Question

Let $f:\{x,y,z\}\to \{1,2,3\}$ be a one-one mapping such that only one of the following three statements and remaining two are false : $f\left(x\right)\ne 2,f\left(y\right)=2,f\left(z\right)\ne 1$, then

• A. $f\left(x\right)>f\left(y\right)>f\left(z\right)$
• B. $f\left(x\right)<f\left(y\right)<f\left(z\right)$
• C. $f\left(y\right)<f\left(y\right)<f\left(z\right)$
• D. $f\left(y\right)<f\left(z\right)<f\left(x\right)$

Answer 1

The correct option is A $f\left(x\right)>f\left(y\right)>f\left(z\right)$

$f:\{x,y,z\}\to \{1,2,3\}$
$I-f\left(x\right)\ne 2$
$II-f\left(y\right)=2$
$III-f\left(z\right)\ne 1$
1. Let statement I is true, $\therefore$ other two are false
$\therefore f\left(x\right)\ne 2,f\left(y\right)\ne 2$
$\therefore f\left(z\right)=1$
$f\left(x\right)\&f\left(y\right)$ can be $2$ or $3$
We don't know exact value of $f\left(x\right)\&f\left(y\right)$
2. Let statement II is true
$f\left(y\right)=2\therefore f\left(z\right)=1$ [$\because f\left(z\right)\ne 1$ is false]
$f\left(x\right)=3$
$f\left(x\right)>f\left(y\right)>f\left(z\right)$