UPSEE Solved Model Paper 7

Given
f : N → Z , f (n) = {

n−1 2	when‌n‌is‌odd
− n 2	when‌n‌is‌even

... (i)
Now. we will consider two cases
1. When n ∊ N is odd
Let n = 2m + 1 m ∊ W
f () =

2m+1−1

2

, m ∊ W [Using (i)]
⇒ f (n) = m , m ∊ W
∴ f (n) is one-one function but not onto function because range
= Non-negative integers c Set of integers = Z
2. When n ∊ N is even
Let n = 2m , m ∊ N.
Then, f (n) = −

2m

2

= - m, m ∊ N [Using (i)]
∴ f (n) Is one-one function but not onto function because range
= Negative integers c Set of integers = Z
In both the cases f (n) , n ∊ N js one-one but not onto.
Hence, option 'A' is correct.