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Problem #542
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542
The function f is defined by \(f : x \mapsto 5 - 3 \sin 2x\) for \(0 \leq x \leq \pi\).
(i) Find the range of \(f\). [2]
(ii) Sketch the graph of \(y = f(x)\). [3]
(iii) State, with a reason, whether \(f\) has an inverse. [1]
Solution
(i) The function \(f(x) = 5 - 3 \sin 2x\) is based on the sine function, which oscillates between -1 and 1. Therefore, \(-3 \leq -3 \sin 2x \leq 3\). Adding 5 to each part of the inequality gives \(2 \leq 5 - 3 \sin 2x \leq 8\). Thus, the range of \(f\) is \(2 \leq f(x) \leq 8\).
(ii) The graph of \(y = 5 - 3 \sin 2x\) will show one complete oscillation from \(x = 0\) to \(x = \pi\). The graph starts at \(y = 5\), decreases to \(y = 2\), and then increases back to \(y = 8\) at \(x = \pi\). It is important to note that the graph does not touch the x-axis and is in the first quadrant.
(iii) The function \(f\) does not have an inverse because it is not one-to-one. The function \(f(x) = 5 - 3 \sin 2x\) repeats values within the interval \(0 \leq x \leq \pi\), which means it fails the horizontal line test for invertibility.
