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```									3/20/2011                                  NCERT CBSE Math Solutions for Class XI…

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Question 2:

Let f: W → W be defined as f(n) = n − 1, if is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the
inverse of f. Here, W is the set of all whole numbers.

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It is given that:

f: W → W is defined as

One-one:

Let f(n) = f(m).

It can be observed that if n is odd and m is even, then we will have n − 1 = m + 1.

⇒n−m=2

However, this is impossible.

Similarly, the possibility of n being even and m being odd can also be ignored under a similar argument.

∴Both n and m must be either odd or even.

Now, if both n and m are odd, then we have:

f(n) = f(m) ⇒ n − 1 = m − 1 ⇒ n = m

Again, if both n and m are even, then we have:

f(n) = f(m) ⇒ n + 1 = m + 1 ⇒ n = m

∴f is one-one.

It is clear that any odd number 2r + 1 in co-domain     is the image of 2r in domain         and any even number 2r in
co-domain is the image of 2r + 1 in domain .

∴f is onto.

Hence, f is an invertible function.

Let us define g: W → W as:
…meritnation.com/…/9LxV5TAnkcU8P6B…                                                                                                1/2
3/20/2011                                   NCERT CBSE Math Solutions for Class XI…

Now, when n is odd:

And, when n is even:

Similarly, when m is odd:

When m is even:

∴

Thus, f is invertible and the inverse of f is given by f—1 = g, which is the same as f.

Hence, the inverse of f is f itself.

…meritnation.com/…/9LxV5TAnkcU8P6B…                                                            2/2

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