Let f(x)=αx(x+1),x≠−1. Then for what value of α is f(f(x))=x?

f(x)=αxx+1,x≠−1 f(f(x))=x or α∞xx+1αxx+1+1=x or α2x(α+1)x+1=x or (α+1)x2+1−α2x=0 or α+1=0 and 1−α2=0 [As it is true ∀x≠−1 , Eq. (1) is an identity] or α=−1

Let f(x)=αx(x+1),x≠−1. Then for what value of α is f(f(x))=x?

f(x)=αxx+1,x≠−1 f(f(x))=x or α∞xx+1αxx+1+1=x or α2x(α+1)x+1=x or (α+1)x2+1−α2x=0 or α+1=0 and 1−α2=0 [As it is true ∀x≠−1 , Eq. (1) is an identity] or α=−1

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

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$\sqrt{2}$

$- \sqrt{2}$

-1

answer is D.

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Detailed Solution

$\begin{array}{l} \begin{aligned} f (x) = \frac{α x}{x + 1}, x \neq - 1 \\ f (f (x)) = x or \frac{α (\frac{\infty x}{x + 1})}{\frac{α x}{x + 1} + 1} = x \\ or \frac{α^{2} x}{(α + 1) x + 1} = x \\ or (α + 1) x^{2} + (1 - α^{2}) x = 0 \\ or α + 1 = 0 and 1 - α^{2} = 0 \end{aligned} \\ [As it is true \forall x \neq - 1, Eq. (1) is an identity] \\ or α = - 1 \end{array}$