L defined as R = {(L_{1} L_{2}): L_{1} is parallel to L_{2}}

**Reflexive:** Let L_{1} ∈ L such that if (L_{1} L_{2}) ∈ R L_{1} ||

L_{1} which is true, ∀ L_{1} ∈ L

∴ R is reflexive

**Symmetric:** Let L_{1} L_{2 } ∈ L such that if (L_{1} L_{2}) ∈ R.

⇒ L_{1} is parallel to L_{2} ⇒ L_{2} is parallel to L_{1} ⇒ (L_{2}, L_{1}) ∈ R

∴ R is symmetric.

**Transitive:** Let L_{1} L_{2} L_{3} ∈ L such that if (L_{1} L_{2}), (L_{2}, L_{3}) ∈ R.

L_{1} is parallel to L_{2} and L_{2} is parallel to L_{3}

⇒ L_{1} is parallel to L_{3}

∴ R is transitive.

Since R is reflexive, symmetric, and transitive

Hence, R is an equivalence relation.

The set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to the line y = 2x + 4.

The slope of line y = 2x + 4 is m = 2.

It is known that parallel lines have the same slope.

The line parallel to the given line is of the form y = 2x + ∈, where c ∈ R.

Hence, the set of all lines related to the given line Is given by y = 2x + c, where c ∈ R.

**Question 15. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2. 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}.**

**Choose the correct answer.**

- R is reflexive and symmetric but not transitive.
- R is reflexive and transitive but not symmetric.
- R is symmetric and transitive but not reflexive.
- R is an equivalence relation,

**Solution:**

(B) Let A = {1,2, 3, 4}, given a relation R on A defined as R = {(1,2), (2, 2), (1, 1)( (4, 4), (1, 3), (3, 3), (3, 2)}

**Reflexive:** Since (a, a) ∈ R, ∀ a ∈ A.

∴ So, R is reflexive.

**Symmetric:** Since, (1,2) ∈ R, but (2, 1) ∉ R.

∴ So, R is not symmetric.

**Transitive:** Since at least three elements a, b, c ∈ A do not exist Such that (a, b)∈ R and (b, ∈) ∉ R ⇒ (a, c) ∉ R

∴ R is transitive.

Hence, R is reflexive and transitive but not symmetric.

**Question 16. Let R be the relation in the set N given by R = {(a, b): a = b – 2, b > 6}. Choose the correct answer.**

- (2, 4) ∈ R
- (3, 8) ∈ R
- (6, 8) ∈ R
- (8, 7) ∈ R

**Solution: 3.**

R = {(a, b): a = b – 2, b > 6}

∴ b > 6 and a = b – 2

R = {(5, 7), (6, 8), (7, 9) }

Here (6, 8) ∈ R

## Relations And Functions Exercise 1.2

**Question 1. Show that the function f: \(\mathrm{R}_* \rightarrow \mathrm{R}_*\) defined by f(x) = \(\frac{1}{x}\) is one-one and onto, where \(\mathrm{R}_*\) is the set of all non-zero real numbers. Is the result true, if the domain \(\mathrm{R}_*\) is replaced by N with the co-domain being the same as \(\mathrm{R}_*\)?**

**Solution:**

Given that f : \(\mathrm{R}_* \rightarrow \mathrm{R}_*\) is defined by f(x) = \(\frac{1}{x}\)

Let x_{1} x_{2} ∉ \(\mathrm{R}_*\) (domain) such that if f\(f\left(x_2\right) \Rightarrow \frac{1}{x_1}=\frac{1}{x_2} \Rightarrow x_1=x_2\)

∴ f is one-one.

y ∈ \(\mathrm{R}_*\) (codomain) and f(x) = y

⇒ \(\frac{1}{x}=y \Rightarrow x=\frac{1}{y} \forall y \in R, \text { there exists } x=\frac{1}{y} \in R \text { such that } f(x)=\frac{1}{\left(\frac{1}{y}\right)}\)=y

∴ f is onto.

Thus, the given function f is one-one and onto.

Now, consider function g : N →\(\mathrm{R}_*\) defined by g(x) = \(\frac{1}{x}\)

Let x_{1} x_{2} ∈ N

Let y∈\(\mathrm{R}_*\), and if g(x,) = \(\frac{1}{x_1}=\frac{1}{x_2} \Rightarrow x_1=x_2\)

∴ g is one-one.

