Question

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

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

Answer 1

The correct option is D $-1$

$f\left(f\left(x\right)\right)=\frac{a\frac{ax}{x+1}}{\frac{ax}{x+1}+1}=\frac{\frac{a^{2}x}{x+1}}{\frac{ax+x+1}{x+1}}=\frac{a^{2}x}{ax+x+1}$

Since, $f\left(f\left(x\right)\right)=x$, we have,

$\frac{a^{2}x}{ax+x+1}=x$.

Simplifying the equation we get,

$a^{2}x=\left(a+1\right)x^2+x$

$\therefore \left(a+1\right)x^2+\left(1-a^2\right)x=0$

or $\left(a+1\right)x\left(x+1-a\right)=0$

Hence the only possible value is $a=-1$