0606 P12 - Nov 2020 - Q3 - 5 marks
8175
The function \(\mathrm{f}\) is given by
\(\mathrm{f}(x)=2\cos\frac{x}{3}-1.\)
(a) Write down the amplitude of \(\mathrm{f}\).
(b) Write down the period of \(\mathrm{f}\).
(c) Sketch the graph of \(y=\mathrm{f}(x)\) for \(-\pi\leq x\leq3\pi\).
Solution
Answer: amplitude \(2\); period \(6\pi\); key points \((-\pi,0)\), \((0,1)\), \((\pi,0)\), \((3\pi,-3)\).
(a) The amplitude is the coefficient of the cosine term, so the amplitude is
\(2.\)
(b) The period of \(\cos x\) is \(2\pi\). Since the angle is \(\frac{x}{3}\), the period is
\(2\pi\times3=6\pi.\)
(c) Use key points in the interval \(-\pi\leq x\leq3\pi\):
\(\begin{aligned} x=-\pi&:\quad y=2\cos\left(-\frac{\pi}{3}\right)-1=0,\\ x=0&:\quad y=2\cos0-1=1,\\ x=\pi&:\quad y=2\cos\frac{\pi}{3}-1=0,\\ x=3\pi&:\quad y=2\cos\pi-1=-3. \end{aligned}\)
The sketch should be a smooth cosine curve through these points, with maximum \(y=1\) at \(x=0\) and minimum \(y=-3\) at \(x=3\pi\).