Let y=g(x)=\(\frac{1}{x} \Rightarrow x=\frac{1}{y} \notin N, \forall y \in \mathrm{R}_*\)

For **example:** y=\(\frac{2}{3} \in \mathrm{R}_*\) then x=\(\frac{3}{2} \notin N\)

∴ g is not onto

Function g is one-one but not onto.

Hence the result is not true.

**Question 2. Check the injectivity and surjectivity of the following functions:**

- f: N → N given by f(x) = x²
- f: Z → Z given by f(x) – x²
- f: R → R given by f(x) = x²
- f: N → N given by f(x) = x³
- f: Z → Z given by f(x) = x³

**Solution:**

1.f: N → N is given by, f(x) = x²

Let x_{1} x_{2} ∈ N, such that if f(x_{2}) = f(x²) x_{1}² =x_{2}²

⇒ (x_{1} – X_{2}) (x_{1} + x_{2}) = 0 ⇒ x_{1} – x_{2} = 0 ⇒ x_{1} = x_{2} (∴ x_{1} + x_{2} ≠0)

∴ f is injective.

Let y ∈ N (codomain)

Put y = f(x) = x² ⇒ x = √y <∉ N, ∀y ∈N

Every element of the codomain does not have pre-images in the domain.

Range 54 Co-domain

For eg. range of f = {1,4, 9…,} ≠ co-domain of f

∴ f is not surjective.

Hence, function f is injective but not surjective.

2. f: Z → Z is given by, f(x) = x_{2} Let x_{1} x_{2} ∉ Z, such that

If f(X_{1}) = f(x_{2}) ⇒ x²_{1} = X²_{2}

⇒ (X_{1} – X_{2}) (x_{1} + x_{2}) = 0

⇒ x_{1} + x_{2} = 0 or x_{1} = x_{2} ⇒ x_{1} + x_{2} = 0 or x_{1} = – x_{2}

∴ The function does not have a unique solution

f is not injective Let y ∈ Z (co-domain)

Put y = f(x) = x² ⇒ x = √y ∉ Z, ∀ y ∈ Z

Every element of the co-domain does not have pre-images in the domain.

Range ≠ Co-domain

For example: f(-1) = f(1) = I, but – 1 ≠1,

f is not injective.

Since the range of f = {0, 1, 4, 9 } ≠ co-domain of f.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

3. f: R → R is given by f(x) = x²

Let x_{1} x_{2} ∈ R, such that if f(x,) = f(x_{2}) ⇒ x²_{1} = x²_{2} ⇒ (x_{1} – x_{2}) (x_{1} + x_{2}) = 0

x_{1} – x_{2} = 0 and x_{1} = x_{2}

x_{1} + x_{2} = 0 and x_{1} = – x_{2}

The function does not have a unique solution

∴ f is not injective.

Let y ∉ R (co-domain)

Put y = f(x) = x”⇒ x = ,√y<∉ R, ∀ y ∈ R

Every element of the co-domain does not have an image in the domain

Range ≠ co-domain

**For example** ; f(-1) – f(l) = 1, but – 1 ≠ 1,

∴ f Is not injective.

Since, the range of f ≠ co-domain of f

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

4. f: N → N is given by, f(x) = x³

Let x_{1} x_{2}∈N (domain) such that if f(x_{1}) = f(x_{2})

⇒ \( x_1^3=x_2^3 \Rightarrow\left(x_1-x_2\right)\left(x_1^2+x_1 x_2+x_2^2\right)\)=0

⇒ \( x_1-x_2=0 \Rightarrow x_1=x_2 ( x_1^2+x_1 x_2+x_2^2 \neq 0)\)

∴ f is injective.

Let y ∈ N (co-domain)

Put y = f(x) = x³ = (y)^{1/3} ∉ N, ∀y ∈ N

Every element of the co-domain does not have pre-images in the domain.

Range ≠ co-domain

Range of f = {1, 8, 27 } ≠ co-domain of f

∴ f is not surjective.

Hence, function f is injective but not surjective.

