Answer :

Given that,

In order to prove that f is one-one, it is sufficient to prove that f(x1)=f(x2) x1=x2 x1, x2 A .

Let f(x1)=f(x2)

(x1-2)(x2-3) = (x2-2)(x1-3)

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

-3x1-2x2 = -3x2-2x1

-3x1+2x1 = -3x2+2x2

(-3+2) x1 = (-3+2)x2

x1 = x2

f is one-one.

f is onto if every element of B is the f-image of some element of A.

let f(x) = y

x-2 = y(x-3)

x-2 = xy-3y

x-xy = -3y+2

x(1-y) = -3y+2

Thus, for each y B there exists such that f(x) = y.

Hence, f is onto.

Since, f is one-one and onto therefore f is bijective.

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