Let $f$  be injective map with domain $\left\{x,y,z\right\}$  and range $\left\{1,2,4\right\}$  such that exactly one of the following statement is correct and the remaining are false $f\left(x\right)=1,f\left(y\right)\ne 1,f\left(z\right)\ne 2$  then $f^{-1}\left(1\right)$  is equal to

Case – I : If 1 true 2 false and 3 false $f\left(x\right)=1,f\left(y\right)=1,f\left(z\right)=2$

$\Rightarrow f$ is not $1-1$ Case – II : If (1) false (2) true and (3) false

$f\left(x\right)\ne 1,f\left(y\right)\ne 1,f\left(z\right)=2$ $\Rightarrow f$ is not $1-1$ Case – III : If (1) and (2) false and (3) true $\Rightarrow f\left(x\right)\ne 1,f\left(y\right)\ne 1,f\left(z\right)\ne 2$

$\Rightarrow f\left(x\right)=2,f\left(y\right)=2,f\left(z\right)=4\Rightarrow f^{-1}\left(1\right)=y\&f^{-1}\left(4\right)=z$