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9709 P12 - Jun 2025 - Q5
4102
The equation of a curve is \(y = 4 \cos 2x + 3\) for \(0 \leq x \leq 2\pi\).
State the greatest and least possible values of \(y\).
Sketch the curve.
Hence determine the number of solutions of the equation \(4 \cos 2x + 3 = 2x - 1\) for \(0 \leq x \leq 2\pi\).
Solution
(a) The function \(y = 4 \cos 2x + 3\) is based on the cosine function, which has a range of \([-1, 1]\). Therefore, the range of \(4 \cos 2x\) is \([-4, 4]\). Adding 3 shifts this range to \([-1, 7]\). Thus, the greatest value is 7 and the least value is -1.
(b) The graph of \(y = 4 \cos 2x + 3\) shows two complete cycles from \(0\) to \(2\pi\), starting at the maximum value of 7 and decreasing to the minimum value of -1, then repeating.
(c) To find the number of solutions to \(4 \cos 2x + 3 = 2x - 1\), we equate the expressions: \(4 \cos 2x + 3 = 2x - 1\). Rearranging gives \(4 \cos 2x = 2x - 4\). The number of intersections of the curve \(y = 4 \cos 2x + 3\) with the line \(y = 2x - 1\) within the interval \(0 \leq x \leq 2\pi\) is 3.
