Question

If $f\left(x\right)=\frac{ax}{x+1}$, $x\ne -1$, then for what value of $a$ is $f\left(f\left(x\right)\right)=x$?

• A. $\sqrt{2}$
• B. $-\sqrt{2}$
• C. $1$
• D. $2$
• E. $-1$

Answer 1

The correct option is E

$-1$

Explanation for the correct option.

Step 1. Form the equation.

For the function $f\left(x\right)=\frac{ax}{x+1}$, $f\left(f\left(x\right)\right)$ is given as:

$\begin{array}{rcl}f\left(f\left(x\right)\right)& =& \frac{a{\displaystyle \frac{ax}{x+1}}}{{\displaystyle \frac{ax}{x+1}}+1}\\ & =& \frac{a^{2}x}{ax+x+1}\end{array}$.

But it is given that $f\left(f\left(x\right)\right)=x$, thus

$$\begin{gathered} \frac{a^{2}x}{ax+x+1}=x \\ \Rightarrow a^{2}x=(a+1)x^2+x \\ \Rightarrow \left(a+1\right)x^2-\left(a^2-1\right)x=0 \end{gathered}$$

Step 2. Solve the equation for $a$.

The equation $\left(a+1\right)x^2-\left(a^2-1\right)x=0$ can be simplified and solve for $a$ as:

$$\begin{gathered} \left(a+1\right)x^2-\left(a^2-1\right)x=0 \\ \Rightarrow \left(a+1\right)x-\left(a+1\right)\left(a-1\right)=0 \\ \Rightarrow \left(a+1\right)\left(x-a+1\right)=0 \end{gathered}$$

So, $a+1=0$ and thus $a=-1$.

Thus the value of $a$ is $-1$.

Hence, the correct option is E.