Algebra

Relations & Functions

Medium

core

The relation R in the set of real numbers, defined as R = {(a, b): 1 + ab > 0}, is:

Reflexive and symmetric but not transitive

Reflexive but neither symmetric nor transitive

Reflexive and transitive but not symmetric

Neither reflexive nor symmetric nor transitive

✅ Correct Option: 1

$1+a^2>0$ always, so R is reflexive, and $1+ab=1+ba$ makes it symmetric. For transitivity take $a=1,b=0,c=-1$ : $(1,0)$ and $(0,-1)$ lie in R, but $1+(1)(-1)=0$ is not positive, so $(1,-1)$ does not. Hence not transitive.

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

2026: 21 May Shift 1

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

2026: 31 May Shift 2

The function $f: R \to [0,1]$ defined by $f(x) = \dfrac{x^2}{x^2+1}$ is: (Where $R$ is set of real numbers)

A. One-one

B. one-one and onto

C. onto

D. Not one-one

Choose the correct answer from the options given below:

2025: 16 May Shift 1

If R and S are two equivalence relations on a set A, then

2026: 14 May Shift 2

Let N be the set of natural numbers and R be the set of all real numbers. If function $f:N\to R$ defined by $f(x)=4x^2+12x+15$ , then $f:N\to range(f)$ 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:

The relation R in the set of real numbers, defined as R = {(a, b): 1 + ab > 0}, is:

Solution

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