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Problem 550
550
The function \(f : x \mapsto 6 - 4\cos\left(\frac{1}{2}x\right)\) is defined for \(0 \leq x \leq 2\pi\).
Find the exact value of \(x\) for which \(f(x) = 4\). [3]
State the range of \(f\). [2]
Sketch the graph of \(y = f(x)\). [2]
Find an expression for \(f^{-1}(x)\). [3]
Solution
(i) Start with the equation \(6 - 4\cos\left(\frac{1}{2}x\right) = 4\). Simplify to \(4\cos\left(\frac{1}{2}x\right) = 2\), giving \(\cos\left(\frac{1}{2}x\right) = \frac{1}{2}\). The solutions for \(\frac{1}{2}x = \frac{1}{3}\pi\) and \(\frac{1}{2}x = \frac{5}{3}\pi\) lead to \(x = \frac{2}{3}\pi\) and \(x = \frac{10}{3}\pi\). However, \(x = \frac{10}{3}\pi\) is outside the given domain, so the valid solutions are \(x = \frac{2}{3}\pi\) and \(x = \frac{4}{3}\pi\).
(ii) The range of \(f(x) = 6 - 4\cos\left(\frac{1}{2}x\right)\) is determined by the range of \(\cos\left(\frac{1}{2}x\right)\), which is \([-1, 1]\). Thus, \(f(x)\) ranges from \(6 - 4(1) = 2\) to \(6 - 4(-1) = 10\), so \(2 \leq f(x) \leq 10\).
(iii) The graph of \(y = f(x)\) is a cosine wave with a point of inflexion at \(x = \pi\).
(iv) To find \(f^{-1}(x)\), start with \(y = 6 - 4\cos\left(\frac{1}{2}x\right)\). Rearrange to \(\cos\left(\frac{1}{2}x\right) = \frac{6-y}{4}\). Then, \(\frac{1}{2}x = \cos^{-1}\left(\frac{6-y}{4}\right)\), leading to \(x = 2\cos^{-1}\left(\frac{6-y}{4}\right)\). Therefore, \(f^{-1}(x) = 2\cos^{-1}\left(\frac{6-x}{4}\right)\).
