# Let f : R → R be defined as f (x) = 10x + 7. Find the function g : R → R such that gof = fog = I_{R}

**Solution:**

f : R → R is defined as f (x) = 10x + 7

For one-one:

f (x) = f (y) where x, y ∈ R

⇒ 10x + 7 = 10y + 7

⇒ x = y

⇒ f is one-one.

For onto:

y ∈ R, Let y = 10x + 7

⇒ x = (y - 7)/10 ∈ R

For any y ∈ R, there exists x = (y - 7)/10 ∈ R such that

f (x) = ((y - 7)/10) + 7 = y - 7 + 7 = y f (x)

= f ((y - 7)/10) = 10 ((y - 7)/10) + 7

= y - 7 + 7 = y

⇒ f is onto.

Thus, f is an invertible function.

Let us define g : R → R as g (y) = (y - 7)/10.

Now,

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

⇒ gof = I_{R} and fog = I_{R}

Hence, the required function

g : R → R as g (y) = (y - 7)/10

NCERT Solutions for Class 12 Maths - Chapter 1 Exercise ME Question 1

## Let f : R → R be defined as f (x) = 10x + 7. Find the function g : R → R such that gof = fog = I_{R}

**Summary:**

For the function f : R → R be defined as f (x) = 10x + 7. The required function g : R → R as g (y) = (y - 7)/10