KCET 2014 Math Solved Paper

Given that, f:N→N defined by
f(n)={

n+1 2	n is odd
n 2	n is even

For n = 1, we have f(1)=

1+1

2

=1
and, if n = 2, we have f(2)=

2

2

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

n+1

2

if n is odd
if y=

n+1

2

, then n=2y−1,∀y
Also, f(x)=

n

2

if n is even. That is,
y=‌

n

2

or n=2y,∀y
Therefore, f(x) is onto.