Question

Let $f\left(x\right)=\frac{ax}{x+1},x\ne -1$. Then for what values of $a$ is $f\left(f\left(x\right)\right)=x$.

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

Answer 1

The correct option is D

$-1$

Explanation of correct answer :

Finding the value of $a$ :

Given, $f\left(x\right)=\frac{ax}{x+1},x\ne -1\dots \left(1\right)$

and, $f\left(f\left(x\right)\right)=x$

Replace $x$with $f\left(x\right)$ in equation $\left(1\right)$., we get

$$\begin{gathered} \Rightarrow f\left(f\left(x\right)\right)=\frac{a\times {\displaystyle \frac{ax}{x+1}}}{{\displaystyle \frac{ax}{x+1}}+1} \\ \Rightarrow x=\frac{a^{2}x}{ax+x+1}(\text{given,}f(f\left(x\right))=x) \end{gathered}$$

Simplifying the equation,

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

Hence, the value of $a$ is $-1$.

Thus, the correct answer is Option(D).