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Q.
Let $X=\left\{x: x=n^3+2 n+1, n \in N \right\}$ and $Y=\left\{x: x=3 n^2+7, n \in N \right\}$ then
Sets
Solution:
If $ n^3+2 n+1=3 n^2+7$
$\Rightarrow n^3-3 n^2+2 n-6=0$
$\Rightarrow (n-3)\left(n^2+2\right)=0$
$\Rightarrow n=3 \text { as } n \in N$
$\text { So, } x=3 \times 3^2+7=34 \in X \cap Y $
In (a) and (b) $x \neq 34$, for any $n \in N$.