... (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.