Question 11 - A-Level Maths

Edexcel C34 2015 June — Question 11 13 marks

Exam Board	Edexcel
Module	C34 (Core Mathematics 3 & 4)
Year	2015
Session	June
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Harmonic Form
Type	Applied context modeling
Difficulty	Standard +0.3 This is a standard harmonic form question with straightforward application to a real-world context. Part (a) uses the routine R sin(θ - α) formula with R = √(1.5² + 1.2²) and tan α = 1.2/1.5. Parts (b) and (c) involve simple substitution and solving basic trigonometric equations. While multi-part, each step follows a well-practiced procedure with no novel insight required, making it slightly easier than the average A-level question.
Spec	1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc

11. (a) Express $1.5 \sin \theta - 1.2 \cos \theta$ in the form $R \sin ( \theta - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$ Give the value of $R$ and the value of $\alpha$ to 3 decimal places. The height, $H$ metres, of sea water at the entrance to a harbour on a particular day, is modelled by the equation $$H = 3 + 1.5 \sin \left( \frac { \pi t } { 6 } \right) - 1.2 \cos \left( \frac { \pi t } { 6 } \right) , \quad 0 \leqslant t < 12$$ where $t$ is the number of hours after midday.
(b) Using your answer to part (a), calculate the minimum value of $H$ predicted by this model and the value of $t$, to 2 decimal places, when this minimum occurs.
(c) Find, to the nearest minute, the times when the height of sea water at the entrance to the harbour is predicted by this model to be 4 metres.

Show mark scheme Show mark scheme source

Question 11:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$R = \sqrt{1.5^2 + 1.2^2}$ = awrt 1.921; accept $\sqrt{3.69}$ or $\frac{3\sqrt{41}}{10}$	B1
$\tan\alpha = \frac{1.2}{1.5}$	M1	$\tan\alpha = \pm\frac{1.2}{1.5}$ or $\tan\alpha = \pm\frac{1.5}{1.2}$
$\alpha = 0.675$ or $0.215\pi$	A1	Must be in radians
(3)

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$H = 3 + 1.921\sin\left(\frac{\pi t}{6} - 0.675\right)$
$H_{\min} = 3 - \text{'1.921'} =$ awrt $1.08$	M1A1	States or attempts $3-R$ with their $R$
$\left(\frac{\pi t}{6} - \text{"0.675"}\right) = \frac{3\pi}{2} \Rightarrow t = 10.29$	M1A1	Attempts $\left(\frac{\pi t}{6} - \alpha'\right) = \frac{3\pi}{2}$; putting equal to $-\frac{\pi}{2}$ is M1A0; putting equal to $\frac{\pi}{2}$ is M0A0
(4)

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$4 = 3 + 1.921\sin\left(\frac{\pi t}{6} - 0.675\right) \Rightarrow \sin\left(\frac{\pi t}{6} - 0.675\right) = \frac{1}{1.921}$	M1	$\sin\left(\frac{\pi t}{6} \pm \alpha'\right) = \frac{4-3}{R}$, where \(\left
$\frac{\pi t}{6} - 0.675 = 0.548 \Rightarrow t =$ awrt $2.33$ or $2.34$	dM1A1	Dependent on previous M; correct order for one value of $t$
$\frac{\pi t}{6} - 0.675 = \pi - 0.548 = 2.594 \Rightarrow t =$ awrt $6.24$ or $6.25$	ddM1A1	Dependent on previous M; correct order for second value of $t$
Times are 2:20pm and 6:15pm or 6:14pm (14:20 and 18:15 or 18:14)	A1	Need both times; extra values in range loses final A mark
(6)
(13 marks)

# Question 11:

**Part (a):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = \sqrt{1.5^2 + 1.2^2}$ = awrt 1.921; accept $\sqrt{3.69}$ or $\frac{3\sqrt{41}}{10}$ | B1 | |
| $\tan\alpha = \frac{1.2}{1.5}$ | M1 | $\tan\alpha = \pm\frac{1.2}{1.5}$ or $\tan\alpha = \pm\frac{1.5}{1.2}$ |
| $\alpha = 0.675$ or $0.215\pi$ | A1 | Must be in radians |
| | **(3)** | |

**Part (b):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $H = 3 + 1.921\sin\left(\frac{\pi t}{6} - 0.675\right)$ | | |
| $H_{\min} = 3 - \text{'1.921'} =$ awrt $1.08$ | M1A1 | States or attempts $3-R$ with their $R$ |
| $\left(\frac{\pi t}{6} - \text{"0.675"}\right) = \frac{3\pi}{2} \Rightarrow t = 10.29$ | M1A1 | Attempts $\left(\frac{\pi t}{6} - \alpha'\right) = \frac{3\pi}{2}$; putting equal to $-\frac{\pi}{2}$ is M1A0; putting equal to $\frac{\pi}{2}$ is M0A0 |
| | **(4)** | |

**Part (c):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4 = 3 + 1.921\sin\left(\frac{\pi t}{6} - 0.675\right) \Rightarrow \sin\left(\frac{\pi t}{6} - 0.675\right) = \frac{1}{1.921}$ | M1 | $\sin\left(\frac{\pi t}{6} \pm \alpha'\right) = \frac{4-3}{R}$, where $\left|\frac{4-3}{R}\right| < 1$ |
| $\frac{\pi t}{6} - 0.675 = 0.548 \Rightarrow t =$ awrt $2.33$ or $2.34$ | dM1A1 | Dependent on previous M; correct order for one value of $t$ |
| $\frac{\pi t}{6} - 0.675 = \pi - 0.548 = 2.594 \Rightarrow t =$ awrt $6.24$ or $6.25$ | ddM1A1 | Dependent on previous M; correct order for second value of $t$ |
| Times are 2:20pm and 6:15pm or 6:14pm (14:20 and 18:15 or 18:14) | A1 | Need both times; extra values in range loses final A mark |
| | **(6)** | |
| | **(13 marks)** | |

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11. (a) Express $1.5 \sin \theta - 1.2 \cos \theta$ in the form $R \sin ( \theta - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$ Give the value of $R$ and the value of $\alpha$ to 3 decimal places.

The height, $H$ metres, of sea water at the entrance to a harbour on a particular day, is modelled by the equation

$$H = 3 + 1.5 \sin \left( \frac { \pi t } { 6 } \right) - 1.2 \cos \left( \frac { \pi t } { 6 } \right) , \quad 0 \leqslant t < 12$$

where $t$ is the number of hours after midday.\\
(b) Using your answer to part (a), calculate the minimum value of $H$ predicted by this model and the value of $t$, to 2 decimal places, when this minimum occurs.\\
(c) Find, to the nearest minute, the times when the height of sea water at the entrance to the harbour is predicted by this model to be 4 metres.\\

\hfill \mbox{\textit{Edexcel C34 2015 Q11 [13]}}

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Q1 9 Q2 9 Q3 10 Q4 7 Q5 8 Q6 8 Q7 8 Q8 10 Q9 13 Q10 6 Q11 13 Q12 10 Q13 14

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(R = \sqrt{1.5^2 + 1.2^2}\) = awrt 1.921; accept \(\sqrt{3.69}\) or \(\frac{3\sqrt{41}}{10}\)	B1
\(\tan\alpha = \frac{1.2}{1.5}\)	M1	\(\tan\alpha = \pm\frac{1.2}{1.5}\) or \(\tan\alpha = \pm\frac{1.5}{1.2}\)
\(\alpha = 0.675\) or \(0.215\pi\)	A1	Must be in radians
(3)