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B
Y
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C
ϕ
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D
{0}
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Solution
The correct option is BY Let us restrict the domain of X to natural numbers,
⟹4n−3n−1⟹(1+3)n−3n−1⟹1+3n+nC232+...+3n−3n−1⟹9k
Where k is some natural number,
So,the elements of X are multiples of 9,
The elements of the set Y are also multiples of 9,
But if we extend the domain of X to the set of real numbers there would be other elements which are not present in Y as the elements of Y are only natural numbers where as the elements of X can also be rational and irrational,
So the elements of Y are contained in X for all natural numbers,