Question

A function f from the set of natural numbers to the set of integers defined by

\[f\left( n \right)\begin{cases}\frac{n - 1}{2}, & \text{when n is odd} \\ - \frac{n}{2}, & \text{when n is even}\end{cases}\]

 

पर्याय

• neither one-one nor onto

• one-one but not onto

• onto but not one-one

• one-one and onto

Answer 1

Injectivity:
Let x and y be any two elements in the domain (N).

\[Case-1: \text{Both  x and y are even}.\] 

\[\text{Let}f\left( x \right) = f\left( y \right)\] 
\[ \Rightarrow \frac{- x}{2} = \frac{- y}{2}\] 
\[ \Rightarrow - x = - y\] 
\[ \Rightarrow x = y\] 
\[Case-2: \text{Both x and y are odd}.\] 
\[\text{Let}f\left( x \right) = f\left( y \right)\] 
\[ \Rightarrow \frac{x - 1}{2} = \frac{y - 1}{2}\] 
\[ \Rightarrow x - 1 = y - 1\] 
\[ \Rightarrow x = y\] 

\[Case-3: \text{ Let  x be even and y be odd}.\] 
\[\text{Then},f\left( x \right) = \frac{- x}{2}\text{and}f\left( y \right) = \frac{y - 1}{2}\] 
\[\text{Then, clearly}\] 
\[x \neq y \] 
\[ \Rightarrow f\left( x \right) \neq f\left( y \right)\] 
\[\text{From all the cases,fis one-one}.\] 

Surjectivity:

\[\text{Co-domain off} = Z = \left\{ . . . , - 3, - 2, - 1, 0, 1, 2, 3, . . . . \right\}\] 

\[\text{Range of f} = \left\{ . . . , \frac{- 3 - 1}{2}, \frac{- \left( - 2 \right)}{2}, \frac{- 1 - 1}{2}, \frac{0}{2}, \frac{1 - 1}{2}, \frac{- 2}{2}, \frac{3 - 1}{2}, . . . \right\}\] 
\[ \Rightarrow \text{Range of f} = \left\{ . . . , - 2, 1, - 1, 0, 0, - 1, 1, . . . \right\}\] 
\[ \Rightarrow\text{Range of f } = \left\{ . . . , - 2, - 1, 0, 1, 2, . . . . \right\}\] 
\[ \Rightarrow \text{Co-domain of f = Range of f}\] 

\[\Rightarrow\] f is onto.

So, the answer is (d).