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June 2017 p11 q9
717
The function f is defined by \(f : x \mapsto \frac{2}{3 - 2x}\) for \(x \in \mathbb{R}, x \neq \frac{3}{2}\).
(i) Find an expression for \(f^{-1}(x)\).
The function g is defined by \(g : x \mapsto 4x + a\) for \(x \in \mathbb{R}\), where \(a\) is a constant.
(ii) Find the value of \(a\) for which \(gf(-1) = 3\).
(iii) Find the possible values of \(a\) given that the equation \(f^{-1}(x) = g^{-1}(x)\) has two equal roots.
Solution
(i) To find \(f^{-1}(x)\), start with \(y = \frac{2}{3 - 2x}\). Rearrange to get \(y(3 - 2x) = 2\), which simplifies to \(3y - 2xy = 2\). Solving for \(x\), we get \(2xy = 3y - 2\), so \(x = \frac{3y - 2}{2y}\). Thus, \(f^{-1}(x) = \frac{3}{2} - \frac{1}{x}\).
(ii) For \(gf(-1) = 3\), calculate \(f(-1) = \frac{2}{3 - 2(-1)} = \frac{2}{5}\). Then \(g(f(-1)) = 4 \times \frac{2}{5} + a = \frac{8}{5} + a\). Set \(\frac{8}{5} + a = 3\) to find \(a = \frac{7}{5}\).
(iii) Set \(f^{-1}(x) = g^{-1}(x)\). From (i), \(f^{-1}(x) = \frac{3}{2} - \frac{1}{x}\). For \(g^{-1}(x) = \frac{x - a}{4}\), equate \(\frac{3}{2} - \frac{1}{x} = \frac{x - a}{4}\). Rearrange to form a quadratic equation: \(x^2 - x(a + 6) + 4 = 0\). For two equal roots, the discriminant must be zero: \((a + 6)^2 = 16\). Solving gives \(a + 6 = 4\) or \(a + 6 = -4\), so \(a = -2\) or \(a = -10\).
