If f(x) = frac{alpha x}{x + 1}, x neq -1. The | vedclass

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(D) Given $f(x) = \frac{\alpha x}{x + 1}$.We need to find $\alpha$ such that $f(f(x)) = x$.First,calculate $f(f(x)) = \frac{\alpha f(x)}{f(x) + 1}$.Su

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$-1$

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(D) Given $f(x) = \frac{\alpha x}{x + 1}$.We need to find $\alpha$ such that $f(f(x)) = x$.First,calculate $f(f(x)) = \frac{\alpha f(x)}{f(x) + 1}$.Substitute $f(x) = \frac{\alpha x}{x + 1}$ into the expression:$f(f(x)) = \frac{\alpha \left( \frac{\alpha x}{x + 1} ight)}{\frac{\alpha x}{x + 1} + 1} = \frac{\frac{\alpha^2 x}{x + 1}}{\frac{\alpha x + x + 1}{x + 1}} = \frac{\alpha^2 x}{(\alpha + 1)x + 1}$.We are given $f(f(x)) = x$,so:$\frac{\alpha^2 x}{(\alpha + 1)x + 1} = x$.Assuming $x 
eq 0$,we have $\frac{\alpha^2}{(\alpha + 1)x + 1} = 1$.$\alpha^2 = (\alpha + 1)x + 1$.For this to hold for all $x$ in the domain,the coefficients of $x$ must be zero and the constant terms must be equal:$\alpha + 1 = 0 \implies \alpha = -1$.Checking the constant term: $\alpha^2 = 1$. If $\alpha = -1$,then $(-1)^2 = 1$,which is true.Thus,the value of $\alpha$ is $-1$.

If $f(x) = \frac{\alpha x}{x + 1}, x \neq -1$. Then,for what value of $\alpha$ is $f(f(x)) = x$?

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