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.
$f(x)=1, f(y) ≠1, f(z)≠2$
The value of $f^{-1}(1)$ is
Answer & explanation
Correct answer: option 2
Following cases arise.
CASE I When $f(x) = 1$ is true
In this case, the remaining two statements are false.
$∴f(y)=1$ and $f(z) = 2$
This means that x and y have the same image. So, f(x) is not an injection, which is a contradiction.
Hence, $f(x) =1$ is not true.
CASE II When $f(y) ≠1$ is true
If $f(y) ≠1$ is true, then the remaining statements are false.
$∴f(x) ≠1$ and $f(z) = 2$
This shows that x and y both are not mapped to 1. So, either both are associated to 3 or both are associated to 2 or one of them is mapped to 3 and other to 2. In all the three cases, f cannot be an injective map.
So, $f(y) ≠1$ is not true.
CASE III When $f(z) ≠ 2$ is true
If $f(z) ≠2$ is true, then the remaining statements are false.
$∴f(x) ≠1$ and $f(y)=1$.
But, f is an injective map.
Thus, we have
$f(y)=1, f (z) = 3$ and $f (x) = 2$
Hence, $f^{-1}(1) = y$.