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November 2020 P12 Q11
525
A curve has equation \(y = 3 \cos 2x + 2\) for \(0 \leq x \leq \pi\).
(a) State the greatest and least values of \(y\).
(b) Sketch the graph of \(y = 3 \cos 2x + 2\) for \(0 \leq x \leq \pi\).
(c) By considering the straight line \(y = kx\), where \(k\) is a constant, state the number of solutions of the equation \(3 \cos 2x + 2 = kx\) for \(0 \leq x \leq \pi\) in each of the following cases.
(i) \(k = -3\)
(ii) \(k = 1\)
(iii) \(k = 3\)
Solution
(a) The function \(y = 3 \cos 2x + 2\) is a cosine function scaled and shifted. The cosine function \(\cos 2x\) has a range of \([-1, 1]\). Therefore, \(3 \cos 2x\) has a range of \([-3, 3]\). Adding 2 shifts this range to \([-1, 5]\). Thus, the greatest value is 5 and the least value is -1.
(b) The graph of \(y = 3 \cos 2x + 2\) is a cosine wave starting at its maximum value of 5 at \(x = 0\), decreasing to its minimum value of -1 at \(x = \frac{\pi}{2}\), and returning to its maximum value of 5 at \(x = \pi\). This completes one full cycle.
(c)(i) For \(k = -3\), the line \(y = -3x\) does not intersect the curve \(y = 3 \cos 2x + 2\) within the given range, so there are 0 solutions.
(c)(ii) For \(k = 1\), the line \(y = x\) intersects the curve \(y = 3 \cos 2x + 2\) at two points within the given range, so there are 2 solutions.
(c)(iii) For \(k = 3\), the line \(y = 3x\) intersects the curve \(y = 3 \cos 2x + 2\) at one point within the given range, so there is 1 solution.
