Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1980 AHSME Problems/Problem 14"

(Added missing navbox)
(Added solution 3 and categorized 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 1==
 
==Solution 1==
−
 
  
 
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>
Line 14: Line 14:
 
==Solution 2==
 
==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 \Rightarrow \boxed{A}</math>
 +
 +
 +
== Solution 3 ==
 +
 +
We are given a function <math>f(x) = \frac{cx}{2x + 3}</math> for <math>x \neq -\frac{3}{2}</math>, and it satisfies the functional equation <math>x = f(f(x))</math> for all real numbers <math>x \neq -\frac{3}{2}</math>. We are tasked with finding the value of <math>c</math>.
 +
 +
'''Step 1: Calculate <math>f(f(x))</math>'''
 +
 +
We begin by calculating <math>f(f(x))</math>, which is the composition of the function <math>f(x)</math> with itself. To do this, we substitute <math>f(x) = \frac{cx}{2x + 3}</math> into itself:
 +
 +
<cmath>f(f(x)) = f\left( \frac{cx}{2x + 3} \right).</cmath>
 +
 +
Substitute <math>\frac{cx}{2x + 3}</math> into the formula for <math>f</math>:
 +
 +
<cmath>f\left( \frac{cx}{2x + 3} \right) = \frac{c \left( \frac{cx}{2x + 3} \right)}{2 \left( \frac{cx}{2x + 3} \right) + 3}.</cmath>
 +
 +
Simplify the numerator:
 +
 +
<cmath>\text{Numerator} = c \times \frac{cx}{2x + 3} = \frac{c^2 x}{2x + 3}.</cmath>
 +
 +
Now simplify the denominator:
 +
 +
<cmath>\text{Denominator} = 2 \times \frac{cx}{2x + 3} + 3 = \frac{2cx}{2x + 3} + 3.</cmath>
 +
 +
To combine the terms in the denominator, express <math>3</math> with a denominator of <math>2x + 3</math>:
 +
 +
<cmath>\frac{2cx}{2x + 3} + 3 = \frac{2cx + 3(2x + 3)}{2x + 3} = \frac{2cx + 6x + 9}{2x + 3} = \frac{(2c + 6)x + 9}{2x + 3}.</cmath>
 +
 +
Thus, we have:
 +
 +
<cmath>f(f(x)) = \frac{\frac{c^2 x}{2x + 3}}{\frac{(2c + 6)x + 9}{2x + 3}} = \frac{c^2 x}{(2c + 6)x + 9}.</cmath>
 +
 +
'''Step 2: Set up the functional equation'''
 +
 +
We are given that <math>f(f(x)) = x</math> for all <math>x \neq -\frac{3}{2}</math>. Therefore, we set the expression for <math>f(f(x))</math> equal to <math>x</math>:
 +
 +
<cmath>\frac{c^2 x}{(2c + 6)x + 9} = x.</cmath>
 +
 +
'''Step 3: Solve the equation'''
 +
 +
To eliminate the fraction, multiply both sides of the equation by <math>(2c + 6)x + 9</math>:
 +
 +
<cmath>c^2 x = x \left( (2c + 6)x + 9 \right).</cmath>
 +
 +
Expand both sides:
 +
 +
<cmath>c^2 x = (2c + 6)x^2 + 9x.</cmath>
 +
 +
Now, move all terms to one side of the equation:
 +
 +
<cmath>0 = (2c + 6)x^2 + 9x - c^2 x.</cmath>
 +
 +
Factor out <math>x</math>:
 +
 +
<cmath>0 = x \left( (2c + 6)x + 9 - c^2 \right).</cmath>
 +
 +
Since this equation must hold for all <math>x \neq 0</math>, the expression in parentheses must be equal to zero:
 +
 +
<cmath>(2c + 6)x + 9 - c^2 = 0.</cmath>
 +
 +
This simplifies to:
 +
 +
<cmath>(2c + 6)x + (9 - c^2) = 0.</cmath>
 +
 +
For this to hold for all <math>x \neq 0</math>, the coefficient of <math>x</math> must be zero, and the constant term must also be zero. Thus, we have the system of equations:
 +
 +
* <math>2c + 6 = 0</math>
 +
* <math>9 - c^2 = 0</math>
 +
 +
'''Step 4: Solve for <math>c</math>'''
 +
 +
From <math>2c + 6 = 0</math>, we solve for <math>c</math>:
 +
 +
<cmath>2c = -6 \quad \Rightarrow \quad c = -3.</cmath>
 +
 +
Substitute <math>c = -3</math> into <math>9 - c^2 = 0</math>:
 +
 +
<cmath>9 - (-3)^2 = 9 - 9 = 0.</cmath>
 +
 +
So, <math>c = -3</math> satisfies both equations.
 +
 +
'''Final Answer:'''
 +
 +
The value of <math>c</math> is <math>\boxed{\textbf{(A) -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>
 
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1980|num-b=13|num-a=15}}   
 
{{AHSME box|year=1980|num-b=13|num-a=15}}   
  
 +
{{MAA Notice}}
  
−
{{MAA Notice}}
+
[[Category:Intermediate Algebra Problems]]

