1952 AHSME Problems/Problem 11

Problem

If $y=f(x)=\frac{x+2}{x-1}$ , then it is incorrect to say:

$\textbf{(A)\ }x=\frac{y+2}{y-1}\qquad\textbf{(B)\ }f(0)=-2\qquad\textbf{(C)\ }f(1)=0\qquad$

$\textbf{(D)\ }f(-2)=0\qquad\textbf{(E)\ }f(y)=x$

Solution

$f(1)=\frac{3}{0}$ , which is undefined. Hence, the (in)correct answer is $\boxed{\textbf{(C)}\ f(1)=0}$ . We can verify that the other statements are valid.

$\textbf{(A).\ } y=\frac{x+2}{x-1}\implies xy-y=x+2\implies x(y-1)=y+2\implies x=\frac{y+2}{y-1}$

$\textbf{(B).\ } f(0)=\frac{(0)+2}{(0)-1}=-2$

$\textbf{(D).\ } f(-2)=\frac{(-2)+2}{(-2)-1}=\frac{0}{-3}=0$

$\textbf{(E).\ } f(y)=\frac{\frac{x+2}{x-1}+2}{\frac{x+2}{x-1}-1}=\frac{3x}{3}=x$

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1952 AHSME Problems/Problem 11

Problem

Solution

See also