JEE Main 2022 — Sets And Relations Question with Solution

Given, $R_1=\{\left(p,p^n\right):p$ is a Prime and $n\ge 0$ is an integer $\}$

and, set $A=\{1,2,3\dots \dots .50\}$

$p$ is a Prime number which can take 15 values $2,3,5,7,11,13,17,19,23,29,31,37,41,43$ and 47

$\therefore$ We can calculate no. of elements in ${\mathrm{R}}_1$

$${\mathrm{R}}_1=\left(2,2^0\right),\left(2,2^1\right),\left(2,2^2\right)\left(2,2^3\right)\dots \dots \left(2,2^5\right)=6$$ number of ordered pairs

$\left(3,3^0\right),\left(3,3^1\right),\left(3,3^2\right)\dots \dots ..\left(3,3^3\right)=4$ number of order paris

$\left(5,5^0\right),\left(5,5^1\right),\left(5,5^2\right)\dots \dots \dots ..=3$ number of order paris

$\left(7,7^0\right)\dots \dots .\left(7,7^2\right)\dots \dots \dots =3$ number of order paris

$\left(11,11^0\right)$ and $\left(11,11^1\right)=2$ number of order paris

$\left(13,13^0\right)$ and $\left(13,13^1\right)=2$ number of order paris

$$\therefore$$ For the 11 prime numbers ($11,13,17,19,23,29,31,37,41,43$ and 47), $n$ can only be 0, 1 (two pairs each).

$$\therefore n\left({\mathrm{R}}_1\right)=6+4+3+3+(2\times 11)=38$$

${\mathrm{R}}_2=\left(p,p^n\right)$, where n = 0 or 1

$$\left(2,2^0\right),\left(2,2^1\right)\left(3,3^0\right)\left(3,3^1\right)\dots ..\left(47,47^0\right)\left(47,47^1\right)$$

Two ordered pairs of each element $n\left(R_2\right)=2\times 15=30$ elements

Hence $R_1-R_2=38-30=8$