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

Revision as of 18:13, 14 August 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

Since $f(x)=\frac{cx}{2x+3}$ , $f(f(x)) = \frac{c^2x}{2cx+6x+9} = x$ for all $x \neq \frac{3}{2}$ .

Multiplying both sides by $\frac{2cx+6x+9}{x}$ yields $c^2=2cx+6x+9$ . That is, $x(2c+6) + (3+c)(3-c) = 0$ . Therefore, both $2c+6 = 0$ and $(3+c)(3-c) = 0$ , so $c= \boxed{(\textbf{A}) -3}$ .

Solution 2

Since $f(x)=\frac{cx}{2x+3}$ , $f(f(x)) = \frac{c^2x}{2cx+6x+9} = x$ . However, $f$ is undefined at $x=-\frac{3}{2}$ .

Substituting $x=-\frac{3}{2}$ into $\frac{c^2x}{2cx+6x+9} = x$ yields $\frac{c}{2} = -\frac{3}{2}$ , so $c= \boxed{(\textbf{A}) -3}$ .

Solution 3

Since $f(x) = \frac{cx}{2x+3} = \frac{\frac{1}{2}cx}{x+\frac{3}{2}} = \frac{\frac{1}{2}cx + \frac{3}{4}c - \frac{3}{4}c}{x+\frac{3}{2}} = \frac{\frac{1}{2}c\left(x+\frac{3}{2}\right) - \frac{3}{4}c}{x+\frac{3}{2}} = \frac{1}{2}c - \frac{\frac{3}{4}c}{x+\frac{3}{2}}$ , the graph of $f(x)$ must be a hyperbola with asymptotes $x=-\frac{3}{2}$ and $y=\frac{1}{2}{c}$ .

Since $f(f(x)) = x$ , $f(x) = f^{-1}(x)$ and the graph of $f(x)$ is symmetrical across the line $y=x$ . The reflections of the asymptotes of $f(x)$ across $y=x$ are $y = -\frac{3}{2}$ and $x=\frac{1}{2}c$ , so $-\frac{3}{2} = \frac{1}{2}c$ and $c= \boxed{(\textbf{A}) -3}$

-j314andrews

Solution 4

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 $\boxed{\textbf{(A) -3}}$ .

@@ Line 19: / Line 19: @@
 ==Solution 3==
-Since <math>f(x) = \frac{cx}{2x+3} = \frac{\frac{1}{2}cx}{x+\frac{3}{2}} = \frac{\frac{1}{2}cx + \frac{3}{4}c - \frac{3}{4}c}{x+\frac{3}{2}} = \frac{\frac{1}{2}c\left(x+\frac{3}{2}\right) - \frac{3}{4}c}{x+\frac{3}{2}} = \frac{1}{2}c - \frac{\frac{3}{4}c}{x+\frac{3}{2}}</math>, the graph of <math>f(x)</math> must be a hyperbola with asymptotes <math>x=\frac{-3}{2}</math> and <math>y=\frac{1}{2}{c}</math>.
+Since <math>f(x) = \frac{cx}{2x+3} = \frac{\frac{1}{2}cx}{x+\frac{3}{2}} = \frac{\frac{1}{2}cx + \frac{3}{4}c - \frac{3}{4}c}{x+\frac{3}{2}} = \frac{\frac{1}{2}c\left(x+\frac{3}{2}\right) - \frac{3}{4}c}{x+\frac{3}{2}} = \frac{1}{2}c - \frac{\frac{3}{4}c}{x+\frac{3}{2}}</math>, the graph of <math>f(x)</math> must be a hyperbola with asymptotes <math>x=-\frac{3}{2}</math> and <math>y=\frac{1}{2}{c}</math>.
-Since <math>f(f(x)) = x</math>, <math>f(x) = f^{-1}(x)</math> and the graph of <math>f(x)</math> is symmetrical across the line <math>y=x</math>.  The reflections of the asymptotes of <math>f(x)</math> across <math>y=x</math> are <math>y = \frac{-3}{2}</math> and <math>x=\frac{1}{2}c</math>, so <math>\frac{-3}{2} = \frac{1}{2}c</math> and <math>c= \boxed{(\textbf{A}) -3}</math>
+Since <math>f(f(x)) = x</math>, <math>f(x) = f^{-1}(x)</math> and the graph of <math>f(x)</math> is symmetrical across the line <math>y=x</math>.  The reflections of the asymptotes of <math>f(x)</math> across <math>y=x</math> are <math>y = -\frac{3}{2}</math> and <math>x=\frac{1}{2}c</math>, so <math>-\frac{3}{2} = \frac{1}{2}c</math> and <math>c= \boxed{(\textbf{A}) -3}</math>
 -j314andrews
 == Solution 4==

1980 AHSME (Problems • Answer Key • Resources)
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