Let f(x)= x+1 ,x -1. Then write the value of satisfying f(f(x))=x for all x -1

We have, f(x)= ax x+1 Now, f(f(x))=x ( ax x+1 )=x ( ax x+1 ) ax x+1 +1 =x ^2x x+1 +x+1 x+1 =x a^2x ax+x+1 =x ^2 ax+x+1 =1 ^2= +x+1 ^2- -(x+1)=0 ^2- (x+1)+ -(x+1)=0 [( )-(x+1) ]+1 [ -(x+1) ]=0 [ -(x+1) ] [ +1 ]=0 +1=0 [ =x+1 does not satisfies f(f(x))=x ] =-1

Let f(x)= x+1 ,x -1. Then write the value of satisfying f(f(x))=x…

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Question

Let $\text{f(x)}=\frac{\alpha\text{x}}{\text{x}+1},\text{x}\neq-1.$ Then write the value of $\alpha$ satisfying $\text{f}(\text{f(x)})=\text{x}$ for all $\text{x}\neq-1$

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Answer

We have,
$\text{f(x)}=\frac{\text{ax}}{\text{x}+1}$
Now,
$\text{f}(\text{f(x)})=\text{x}$
$\Rightarrow\text{f}\Big(\frac{\text{ax}}{\text{x}+1}\Big)=\text{x}$
$\Rightarrow\frac{\alpha\big(\frac{\text{ax}}{\text{x}+1}\big)}{\frac{\text{ax}}{\text{x}+1}+1}=\text{x}$
$\Rightarrow\frac{\frac{\alpha^2\text{x}}{\text{x}+1}}{\frac{\alpha\text{x}+\text{x}+1}{\text{x}+1}}=\text{x}$
$\Rightarrow\frac{\text{a}^2\text{x}}{\text{ax}+\text{x}+1}=\text{x}$
$\Rightarrow\frac{\alpha^2}{\text{ax}+\text{x}+1}=1$
$\Rightarrow\alpha^2=\alpha\text{x}+\text{x}+1$
$\Rightarrow\alpha^2-\alpha\text{x}-(\text{x}+1)=0$
$\Rightarrow\alpha^2-\alpha(\text{x}+1)+\alpha-(\text{x}+1)=0$
$\Rightarrow\alpha\big[(\alpha)-(\text{x}+1)\big]+1\big[\alpha-(\text{x}+1)\big]=0$
$\Rightarrow\big[\alpha-(\text{x}+1)\big]\big[\alpha+1\big]=0$
$\Rightarrow\alpha+1=0$ $\big[\because\ \alpha=\text{x}+1$ does not satisfies $\text{f}(\text{f(x)})=\text{x}\big]$
$\Rightarrow\alpha=-1$

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