Question

Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by f(x) = (x-2) / (x-3) . Is f one-one and onto? Justify your answer.

Answer 1

it is given that A = R - {3} and B = R - {1}
f : A → B defined by f(x) = (x -2)/(x -3)
Now, let x, y ∈ A such that f(x) = f(y),
e.g., f(x) = (x -2)/(x - 3) , f(y) = (y - 2)/(y - 3)
=> (x - 2)/(x - 3) = (y - 2)/(y - 3)
=> (x - 2)(y - 3) = (y - 2)(x - 3)
=> xy - 3x - 2y + 6 = xy - 3y - 2x + 6
=> -3x + 2x = -3y + 2y
=> - x = -y
=> x = y
hence, f is one - one function.

Let's find range of function f
f(x) = y = (x - 2)/(x - 3)
=> yx - 3y = x - 2
=> (y - 1)x = (3y - 2)
=> x = f(y) = (3y - 2)/(y - 1)
here you can see that domain of f(y) = range of f(x)
so, range of f(x) = R - { 1}
also given co - domain = B = R - { 1}
here, co-domain = range
therefore, f is onto.

hence f is one - one and onto.