Discussion
The inverse function of \(f(x) = \lambda\ x + 1, x \in \mathbb{R}\) is itself. Then the value of \(\lambda\) is
Answer: B
The value is \(\lambda = -1\).
A function is its own inverse when \(f(f(x)) = x\) for every x.
\[ \begin{aligned} f(f(x)) &= \lambda(\lambda x + 1) + 1 \\ &= \lambda^{2}x + \lambda + 1 \end{aligned} \]
For this to equal x we need \(\lambda^{2} = 1\) and \(\lambda + 1 = 0\), so \(\lambda = -1\), giving \(f(x) = 1 - x\), which is indeed its own inverse. \(\lambda = 1\) satisfies the first condition but gives \(f(f(x)) = x + 2\), and the fractional values fail both.
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