21isnt it 0?????????
If f(x)=\frac{ax}{x+1} , x\neq -1
then for what value of a is f(f(x))=x ?
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4 Answers
1708$\textbf{Here $\mathbf{f(x)=\frac{ax}{x+1}}$}$\\\\ $\textbf{Now $\mathbf{f(f(x))=x\Leftrightarrow f(x)=f^{-1}(x).................................(1)}$}$\\\\ $\textbf{Now We have to Calculate $\mathbf{f^{-1}(x)}$ Using $\mathbf{f(x)}$}$\\\\ $\textbf{So $\mathbf{f^{-1}(x)=\frac{x}{a-x}}$ (Calculate yourself.)}$\\\\ $\textbf{Now Put value of $\mathbf{f^{-1}(x)}$ in equa.......(1),we Get $}$\\\\ $\mathbf{\frac{ax}{x+1}=\frac{x}{a-x}\Leftrightarrow x^2.(1+a)+x.(1-a^2)=0=0.x^2+0.x}$\\\\ \textbf{So We Get $\mathbf{1+a=0\Leftrightarrow a=-1}$ and $\mathbf{1-a^2=0\Leftrightarrow a=\pm 1}$}$\\\\ $\boxed{\boxed{\mathbf{a=-1}}}$
1x2(a+1)+x.(1-a2)=0
How can it be this 0.x2+0.x?Are those both the positive numbers that you are applying this rule?
=>x[x(1+a)+1-a2]=0
=>x=0 or x(1+a)+1-a2=0
x(1+a)+1-a2=0
=>x=a2-11+a=a-1
If x is known then a can be calculated.But there is no x.