Q. If $y = f(x) = \frac{x + 2}{x - 1}$, then
A) $x = f(y)$
B) $f(1) = 3$
C) $y$ increase with $x$ for $x < 1$
D) $f$ is a rational function of $x$
Explanation
(a)
$\because f(y) = f\left( \frac{x + 2}{x - 1} \right) = \frac{\frac{x + 2}{x - 1} + 2}{\frac{x + 2}{x - 1} - 1}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(y) = x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.