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Let $R_{1}$ and $R_{2}$ be relations on the set $\{1,2, \ldots, 50\}$ such that $R _{1}=\left\{\left( p , p ^{ n }\right)\right.$ : $p$ is a prime and $n \geq 0$ is an integer $\}$ and $R _{2}=\left\{\left( p , p ^{ n }\right)\right.$ : $p$ is a prime and $n =0$ or $1\}$. Then, the number of elements in $R _{1}- R _{2}$ is........
$90$
$3$
$9$
$8$
Solution
Here, $p , p ^{ n } \in\{1,2, \ldots 50\}$
Now p can take values
$2,3,5,7,11,13,17,23,29,31,37,41,43$ and $47 .$
we can calculate no. of elements in $R$, as
$\left(2,2^{\circ}\right),\left(2,2^{1}\right) \ldots\left(2,2^{5}\right)$
$\left(3,3^{\circ}\right), \ldots\left(3,3^{3}\right)$
$\left(5,5^{\circ}\right), \ldots\left(5,5^{2}\right)$
$\left(7,7^{\circ}\right), \ldots\left(7,7^{2}\right)$
$\left(11,11^{\circ}\right), \ldots\left(11,11^{1}\right)$
And rest for all other two elements each
$n \left( R _{1}\right)=6+4+3+3+(2 \times 10)=36$
Similarly for $R _{2}$
$\left(2,2^{\circ}\right),\left(2,2^{1}\right)$
$\left(47,47^{\circ}\right),\left(47,47^{1}\right)$
$n \left( R _{2}\right)=2 \times 14=28$
$n \left( R _{1}\right)- n \left( R _{2}\right)=36-28=8$
