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

Given function:

f (x) = (x- 2)/(x -3)

Checking for one-one function:

f (x₁) = (x₁– 2)/ (x₁– 3)

f (x₂) = (x₂-2)/ (x₂-3)

Putting f (x₁) = f (x₂)

(x₁-2)/(x₁-3)= (x₂-2 )/(x₂ -3)

(x₁-2) (x₂– 3) = (x₁– 3) (x₂-2)

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

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

-3x₁– 2x₂ =- 2x₁ -3x₂

3x₂ -2x₂ = – 2x₁ + 3x₁

x₁= x₂

Hence, if f (x₁) = f (x₂), then x₁ = x₂

Thus, the function f is one-one  function.

Checking for onto function:

f (x) = (x-2)/(x-3)

Let f(x) = y such that y B i.e. y ∈ R – {1}

So, y = (x -2)/(x- 3)
y(x -3) = x- 2

xy -3y = x -2

xy – x = 3y-2

x (y -1) = 3y- 2

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

For y = 1, x is not defined But it is given that. y ∈ R – {1}

Hence, x = (3y- 2)/(y- 1) ∈ R -{3} Hence, f is onto.