Question 7 - A-Level Maths

Edexcel C3 2010 June — Question 7 15 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Year	2010
Session	June
Marks	15
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Harmonic Form
Type	Applied context modeling
Difficulty	Standard +0.3 This is a standard C3 harmonic form question with routine application to a real-world context. Part (a) uses the standard R sin(θ-α) technique, parts (b) and (c) apply this directly to find maxima, and part (d) requires solving a trigonometric equation. All steps follow well-practiced procedures with no novel insight required, making it slightly easier than average.
Spec	1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc 1.05o Trigonometric equations: solve in given intervals

7. (a) Express $2 \sin \theta - 1.5 \cos \theta$ in the form $R \sin ( \theta - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$. Give the value of $\alpha$ to 4 decimal places.
(b) (i) Find the maximum value of $2 \sin \theta - 1.5 \cos \theta$.
(ii) Find the value of $\theta$, for $0 \leqslant \theta < \pi$, at which this maximum occurs. Tom models the height of sea water, $H$ metres, on a particular day by the equation $$H = 6 + 2 \sin \left( \frac { 4 \pi t } { 25 } \right) - 1.5 \cos \left( \frac { 4 \pi t } { 25 } \right) , \quad 0 \leqslant t < 12$$ where $t$ hours is the number of hours after midday.
(c) Calculate the maximum value of $H$ predicted by this model and the value of $t$, to 2 decimal places, when this maximum occurs.
(d) Calculate, to the nearest minute, the times when the height of sea water is predicted, by this model, to be 7 metres.

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Question 7:

Part (a)

Answer	Marks	Guidance
$R = \sqrt{6.25}$ or $2.5$	B1	$R = \pm 2.5$ award B0
$\tan\alpha = \frac{1.5}{2} = \frac{3}{4} \Rightarrow \alpha =$ awrt $0.6435$	M1A1	M1: $\tan\alpha = \pm\frac{1.5}{2}$ or $\pm\frac{2}{1.5}$

Part (b)(i)

Answer	Marks	Guidance
Max Value $= 2.5$	B1$\checkmark$	Follow through their $R$

Part (b)(ii)

Answer	Marks	Guidance
$\sin(\theta - 0.6435) = 1$ or $\theta - \text{their } \alpha = \frac{\pi}{2} \Rightarrow \theta =$ awrt $2.21$	M1; A1$\checkmark$	A1$\checkmark$: awrt 2.21 or $\frac{\pi}{2}$ + their $\alpha$ rounding correctly to 3 sf

Part (c)

Answer	Marks	Guidance
$H_{\max} = 8.5$ (m)	B1$\checkmark$	8.5 or $6 +$ their $R$ as long as answer $> 6$
$\sin\!\left(\frac{4\pi t}{25} - 0.6435\right) = 1$ or $\frac{4\pi t}{25} =$ their (b) answer $\Rightarrow t =$ awrt $4.41$	M1; A1	A1 implied by awrt 4.41 or 4.40

Part (d)

Answer	Marks	Guidance
$6 + 2.5\sin\!\left(\frac{4\pi t}{25} - 0.6435\right) = 7 \Rightarrow \sin\!\left(\frac{4\pi t}{25} - 0.6435\right) = \frac{1}{2.5} = 0.4$	M1; M1
$\left\{\frac{4\pi t}{25} - 0.6435\right\} = \sin^{-1}(0.4)$ or awrt $0.41$	A1	Implied by awrt 0.41 or 2.73
Either $t =$ awrt $2.1$ or awrt $6.7$	A1
$\left\{\frac{4\pi t}{25} - 0.6435\right\} = \left\{\pi - 0.411517\ldots \text{ or } 2.730076\ldots\right\}$	ddM1	Dependent on both M marks; implied by seeing 2.730… or 3.373…
Times $= \{14:06,\ 18:43\}$	A1	Both times required

## Question 7:

### Part (a)
| $R = \sqrt{6.25}$ or $2.5$ | B1 | $R = \pm 2.5$ award B0 |
| $\tan\alpha = \frac{1.5}{2} = \frac{3}{4} \Rightarrow \alpha =$ awrt $0.6435$ | M1A1 | M1: $\tan\alpha = \pm\frac{1.5}{2}$ or $\pm\frac{2}{1.5}$ |

### Part (b)(i)
| Max Value $= 2.5$ | B1$\checkmark$ | Follow through their $R$ |

### Part (b)(ii)
| $\sin(\theta - 0.6435) = 1$ or $\theta - \text{their } \alpha = \frac{\pi}{2} \Rightarrow \theta =$ awrt $2.21$ | M1; A1$\checkmark$ | A1$\checkmark$: awrt 2.21 or $\frac{\pi}{2}$ + their $\alpha$ rounding correctly to 3 sf |

