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If  f : X -> Y and a, b  ⊆ X, then f (a  ∩ b) is equal to

  • a)
    f(a) - f(b)

  • b)
    f(a) ∩ f(b)

  • c)
    a proper subset of f(a) ∩ f(b)

  • d)
    f(b) - f(a)

Correct answer is option 'C'. Can you explain this answer?
Verified Answer
If f : X -> Y and a, b ⊆ X, then f (a ∩ b) is equal toa)f(...
The only requirement to answer the above question is to know the definition of function- a relation becomes a function if every element in domain is mapped to some element in co-domain and no element is mapped to more than one element.



Now, we have a,b⊆ X. Their intersection can be even empty set. So, lets try out options:



Options a and d don't even need a check.



Lets take a case where a∩b = φ. Now, f(a∩b)=φ, but f(a) ∩ f(b) can be non empty. So, option B can be false.



Option C is always true provided "proper subset" is replaced by "subset". This is because no element in domain of a function can be mapped to more than one element. And the subset needn't be "proper" as for a one-one mapping, we get


f(a∩b) = f(a) ∩ f(b) ,Hence option C
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Most Upvoted Answer
If f : X -> Y and a, b ⊆ X, then f (a ∩ b) is equal toa)f(...
> Y is a function from a set X to a set Y, then f is a surjective function if and only if every element of Y has at least one pre-image in X.

In other words, for every y in Y, there exists at least one x in X such that f(x) = y.

This is also known as the onto property, since f maps every element in X onto some element in Y. A surjective function is also called a surjection or a onto function.

Note that a function can be surjective even if it is not injective (i.e., one-to-one). In this case, multiple elements in X may map to the same element in Y.

For example, the function f : R -> R given by f(x) = x^2 is surjective, since every real number y has a pre-image in R (namely, either sqrt(y) or -sqrt(y)). However, f is not injective, since multiple values of x (e.g., x and -x) can map to the same value of y.
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