Question

Consider the given function \[f:N\to z:f\left( n \right)\left\{ \begin{matrix}
   \dfrac{n}{2},n-even \\
   -\left( \dfrac{n-1}{2} \right),n-odd \\
\end{matrix} \right.\] , what type of function is this?
a.Not one-one but onto
b.One-one but onto
c.One-one and onto
d.Not one-one and onto.

Answer 1

Hint: Consider the case of odd by putting n = 2m+1. For the case of even take n = 2m. Substitute these values for n-even and n-odd functions by considering function \[f\left( {{m}_{1}} \right)\]and \[f\left( {{m}_{2}} \right)\].

Complete Step-by-step answer:
Given a function \[f\left( n \right)=\left\{ \begin{matrix}
   \dfrac{n}{2},n-even \\
   -\left( \dfrac{n-1}{2} \right),n-odd \\
\end{matrix} \right.\]
Now, let us consider the case where n is odd.
Let, \[n=2{{m}_{1}}+1\], is odd.
\[\therefore f\left( {{m}_{1}} \right)=f\left( {{m}_{2}} \right)\]can be taken as one-one so, \[{{m}_{1}}\ne {{m}_{2}}\].
Now checking the condition of \[f\left( {{m}_{1}} \right)=f\left( {{m}_{2}} \right)\], substituting the value of n in the function for n-odd.
\[\dfrac{-\left( 2{{m}_{1}}+1 \right)-1}{2}=\dfrac{-\left( 2{{m}_{2}}+1 \right)-1}{2}\]
Cancelling out the like terms, we get
\[{{m}_{1}}={{m}_{2}}\]
Which means that \[f\left( {{m}_{1}} \right)\]will only be equal to \[f\left( {{m}_{2}} \right)\]when \[{{m}_{1}}={{m}_{2}}\].
\[\therefore f\left( {{m}_{1}} \right)=f\left( {{m}_{2}} \right)\]
As it is odd, the function is one-one.
\[\because \]One-one function is a function that comprises individually that never discrete elements of its co-domain (z).
Now let us consider the case where n is even.
\[\therefore \]Let \[n=2{{m}_{1}}\]
\[f\left( 2{{m}_{1}} \right)=f\left( 2{{m}_{2}} \right)\]
Now, checking this condition by putting the value of n in the function for n-even
 \[\begin{align}
  & \Rightarrow \dfrac{-2{{m}_{1}}}{2}=\dfrac{-2{{m}_{2}}}{2} \\
 & \Rightarrow {{m}_{1}}={{m}_{2}} \\
\end{align}\]
\[\therefore \]The function is one-one.
Now let us consider giving values for \[{{m}_{1}}\]and \[{{m}_{2}}\].
\[\therefore \]Let \[{{m}_{1}}\]be even which becomes \[2{{m}_{1}}\].
\[\therefore \]\[f\left( n \right)=\dfrac{n}{2}=\dfrac{2{{m}_{1}}}{2}={{m}_{1}}\]
If taking odd \[\left( 2{{m}_{1}}+1 \right)\].
\[\begin{align}
  & \Rightarrow f\left( n \right)=\dfrac{-\left( n-1 \right)}{2}=\dfrac{-\left( 2{{m}_{1}}+1-1 \right)}{2} \\
 & f\left( n \right)=-{{m}_{1}} \\
\end{align}\]
Now, if we are considering natural numbers, N = 1, 2, 3, ……
If we are putting them in the function, we get,
For N = 1, we get (1, -1).
       N = 2, we get (2, -2) etc.
 \[\therefore \]The range of the function becomes\[\Rightarrow \left\{ +1,+2.....-1,-2.... \right\}\in \]integers.
In the question, z is given as a co-domain.
\[\therefore \]Range becomes equal to co-domain, so it becomes onto.
\[\therefore \]The function is one-one onto.
Hence, the correct option is (c).

Note: Check for one-one.
\[f\left( 1 \right)=\dfrac{-\left( n-1 \right)}{2}=\dfrac{-\left( 1-1 \right)}{2}=0\](since 1 is odd)
\[f\left( 2 \right)=\dfrac{n}{2}=\dfrac{2}{2}=1\] (since 2 is even)
\[f\left( 1 \right)=0\]and \[f\left( 2 \right)=1\]but \[1\ne 2\].
\[\therefore \]f is one-one, as both \[f\left( 1 \right)\]and \[f\left( 2 \right)\]don't have the same image.
Check for onto
\[f\left( n \right)=\left\{ \begin{matrix}
   \dfrac{n}{2},n-even \\
   -\left( \dfrac{n-1}{2} \right),n-odd \\
\end{matrix} \right.\]
Let \[f\left( x \right)=y\], such that \[y\in N\].
When n is odd,
\[\begin{align}
  & y=\dfrac{-\left( n-1 \right)}{2}\Rightarrow 2y=-n+1 \\
 & \therefore n=1-2y \\
\end{align}\]
Hence, for y is integer number, n = 1 – 2y is an integer.
When n is even,
\[y=\dfrac{n}{2}\Rightarrow 2y=n\]
Hence, for y is a natural number, n = 2y is also a natural number.
Thus, for every \[y\in N\], there exists \[x\in N\]such that,
\[f\left( n \right)=y\]
\[\therefore \]f is onto.
\[\therefore \]Function \[f\left( n \right)\]is both one-one and onto.