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Show that the function f on A = R – {2/3}, defined as f(x) = (4x + 3)/ (6x – 4) in one-one and onto. Hence, find f -1.

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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 xwhere 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

-1 (y) = (3 + 4y)/ (6y – 4) where y ≠ 2/3

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