Algebra

Relations & Functions

Medium

core

A relation $f: N \rightarrow N$ be defined by $f(x) = x^2$ , $x \in N$ (Set of Natural numbers). Then $f(x)$ is

neither injective nor surjective

injective only

surjective only

bijective

✅ Correct Option: 2

A function $f: N \rightarrow N$ is defined by $f(x) = x^2$ , where $x \in N$ (the set of natural numbers).

To determine the nature of this function, we check if it is injective (one-to-one) and surjective (onto).

For injectivity, assume $f(a) = f(b)$ for $a, b \in N$ .

$a^2 = b^2$

Since $a, b$ are natural numbers (positive), this implies $a = b$ .

Different inputs produce different outputs:

$f(1) = 1$
$f(2) = 4$
$f(3) = 9$
$f(4) = 16$

Therefore, $f(x)$ is injective.

For surjectivity, every natural number in the codomain must have a pre-image.

Checking which natural numbers can be obtained:

$1 = f(1)$ ✓
$2$ cannot be obtained (no natural number squares to give $2$ )
$3$ cannot be obtained (no natural number squares to give $3$ )
$4 = f(2)$ ✓
$5$ cannot be obtained (no natural number squares to give $5$ )

Only perfect squares $(1, 4, 9, 16, 25, ...)$ can be obtained, not all natural numbers.

Therefore, $f(x)$ is not surjective.

The function $f(x) = x^2$ from $N$ to $N$ is injective only.

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Which of the following relations on the set $A = \{1, 2, 3\}$ are equivalence ?

(A) $R = \{(1,1), (2,2), (1,2), (2,1)\}$

(B) $R = \{(1,1), (2,2), (3,3)\}$

(D) $R = \{(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2), (1,3), (3,1)\}$

(E) $R = \{(1,1), (2,2), (3,3), (1,2)\}$

Choose the correct answer from the options given below :

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2026: 21 May Shift 1

The number of symmetric relations on set $A = \{1,2,3\}$ is

A relation f:N→Nf: N \rightarrow Nf:N→N be defined by f(x)=x2f(x) = x^2f(x)=x2, x∈Nx \in Nx∈N (Set of Natural numbers). Then f(x)f(x)f(x) is

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CUET Mathematics Past Year Questions (Topic-Wise)

A relation $f: N \rightarrow N$ be defined by $f(x) = x^2$ , $x \in N$ (Set of Natural numbers). Then $f(x)$ is