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Difference between revisions of "1980 AHSME Problems/Problem 14"
(→‎Solution 1)
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==Solution 1==
 
==Solution 1==
  
−
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}.  
+
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>.  
  
−
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>.
+
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==
  
−
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>.  
+
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>.  
  
−
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>.
+
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>.
  
  
 
== Solution 3 ==
 
== 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>.
+
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>'''
+
'''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:
+
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>
 
<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>:
+
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>
 
<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>
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<cmath>\text{Denominator} = 2 \times \frac{cx}{2x + 3} + 3 = \frac{2cx}{2x + 3} + 3.</cmath>
 
<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>:
+
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>
 
<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>
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'''Step 2: Set up the functional equation'''
 
'''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>:
+
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>
 
<cmath>\frac{c^2 x}{(2c + 6)x + 9} = x.</cmath>
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'''Step 3: Solve the equation'''
 
'''Step 3: Solve the equation'''
  
−
To eliminate the fraction, multiply both sides of the equation by </math>(2c + 6)x + 9<math>:
+
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>
 
<cmath>c^2 x = x \left( (2c + 6)x + 9 \right).</cmath>
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<cmath>0 = (2c + 6)x^2 + 9x - c^2 x.</cmath>
 
<cmath>0 = (2c + 6)x^2 + 9x - c^2 x.</cmath>
  
−
Factor out </math>x<math>:
+
Factor out <math>x</math>:
  
 
<cmath>0 = x \left( (2c + 6)x + 9 - c^2 \right).</cmath>
 
<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:
+
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>
 
<cmath>(2c + 6)x + 9 - c^2 = 0.</cmath>
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<cmath>(2c + 6)x + (9 - c^2) = 0.</cmath>
 
<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:
+
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>2c + 6 = 0</math>
−
* </math>9 - c^2 = 0<math>
+
* <math>9 - c^2 = 0</math>
  
−
'''Step 4: Solve for </math>c<math>'''
+
'''Step 4: Solve for <math>c</math>'''
  
−
From </math>2c + 6 = 0<math>, we 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>
 
<cmath>2c = -6 \quad \Rightarrow \quad c = -3.</cmath>
  
−
Substitute </math>c = -3<math> into </math>9 - c^2 = 0<math>:
+
Substitute <math>c = -3</math> into <math>9 - c^2 = 0</math>:
  
 
<cmath>9 - (-3)^2 = 9 - 9 = 0.</cmath>
 
<cmath>9 - (-3)^2 = 9 - 9 = 0.</cmath>
  
−
So, </math>c = -3<math> satisfies both equations.
+
So, <math>c = -3</math> satisfies both equations.
  
 
'''Final Answer:'''
 
'''Final Answer:'''
  
−
The value of </math>c<math> is </math>\boxed{\textbf{(A) -3}}$.
+
The value of <math>c</math> is <math>\boxed{\textbf{(A) -3}}</math>.
−
 
 
  
 
== See also ==
 
== See also ==

Revision as of 17:56, 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

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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