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Question: Answered & Verified by Expert
Let $f: X \rightarrow Y$ be a function and $A_y=\left\{\frac{f^{-1}(y)}{y \in Y}\right\}$. Then $A_i \cap A_j=\dot{\phi}(i \neq j) \forall i, j \in Y$ and $\bigcup_{y \in Y} A_y=X$, if
MathematicsFunctionsTS EAMCETTS EAMCET 2019 (06 May Shift 1)
Options:
  • A $f$ is onto function only
  • B $f$ is one-one function only
  • C $f$ is any function
  • D $X$ and $Y$ are finite sets only
Solution:
1125 Upvotes Verified Answer
The correct answer is: $f$ is any function
It is given that function $f: X \rightarrow Y$ and
$A_y=\left\{\frac{f^{-1}(y)}{y \in Y}\right\}$
$\because$ Inverse of the function exists, so function $f$ is one-one and onto.
then $A_i \cap A_j=\phi,(i \neq j) \forall i, j \in Y$,
and $\bigcup A_y=X$ for every function $f$.
Hence, ${ }^Y$ option (c) is correct.

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