Question

Let $f\left(x\right)=\frac{\mathrm{kx}}{x+1},x\ne -1.$If $f\left(f\right(x\left)\right)=x$ for all $x$ $\ne$

$-1$, then the value of $k$ is

• A. $k=1$
• B. $k=0$
• C. $k=-1$
• D. $k=2$

Answer 1

The correct option is C $k=-1$

$f\left(x\right)=\frac{\mathrm{kx}}{x+1},x\ne -1.f\left(f\left(x\right)\right)=f\left(\frac{\mathrm{kx}}{x+1}\right)=\frac{k\left(\frac{\mathrm{kx}}{x+1}\right)}{\frac{\mathrm{kx}}{x+1}+1}=\frac{\frac{k^{2}x}{x+1}}{\frac{\mathrm{kx}+x+1}{x+1}}=\frac{k^{2}x}{x(k+1)+1}\because f\left(f\left(x\right)\right)=x\Rightarrow \frac{k^{2}x}{x(k+1)+1}=x\Rightarrow k^{2}x=x^2\left(k+1\right)+x\Rightarrow x^2\left(k+1\right)+x\left(1-k^2\right)=0\Rightarrow x^2\left(k+1\right)+x\left(1-k\right)\left(1+k\right)=0\Rightarrow x\left(k+1\right)\left(x+1-k\right]=0\Rightarrow k+1=0\Rightarrow k=-1$