Question

\[f : Z \to Z\]  be given by

 ` f (x) = {(x/2, ", if  x is even" ) ,(0 , ", if  x  is  odd "):}`

Then,  f is

पर्याय

• onto but not one-one

• one-one but not onto

• one-one and onto

• neither one-one nor onto

Answer 1

Injectivity:
Let x and y be two elements in the domain (Z), such that

LaTeX
\[f\left( x \right) = f\left( y \right)\] 
\[Case-1: \text{Let both x andybe even}.\] 
\[\text{Then},\] 
\[f\left( x \right) = f\left( y \right)\] 
\[ \Rightarrow \frac{x}{2} = \frac{y}{2}\] 
\[ \Rightarrow x = y\] 
\[Case-2: \text{Let bothx andybe odd}.\] 
\[\text{Then},\] 
\[f\left( x \right) = f\left( y \right)\] 
\[ \Rightarrow 0 = 0\] 
\[\text{Here, we cannot determine whether } x = y.\] So, f is not one-one.
Surjectivity:
Let y be an element in the co-domain (Z), such that

\[\text{Co-domain of f} = Z = \left\{ 0, \pm 1, \pm 2, \pm 3, \pm 4, . . . \right\} \]

\[\text{Range of f} = \left\{ 0, 0, \frac{\pm 2}{2}, 0, \frac{\pm 4}{2} , . . . \right\} = \left\{ 0, \pm 1, \pm 2, . . . \right\} \]

\[ \Rightarrow \text{Co-domain of f} = \text{Range of f}\]

\[\Rightarrow\] f is onto.
So, the answer is (a).