Question

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Answer 1

Answer:

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Answer 2

$\bf \to \: { \underline{given \div }}$

$\rm \to \: f(x) = \dfrac{ \alpha x}{x + 1} \: , x \ne \: - 1$

$\rm \to \: then \: write \: the \: value \: of \: \alpha \: satisfying \: \\ \\ \rm \to \: f(f(x)) \: = x \: , \forall \: x \: \ne \: - 1$

$\bf \to \:{ \underline{solution \div }}$

$\rm \to \: f(x) \: = \dfrac{ \alpha x}{x + 1} , \: x \ne \: 1$

$\rm \to \: f(f(x)) = x$

$\rm \to \: put \: the \: value \: of \: f(x) \:$

$\rm \to \: \dfrac{ \alpha f(x)}{f(x) + 1} = x$

$\rm \to \: \frac{ \alpha ( \dfrac{ \alpha x}{x + 1}) }{ (\dfrac{ \alpha x}{x + 1}) + 1} = x$

$\rm \to \: \dfrac{ \frac{ { \alpha }^{2}x }{x + 1} }{ \dfrac{ \alpha x + x + 1}{x + 1} } = x$

$\rm \to \: \dfrac{ \alpha {}^{2}x }{ \alpha x + x + 1} = x$

$\rm \to \: { \alpha }^{2} = \alpha x + x + 1$

$\rm \to \: { \alpha }^{2} - \alpha x - x - 1 = 0$

$\rm \to \: { \alpha }^{2} - \alpha x - (x \: + \: 1) = 0$

$\rm \to \: d \: = {b}^{2} - 4ac$

$\rm \to \: ( - x) {}^{2} - 4(1)(x + 1) = 0$

$\rm \to \: {x}^{2} + 4x + 4 = 0$

$\rm \to \: {x}^{2} + 2x + 2x + 4 = 0$

$\rm \to \: x(x + 2) + 2(x + 2) = 0$

$\rm \to \: (x + 2) {}^{2} = 0$

$\rm \to \: x \: = \dfrac{ - b \pm \: \sqrt{d} }{2a}$

$\rm \to \: x \: = \dfrac{x + \sqrt{(x + 2) {}^{2} } }{2}$

$\rm \to \: x \: = \dfrac{x + x + 2}{2}$

$\rm \to \: x \: = x + 1$

$\rm \to \: x \: = \dfrac{x - x - 2}{2}$

$\rm \to \: x \: = - 1$

Therefore,

value of a = x + 1 , -1