Q22 of 45 Page 1

Let A = R - {3} and B = R - {1}. Consider the function f : A → B defined by Show that f is one - one and onto and hence find f - 1.

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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