Question

Let $f$ be an injective function with domain $\{x,y,z\}$and range $\{1,2,3\}$ such that exactly one of the follwowing statements is correct and the remaining are false :

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

$f\left(z\right)\ne 2$,

then the value of $f^{-1}\left(1\right)$ is

• A. $x$
• B. $y$
• C. $z$
• D. none

Answer 1

The correct option is B $y$

It gives three cases

Case (1) When $f\left(x\right)=1$ is true.

In this case remaining two are false

$f\left(y\right)=1$ and $f\left(z\right)=2$

This means $x$ and $y$ have the same image, so $f\left(x\right)$ is not an injective, which is a contradiction.

Case (2) when $f\left(y\right)\ne 1$ is true

If $f\left(y\right)\ne 1$is true then the remaining statements are false.

$\therefore f\left(x\right)\ne 1$ and $f\left(z\right)=2$

i.e., both $x$ and $y$ are not mapped to $1.$

So, either both associate to $2$ or $3,$

Thus, it is not injective.

Case (3) When $f\left(z\right)\ne 2$ is true

If $f\left(z\right)\ne 2$ is true then remaining statements are fales

$\therefore$ If $f\left(x\right)\ne 1$ and $f\left(y\right)=1$

But $f$ is injective

Thus, we have $f\left(x\right)=2,f\left(y\right)=1$ and $f\left(z\right)=3$

Hence, $f^{-1}\left(1\right)=y$