Comprehensive assessment covering both lessons: transformations of the trigonometric functions, amplitude, period and phase shift, sketching transformed graphs, solving trigonometric equations, modelling periodic phenomena, and the logarithmic scales used for decibels, Richter magnitude, stellar magnitude and pH.
Assessment
Select the best answer for each question. 1 mark each.
The amplitude of $y = 4\sin(2x) - 3$ is:
The period of $y = 3\cos(4x)$ is:
The period of $y = \tan(2x)$ is:
The graph of $y = \sin\!\left(x - \dfrac{\pi}{3}\right)$ is the graph of $y = \sin x$ shifted:
The equation of the midline of $y = -2\sin(x) + 5$ is:
The maximum value of $y = 3\cos(x) - 1$ is:
Compared with $y = 5\cos(2x)$, the graph of $y = -5\cos(2x)$ is:
How many solutions does $2\sin x = 1$ have for $0 \le x \le 2\pi$?
Which equation has amplitude $3$ and period $\pi$?
The height of the tide is modelled by $h(t) = a\cos(bt) + d$. High tide is $6$ m at $t = 0$ and the next low tide is $2$ m at $t = 6$ hours. The model is:
A Ferris wheel of radius $8$ m has its centre $10$ m above the ground and completes one revolution every $20$ seconds. A rider starts at the bottom. The correct height model $h(t)$ in metres is:
A sound has intensity $I = 10^{-4}$ W/m². Using $L = 10\log_{10}\!\left(\dfrac{I}{I_0}\right)$ with $I_0 = 10^{-12}$ W/m², the sound level is:
A solution has hydrogen ion concentration $[\text{H}^+] = 10^{-9}$ mol/L. Using $\text{pH} = -\log_{10}[\text{H}^+]$, its pH is:
On the Richter scale, $M = \log_{10}\!\left(\dfrac{A}{A_0}\right)$. An earthquake of magnitude $8$ has ground-motion amplitude how many times greater than one of magnitude $5$?
Using $m_1 - m_2 = -2.5\log_{10}\!\left(\dfrac{F_1}{F_2}\right)$, two stars differ in apparent magnitude by $5$. The ratio of their brightness (flux) is:
Short Answer
For $y = -3\cos\!\left(2x - \dfrac{\pi}{2}\right) - 1$, state: (a) the amplitude, (b) the period, (c) the phase shift, (d) the equation of the midline, and (e) the maximum and minimum values. Show all working.
Sketch one complete cycle of $y = 2\sin\!\left(x + \dfrac{\pi}{4}\right)$, starting from the phase-shifted position. Label the coordinates of the maximum, the minimum, and the midline crossings. Show all working.
Solve $2\cos\!\left(x - \dfrac{\pi}{4}\right) = \sqrt{2}$ for $0 \le x \le 2\pi$. Show all working and justify the number of solutions in the domain.
The temperature in a greenhouse over one day is modelled by $T(t) = 6\sin\!\left(\dfrac{\pi t}{12}\right) + 22$ degrees Celsius, where $t$ is hours after 6 am. Find: (a) the maximum and minimum temperatures, (b) the period of the model, and (c) the first time after 6 am at which the temperature reaches $25^\circ$C. Show all working.
A buoy oscillates with a period of $8$ seconds. Its height above the seabed ranges between a maximum of $5$ m and a minimum of $1$ m, and it is at its maximum height at $t = 0$. Find a model of the form $h(t) = a\cos(bt) + d$, and use it to find the height of the buoy at $t = 3$ seconds. Show all working.
Two sounds are recorded: a whisper at $20$ dB and a vacuum cleaner at $70$ dB. Using $L = 10\log_{10}\!\left(\dfrac{I}{I_0}\right)$ with $I_0 = 10^{-12}$ W/m²: (a) find the intensity of each sound in W/m², and (b) determine how many times more intense the vacuum cleaner is than the whisper. Show all working.
An earthquake registers $M = 6.0$ on the Richter scale, where $M = \log_{10}\!\left(\dfrac{A}{A_0}\right)$. A second earthquake has ground-motion amplitude $40$ times greater. (a) Show that its magnitude is $6.0 + \log_{10}(40)$. (b) Find this magnitude correct to two decimal places. Show all working.
Two solutions are tested: solution X has pH $3$ and solution Y has pH $6$, where $\text{pH} = -\log_{10}[\text{H}^+]$. (a) Find the hydrogen ion concentration $[\text{H}^+]$ of each solution. (b) Determine how many times more acidic solution X is than solution Y. Show all working.
Q1: B, amplitude $= |a| = |4| = 4$. The $-3$ is a vertical shift, not the amplitude, and $b = 2$ affects the period, not the amplitude.
