Expression Boils down to:
$f1.f2 + f3 = f$
$(f1 ⋂ f2) U f3 = f$
we use intersection for AND because if a f1 and f2 both are true for a minterm then only output will be true.
we use Union for OR because if one among f1 and f2 is true for a minterm then output will be true.
${4,5,6,7,8}⋂ f2 U {1,6,15} = {1,6,8,15}$
${4,5,6,7,8}⋂ f2 = {8}$
$f2 = {8,any other elements preset in f }$
check for options c contains {6,8} 6 is in f
option (C)