Question

Let f be a one-one function with domain {x,y,z} and range {1,2,3}. It is given that exactly one of the following statements is true and remaining two are false, $f\left(x\right)=1,f\left(y\right)=1,f\left(z\right)\ne 2$, Determine $f^{-1}\left(1\right)$.

• A. $x$
• B. $y$
• C. $z$
• D. $x+z$

Answer 1

Answer: (B)

$y$

It gives three cases

Case I:

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

In this case remaining two are false i.e. $f\left(y\right)=1$ 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 as the function is injective.

Case II:

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

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

$⟹$ Either of the $x,y$ mapped to $2$

Or both $x$ and $y$ are not mapped to $1.$

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

Thus, it is not injective which is a contradiction.

Thus, both the cases I and II are not true.

Case III:

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

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