Answer :

**TIP:** – __One – One Function__: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

__Onto Function__: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

__Bijection Function__: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, suppose

f(n_{1}) = f(n_{2})

If n_{1} is odd and n_{2} is even, then we have

⇒ n_{1} + 1 = n_{2} – 2

⇒ n_{2} – n_{1} = 2

Not possible

Suppose both n_{1} even and n_{2} is odd.

Then, f(n_{1}) = f(n_{2})

⇒ n_{1} – 1 = n_{2} + 1

⇒ n_{1} – n_{2} = 2

Not possible

Therefore, both n_{1} and n_{2} must be either odd or even

Suppose both n_{1} and n_{2} are odd.

Then, f(n_{1}) = f(n_{2})

⇒ n_{1} + 1 = n_{2} + 1

⇒ n_{1} = n_{2}

Suppose both n_{1} and n_{2} are even.

Then, f(n_{1}) = f(n_{2})

⇒ n_{1} – 1 = n_{2} – 1

⇒ n_{1} = n_{2}

Then, f is One – One

Also, any odd number 2r + 1 in the co – domain N will have an even number as image in domain N which is

⇒ f(n) = 2r + 1

⇒ n – 1 = 2r + 1

⇒ n = 2r + 2

Any even number 2r in the co – domain N will have an odd number as image in domain N which is

⇒ f(n) = 2r

⇒ n + 1 = 2r

⇒ n = 2r – 1

Thus f is Onto function.

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