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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>
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==Solution 1==
 
==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>
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Since <math>f(x)=\frac{cx}{2x+3}</math>, <math>f(f(x)) = \frac{c^2x}{2cx+6x+9} = x</math> for all <math>x \neq \frac{3}{2}</math>.  
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. 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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Multiplying both sides by <math>\frac{2cx+6x+9}{x}</math> yields <math>c^2=2cx+6x+9</math>. That is, <math>x(2c+6) + (3+c)(3-c) = 0</math>. Therefore, both <math>2c+6 = 0</math> and <math>(3+c)(3-c) = 0</math>, so <math>c= \boxed{(\textbf{A}) -3}</math>.
  
 
==Solution 2==
 
==Solution 2==
  
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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>
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Since <math>f(x)=\frac{cx}{2x+3}</math>, <math>f(f(x)) = \frac{c^2x}{2cx+6x+9} = x</math>.  However, <math>f</math> is undefined at <math>x=-\frac{3}{2}</math>.  
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Substituting <math>x=-\frac{3}{2}</math> into <math>\frac{c^2x}{2cx+6x+9} = x</math> yields <math>\frac{c}{2} = -\frac{3}{2}</math>, so <math>c= \boxed{(\textbf{A}) -3}</math>.
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==Solution 3==
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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 thus <math>c= \boxed{(\textbf{A}) -3}</math>.
  
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== Solution 3 ==
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-j314andrews
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== Solution 4==
  
 
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>.
 
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>.
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The value of <math>c</math> is <math>\boxed{\textbf{(A) -3}}</math>.
 
The value of <math>c</math> is <math>\boxed{\textbf{(A) -3}}</math>.
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== See also ==
 
== See also ==

Latest 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 thus $c= \boxed{(\textbf{A}) -3}$.

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

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


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