Q. 223.9( 9 Votes )

Let A = R - {3} a

Answer :

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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