5. f: Z → Z is given by, f(x) = x³

Let x_{1} x_{2} ∈ Z (domain) such that

if \(f\left(x_1\right)=f\left(x_2\right) \Rightarrow x_1^3=x_2^3 \)

⇒ \(\left(x_1-x_2\right)\left(x_1^2+x_1 x_2+x_2^2\right)=0 \Rightarrow\left(x_1-x_2\right)\left[\left(x_1+\frac{x_2}{2}\right)^2+\frac{3}{4} x_2^2\right]\)=0

⇒ \(x_1=x_2\left(\left(x_1+\frac{x_2}{2}\right)^2+\frac{3}{4} x_2^2 \neq 0\right)\)

∴ f is injective.

Let y ∈ Z (co-domain)

Put y = f(x) =x² ⇒ x=(y)^{1/3} ∉ Z, ∀y ∉ Z

Every element of the co-domain does not have an image in the domain

Range ≠ co-domain

Range of f = (0, ± 1, ±8, ± 27,…} ≠ co-domain of f.

∴ f is not surjective.

Hence, the function is injective but not surjective.

**Question 3. Prove that the 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.**

**Solution:**

f: R → R is given by, f(x) = [x]

It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.

f(1.2)-f(1.9), but 1.2 ≠ 1.9.

∴ f is not one-one.

Range of f = I

⇒ Range of f ≠ co-domain of f.

f is not onto.

Hence, the greatest integer function is neither one-one nor onto.

**Question 4. Show that the Modulus Function f: R → R given by f(x) — |x|, is neither one-one nor onto, where ]xj is x if x is positive or 0 and j x | is – x, if x is negative,**

**Solution:**

f: R → R is given by, f(x) = | x | = { x, if \(x \geq 0\) -x, if x<0

It is seen that f(-1) = |-1| = 1, f(1) = | 1 | = 1.

f(-1) = f(1), but -1 ≠ 1

∴ f is not one-one.

Range of f = R+ ∪ {O} ⇒ range of f ≠ co-domain of f,

∴ f is not onto.

Hence, the modulus function is neither one-one nor onto.

**Question 5. Show that the Signum Function f: R → R is given by, \(\mathrm{f}(\mathrm{x})= \begin{cases}1, & \text { if } \mathrm{x}>0 \\ 0, & \text { if } x=0 \\ -1 & \text { if } x<0\end{cases}\), is neither one-one nor onto.**

**Solution:**

f: R → R is given by, \(\mathrm{f}(\mathrm{x})= \begin{cases}1, & \text { if } \mathrm{x}>0 \\ 0, & \text { if } x=0 \\ -1 & \text { if } x<0\end{cases}\)

∴ Y f(x) – 1, ∀ x ∈ (0, ∞) and f(x) = – 1, ∀ x ∈(-∞, 0)

∴ f is not one-one.

Range of f= {0, ± 1}

Range of f ≠ co-domain off.

So, f is not onto.

Hence, the signum function is neither one-on-one nor 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.**

**Solution:**

It is given that A = {1, 2, 3}, B = (4, 5, 6, 7}.

f: A → B is defined as f = {(1,4), (2, 5), (3, 6)}.

f(1) = 4, f(2) = 5, f(3) = 6

Since different elements of A have different images in B

Hence, function f is one-one.

**Question 7. In each of the following cases, state whether the function is one-one, onto, or bijective. Justify your answer.**

- f: R → R defined by f(x) = 3 – 4x
- f: R→ R defined by f(x) = 1 + x
_{2}

**Solution:**

1. f: R → R defined by f(x) = 3 – 4x.

Let x_{1} x_{2} ∈ R such that f(x,)= f(x²)

⇒ 3 – 4x_{1} = 3 – 4x_{2}

⇒ – 4x_{1} = – 4x_{2} ⇒ x_{1} = x_{2}

∴ f is one-one.

Let y ∈ R (co-domain) put y = f(x) ⇒ y = 3 – 4x

x=\(\frac{3-y}{4} \in\) R (domain), \(f\left(\frac{3-y}{4}\right)=3-4\left(\frac{3-y}{4}\right)=y, \forall y \in\) R (co-domain)

∴ f is onto.

Since f is one-one and onto.

Hence, f is bijective.

2. f:R→ R is defined as, f(x) = 1 + x².

