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

1980 AHSME Problems/Problem 14

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"