Find \(\alpha\) such that \(f(f(x))=x\)

🎥 Video Explanation

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📝 Question

Given:

\[ f(x)=\frac{\alpha x}{x+1}, \quad x\ne -1 \]

Find \(\alpha\) such that:

\[ f(f(x))=x \]

• (a) \(\sqrt{2}\)
• (b) \(-\sqrt{2}\)
• (c) \(1\)
• (d) \(-1\)

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✅ Solution

🔹 Step 1: Compute \(f(f(x))\)

\[ f(f(x)) = f\!\left(\frac{\alpha x}{x+1}\right) \]

\[ = \frac{\alpha \cdot \frac{\alpha x}{x+1}}{\frac{\alpha x}{x+1}+1} \] —

🔹 Step 2: Simplify

Denominator:

\[ \frac{\alpha x}{x+1}+1 = \frac{\alpha x + x +1}{x+1} \]

So:

\[ f(f(x))=\frac{\alpha^2 x}{\alpha x + x +1} \]

\[ =\frac{\alpha^2 x}{x(\alpha+1)+1} \] —

🔹 Step 3: Set Equal to \(x\)

\[ \frac{\alpha^2 x}{x(\alpha+1)+1}=x \] —

🔹 Step 4: Solve

\[ \alpha^2 x = x\big(x(\alpha+1)+1\big) \]

\[ \alpha^2 x = x^2(\alpha+1)+x \]

Divide by \(x\) (for \(x\ne0\)):

\[ \alpha^2 = x(\alpha+1)+1 \]

For all \(x\), coefficient of \(x\) must be zero:

\[ \alpha+1=0 \Rightarrow \alpha=-1 \] —

🔹 Step 5: Verify

\[ \alpha^2=1 \] ✔️ satisfies equation

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🔹 Final Answer

\[ \boxed{\text{Option (d): } -1} \]