Test your understanding of graphing sine, cosine, and tangent, transformations, phase shifts, solving equations graphically, and modelling with trig functions.
8 random questions from a replayable checkpoint bank
✍️ Short Answer
9. For $y = 2\sin\left(3x - \frac{\pi}{2}\right) + 1$, state the amplitude, period, phase shift, and range. 4 MARKS
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10. Sketch $y = \tan(2x)$ for $0 \leq x \leq \pi$, showing all asymptotes and intercepts. 3 MARKS
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11. The temperature in a greenhouse is modelled by $T = 6\cos\left(\frac{\pi}{12}t\right) + 18$, where $T$ is in $^\circ$C and $t$ is hours after 6:00 am. (a) Find the maximum and minimum temperatures. (b) Find the first time after 6:00 am when the temperature is $21^\circ$C. (c) Explain why a cosine model is appropriate if the maximum temperature occurs at 6:00 am. 5 MARKS
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12. By considering the graphs of $y = \sin x$ and $y = \frac{1}{2}$, find the number of solutions to $\sin x = \frac{1}{2}$ in $0 \leq x \leq 6\pi$. Explain your reasoning without listing every solution. 3 MARKS
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1. A — Period of $\sin(2x)$ is $\pi$.
2. A — Amplitude of $3\cos x$ is $3$.
3. A — Tangent asymptotes at odd multiples of $\frac{\pi}{2}$; $\tan(2x)$ has asymptotes at $\frac{\pi}{4}, \frac{3\pi}{4}, \dots$
4. A — Phase shift = $\frac{\pi}{6}$ right.
5. A — Range of $2\sin x + 1$ is $[-1, 3]$.
6. A — $\cos x = 0.5$ at $\frac{\pi}{3}$ and $\frac{5\pi}{3}$.
7. A — $a = \frac{16 - 4}{2} = 6$.
8. A — Cotangent period is $\pi$.
Q9 (4 marks): Rewrite: $y = 2\sin\left(3\left(x - \frac{\pi}{6}\right)\right) + 1$ [1]. Amplitude = 2 [0.5], Period = $\frac{2\pi}{3}$ [1], Phase shift = $\frac{\pi}{6}$ right [1], Range = $[-1, 3]$ [0.5].
Q10 (3 marks): Period = $\frac{\pi}{2}$ [0.5]. Asymptotes at $x = \frac{\pi}{4}, \frac{3\pi}{4}$ [1]. Intercepts at $x = 0, \frac{\pi}{2}, \pi$ [1]. Increasing branches [0.5].
Q11 (5 marks): (a) Max = $6 + 18 = 24^\circ$C, Min = $-6 + 18 = 12^\circ$C [2]. (b) $6\cos\left(\frac{\pi}{12}t\right) + 18 = 21 \Rightarrow \cos\left(\frac{\pi}{12}t\right) = 0.5 \Rightarrow \frac{\pi}{12}t = \frac{\pi}{3} \Rightarrow t = 4$ hours [2]. (c) At $t = 0$, $\cos(0) = 1$ gives the maximum value, matching the condition [1].
Q12 (3 marks): Sine has period $2\pi$, so $6\pi$ contains 3 complete cycles [1]. In each cycle, $y = \frac{1}{2}$ intersects $y = \sin x$ twice [1]. Therefore, there are $3 \times 2 = 6$ solutions [1].