Question

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

• A. $f\left(x\right)-f\left(y\right)+f\left(z\right)=5$
• B. $f\left(x\right)+f\left(y\right)-f\left(z\right)=0$
• C. $f\left(x\right)-f\left(y\right)-f\left(z\right)=0$
• D. $-f\left(x\right)-2f\left(y\right)+f\left(z\right)=3$

Answer 1

The correct option is B

$f\left(x\right)+f\left(y\right)-f\left(z\right)=0$

Only possible case for given data is:
$f\left(x\right)=1$ is false $\Rightarrow f\left(x\right)=2\text{or}3$
$f\left(y\right)\ne 1$ is false $\Rightarrow f\left(y\right)=1$
$f\left(z\right)\ne 2$ is true $\Rightarrow f\left(z\right)=1\text{or}3$
$f\left(y\right)=1,f\left(x\right)=2,f\left(z\right)=3$
In this case, $f$ is one-one.

In all other cases, we won't be getting a one one function.
So, clearly from the above analogy we have $f\left(x\right)+f\left(y\right)-f\left(z\right)=0$