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 $c=-3 \Rightarrow \boxed{A}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1980 AHSME Problems/Problem 14

Problem

Solution 1

Solution 2