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Problem 520
520
The diagram shows part of the graph of \(y = a \cos(bx) + c\).
(a) Find the values of the positive integers \(a\), \(b\) and \(c\).
(b) For these values of \(a\), \(b\) and \(c\), use the given diagram to determine the number of solutions in the interval \(0 \leq x \leq 2\pi\) for each of the following equations.
(i) \(a \cos(bx) + c = \frac{6}{\pi} x\)
(ii) \(a \cos(bx) + c = 6 - \frac{6}{\pi} x\)
Solution
(a) The graph of \(y = a \cos(bx) + c\) shows a cosine wave with a maximum value of 8 and a minimum value of -2. The amplitude \(a\) is half the distance between the maximum and minimum values, so \(a = \frac{8 - (-2)}{2} = 5\).
The period of the cosine function is \(\frac{2\pi}{b}\). From the graph, the period is \(\pi\), so \(\frac{2\pi}{b} = \pi\), giving \(b = 2\).
The vertical shift \(c\) is the average of the maximum and minimum values, so \(c = \frac{8 + (-2)}{2} = 3\).
(b)(i) For \(a \cos(bx) + c = \frac{6}{\pi} x\), substitute \(a = 5\), \(b = 2\), \(c = 3\) to get \(5 \cos(2x) + 3 = \frac{6}{\pi} x\). The number of solutions is the number of intersections of the line \(y = \frac{6}{\pi} x\) with the cosine graph in the interval \(0 \leq x \leq 2\pi\), which is 3.
(b)(ii) For \(a \cos(bx) + c = 6 - \frac{6}{\pi} x\), substitute \(a = 5\), \(b = 2\), \(c = 3\) to get \(5 \cos(2x) + 3 = 6 - \frac{6}{\pi} x\). The number of solutions is the number of intersections of the line \(y = 6 - \frac{6}{\pi} x\) with the cosine graph in the interval \(0 \leq x \leq 2\pi\), which is 2.