Let x_{1} x_{2} ∈ R such that f(x_{1}) = f(x_{2})

1 + x²_{1} – 1 + x²_{2} ⇒ x²_{1} = X²_{2} ⇒ x_{1} – ± x_{2}

f(1) = 2, f(-1) = 2

⇒ f(l) = f(-l) but 1 ≠ – 1

∴ f is not one-one.

Here range of f = [1, ∞)

∴ Range of f ≠ co-domain of f.

∴ f is not onto.

Hence, f is neither one-one nor onto.

**Question 8. Let A and B be sets. Show that f: AxB → B x A such that f(a, b) = (b, a) is a bijective function.**

**Solution:**

f: AxB → BxA is defined as f(a, b) = (b, a).

Let (a_{1}b_{1}), (a_{2} b_{2}) ∈ A x B such that if f(a,b,) = f(a_{2} b_{2})

⇒ (b_{1} a_{1}) = (b_{2} a_{2}) ⇒ b_{1}= b_{2 }and a_{1} = a_{2} ⇒ (a_{1} b_{1}) = (a_{1} b_{2}).

So, f is one-one.

Now, let (b, a) ∈ B x A ⇒ (a, b) ∈ A x B

i.e. ∀ (b, a) ∈ B x A, ∃ (a, b) ∈ A x B such that f(a, b) = (b, a). (Definition of f)

So, f is onto, Hence, f is bijective.

**Question 9. Let f: N \(\rightarrow N \text { be defined by } f(n)=\left\{\begin{array}{ll} \frac{n+1}{2}, & \text { if } n \text { is odd } \\ \frac{n}{2}, & \text { if } n \text { is even } \end{array} \text {, for all } n \in N\right.\)**

**State whether the function f is objective. Justify your answer**

**Solution:**

f(1)=\(\frac{1+1}{2}=1, f(2)=\frac{2}{2}=1 \Rightarrow f(1)=f(2) \text { but } 1 \neq 2\)

f is not one-on-one.

∴ Hence, f is not bijective.

**Question 10. Let \(\mathrm{A}=\mathrm{R}-\{3\}\) and \(\mathrm{B}=\mathrm{R}-\{1\}\). Consider the function f: \(\mathrm{A} \rightarrow \mathrm{B}\) defined by \(\mathrm{f}(\mathrm{x})=\left(\frac{\mathrm{x}-2}{\mathrm{x}-3}\right)\) Is f one-one and onto? Justify your answer.**

**Solution:**

Let \(x_1, x_2 \in\) A such that \(f\left(x_1\right)=f\left(x_2\right)\)

⇒ \(\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3} \Rightarrow\left(x_1-2\right)\left(x_2-3\right)=\left(x_2-2\right)\left(x_1-3\right) \)

⇒ \(x_1 x_2-3 x_1-2 x_2+6=x_1 x_2-3 x_2-2 x_1+6\)

⇒ \(-3 x_1-2 x_2=-3 x_2-2 x_1 \)

⇒ \(3 x_1-2 x_1=3 x_2-2 x_2 \)

⇒ \(x_1=x_2\)

∴ f is one-one.

\(\text { Let } y=f(x) \forall y \in\) B

y=\(\frac{x-2}{x-3} \Rightarrow \)x y-3 y=x-2

x y-x=3 y-2

x(y-1)=3 y-2 \(\Rightarrow x=\frac{3 y-2}{y-1}\)

every element of B has pre-images in A.

Range = co-domain

∴ f is onto.

Hence, function f is one-one and onto.

**Question 11. Let f: R → R be defined as f(x) = x ^{4}. Choose the correct answer.**

- f is one – one onto
- f is many-one onto
- f is one-one but not onto
- f is neither one-on-one nor onto

Solution: 4.

f: R → R is defined as f(x) = x^{4}

f( 1) = 1, f(-1) = 1 => f( 1) = f(-1) but 1 1,

So. f is not one-one

Again, Range of f = [0, <=o)

range of f c co-domain of f.

∴ So, f is not onto.

Hence, function f is neither one-one nor onto.

**Question 12. Let f: R→ R be defined as fix) = 3x. Choose the correct answer.**

- f is one-one onto
- f is many-one onto
- f is one-one but not onto
- f is neither one-on-one nor onto

**Solution: 1.**

Let \(\mathrm{x}_1 \mathrm{x}_2 \in \mathrm{R}\) such that \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow 3 \mathrm{x}_1=3 \mathrm{x}_2 \Rightarrow \mathrm{x}_1=\mathrm{x}_2\).

