Difference between revisions of "1980 AHSME Problems/Problem 14"
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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 < | + | 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 < | + | 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 < | + | 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== | ||
− | == | + | 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>. | |
− | + | -j314andrews | |
− | We begin by calculating < | + | == 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>. | ||
+ | |||
+ | '''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> | <cmath>f(f(x)) = f\left( \frac{cx}{2x + 3} \right).</cmath> | ||
− | Substitute < | + | 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 < | + | 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 < | + | 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 < | + | 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 < | + | 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 < | + | 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 < | + | 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 < | + | '''Step 4: Solve for <math>c</math>''' |
− | From < | + | 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 < | + | 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, < | + | So, <math>c = -3</math> satisfies both equations. |
'''Final Answer:''' | '''Final Answer:''' | ||
− | The value of < | + | The value of <math>c</math> is <math>\boxed{\textbf{(A) -3}}</math>. |
− | |||
== See also == | == See also == |
Latest revision as of 18:13, 14 August 2025
Problem
If the function is defined by
satisfies
for all real numbers
except
, then
is
Solution 1
Since ,
for all
.
Multiplying both sides by yields
. That is,
. Therefore, both
and
, so
.
Solution 2
Since ,
. However,
is undefined at
.
Substituting into
yields
, so
.
Solution 3
Since , the graph of
must be a hyperbola with asymptotes
and
.
Since ,
and the graph of
is symmetrical across the line
. The reflections of the asymptotes of
across
are
and
, so
and thus
.
-j314andrews
Solution 4
We are given a function for
, and it satisfies the functional equation
for all real numbers
. We are tasked with finding the value of
.
Step 1: Calculate
We begin by calculating , which is the composition of the function
with itself. To do this, we substitute
into itself:
Substitute into the formula for
:
Simplify the numerator:
Now simplify the denominator:
To combine the terms in the denominator, express with a denominator of
:
Thus, we have:
Step 2: Set up the functional equation
We are given that for all
. Therefore, we set the expression for
equal to
:
Step 3: Solve the equation
To eliminate the fraction, multiply both sides of the equation by :
Expand both sides:
Now, move all terms to one side of the equation:
Factor out :
Since this equation must hold for all , the expression in parentheses must be equal to zero:
This simplifies to:
For this to hold for all , the coefficient of
must be zero, and the constant term must also be zero. Thus, we have the system of equations:
Step 4: Solve for
From , we solve for
:
Substitute into
:
So, satisfies both equations.
Final Answer:
The value of is
.
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.