Question
Let $$f:\left\{ {x,\,y,\,z} \right\} \to \left\{ {1,\,2,\,3} \right\}$$ be a one-one mapping such that only one of the following three statements is true 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.
$$f\left( y \right) < f\left( x \right) < f\left( z \right)$$
D.
$$f\left( y \right) < f\left( z \right) < f\left( z \right)$$
Answer :
$$f\left( y \right) < f\left( x \right) < f\left( z \right)$$
Solution :
$$\eqalign{
& {\text{Let }}f\left( x \right) \ne 2{\text{ be true }} \cr
& {\text{and }}f\left( y \right) = 2,\,f\left( z \right) \ne 1{\text{ are false}} \cr
& \Rightarrow f\left( x \right) \ne 2,\,f\left( y \right) \ne 2,\,f\left( z \right) = 1 \cr
& \Rightarrow f\left( x \right) = 3,\,f\left( y \right) = 3,\,f\left( z \right) = 1 \cr} $$
but the function is many one, similarly to other cases.