June 2025 Paper 3 Q10
10
(a) Express\[2\sin x + 5\cos x\]in the form\[R\sin(x + \alpha)\]
where \(R \gt 0\) and \(0^\circ \leqslant \alpha \leqslant 90^\circ\)
[3 marks](b) In 2010, the water temperature at a beach could be modelled by the formula\[T = 14.2 - (2\sin d^\circ + 5\cos d^\circ)\]
where:
- \(T\) is the temperature of the water in °C
- \(d\) is the number of days after 1 January 2010
(i) Find the value of \(d\) when the temperature of the water reached its lowest value in 2010 [1 mark]
(ii) Find the maximum temperature of the water predicted by the model in 2010
Give your answer to one decimal place.
[1 mark](iii) Find the number of weeks in 2010 for which the temperature of the water was higher than 15 °C
Give your answer to the nearest whole number.
[5 marks]| Scheme | Marks | AO |
|---|---|---|
| Obtains \(R = \sqrt{29}\) AWFW [5.38, 5.4] | B1 | 1.1b |
| Uses compound angle identity to obtain \(R\sin\alpha = 5\) or \(R\cos\alpha = 2\) or obtains \(\tan\alpha = \dfrac{5}{2}\) or \(\alpha\) = AWFW [68, 68.2] Allow angle in radians | M1 | 1.1a |
| Obtains \(\sqrt{29}\sin(x + 68.2)\) For \(\sqrt{29}\), AWFW [5.38, 5.4] and for 68.2, AWFW [68, 68.2] | R1 | 2.1 |
| (3) |
Typical solution
\[R = \sqrt{29}\]\[\tan\alpha = \frac{5}{2}\]\[\alpha = 68.2\]Hence \(\sqrt{29}\sin(x + 68.2)\)
| Scheme | Marks | AO |
|---|---|---|
| (i) Deduces 90 − their 68.2 Ignore rounding after value of 90 − their 68.2 seen | B1F | 2.2a |
| (1) | ||
| (ii) Deduces 14.2 + their \(\sqrt{29}\) Condone missing or wrong units Accept answer with more than 1 decimal place | B1F | 2.2a |
| (1) | ||
| (iii) Equates a complete model for the temperature to 15 eg \(14.2 - \text{their }\sqrt{29}\sin(d + 68.2) = 15\) or \(14.2 - (2\sin d^\circ + 5\cos d^\circ) = 15\) Allow use of inequality Condone use of another letter for \(d\) throughout PI by correct answer or angles | M1 | 3.4 |
| Rearranges their equation correctly to obtain \(\sin(d + \text{their } 68.2) = \ldots\) Allow use of inequality PI by correct answer or angles | M1 | 1.1a |
| Obtains AWFW [188.2, 189] and AWFW [351, 351.5] or obtains AWFW [120, 120.8] and AWFW [282.8, 283.3] Ignore additional values PI by correct answer | A1 | 1.1b |
| Subtracts their two positive values of \(d\) or their two positive values of \(d\) + their 68.2 and divides by 7 PI by correct answer | M1 | 3.1b |
| Obtains AWFW [23.1, 23.4] or 23 ISW | A1 | 1.1b |
| (5) | ||
| (10 marks) |
Typical solution
(b)(i)
\[90 - 68.2 = 21.8\](b)(ii)
\[14.2 + \sqrt{29} = 19.6\,{}^\circ\text{C}\](b)(iii)
\[14.2 - \sqrt{29}\sin(d + 68.2) = 15\]\[\sin(d + 68.2) = -\frac{0.8}{\sqrt{29}}\]\[d + 68.2 = -8.54,\ 188.54,\ 351.46\]\[d = 120.34,\ 283.26\]\[\frac{283.26 - 120.34}{7} = 23 \text{ weeks}\]