f(x) = `(alpha x)/(x + 1) (x ne -1)`, then for what value of `alpha , f{f(x)} = x?`

Answer

-1

Explanation

`f(x)=(alpha x)/(x+1), x ne -1`
Now `f(f(x))=x`
`implies(alpha((alpha x)/(x+1)))/((alpha x)/(x+1)+1)=x`
`implies(alpha^(2)x)/((alpha +1)x+1)=x`
`implies (alpha+1)x^(2)+(1-alpha^(2))x=0 " ...(1)" `
Now this is true for all real x.
`implies alpha +1=0 " and " 1-alpha^(2)=0`
`implies alpha = -1 ` (common value)