Let f : N → N defined by f(n) = {cases} n+1/2, & {if } n { is odd} \\ n/2, & {if } n { is even} {cases} then f is

Correct Answer is : onto but not one-one

Correct Answer is : onto but not one-one

Test Index

KCET 2014 Math Solved Paper

Question : 3 of 60

Marks: +1, -0

\begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases}

then f is

Solution:

Given that,

f:N\rightarrow N

defined by

f(n)=\begin{cases} \frac{n+1}{2}, & n \text{ is odd} \\ \frac{n}{2}, & n \text{ is even} \end{cases}

For n = 1, we have

f(1)=\frac{1+1}{2}=1

and, if n = 2, we have

f(2)=\frac{2}{2}=1

So, f(1) = f(2) but 1 ≠ 2 . Therefore, f(x) is not one-one.

Now,

f(x)=\frac{n+1}{2}

if n is odd

y=\frac{n+1}{2},

then

n=2y-1,\forall y

Also,

f(x)=\frac{n}{2}

if n is even. That is,

y=\frac{n}{2}

n=2y,\forall y

Therefore, f(x) is onto.

Go to Question:

Prev Question

Next Question