One-one f:
Consider x1 and x2 ∈ dom (f)
We know that
f(x1) = f(x2)
It can be written as
(4x1 + 3)/ (6x1 – 4) = (4x2 + 3)/ (6x2 – 4)
So we get
(4x1 + 3) (6x2 – 4) = (4x2 + 3) (6x1 – 4)
On further calculation
24 x1 x2 – 16x1 + 18 x2 – 12 = 24 x1 x2 – 16x2 + 18x1 – 12
We get
34 x1 = 34 x2 where x1 = x2
f is one-one
Onto f:
Consider y ∈ co domain (f)
We know that y = f(x)
y = (4x + 3)/ (6x – 4)
On further calculation
6xy – 4y = 4x + 3
So we get
6xy – 4x = 3 + 4y
It can be written as
x (6y – 4) = 3 + 4y
So x = (3 + 4y)/ (6y – 4) ∈ domain Ɐ y ∈ co-domain
f is an onto function
Here, x = (3 + 4y)/ (6y – 4) where y ≠ 2/3
We get
f -1 (y) = (3 + 4y)/ (6y – 4) where y ≠ 2/3