Question

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

• A. $\left\{(1,1),(2,1),(3,1),(4,1),(2,3)\right\}$
• B. $\left\{(2,2),(3,2),(4,2),(2,4)\right\}$
• C. $\left\{(3,3),(3,4),(5,4),(4,3),(3,1)\right\}$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for correct option

Given that $\mathrm{A}=\left\{1,2,3,4,5\right\}$ and $\mathrm{R}=\left\{(\mathrm{x},\mathrm{y}):|{\mathrm{x}}^2–{\mathrm{y}}^2|<16\right\}$

$\bullet$For, $\mathrm{x}=1$, we have;

$|{\left(1\right)}^2–{\mathrm{y}}^2|<16$

$\Rightarrow$$|{\mathrm{y}}^2-1|<16$

$\Rightarrow$$\mathrm{y}=1,2,3,4$

$\bullet$For, $\mathrm{x}=2$, we have;

$|{\left(2\right)}^2–{\mathrm{y}}^2|<16$

$\Rightarrow$$|{\mathrm{y}}^2-4|<16$

$\Rightarrow$$\mathrm{y}=1,2,3,4$

$\bullet$For, $\mathrm{x}=3$, we have;

$|{\left(3\right)}^2–{\mathrm{y}}^2|<16$

$\Rightarrow$$|{\mathrm{y}}^2-9|<16$

$\Rightarrow$$\mathrm{y}=1,2,3,4$

$\bullet$For, $\mathrm{x}=4$, we have;

$|{\left(4\right)}^2–{\mathrm{y}}^2|<16$

$\Rightarrow$$|{\mathrm{y}}^2-16|<16$

$\Rightarrow$$\mathrm{y}=1,2,3,4,5$

$\bullet$For, $\mathrm{x}=5$, we have;

$|{\left(5\right)}^2–{\mathrm{y}}^2|<16$

$\Rightarrow$$|{\mathrm{y}}^2-25|<16$

$\Rightarrow$$\mathrm{y}=4,5$

Thus, the relation $\mathrm{R}$ is given by $\left\{\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(5,4\right),\left(5,5\right)\right\}$

Hence, the correct option is option(D) i.e. None of these