Algebra

Relations & Functions

Medium

core

Given relation $R = \{(x, y) : y = x + 5, x < 4, x, y \in N\}$ . Where N is a set of natural numbers then :

R is an equivalence relation.

R is transitive but neither reflexive nor symmetric.

R is reflexive but neither symmetric nor transitive.

R is symmetric & transitive but not reflexive.

✅ Correct Option: 2

$R = \{(1,6), (2,7), (3,8)\}$ .

Not reflexive: $(1,1) \notin R$ .

Not symmetric: $(1,6) \in R$ but $(6,1) \notin R$ .

Transitive vacuously: no $(a,b),(b,c)$ pairs exist where $b$ is the first element of another pair.

2026: 26 May Shift 2

Let $A = \{ 1,2,3 \}$ , then Match List-I with List-II

LIST-I	LIST-II
A. $R_1 = \{(1,1), (2,2), (3,3), (1,2)\}$	I. Symmetric but neither reflexive nor transitive
B. $R_2 = \{(1,1), (3,3), (1,3), (3,1)\}$	II. Reflexive and transitive but not symmetric
C. $R_3 = \{(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)\}$	III. Symmetric and transitive but not reflexive
D. $R_4 = \{(2,2), (3,3), (1,2), (2,1)\}$	IV. Reflexive and symmetric but not transitive

Choose the correct answer from the options given below:

2025: 26 May Shift 2

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

2026: 31 May Shift 2

A relation $R$ on set of real numbers $(\mathbb{R})$ is defined as $R = \{(a,b): a \le b, a,b \in \mathbb{R}\}$ . The relation is :

2025: 30 May Shift 2

The relation R on the set of real numbers defined by $R = \{(a, b): a \leq b^2\}$ is

(A) Reflexive

(B) Not symmetric

(D) Transitive

Choose the correct answer from the options given below:

2024: 16 May Shift 1

If the function $f: \mathbb{N} \rightarrow \mathbb{N}$ is defined as $f(n)=\left\{\begin{array}{ll}n-1, & \text { if } n \text { is even } \\ n+1, & \text { if } n \text { is odd }\end{array}\right.$ , then

(A) f is injective

(B) f is into

(D) f is invertible

2022: 16 July Shift 2

A relation R = {(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(2,3)} on A = {1,2,3} will be an equivalence relation, if we delete:

Choose the correct answer from the options given below:

Given relation R={(x,y):y=x+5,x<4,x,y∈N}R = \{(x, y) : y = x + 5, x < 4, x, y \in N\}R={(x,y):y=x+5,x<4,x,y∈N}. Where N is a set of natural numbers then :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Given relation $R = \{(x, y) : y = x + 5, x < 4, x, y \in N\}$ . Where N is a set of natural numbers then :