Difference between revisions of "1980 AHSME Problems/Problem 14"
(→Solution) |
(→Problem) |
||
| Line 4: | Line 4: | ||
<math>\text{(A)} \ -3 \qquad \text{(B)} \ - \frac{3}{2} \qquad \text{(C)} \ \frac{3}{2} \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely determined}</math> | <math>\text{(A)} \ -3 \qquad \text{(B)} \ - \frac{3}{2} \qquad \text{(C)} \ \frac{3}{2} \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely determined}</math> | ||
| − | |||
| − | |||
==Solution== | ==Solution== | ||
Revision as of 20:29, 24 April 2016
Problem
If the function 
is defined by ![\[f(x)=\frac{cx}{2x+3} ,\quad x\neq -\frac{3}{2} ,\]](http://latex.artofproblemsolving.com/5/7/7/5773326d253805644b66a95dd0cf5009bfd69085.png)
satisfies 
for all real numbers 
except 
, then 
is
Solution
As 
, we can plug that into 
and simplify to get 
. However, we have a restriction on x such that if 
we have an undefined function. We can use this to our advantage. Plugging that value for x into yields 
, as the left hand side simplifies and the right hand side is simply the value we have chosen. This means that 
, which is answer choice 
.
Alternatively, after simplifying the function to , multiply both sides by 
and divide by to yield 
. This can be factored to 
. This means that both 
and either one of 
or 
are equal to 0. 
yields and the other two yield 
. The clear solution is .
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