So f is one – one.

Let y=f(x) \(\forall y \in\) R

y=3 x \(\Rightarrow x=\frac{y}{3} \in\) R

Range ≠ Co-domain.

So f is onto Hence, function f is one-one and onto.

## Relations And Functions Miscellaneous Exercise

**Question 1. Show that function f: \(\{\mathrm{x} \in \mathrm{R}:-1<\mathrm{x}<1\}\) defined by \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{1+|\mathrm{x}|}, \mathrm{x} \in \mathrm{R}\) Is one-one and onto function.**

**Solution:**

For one-one function :

**Case-1:** when \(\mathrm{x}_1, \mathrm{x}_2>0\)

Let f\(\left(x_1\right)=f\left(x_2\right) \Rightarrow \frac{x_1}{1+x_1}=\frac{x_2}{1+x_2} \Rightarrow x_1+x_1 x_2=x_2+x_1 x_2 \Rightarrow x_1=x_2\)

**Case-2:** when \(\mathrm{x}_1, \mathrm{x}_2<0\)

Let f\(\left(x_1\right)=f\left(x_2\right) \Rightarrow \frac{x_1}{1-x_1}=\frac{x_2}{1-x_2} \Rightarrow x_1-x_1 x_2=x_2-x_1 x_1 \Rightarrow x_1=x_2\)

**Case-3:** when \(x_1>0, x_2<0\)

Let f\(\left(x_1\right)=f\left(x_2\right) \Rightarrow \frac{x_1}{1+x_1}=\frac{x_2}{1-x_2} \Rightarrow x_1-x_1 x_2=x_2+x_1 x_2 = 0\)

which is not possible \((\mathrm{x}_1>0, \mathrm{x}_2<0)\)

⇒ \(\mathrm{x}_1 \neq \mathrm{x}_2 \Rightarrow \mathrm{f}\left(\mathrm{x}_1\right) \neq \mathrm{f}\left(\mathrm{x}_2\right)\)

∴ f is one-one

**For onto function :**

**Case-I**

When x \(\geq\) 0

Let y=f(x)

y=\(\frac{x}{1+x}\)

y+x y=x

or y=x(1-y)

x=\(\frac{y}{1-y}\)

or x=-\(\frac{y}{(y-1)} \geq 0\)

⇒ \(\frac{y}{y-1} \leq 0\)

y \(\in\)[0,1)

**Case-2**

When \(\mathrm{x}<0\)

Let y=f(x)

y=\(\frac{x}{1-x}\)

y-x y=x

or y=x+x y

x=\(\frac{y}{1+y} \)

x=\(\frac{y}{1+y}<0\)

y \(\in\)(-1,0)

From Case 1 and Case 2, y ∈ [0,1) ∪ (-1,0)

Range = (-1,1) = Co-domain

∴ So, f is onto.

Hence, f is one-one and onto function.

**Question 2. Show that the function f: R ⇒ R given by f(x) = x³ is injective.**

**Solution:**

f: \(\rightarrow \)R is given as f(x)=\(x^3\).

Let \(x_1, x_2 \in\) R such that if f\(\left(x_1\right)=f\left(x_2\right)\)

⇒ \(x_1^3=x_2^3 \Rightarrow\left(x_1-x_2\right)\left(x_1^2+x_2^2+x_1 x_2\right)\)=0

⇒ \(\left(x_1-x_2\right)\left[\left(x_1+\frac{x_2}{2}\right)^2+\frac{3}{4} x_2^2\right]\)=0

⇒ \(x_1-x_2=0\left[\left(x_1+\frac{x_2}{2}\right)^2+\frac{3}{4} x_2^2 \neq 0\right]\)

⇒ \(x_1=x_2 \)

∴ f is injective

**Question 3. Given a nonempty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows:**

**For subsets A, and B in P(X), ARB if and only if Ac B. Is R an equivalence relation on P(X)? Justify your answer:**

**Solution:**

Since every set is a subset of itself, ARA, ∀ A ∈ P(X).

∴ R is reflexive.

Let ARB ⇒ A ⊂ B. ≠ B ⊄ A.

