First slide

Functions (XII)

Question

Let $f : (1, \infty) \to (1, \infty)$ be defined by $f (x) = \frac{x + 2}{x - 1} .$
Then

Moderate

A

$f$ is 1 – 1 and onto
B

$f$ is 1 – 1 but not onto
C

$f$ is not 1 – 1 but onto
D

$f$ is neither 1 – 1nor onto

Solution

If $\begin{array}{l} \frac{x_{1} + 2}{x_{1} - 1} = \frac{x_{2} + 2}{x_{2} - 1} \Rightarrow x_{1} x_{2} + 2 x_{2} - x_{1} - 2 \\ = x_{1} x_{2} + 2 x_{1} - x_{2} - 2 \end{array}$
$\Rightarrow 3 x_{2} = 3 x_{1} \Rightarrow x_{1} = x_{2}$
So $f$ in one - one
$\begin{array}{l} If y = f (x) = \frac{x + 2}{x - 1} \Rightarrow y x - y = x + 2 \\ \Rightarrow (y - 1) x = y + 2 \\ \Rightarrow x = \frac{y + 2}{y - 1} \in (1, \infty) \end{array}$
Thus $f$ is onto

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