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Problem 547
547
A function \(f\) is defined by \(f : x \mapsto 5 - 2 \sin 2x\) for \(0 \leq x \leq \pi\).
(i) Find the range of \(f\). [2]
(ii) Sketch the graph of \(y = f(x)\). [2]
(iii) Solve the equation \(f(x) = 6\), giving answers in terms of \(\pi\). [3]
The function \(g\) is defined by \(g : x \mapsto 5 - 2 \sin 2x\) for \(0 \leq x \leq k\), where \(k\) is a constant.
(iv) State the largest value of \(k\) for which \(g\) has an inverse. [1]
(v) For this value of \(k\), find an expression for \(g^{-1}(x)\). [3]
Solution
(i) The function \(f(x) = 5 - 2 \sin 2x\) has a range determined by the range of \(\sin 2x\), which is \([-1, 1]\). Thus, \(f(x)\) ranges from \(5 - 2(1) = 3\) to \(5 - 2(-1) = 7\). Therefore, the range is \(3 \leq f(x) < 7\).
(ii) The graph of \(y = f(x)\) is a sinusoidal curve with one complete oscillation between \(0\) and \(\pi\), initially going downwards, all above \(f(x) = 0\).
(iii) To solve \(f(x) = 6\), set \(5 - 2 \sin 2x = 6\), giving \(\sin 2x = -\frac{1}{2}\). Solving \(2x = \frac{7\pi}{6}\) or \(2x = \frac{11\pi}{6}\), we find \(x = \frac{7\pi}{12}\) or \(x = \frac{11\pi}{12}\).
(iv) For \(g\) to have an inverse, it must be one-to-one. The largest \(k\) for which this is true is \(\frac{\pi}{4}\).
(v) To find \(g^{-1}(x)\), solve \(5 - 2 \sin 2x = y\) for \(x\). This gives \(\sin 2x = \frac{5-y}{2}\), so \(2x = \sin^{-1} \left( \frac{5-y}{2} \right)\) and \(x = \frac{1}{2} \sin^{-1} \left( \frac{5-x}{2} \right)\).
