# Let \[f:X \to Y\]and A, B are non-void subsets of Y, then? (where the symbols have their usual interpretation)

A.\[{f^{ - 1}}(A) - {f^{ - 1}}(B) \supset {f^{ - 1}}(A - B)\] but the opposite does not hold

B.\[{f^{ - 1}}(A) - {f^{ - 1}}(B) \subset {f^{ - 1}}(A - B)\] but the opposite does not hold

C.\[{f^{ - 1}}(A - B) = {f^{ - 1}}(A) - {f^{ - 1}}(B)\]

D.\[{f^{ - 1}}(A - B) = {f^{ - 1}}(A) \cap {f^{ - 1}}(B)\]

Answer

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**Hint**: A function takes input and performs specific operations on it before returning an output. The inverse function agrees with the resultant, performs its function, and returns to the original function.

The inverse function returns the original value of the output of a function.

**Complete step-by-step answer**:

The given question is explained as \[f:X \to Y\] where \[f\] is a solution defined between \[X\] and \[Y\]

And similarly, A and B are two non-void subsets of \[X\]and \[Y\].

Based on the given question.

The Correct option is C \[{f^{ - 1}}(A - B) = {f^{ - 1}}(A) - {f^{ - 1}}(B)\] which defines the difference operation between them and exists same in both LHS and RHS and they exist.

Through the given options we need to find the correct form of interpretation of the result

Option A - \[{f^{ - 1}}(A) - {f^{ - 1}}(B) \supset {f^{ - 1}}(A - B)\]

Here for the two non-void subsets A and B with regarded of the function \[f:X \to Y\] it tempts to find the inverse image of the function as \[{f^{ - 1}}:X \to Y\] hence during this function format avail it forms another two non-void subsets as \[{A^{ - 1}}\] and \[{B^{ - 1}}\]. Thus, we formatting the given function \[{f^{ - 1}}(A) - {f^{ - 1}}(B)\] it exists separately and similarly \[{f^{ - 1}}(A - B)\] it also exits but these are not the subsets of each other.

Hence the option A is not the correct one.

Option B - \[{f^{ - 1}}(A) - {f^{ - 1}}(B) \subset {f^{ - 1}}(A - B)\]

As mentioned in the above option A these two-inverse images of the given function \[f:X \to Y\] will exist but do not form the subsets between them.

Hence Option B is also not the correct one.

Option C - \[{f^{ - 1}}(A - B) = {f^{ - 1}}(A) - {f^{ - 1}}(B)\]

As per the given option based on the explanation which is detailed in Option A this A and B forms the inverse image through the function of \[{f^{ - 1}}:X \to Y\] which contributes on forming the two non-void subsets \[{A^{ - 1}}\] and \[{B^{ - 1}}\]. While the difference operation is operated on this functional form \[{f^{ - 1}}(A - B)\] which definitely exists and \[{f^{ - 1}}(A) - {f^{ - 1}}(B)\] it also exists and with this they are equal to each other. The reason is that the difference operation exists the same in the inverse function.

Hence the Option C is the correct one.

Option D - \[{f^{ - 1}}(A - B) = {f^{ - 1}}(A) \cap {f^{ - 1}}(B)\]

In this option the LHS is defined with the difference operation and the RHS is defined with the intersection operation. Hence, they will not be equal to each other.

Thus, the option D is not the correct one.

**So, the correct answer is “Option C”.**

**Note**: An inverse function is a mathematical concept. Given some input x, a function can measure some output \[Y\].\[f(X) = Y\] is a common notation for this. The inverse function does exactly as it says on the function. If \[g\] is the inverse function of f, then \[g(Y) = X\] is the result. Alternatively, \[g(f(X)) = X\]

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