smc

$f(1)=\frac{3}{0}$, which is undefined. Hence, the (in)correct answer is $\boxed{f(1)=0}$. We can verify that the other statements are valid. $\text{(A). }y=\frac{x+2}{x-1}⟹xy-y=x+2⟹x(y-1)=y+2⟹x=\frac{y+2}{y-1}$ $\text{(B). }f(0)=\frac{(0)+2}{(0)-1}=-2$ $\text{(D). }f(-2)=\frac{(-2)+2}{(-2)-1}=\frac{0}{-3}=0$ $\text{(E). }f(y)=\frac{\frac{x+2}{x-1}+2}{\frac{x+2}{x-1}-1}=\frac{3x}{3}=x$