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Answer the following.
Let A = mathbbR - {2} and B = mathbbR - {1}. If f : A → B is a function defined by f(x) = (x -…
Let $A = \mathbb{R} - \{2\}$ and $B = \mathbb{R} - \{1\}$. If $f : A \to B$ is a function defined by $f(x) = \frac{x - 1}{x - 2}$ then show that $f$ is a one-one and an onto function.
Given: $f : A \to B \text{ defined by } f(x) = \frac{x - 1}{x - 2}$
To show: $f \text{ is a one-one and an onto function}$
Answer
Answer
One-one: let $f(x_1) = f(x_2)$ for $x_1, x_2 \in A$.
$\frac{x_1-1}{x_1-2} = \frac{x_2-1}{x_2-2}$
$(x_1-1)(x_2-2) = (x_2-1)(x_1-2)$
$x_1x_2 - 2x_1 - x_2 + 2 = x_1x_2 - 2x_2 - x_1 + 2$
$x_1 = x_2$, so $f$ is one-one.
Onto: let $y \in B$, $y \neq 1$, and $y = \frac{x-1}{x-2}$.
$y(x-2) = x-1 \Rightarrow x(y-1) = 2y-1 \Rightarrow x = \frac{2y-1}{y-1}$, which is defined as $y \neq 1$.
$x = 2$ would need $2y-1 = 2y-2$, which is impossible, so $x \in A$ and $f(x) = y$.
Every element of $B$ has a pre-image in $A$, so $f$ is onto.
Hence $f$ is a one-one and an onto function.
- Step 1: Let $f(x_1) = f(x_2)$, i.e. $\frac{x_1-1}{x_1-2} = \frac{x_2-1}{x_2-2}$ (definition of one-one).
- Step 2: Cross-multiplying: $(x_1-1)(x_2-2) = (x_2-1)(x_1-2)$.
- Step 3: Expanding: $x_1x_2 - 2x_1 - x_2 + 2 = x_1x_2 - 2x_2 - x_1 + 2$.
- Step 4: Simplifying: $-x_1 = -x_2$, so $x_1 = x_2$; hence $f$ is one-one.
- Step 5: For onto, let $y \in B$ and put $y = \frac{x-1}{x-2}$; then $y(x-2) = x-1$.
- Step 6: Solving for $x$: $x(y-1) = 2y-1$, so $x = \frac{2y-1}{y-1}$, defined since $y \neq 1$.
- Step 7: $x \neq 2$, because $x = 2$ gives $2y-1 = 2y-2$, which is false; so $x \in A$ and $f(x) = y$.
- Step 8: Every $y \in B$ has a pre-image in $A$, so $f$ is onto. Hence $f$ is a one-one and an onto function.