Functions 2 Question 15
16.
If $y=f(x)=\frac{x+2}{x-1}$, then
(1984, 3M)
(a) $x=f(y)$
(b) $f(1)=3$
(c) $y$ increases with $x$ for $x<1$
(d) $f$ is a rational function of $x$
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Answer:
Correct Answer: 16. (a,d)
Solution:
- Given, $y=f(x)=\frac{x+2}{x-1}$
$ \begin{array}{lrll} \Rightarrow & y x-y=x+2 & \Rightarrow & x(y-1)=y+2 \\ \Rightarrow & x=\frac{y+2}{y-1} & \Rightarrow & x=f(y) \end{array} $
Here, $f(1)$ does not exist, so domain $\in R-{1}$
$ \begin{aligned} \frac{d y}{d x} & =\frac{(x-1) \cdot 1-(x+2) \cdot 1}{(x-1)^{2}} \\ & =-\frac{3}{(x-1)^{2}} \end{aligned} $
$\Rightarrow f(x)$ is decreasing for all $x \in R-{1}$.
Also, $f$ is rational function of $x$.
Hence, (a) and (d) are correct options.
