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Question
Let f be an injective map with domain {x, y, z} and range {1, 2, 3}, such that exactly one of the following statements is correct and the remaining are false.
\[f\left( x \right) = 1, f\left( y \right) \neq 1, f\left( z \right) \neq 2 .\]
The value of
\[f^{- 1} \left( 1 \right)\] is
Options
x
y
z
none of these
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Solution
\[\text{Case}-1: Letf\left( x \right) = 1 \text{ P be true}.\]
\[\text{Then,f } \left( y \right)\neq1 \text{ and f }\left( z \right) \neq 2\text{ are false}.\]
\[\text{So,f } (y) = 1 \text{ and } f \left( z \right) = 2\]
\[\Rightarrow f\left( x \right) = 1, f\left( y \right) = 1\]
\[ \Rightarrow \text{ x and y have the same images}.\]
\[\text{This contradicts the fact that fis one-one}.\]
\[\text{Case}-2: \text{Letf}\left( y \right) \neq1 \text{be true}.\]
\[\text{Then},f\left( x \right) = 1 \text{and}f\left( z \right) \neq 2 \text{ are false}.\]
\[So, f\left( x \right) \neq1 \text{and f}\left( z \right) = 2\]
\[\Rightarrow f\left( x \right) \neq 1, f\left( y \right) \neq 1 andf\left( z \right) = 2\]
\[\Rightarrow\text{There is no pre-image for 1}.\]
\[\text{This contradicts the fact that range is}\left\{ 1, 2, 3 \right\}.\]
\[\text{Case}-3: Letf\left( z \right) \neq 2\text{ be true}.\]
\[\text{Then},f\left( x \right) = 1\text{and}f\left( y \right) \neq 1 \text{are false}.\]
\[So, f\left( x \right) \neq1 \text {and} f\left( y \right) = 1\]
\[\Rightarrow f\left( x \right) = 2, f\left( y \right) = 1 \text{and }f\left( z \right) = 3\]
\[ \Rightarrow f \left( y \right) = 1\]
\[ \Rightarrow f^{- 1} \left( 1 \right) = y\]
So, the answer is (b).
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