Question

Let $A=\{x:x\in Z,|x|\le 3\}$ and $B=\{y;y\in N,-1<y+2\le 4\}$. If $R$ be the relation from $A$ to $B$, then

• A. Co domain of $R$ is $\{1,2\}$
• B. Range of $R$ is $\{0,1,2\}$
• C. $R^{-1}$ can be $\left\{\right(1,1),(1,2),(1,3),(2,1),(2,2),(2,3\left)\right\}$
• D. Co domain of $R^{-1}$ is $\{-3,-2,-1.0,1,2,3\}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A Co domain of $R$ is $\{1,2\}$
C $R^{-1}$ can be $\left\{\right(1,1),(1,2),(1,3),(2,1),(2,2),(2,3\left)\right\}$
D Co domain of $R^{-1}$ is $\{-3,-2,-1.0,1,2,3\}$

Here, $A=\{x:x\in Z,|x|\le 3\}$
$|x|\le 3$
$\Rightarrow -3\le x\le 3$
Since, $x\in Z$
$A=\{-3,-2,-1,0,1,2,3\}$

$B=\{y;y\in N,-1<y+2\le 4\}$
$-1<y+2\le 4$
$\Rightarrow -3<y\le 2$
Since, $y\in N$
$B=\{1,2\}$

Here, $R$ is a relation from $A$ to $B$
$\therefore$ Codomain of $R$ is $B=\{1,2\}$
Since Range of a relation is $\subset$ of Co domain
Range $\ne \{0,1,2\}$

$R^{-1}$ will be a relation from $B$ to $A$
$R^{-1}\subset B\times A$
$\therefore R^{-1}$ can be
$\left\{\right(1,1),(1,2),(1,3),(2,1),(2,2),(2,3\left)\right\}$
$\therefore$ Codomain of $R^{-1}$ is $A=\{-3,-2,-1,0,1,2,3\}$