NCERT Solutions
Class 12 Maths
Relations and Functions

Misc. Ex. Q1
Let f: R → R be defined as f(x) = 10x + 7. Find the function g: R → R such that
g o f = f o g = IR.
It is given that f: R → R is defined as f(x) = 10x + 7
For one – one:
Let f(x) = f(y), where x, y ∈ R.
⇒ 10x + 7 = 10y + 7
⇒ x = y
So, f is a one – one function.
For onto:
For y ∈ R, let y = 10x + 7
⇒ x = (y – 7)/10 ∈ R
Therefore, for any y ∈ R, there exists x = (y – 7)/10 ∈ R such that
f(x) = f((y−7)/10) = 10(y−7)/10 + 7 = y − 7 + 7 = y
So, f is onto.
Therefore, f is one – one and onto.
Thus, f is an invertible function.
Let us define g: R → R as g(y) = (y – 7)/10
Now, we have
gof(x) = g(f(x)) = g(10x + 7) = {(10x + 7) – 7}/10 = 10x/10 = x and
fog(y) = f(g(y)) = f ((y – 7)/10) = 10(y – 7)/10 + 7 = y − 7 + 7 = y
So, g o f = IR and f o g = IR
Hence, the required function g: R → R is defined as g(y) = (y – 7)/10