# Let A = R – {3} and B = R – {1}. Consider the function f: A → B defined by . Is f one-one and onto? Justify your answer.

It is given that A = R – {3} and B = R – {1}

f: A B defined by Now, let x, y ϵ A such that f(x) = f(y) (x-2)(y-3) = (y-2)(x-3)

xy 3x -2y + 6 = xy -3y -2x +6

-3x -2y = -3y -2x

x = y

f is one one.

Let y ϵ B = R – {1}

Then, y ≠ 1.

The function f is onto if there exist x ϵ A such that f(x) = y

Now, f(x) = y x 2 = xy -3y

x(1-y) = -3y + 2 y ϵ B, there exists ϵ A such that f is onto.

Therefore, function f is one- one and onto.

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