Q42 / 75 · VITEEE 2021 · Mathematics
The relation $\mathrm{R}$ defined on the set $\mathrm{A}=\{1,2,3$, $4,5\}$ by $R=\left\{(x, y):\left|x^2-y^2\right| < 16\right\}$ is given by
- A$\{(1,1),(2,1),(3,1),(4,1),(2,3)\}$
- B$\{(2,2),(3,2),(4,2),(2,4)\}$
- C$\{(3,3),(4,3),(5,4),(3,4)\}$
- Correct answerDNone of these
Solution
We have $\mathrm{R}=\left\{(\mathrm{x}, \mathrm{y}):\left|\mathrm{x}^2-\mathrm{y}^2\right| < 16\right\}$
Let $\mathrm{x}=1,\left|\mathrm{x}^2-\mathrm{y}^2\right| < 16 \Rightarrow\left|1-\mathrm{y}^2\right| < 16$
$\Rightarrow\left|y^2-1\right| < 16 \Rightarrow y=1,2,3,4$
Let $x=2,\left|x^2-y^2\right| < 16 \Rightarrow\left|4-y^2\right| < 16$
$\Rightarrow\left|y^2-4\right| < 16 \Rightarrow y=1,2,3,4$
Let $x=3,\left|x^2-y^2\right| < 16 \Rightarrow\left|9-y^2\right| < 16$
$\Rightarrow\left|y^2-9\right| < 16 \Rightarrow y=1,2,3,4$
Let $x=4,\left|x^2-y^2\right| < 16 \Rightarrow\left|16-y^2\right| < 16$
$\Rightarrow\left|y^2-16\right| < 16 \Rightarrow y=1,2,3,4,5$
Let $\mathrm{x}=5,\left|\mathrm{x}^2-\mathrm{y}^2\right| < 16 \Rightarrow\left|25-\mathrm{y}^2\right| < 16$
$\Rightarrow\left|\mathrm{y}^2-25\right| < 16 \Rightarrow \mathrm{y}=4,5$
$\therefore \quad \mathrm{R}=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2$,
$2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1)$,
$(4,2),(4,3),(4,4),(4,5),(5,4),(5,5)\}$
Hence, the correct answer is option: 'D'.