# Let A = R - {3} a

Given; A = R - {3} and B = R - {1}

function f: A B defined by  Let f(x1) = f(x2) (x1 − 2)(x2 − 3) = (x2 − 2)(x1 − 3)

x1x2 − 2x2 − 3x1 + 6 = x1x2 − 2x1 − 3x2 + 6

x1 = x2

Hence, if f(x1) = f(x2), then x1 = x2

f is one - one

Let f(x) = y such that y ϵ B i.e. y ϵ R − {1}

So, y(x − 3) = x − 2

xy − 3y = x − 2

xy − x = 3y − 2 x = f(y)

Hence f is onto

Hence f - 1 exists.

Now,

fof - 1(x) = x

f(f - 1(x)) = x f - 1(x) – 2 = x[f - 1(x)) – 3]

f - 1(x) – 2 = xf - 1(x)) – 3x

3x – 2 = f - 1(x) [x – 1] Hence f - 1(x): R R is given by for all x R.

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