Q2: A, period $= \dfrac{2\pi}{|b|} = \dfrac{2\pi}{4} = \dfrac{\pi}{2}$.
Q3: D, tangent has natural period $\pi$, so period $= \dfrac{\pi}{|b|} = \dfrac{\pi}{2}$.
Q4: C, phase shift $= -\dfrac{c}{b} = -\dfrac{-\pi/3}{1} = +\dfrac{\pi}{3}$, a shift to the right by $\dfrac{\pi}{3}$.
Q5: A, the midline is $y = d$. Here $d = 5$, so the midline is $y = 5$. The $-2$ is the amplitude coefficient (reflection), not the midline.
Q6: D, maximum $= d + |a| = -1 + 3 = 2$.
Q7: B, the negative sign on $a$ reflects the graph over the $x$-axis. The amplitude remains $|-5| = 5$ (always positive).
Q8: C, $2\sin x = 1 \Rightarrow \sin x = \dfrac{1}{2}$, giving $x = \dfrac{\pi}{6}$ and $x = \dfrac{5\pi}{6}$ in $[0, 2\pi]$, so two solutions.
Q9: A, need $|a| = 3$ and $\dfrac{2\pi}{|b|} = \pi$, so $|b| = 2$. Only $y = 3\sin(2x)$ satisfies both. Option B has period $2\pi$; option C has amplitude $2$; option D has period $\dfrac{\pi}{2}$.
Q10: B, midline $d = \dfrac{6 + 2}{2} = 4$; amplitude $a = \dfrac{6 - 2}{2} = 2$; high to low is half a cycle, so $T = 12$ and $b = \dfrac{2\pi}{12} = \dfrac{\pi}{6}$. Since the maximum occurs at $t = 0$, use cosine: $h(t) = 2\cos\!\left(\dfrac{\pi t}{6}\right) + 4$.
Q11: D, amplitude $= 8$ (radius), midline $d = 10$ (centre height), period $= 20$ so $b = \dfrac{2\pi}{20} = \dfrac{\pi}{10}$. Starting at the bottom means the minimum is at $t = 0$, which requires a negative cosine: $h(t) = -8\cos\!\left(\dfrac{\pi t}{10}\right) + 10$. At $t = 0$: $-8 + 10 = 2$ m (the lowest point). Option C uses the wrong $b$ ($\pi/20$ gives period $40$).
Q12: C, $L = 10\log_{10}\!\left(\dfrac{10^{-4}}{10^{-12}}\right) = 10\log_{10}(10^{8}) = 10 \times 8 = 80$ dB.
Q13: A, $\text{pH} = -\log_{10}(10^{-9}) = -(-9) = 9$. Since pH $> 7$, the solution is basic.
Q14: D, a difference of $8 - 5 = 3$ magnitudes gives an amplitude ratio of $10^{3} = 1000$.
Q15: B, a difference of $5$ magnitudes gives a flux ratio of $10^{5/2.5} = 10^{2} = 100$ (Pogson's ratio).
Q16 (3 marks): $a = -3$, $b = 2$, $c = -\dfrac{\pi}{2}$, $d = -1$ [identify]. (a) Amplitude $= |-3| = 3$ [1]. (b) Period $= \dfrac{2\pi}{|2|} = \pi$. (c) Factor: $y = -3\cos\!\left(2\!\left(x - \dfrac{\pi}{4}\right)\!\right) - 1$, so phase shift $= -\dfrac{c}{b} = -\dfrac{-\pi/2}{2} = +\dfrac{\pi}{4}$ (right $\dfrac{\pi}{4}$) [1]. (d) Midline $y = -1$. (e) Max $= d + |a| = -1 + 3 = 2$; min $= d - |a| = -1 - 3 = -4$ [1].
Q17 (3 marks): $a = 2$, $b = 1$, $c = \dfrac{\pi}{4}$, $d = 0$. Amplitude $= 2$, period $= 2\pi$, phase shift $= -\dfrac{\pi}{4}$ (left $\dfrac{\pi}{4}$) [1]. Five key $x$-values start at the phase shift $x = -\dfrac{\pi}{4}$, stepping by $T/4 = \dfrac{\pi}{2}$: $x = -\dfrac{\pi}{4}, \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}$ [1]. The $y$-values follow the sine pattern midline $\to$ max $\to$ midline $\to$ min $\to$ midline: $\left(-\dfrac{\pi}{4}, 0\right)$, maximum $\left(\dfrac{\pi}{4}, 2\right)$, $\left(\dfrac{3\pi}{4}, 0\right)$, minimum $\left(\dfrac{5\pi}{4}, -2\right)$, $\left(\dfrac{7\pi}{4}, 0\right)$. A smooth sine curve joins these five labelled points [1].
