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**NCERT mathematics class 12 chapter Relations and Functions Exercise 1.2 Problem 9 Solution**. Solutions for other problems are available at Exercise 1.2 SolutionsExercise 1.2 Problem 9 Solution

9. Let f : N → N be defined by \begin{cases}\dfrac{n + 1}{2} & ,\ if\ n\ is\ odd\\\dfrac{n}{2} & ,\ if\ n\ is\ even\end{cases} for all n ∈ N. State whether the function f is bijective. Justify your answer.

To check whether f is one-one

Let k be an odd natural number.

Then {k + 1} will be an even natural number.

Now {f(k) = \dfrac{k + 1}{2}} (∵ k is odd, we’ve to use {f(n) = \dfrac{n + 1}{2}})

Now {f(k + 1) = \dfrac{k + 1}{2}} (∵ {k + 1} is even, we’ve to use {f(n) = \dfrac{n}{2}})

⇒ Two different consecutive natural numbers have the same image.

⇒ f is not one-one

To check whether f is onto

To be onto, every natural number y ∈ N in the co-domain should be an image of an element in the domain N.

When x is odd {y = f(x) = \dfrac{x + 1}{2}}

⇒ {2y - 1 = x}

As we know, for any value of y ∈ N, {2y - 1} is always odd and also {2y - 1} ∈ N

⇒ Every y ∈ N in the co-domain will be an image of an element in the domain.

Now, when x is even {y = f(x) = \dfrac{x}{2}}

⇒ {2y = x}

As we know, for any value of y ∈ N, 2y is always even and also 2y ∈ N

⇒ Every y ∈ N in the co-domain will be an image of an element in the domain.

Thus we see that in all the cases, every element y ∈ N in the co-domain will be an image of an element in the domain. (to be specific every y ∈ N will be an image of two elements in the domain)

In otherwords, all the elements in the co-domain N will have an image in the domain

∴ f is onto(surjective) but not one-one(injective).

∴ f is not bijective.

Note: For short answer/MCQ, as soon as you discover that f is not one-one, you can conclude that f is not bijective. In otherwords, you don’t have to check whether f is onto. This might save some time.