Difference between revisions of "1980 AHSME Problems/Problem 14"
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<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 1== |
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As <math>f(x)=cx/2x+3</math>, we can plug that into <math>f(f(x))</math> and simplify to get <math>c^2x/2cx+6x+9 = x</math> | As <math>f(x)=cx/2x+3</math>, we can plug that into <math>f(f(x))</math> and simplify to get <math>c^2x/2cx+6x+9 = x</math> | ||
| − | . However, we have a restriction on x such that if <math>x=-3/2</math> we have an undefined function. We can use this to our advantage. Plugging that value for x into <math>c^2x/2cx+6x+9 = x</math> yields <math>c/2 = -3/2</math>, as the left hand side simplifies and the right hand side is simply the value we have chosen. This means that <math>c=-3 | + | . However, we have a restriction on x such that if <math>x=-3/2</math> we have an undefined function. We can use this to our advantage. Plugging that value for x into <math>c^2x/2cx+6x+9 = x</math> yields <math>c/2 = -3/2</math>, as the left hand side simplifies and the right hand side is simply the value we have chosen. This means that <math>c=-3 \Rightarrow \boxed{A}</math>. |
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| + | ==Solution 2== | ||
| − | 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</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> |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:31, 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 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
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
