written 3.7 years ago by
yashbeer
★ 11k
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Solution:
1] {$A_1, A_2$}
$A_1$ = { a,b,c,d} and $A_2$ = {a,c,e,g,h}
$A_1 V A_2$ = { a,b,c,d,e,g,h}
$\because$ f E A but f e A, V$A_2$ and $A_1 \ n \ A_2 \neq \phi $
$\therefore$ {$A_1, A_2$} is not a perfect partition of A.
2] { $A_3, A_4, A_5$ }
$A_3$ = {a,c,e,g}
$A_4$ = { b,d }
$A_5$ = { f, d }
$A_3 \ n \ A_4 N A_5 = \phi$
$\because$ All the elements of A is either present in $A_3$ OR $A_4$ OR $A_5$
and $A_3, A_4, A_5$ are mutually disjoint sets.
$\therefore$ { $A_3, A_4, A_5$ } are perfect partitions of A.