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