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Problem 544
544
A curve has equation \(y = 2 + 3 \, \sin \frac{1}{2}x\) for \(0 \leq x \leq 4\pi\).
(a) State greatest and least values of \(y\). [2]
(b) Sketch the curve. [2]
(c) State the number of solutions of the equation \(2 + 3 \, \sin \frac{1}{2}x = 5 - 2x\) for \(0 \leq x \leq 4\pi\). [1]
Solution
(a) The equation of the curve is \(y = 2 + 3 \, \sin \frac{1}{2}x\). The sine function \(\sin \frac{1}{2}x\) oscillates between -1 and 1. Therefore, the maximum value of \(y\) is when \(\sin \frac{1}{2}x = 1\), giving \(y = 2 + 3 \times 1 = 5\). The minimum value of \(y\) is when \(\sin \frac{1}{2}x = -1\), giving \(y = 2 + 3 \times (-1) = -1\).
(b) The sketch of the curve should show one complete cycle starting and finishing at \(y = 2\), with maximum at \(y = 5\) and minimum at \(y = -1\). The curve should have the correct shape and curvature.
(c) To find the number of solutions for \(2 + 3 \, \sin \frac{1}{2}x = 5 - 2x\), rearrange to \(3 \, \sin \frac{1}{2}x = 3 - 2x\). The left side oscillates between -3 and 3, while the right side is a linear function. There is one intersection point within the given range \(0 \leq x \leq 4\pi\).
