Question

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

Answer 1

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Given function:

$\tt f (x) = (x- 2)/(x -3)$

Checking for one-one function:

$\tt f (x^1) = (x^1– 2)/ (x^1– 3)$

$\tt f (x^2) = (x^2-2)/ (x^2-3)$

Putting $\tt f (x^1) = f (x^2)$

$\tt (x^1-2)/(x^1-3)= (x^2-2 )/(x^2 -3)$

$\tt (x^1-2) (x^2– 3) = (x^1– 3) (x^2-2)$

$\tt x^1 (x^2– 3)- 2 (x^2 -3) = x1 (x^2– 2) – 3 (x^2– 2)$

$\tt x^1 x^2 -3x^1 -2x^2 + 6 = x^1 x^2 – 2x^1 -3x^2 + 6$

$\tt -3x^1– 2x^2 =- 2x^1 -3x^2$

$\tt 3x^2 -2x^2 = – 2x^1 + 3x^1$

$\tt x^1= x^2$

Hence, if $\tt f (x^1) = f (x^2), then x^1 = x^2$

Thus, the function f is one-one function.

Checking for onto function:

$\tt f (x) = (x-2)/(x-3)$

Let f(x) = y such that y B i.e. y ∈ R

$\tt So, y = (x -2)/(x- 3)$

$\tt y(x -3) = x- 2$

$\tt xy -3y = x -2$

$\tt xy – x = 3y-2$

$\tt x (y -1) = 3y- 2$

$\tt x = (3y -2) /(y-1)$

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

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

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Hope It's Helpful.....:)

Answer 2

Answer:

Hope this helped u!!!~