Revision as of 19:18, 25 June 2025

Problem

If the function $f$ is defined by \[f(x)=\frac{cx}{2x+3} ,\quad x\neq -\frac{3}{2} ,\] satisfies $x=f(f(x))$ for all real numbers $x$ except $-\frac{3}{2}$, then $c$ is

$\text{(A)} \ -3 \qquad \text{(B)} \ - \frac{3}{2} \qquad \text{(C)} \ \frac{3}{2} \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely determined}$


Solution 1

As $f(x)=cx/2x+3$, we can plug that into $f(f(x))$ and simplify to get $c^2x/2cx+6x+9 = x$ . However, we have a restriction on x such that if $x=-3/2$ we have an undefined function. We can use this to our advantage. Plugging that value for x into $c^2x/2cx+6x+9 = x$ yields $c/2 = -3/2$, as the left hand side simplifies and the right hand side is simply the value we have chosen. This means that $c=-3 \Rightarrow \boxed{A}$.


Solution 2

Alternatively, after simplifying the function to $c^2x/2cx+6x+9 = x$, multiply both sides by $2cx+6x+9$ and divide by $x$ to yield $c^2=2cx+6x+9$. This can be factored to $x(2c+6) + (3+c)(3-c) = 0$. This means that both $2c+6$ and either one of $3+c$ or $3-c$ are equal to 0. $2c+6=0$ yields $c=-3$ and the other two yield $c=3,-3$. The clear solution is $c=-3 \Rightarrow \boxed{A}$


Solution 3

We are given a function $f(x) = \frac{cx}{2x + 3}$ for $x \neq -\frac{3}{2}$, and it satisfies the functional equation $x = f(f(x))$ for all real numbers $x \neq -\frac{3}{2}$. We are tasked with finding the value of $c$.

Step 1: Calculate $f(f(x))$

We begin by calculating $f(f(x))$, which is the composition of the function $f(x)$ with itself. To do this, we substitute $f(x) = \frac{cx}{2x + 3}$ into itself:

\[f(f(x)) = f\left( \frac{cx}{2x + 3} \right).\]

Substitute $\frac{cx}{2x + 3}$ into the formula for $f$:

\[f\left( \frac{cx}{2x + 3} \right) = \frac{c \left( \frac{cx}{2x + 3} \right)}{2 \left( \frac{cx}{2x + 3} \right) + 3}.\]

Simplify the numerator:

\[\text{Numerator} = c \times \frac{cx}{2x + 3} = \frac{c^2 x}{2x + 3}.\]

Now simplify the denominator:

\[\text{Denominator} = 2 \times \frac{cx}{2x + 3} + 3 = \frac{2cx}{2x + 3} + 3.\]

To combine the terms in the denominator, express $3$ with a denominator of $2x + 3$:

\[\frac{2cx}{2x + 3} + 3 = \frac{2cx + 3(2x + 3)}{2x + 3} = \frac{2cx + 6x + 9}{2x + 3} = \frac{(2c + 6)x + 9}{2x + 3}.\]

Thus, we have:

\[f(f(x)) = \frac{\frac{c^2 x}{2x + 3}}{\frac{(2c + 6)x + 9}{2x + 3}} = \frac{c^2 x}{(2c + 6)x + 9}.\]

Step 2: Set up the functional equation

We are given that $f(f(x)) = x$ for all $x \neq -\frac{3}{2}$. Therefore, we set the expression for $f(f(x))$ equal to $x$:

\[\frac{c^2 x}{(2c + 6)x + 9} = x.\]

Step 3: Solve the equation

To eliminate the fraction, multiply both sides of the equation by $(2c + 6)x + 9$:

\[c^2 x = x \left( (2c + 6)x + 9 \right).\]

Expand both sides:

\[c^2 x = (2c + 6)x^2 + 9x.\]

Now, move all terms to one side of the equation:

\[0 = (2c + 6)x^2 + 9x - c^2 x.\]

Factor out $x$:

\[0 = x \left( (2c + 6)x + 9 - c^2 \right).\]

Since this equation must hold for all $x \neq 0$, the expression in parentheses must be equal to zero:

\[(2c + 6)x + 9 - c^2 = 0.\]

This simplifies to:

\[(2c + 6)x + (9 - c^2) = 0.\]

For this to hold for all $x \neq 0$, the coefficient of $x$ must be zero, and the constant term must also be zero. Thus, we have the system of equations:

  • $2c + 6 = 0$
  • $9 - c^2 = 0$

Step 4: Solve for $c$

From $2c + 6 = 0$, we solve for $c$:

\[2c = -6 \quad \Rightarrow \quad c = -3.\]

Substitute $c = -3$ into $9 - c^2 = 0$:

\[9 - (-3)^2 = 9 - 9 = 0.\]

So, $c = -3$ satisfies both equations.

Final Answer:

The value of $c$ is $\boxed{\textbf{(A) -3}}$.


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. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_14&oldid=252348"