Q18 (4 marks): Isolate: $\cos\!\left(x - \dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}$ [1]. Let $u = x - \dfrac{\pi}{4}$; for $x \in [0, 2\pi]$, $u \in \left[-\dfrac{\pi}{4}, \dfrac{7\pi}{4}\right]$ [1]. Reference angle $= \dfrac{\pi}{4}$; cosine is positive in Q1 and Q4, so $u = \dfrac{\pi}{4}$ and $u = 2\pi - \dfrac{\pi}{4} = \dfrac{7\pi}{4}$, both within the adjusted domain [1]. Back-substitute $x = u + \dfrac{\pi}{4}$: $x = \dfrac{\pi}{2}$ and $x = 2\pi$. The domain spans exactly one period ($2\pi$), so $\cos u = \dfrac{\sqrt{2}}{2}$ is met exactly twice: two solutions, $x = \dfrac{\pi}{2}$ and $x = 2\pi$ [1].
Q19 (4 marks): (a) Amplitude $= 6$, midline $= 22$, so max $= 22 + 6 = 28^\circ$C and min $= 22 - 6 = 16^\circ$C [1]. (b) Period $= \dfrac{2\pi}{\pi/12} = 24$ hours [1]. (c) $6\sin\!\left(\dfrac{\pi t}{12}\right) + 22 = 25 \Rightarrow \sin\!\left(\dfrac{\pi t}{12}\right) = \dfrac{3}{6} = \dfrac{1}{2}$ [1]. Principal value: $\dfrac{\pi t}{12} = \dfrac{\pi}{6}$, so $t = \dfrac{12}{6} = 2$. The first time after 6 am is $t = 2$ hours, i.e. 8 am [1].
Q20 (4 marks): Amplitude $a = \dfrac{5 - 1}{2} = 2$; midline $d = \dfrac{5 + 1}{2} = 3$ [1]. $T = 8$, so $b = \dfrac{2\pi}{8} = \dfrac{\pi}{4}$. Maximum at $t = 0$ means cosine: $h(t) = 2\cos\!\left(\dfrac{\pi t}{4}\right) + 3$ [1]. At $t = 3$: $h(3) = 2\cos\!\left(\dfrac{3\pi}{4}\right) + 3$ [1]. Since $\cos\!\left(\dfrac{3\pi}{4}\right) = -\dfrac{\sqrt{2}}{2}$, $h(3) = 2\left(-\dfrac{\sqrt{2}}{2}\right) + 3 = 3 - \sqrt{2} \approx 1.59$ m [1].
Q21 (4 marks): (a) Whisper (20 dB): $\log_{10}\!\left(\dfrac{I}{10^{-12}}\right) = 2$, so $I = 10^{2} \times 10^{-12} = 10^{-10}$ W/m². Vacuum (70 dB): $\log_{10}\!\left(\dfrac{I}{10^{-12}}\right) = 7$, so $I = 10^{7} \times 10^{-12} = 10^{-5}$ W/m² [2]. (b) Ratio $= \dfrac{10^{-5}}{10^{-10}} = 10^{5} = 100\,000$. (Equivalently, $\Delta L = 70 - 20 = 50$ dB, so the ratio is $10^{50/10} = 10^{5}$.) The vacuum cleaner is $100\,000$ times more intense [2].
Q22 (4 marks): (a) Let $A_1$ be the first amplitude: $M_1 = \log_{10}\!\left(\dfrac{A_1}{A_0}\right) = 6.0$, so $\dfrac{A_1}{A_0} = 10^{6}$. The second amplitude is $A_2 = 40 A_1$, so $M_2 = \log_{10}\!\left(\dfrac{40 A_1}{A_0}\right) = \log_{10}(40) + \log_{10}\!\left(\dfrac{A_1}{A_0}\right) = \log_{10}(40) + 6.0$ [2]. (b) $\log_{10}(40) \approx 1.6021$, so $M_2 \approx 6.0 + 1.6021 = 7.60$ (to 2 d.p.) [2].
Q23 (3 marks): (a) $\text{pH} = -\log_{10}[\text{H}^+] \Rightarrow [\text{H}^+] = 10^{-\text{pH}}$. Solution X: $[\text{H}^+] = 10^{-3}$ mol/L; solution Y: $[\text{H}^+] = 10^{-6}$ mol/L [2]. (b) Ratio $= \dfrac{10^{-3}}{10^{-6}} = 10^{3} = 1000$. Solution X is $1000$ times more acidic than solution Y [1].