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प्रश्न
Let \[f\left( x \right) = \frac{\alpha x}{x + 1}, x \neq - 1\] . Then write the value of α satisfying f(f(x)) = x for all x ≠ −1.
उत्तर
Given:
\[Since f(f(x)) = x, \]
\[\frac{\alpha\left( \frac{\alpha x}{x + 1} \right)}{\frac{\alpha x}{x + 1} + 1} = x\]
\[ \Rightarrow \frac{\alpha^2 x}{\alpha x + x + 1} = x\]
\[ \Rightarrow \alpha^2 x - \alpha x^2 - ( x^2 + x) = 0\]
\[\text{ Solving the quadratic equation in } \alpha: \]
\[\alpha = \frac{x^2 \pm \sqrt{x^4 + 4x( x^2 + x)}}{2x} \]
\[ \Rightarrow \alpha = x + 1 \text{ or } - 1\]
\[\text{ Since } , \alpha \neq x + 1, \]
\[\alpha = - 1 . \]
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