Difference between revisions of "1980 AHSME Problems/Problem 14"
(→Solution) |
J314andrews (talk | contribs) (Added missing navbox) |
||
| Line 16: | Line 16: | ||
Alternatively, after simplifying the function to <math>c^2x/2cx+6x+9 = x</math>, multiply both sides by <math>2cx+6x+9</math> and divide by <math>x</math> to yield <math>c^2=2cx+6x+9</math>. This can be factored to <math>x(2c+6) + (3+c)(3-c) = 0</math>. This means that both <math>2c+6</math> and either one of <math>3+c</math> or <math>3-c</math> are equal to 0. <math>2c+6=0</math> yields <math>c=-3</math> and the other two yield <math>c=3,-3</math>. The clear solution is <math>c=-3 \Rightarrow \boxed{A}</math> | Alternatively, after simplifying the function to <math>c^2x/2cx+6x+9 = x</math>, multiply both sides by <math>2cx+6x+9</math> and divide by <math>x</math> to yield <math>c^2=2cx+6x+9</math>. This can be factored to <math>x(2c+6) + (3+c)(3-c) = 0</math>. This means that both <math>2c+6</math> and either one of <math>3+c</math> or <math>3-c</math> are equal to 0. <math>2c+6=0</math> yields <math>c=-3</math> and the other two yield <math>c=3,-3</math>. The clear solution is <math>c=-3 \Rightarrow \boxed{A}</math> | ||
| + | |||
| + | == See also == | ||
| + | {{AHSME box|year=1980|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:01, 25 June 2025
Contents
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 1
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 
.
Solution 2
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
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
