Solve this


If $f(x)=\frac{k x}{x+1}$, where $x \neq-1$ and $f\{f(x)\}=x$ for $x \neq-1$ then find the value of k.




Given. $f(x)=\frac{k x}{x+1}, x \neq-1$

$F(f(x))=f\left(\frac{k x}{(x+1)}\right.$

$=\frac{k \frac{k x}{x+1}}{\frac{k x}{x+1}+1}$

$=\frac{k^{2} x}{k x+x+1}$

Given that f(f(x)) = x

$\mathrm{x}=\frac{k^{2} \mathrm{x}}{k x+x+1}$

Dividing both sides by x

$1=\frac{k^{2}}{k x+x+1}$


$1^{k^{2}}-k x-(x+1)=0$

$\mathrm{k}=\frac{\frac{x+\sqrt{x^{2}+4 x+4}}{2}}{2}$ or $\mathrm{k}=\frac{x-\sqrt{x^{2}+4 x+4}}{2}$

$\mathrm{k}=\frac{\frac{x+x+2}{2}}{2}$ or $\mathrm{k}=\frac{x-x-2}{2}$

$k=x+1$ or $k=-1$

As value of x is variable we take k = -1.

Therefore, k= -1


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