Algebra

Relations & Functions

Easy

core

Let $A = \{0, 1, 2, 3\}$ and R be a relation on set A defined as $R = \{(0,0), (0,1), (0,3), (1,0), (1,1), (2,2), (3,0), (3,3)\}$ , then the relation R is

reflexive and symmetric but not transitive.

reflexive and transitive but not symmetric.

Symmetric and transitive but not reflexive.

equivalence relation

✅ Correct Option: 1

All four pairs $(0,0),(1,1),(2,2),(3,3)$ are present, so R is reflexive, and every off-diagonal pair appears with its reverse, so R is symmetric. But $(1,0)$ and $(0,3)$ are in R while $(1,3)$ is not, so R is not transitive.

2026: 18 May Shift 1

Let a relation $R$ be defined on a set $A = \{1,2,3,4,5\}$ as $R = \{(a,b): |a-b| \text{ is even or zero}\}$ . Then which of the following statements is not TRUE?

2026: 29 May Shift 2

Match the LIST-I with LIST-II (Consider N and R as set of natural numbers and set of real numbers respectively)

	LIST-I Function		LIST-II
A.	$f: N \to N, f(x) = x^2$ is	I.	Neither one-one nor onto
B.	$f: R \to R, f(x) = 3 - 2x$ is	II.	one-one but not onto
C.	$f: R \to R, f(x) = x^2$ is	III.	onto but not one-one
D.	$f: \{-1,1\} \to \{1\}, f(x) = \vert x \vert$ is	IV.	one-one and onto

Choose the correct answer from the options given below:

2025: 21 May Shift 1

Let N be set of natural numbers, then the function $f: \mathbb{N} \rightarrow \mathbb{N}$ , defined by

$f(x) = \begin{cases} \frac{n+1}{2} & \text{, if } n \text{ is odd} \\ \frac{n}{2} & \text{, if } n \text{ is even} \end{cases}$ , is

2026: 12 May Shift 2

Let $A=R-\{2\}$ and $B=R-\{1\}$ and a function $f:A\to B$ defined by $f(x)=\dfrac{x-1}{x-2}\ \forall x\in A$ , then the function $f(x)$ 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: 27 May Shift 1

The domain of the function $y = \sin^{-1}(x-1) + \cos^{-1}\sqrt{x-1}$ is:

Let A={0,1,2,3}A = \{0, 1, 2, 3\}A={0,1,2,3} and R be a relation on set A defined as R={(0,0),(0,1),(0,3),(1,0),(1,1),(2,2),(3,0),(3,3)}R = \{(0,0), (0,1), (0,3), (1,0), (1,1), (2,2), (3,0), (3,3)\}R={(0,0),(0,1),(0,3),(1,0),(1,1),(2,2),(3,0),(3,3)}, then the relation R is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Let $A = \{0, 1, 2, 3\}$ and R be a relation on set A defined as $R = \{(0,0), (0,1), (0,3), (1,0), (1,1), (2,2), (3,0), (3,3)\}$ , then the relation R is