NCERT Solutions
Class 12 Maths
Relations and Functions
Ex. 1.4 Q2
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.
It is given that:
F : W → W is defined as f(n) = n − 1, if n is odd
n + 1, if n is even
For 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.
So, 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
So, f is one – one.
For onto:
It is clear that any odd number 2r + 1 in co-domain N is the image of 2r in domain N and any
even number 2r in co-domain N is the image of 2r + 1 in domain N.
So, f is onto.
Hence, f is an invertible function.Let us define g: W → W as g(m) = 
Now, when n is odd
gof(n) = g(f(n)) = g(n − 1) = n − 1 + 1 = n and
When n is even
gof(n) = g(f(n)) = g(n + 1) = n + 1 − 1 = n
Similarly,
When m is odd
fog(m) = f(g(m)) = f(m − 1) = m − 1 + 1 = m and
When m is even
fog(m) = f(g(m)) = f(m + 1) = m + 1 − 1 = m
So, gof = IW and fog = IW
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.