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Problem 543
543
A function f is defined by \(f : x \mapsto 3 - 2 \sin x\), for \(0^\circ \leq x \leq 360^\circ\).
(i) Find the range of \(f\). [2]
(ii) Sketch the graph of \(y = f(x)\). [2]
A function \(g\) is defined by \(g : x \mapsto 3 - 2 \sin x\), for \(0^\circ \leq x \leq A^\circ\), where \(A\) is a constant.
(iii) State the largest value of \(A\) for which \(g\) has an inverse. [1]
(iv) When \(A\) has this value, obtain an expression, in terms of \(x\), for \(g^{-1}(x)\). [2]
Solution
(i) The function \(f(x) = 3 - 2 \sin x\) has a range determined by the range of \(\sin x\), which is \([-1, 1]\). Therefore, \(3 - 2(-1) = 5\) and \(3 - 2(1) = 1\). Thus, the range of \(f\) is \(1 \leq f(x) \leq 5\).
(ii) The graph of \(y = 3 - 2 \sin x\) is a sinusoidal wave starting at \(y = 3\), reaching a minimum of \(y = 1\) and a maximum of \(y = 5\), completing one full oscillation from \(x = 0^\circ\) to \(x = 360^\circ\).
(iii) For \(g\) to have an inverse, it must be one-to-one. The largest interval for \(x\) where \(\sin x\) is one-to-one is \(0^\circ \leq x \leq 90^\circ\) or \(0 \leq x \leq \frac{\pi}{2}\).
(iv) To find \(g^{-1}(x)\), set \(y = 3 - 2 \sin x\) and solve for \(x\):
