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Question
Let \[f\left( x \right) = \frac{\alpha x}{x + 1}, x \neq - 1\] Then, for what value of α is \[f \left( f\left( x \right) \right) = x?\]
Options
\[\sqrt{2}\]
\[- \sqrt{2}\]
1
-1
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Solution
(d) −1
\[f\left( f\left( x \right) \right) = x\]
\[ \Rightarrow f\left( \frac{\alpha x}{x + 1} \right) = x\]
\[ \Rightarrow \frac{\alpha\left( \frac{\alpha x}{x + 1} \right)}{\left( \frac{\alpha x}{x + 1} \right) + 1} = x\]
\[ \Rightarrow \frac{\alpha^2 x}{\alpha x + x + 1} = x\]
\[ \Rightarrow \alpha^2 x = \alpha x^2 + x^2 + x\]
\[ \Rightarrow \alpha^2 x - \alpha x^2 - x^2 - x = 0\]
\[ \Rightarrow \alpha^2 x - \alpha x^2 - \left( x^2 + x \right) = 0\]
\[\text{Solving } \text{for } \text{ the } \alpha \text{ we get}, \]
\[\alpha = \frac{- \left( - x^2 \right) \pm \sqrt{\left( - x^2 \right)^2 - 4 \times x \times \left[ - \left( x^2 + x \right) \right]}}{2x}\]
\[ = \frac{x^2 \pm \sqrt{x^4 + 4 x^3 + 4 x^2}}{2x}\]
\[ = x + 1, - 1\]
\[\text{Here, - 1 is independent of } x, \]
\[ \therefore for, \alpha = - 1, f\left( f\left( x \right) \right) = x\]
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