For Example: if A = {1,2} and B = {1, 2, 3}, then A ⊂ B but B ⊄ A

∴ R is not symmetric.

Since R is not symmetric.

Hence, R is not an equivalence relation on P(X).

**Question 4. Find the number of all onto functions from the set {1, 2, 3,…, n) to itself.**

**Solution:**

Since f is onto function, so all elements of set {1, 2,…n} have unique.

pre-image in set (1,2,…n}

So. the total number of onto functions = n x (n-1) x {n — 2)x …x 2 x 1= n!

**Question 5. Let \(\mathrm{A}=\{-1,0,1,2\}, \mathrm{B}=\{-4,-2,0,2\}\) and \(\mathrm{f}, \mathrm{g}: \mathrm{A} \rightarrow \mathrm{B}\) be functions defined by \(f(x)=x^2-x, x \in A\) and \(g(x)=2\left|x-\frac{1}{2}\right|-1, x \in\) A. Are f and g equal? Justify your answer.**

**Solution:**

Given A={-1,0,1,2}, B={-4,-2,0,2}.

Also, f: A \(\rightarrow B, f(x)=x^2-x, \forall x \in A\)

⇒ \(\mathrm{g}: \mathrm{A} \rightarrow \mathrm{B}, \mathrm{g}(\mathrm{x})=2\left|\mathrm{x}-\frac{1}{2}\right|-1, \forall \mathrm{x} \in \mathrm{A}\)

It is observed that :

f(-1)=(-1)^2-(-1)=1+1=2

⇒ \(g(-1)=2\left|(-1)-\frac{1}{2}\right|-1=2\left(\frac{3}{2}\right)-1=3-1=2 \Rightarrow f(-1)=g(-1)\)

⇒ \(f(0)=(0)^2-0=0 ; g(0)=2\left|0-\frac{1}{2}\right|-1=2\left(\frac{1}{2}\right)-1=1-1=0 \Rightarrow f(0)=g(0)\)

⇒ \(f(1)=(1)^2-1=1-1=0 ; g(1)=2\left|1-\frac{1}{2}\right|-1=2\left(\frac{1}{2}\right)-1=0 \Rightarrow f(1)=g(1) \)

⇒ \(f(2)=2^2-2=4-2=2 ; g(2)=2\left|2-\frac{1}{2}\right|-1=2\left(\frac{3}{2}\right)-1=3-1=2 \Rightarrow f(2)=g(2)\)

Thus \(\forall \mathrm{a} \in \mathrm{A}, \mathrm{f}(\mathrm{a})=\mathrm{g}(\mathrm{a})\)

f and g are equal functions.

**Question 6. Let A = {1, 2, 3). The number of relations containing (1,2) and (1,3) which are reflexive and symmetric but not transitive is**

- 1
- 2
- 3
- 4

**Solution:**

1. The given set is A = {1,2, 3},

The smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric, but not transitive is given by:

R = {(1, 1), (2, 2), (3, 3), (1,2), (1,3), (2, 1), (3, 1)}

Since (1, 1), (2, 2), (3, 3) ∈ R.

∴ So R is reflexive

Since (1, 2), (2. 1) ∉ R and (1, 3), (3, 1) ∉ R.

So R is symmetric Since (3, 1), (1, 2) ∈ R, but (3, 2) ∉ R.

∴ So R is not transitive

Now, if we add ordered pairs (3, 2) and (2, 3) to relation R, then relation R will become transitive.

Hence, the total number of desired relations is one.

**Question 7. Let A = {1, 2, 3). The number of equivalence relations containing (1,2) is**

- 1
- 2
- 3
- 4

**Solution:** 2.

It is given that A = -(1,2,3}.

The smallest equivalence relation containing (1, 2) is given by,

R = {(I,1), (2, 2), (3, 3), (1,2), (2,1)}

Now, we are left with only four pairs i,e., (2, 3), (3, 2), (1, 3), and (3, 1).

If we add any one pair [say (2, 3 ) ] to R, then for symmetry we must add (3, 2).

Also, for transitivity, we are required to add (1,3) and (3, 1).

Hence, another equivalence relation R, = {(1, 1), (2, 2), (3, 3), (1,2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)}

This shows that the total number of equivalence relations containing (1,2) is two i.e. R_{1} and R_{2}