Question

If $y=f\left(x\right)=\frac{[x+2]}{[x-1]}$, then :

• A. $x=f\left(y\right)$
• B. $f\left(1\right)=3$
• C. $y$ increases with $x$ for $x<1$
• D. $f$ is a rational function of $x$

Answer 1

Answer: (A), (D)

$f$ is a rational function of $x$

Finding the correct condition:

Given that, $y=f\left(x\right)=\frac{[x+2]}{[x-1]}$

$\begin{array}{rcl}y& =& \frac{[x+2]}{[x-1]}\\ y(x-1)& =& x+2\\ yx-y& =& x+2\\ yx-x& =& y+2\\ x(y-1)& =& y+2\\ x& =& \frac{y+2}{y-1}\end{array}$

$\Rightarrow x=f\left(y\right)$

Hence, option A is correct.

We have, $y=f\left(x\right)=\frac{[x+2]}{[x-1]}$, clearly $f$ is a rational function of $x$ as it is in the form $\frac{p}{q};q\ne 0,p\in I$.

Hence, option D is correct.

Explanation for the incorrect options:

Option B: $f\left(1\right)=3$

$f\left(1\right)=3$ is not possible as $y=f\left(x\right)=\frac{[x+2]}{[x-1]}$ is not defined at $x=1$.

Hence, option B is incorrect.

Option C: $y$ increases with $x$ for $x<1$
$\begin{array}{rcl}f\left(x\right)& =& \frac{[x+2]}{[x-1]}\\ f^{\text{'}}\left(x\right)& =& \frac{\left\{\right[x-1]-[x+2\left]\right\}}{(x-1{)}^2}\\ & =& \frac{-3}{(x-1{)}^2}<0\end{array}$

Which means $y$ decreases with $x$ for $x<1$.

Therefore, option C is incorrect.

Hence, (A), (D) are the correct option.