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Relations and Functions Ncert Solutions Maths class 12th Maths Class 12th Relations and Functions

57. 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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Vishal Baghel | Contributor-Level 10

4 months ago

It is given that:

$f : W \to W$ is defined as $f (n) = {\begin{cases} n - 1, i f . n, o d d \\ n + 1, i f . n, e v e n \end{cases}$

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.

$\Rightarrow n - m = 2$

However, 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:

$f (n) = f (m) \Rightarrow n - 1 = m - 1 \Rightarrow n = m$

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

$f (n) = f (m) \Rightarrow n + 1 = m + 1 \Rightarrow n = m$

$∴ f$ is one-one.

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

$∴ f$ is onto.

Hence, f is an invertible function.

Let us define $g : W \to W$ as:

$g (m) = {\begin{cases} m + 1, i f . n, e v e n \\ m - 1, i f . n, o d d \end{cases}$

Now, when n is odd:

$g o f (n) = g (f (n)) = g (n - 1) = n - 1 + 1 = n$

And, when

...more

It is given that:

$f : W \to W$ is defined as $f (n) = {\begin{cases} n - 1, i f . n, o d d \\ n + 1, i f . n, e v e n \end{cases}$

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.

$\Rightarrow n - m = 2$

However, 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:

$f (n) = f (m) \Rightarrow n - 1 = m - 1 \Rightarrow n = m$

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

$f (n) = f (m) \Rightarrow n + 1 = m + 1 \Rightarrow n = m$

$∴ f$ is one-one.

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

$∴ f$ is onto.

Hence, f is an invertible function.

Let us define $g : W \to W$ as:

$g (m) = {\begin{cases} m + 1, i f . n, e v e n \\ m - 1, i f . n, o d d \end{cases}$

Now, when n is odd:

$g o f (n) = g (f (n)) = g (n - 1) = n - 1 + 1 = n$

And, when n is even:

$g o f (n) = g (f (n)) = g (n + 1) = n + 1 - 1 = n$

Similarly, when m is odd:

$f o g (m) = f (g (m)) = f (m - 1) = m - 1 + 1 = m$

When m is even:

$\begin{array}{l} f o g (m) = f (g (m)) = f (m + 1) = m + 1 - 1 = m \\ ∴ g o f = I_{W}, a n d, f o g = I_{W} \end{array}$

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.

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$?$ R is symmetric relation

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So total no. of functions = 10⁵ × 1 = 10⁵

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57. 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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