R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x - 3 . Then {R^{ - 1}} is

English

Gujarati

Hindi

1.Relation and Function

normal

$R$ is a relation from $\{11, 12, 13\}$ to $\{8, 10, 12\}$ defined by $y = x - 3$. Then ${R^{ - 1}}$ is

A

$\{(8, 11), (10, 13)\}$

B

$\{(11, 18), (13, 10)\}$

C

$\{(10, 13), (8, 11)\}$

D

None of these

Solution

(a) $R$ is a relation from $\{11, 12, 13\}$ to $\{8, 10, 12\}$ defined by $y = x – 3 \Rightarrow x – y = 3$

$\therefore R = \{11, 8\},\{13, 10\}.$

Hence, ${R^{ – 1}} = \{ 8,\,11\} ;\{ 10,\,13\} $.

Standard 12

Mathematics

Share

Similar Questions

Let $\mathrm{T}$ be the set of all triangles in a plane with $\mathrm{R}$ a relation in $\mathrm{T}$ given by $\mathrm{R} =\left\{\left( \mathrm{T} _{1}, \mathrm{T} _{2}\right): \mathrm{T} _{1}\right.$ is congruent to $\left. \mathrm{T} _{2}\right\}$ . Show that $\mathrm{R}$ is an equivalence relation.

easy

The relation "less than" in the set of natural numbers is

easy

Let $L$ denote the set of all straight lines in a plane. Let a relation $R$ be defined by $\alpha R\beta \Leftrightarrow \alpha \bot \beta ,\,\alpha ,\,\beta \in L$. Then $R$ is

easy

Show that each of the relation $R$ in the set $A=\{x \in Z: 0 \leq x \leq 12\},$ given by $R =\{( a , b ): a = b \}$ is an equivalence relation. Find the set of all elements related to $1$ in each case.

medium

The relation $R$ defined on the set $A = \{1, 2, 3, 4, 5\}$ by $R = \{(x, y)$ : $|{x^2} – {y^2}| < 16\} $ is given by

easy