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.

$f\left(x\right)=1,f\left(y\right)\ne 1,f\left(z\right)\ne 2.$

The value of $f^{-1}\left(1\right)$ is
(a) x
(b) y
(c) z
(d) none of these

Answer 1

$$\begin{gathered} \text{Case-1: Let}f\left(x\right)=1\text{be true.} \\ \text{Then,}f\left(y\right)\text{≠1 and}f\left(z\right)\ne 2\text{are false.} \\ \text{So,}f\left(y\right)=1\text{and}f\left(z\right)=2 \\ \text{⇒}f\left(x\right)=1,f\left(y\right)=1 \\ \Rightarrow x\text{and}y\text{have the same images.} \\ \text{This contradicts the fact that}f\text{is one-one.} \\ \text{Case-2: Let}f\left(y\right)\ne 1\text{be true.} \\ \text{Then,}f\left(x\right)=1\text{and}f\left(z\right)\ne 2\text{are false.} \\ \text{So,}f\left(x\right)\ne 1\text{and}f\left(z\right)=2 \\ \text{⇒}f\left(x\right)\ne 1,f\left(y\right)\ne 1\text{and}f\left(z\right)=2 \\ \text{⇒There is no pre-image for 1.} \\ \text{This contradicts the fact that range is}\left\{1,2,3\right\}\text{.} \\ \text{Case-3: Let}f\left(z\right)\ne 2\text{be true.} \\ \text{Then,}f\left(x\right)=1\text{and}f\left(y\right)\ne 1\text{are false.} \\ \text{So,}f\left(x\right)\ne 1\text{and}f\left(y\right)=1 \\ \text{⇒}f\left(x\right)=2,f\left(y\right)=1\text{and}f\left(z\right)=3 \\ \Rightarrow f\left(y\right)=1 \\ \Rightarrow f^{-1}\left(1\right)=y \end{gathered}$$
So, the answer is (b).