If the function f(x) = 1 + kx , x is not equal to 0, is the inverse of itself, then the value of k is

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Solution

$Given f is the inverse of itself . \Rightarrow {fof}^{- 1} (x) = x \Rightarrow 1 + {kf}^{- 1} (x) = x \Rightarrow f^{- 1} (x) = \frac{x - 1}{k} Since f is the inverse of itself, f (x) = f^{- 1} (x) for all x \in R \Rightarrow 1 + kx = \frac{x - 1}{k} for all x \in R \Rightarrow kx + 1 = (\frac{1}{k}) x + (- \frac{1}{k}) for all x \in R Comparing both sides, we get k = \frac{1}{k} and 1 = - \frac{1}{k} \Rightarrow k^{2} = 1 and k = - 1 \Rightarrow k = - 1$

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