### Part (c)
| $H_{\max} = 8.5$ (m) | B1$\checkmark$ | 8.5 or $6 +$ their $R$ as long as answer $> 6$ |
| $\sin\!\left(\frac{4\pi t}{25} - 0.6435\right) = 1$ or $\frac{4\pi t}{25} =$ their (b) answer $\Rightarrow t =$ awrt $4.41$ | M1; A1 | A1 implied by awrt 4.41 or 4.40 |

### Part (d)
| $6 + 2.5\sin\!\left(\frac{4\pi t}{25} - 0.6435\right) = 7 \Rightarrow \sin\!\left(\frac{4\pi t}{25} - 0.6435\right) = \frac{1}{2.5} = 0.4$ | M1; M1 | |
| $\left\{\frac{4\pi t}{25} - 0.6435\right\} = \sin^{-1}(0.4)$ or awrt $0.41$ | A1 | Implied by awrt 0.41 or 2.73 |
| Either $t =$ awrt $2.1$ or awrt $6.7$ | A1 | |
| $\left\{\frac{4\pi t}{25} - 0.6435\right\} = \left\{\pi - 0.411517\ldots \text{ or } 2.730076\ldots\right\}$ | ddM1 | Dependent on both M marks; implied by seeing 2.730… or 3.373… |
| Times $= \{14:06,\ 18:43\}$ | A1 | Both times required |

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7. (a) Express $2 \sin \theta - 1.5 \cos \theta$ in the form $R \sin ( \theta - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$.

Give the value of $\alpha$ to 4 decimal places.\\
(b) (i) Find the maximum value of $2 \sin \theta - 1.5 \cos \theta$.\\
(ii) Find the value of $\theta$, for $0 \leqslant \theta < \pi$, at which this maximum occurs.

Tom models the height of sea water, $H$ metres, on a particular day by the equation

$$H = 6 + 2 \sin \left( \frac { 4 \pi t } { 25 } \right) - 1.5 \cos \left( \frac { 4 \pi t } { 25 } \right) , \quad 0 \leqslant t < 12$$

where $t$ hours is the number of hours after midday.\\
(c) Calculate the maximum value of $H$ predicted by this model and the value of $t$, to 2 decimal places, when this maximum occurs.\\
(d) Calculate, to the nearest minute, the times when the height of sea water is predicted, by this model, to be 7 metres.

\hfill \mbox{\textit{Edexcel C3 2010 Q7 [15]}}

This paper (8 questions)

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Q1 5 Q2 7 Q3 9 Q4 10 Q5 12 Q6 10 Q7 15 Q8 7

Answer	Marks	Guidance
\(R = \sqrt{6.25}\) or \(2.5\)	B1	\(R = \pm 2.5\) award B0
\(\tan\alpha = \frac{1.5}{2} = \frac{3}{4} \Rightarrow \alpha =\) awrt \(0.6435\)	M1A1	M1: \(\tan\alpha = \pm\frac{1.5}{2}\) or \(\pm\frac{2}{1.5}\)

Answer	Marks	Guidance
\(H_{\max} = 8.5\) (m)	B1\(\checkmark\)	8.5 or \(6 +\) their \(R\) as long as answer \(> 6\)
\(\sin\!\left(\frac{4\pi t}{25} - 0.6435\right) = 1\) or \(\frac{4\pi t}{25} =\) their (b) answer \(\Rightarrow t =\) awrt \(4.41\)	M1; A1	A1 implied by awrt 4.41 or 4.40

Answer	Marks	Guidance
\(6 + 2.5\sin\!\left(\frac{4\pi t}{25} - 0.6435\right) = 7 \Rightarrow \sin\!\left(\frac{4\pi t}{25} - 0.6435\right) = \frac{1}{2.5} = 0.4\)	M1; M1
\(\left\{\frac{4\pi t}{25} - 0.6435\right\} = \sin^{-1}(0.4)\) or awrt \(0.41\)	A1	Implied by awrt 0.41 or 2.73
Either \(t =\) awrt \(2.1\) or awrt \(6.7\)	A1
\(\left\{\frac{4\pi t}{25} - 0.6435\right\} = \left\{\pi - 0.411517\ldots \text{ or } 2.730076\ldots\right\}\)	ddM1	Dependent on both M marks; implied by seeing 2.730… or 3.373…
Times \(= \{14:06,\ 18:43\}\)	A1	Both times required