AS June 2025 Q1

EdexcelAS paperCurrent spec9 markst-Formulae

In this question you must show all stages of your working.

Solutions based entirely on calculator technology are not acceptable.

The surface temperature of the water in a lake during a particular year is modelled by the equation

\[S = 12 - \frac{15}{2}\cos x^\circ - \frac{27}{10}\sin x^\circ \qquad (\text{I})\]

where \(S\) is the temperature in degrees Celsius and \(x\) is the number of days after the start of the year.

(a) Use the model to write down the surface temperature of the water in the lake at the start of the year. (1)

Using the substitution \(t = \tan\left(\dfrac{x}{2}\right)\)

(b) show that equation \((\text{I})\) can be rewritten as\[S = \frac{At^2 + Bt + C}{10(1 + t^2)}\]where \(A\), \(B\) and \(C\) are integers to be determined. (3)

(c) Hence determine, according to the model, the number of days after the start of the year when the surface temperature of the water in the lake is \(10^\circ\mathrm{C}\) for the second time that year. Give your answer to the nearest day. (5)

Scheme	Marks	AO
\(x = 0 \Rightarrow 12 - 7.5 = 4.5^\circ\mathrm{C}\)	B1	3.4
	(1)

Mark scheme (b)

Scheme	Marks	AO
\(S = 12 - \dfrac{15}{2}\left(\dfrac{1 - t^2}{1 + t^2}\right) - \dfrac{27}{10}\left(\dfrac{2t}{1 + t^2}\right)\)	M1	1.1b
\(= \dfrac{120(1 + t^2) - 75(1 - t^2) - 54t}{10(1 + t^2)}\)	M1	1.1b
\(= \dfrac{195t^2 - 54t + 45}{10(1 + t^2)}\)	A1	2.1
	(3)

Notes

M1: Uses the correct t-formulae to obtain \(S\) in terms of \(t\)

M1: Correct method to obtain a common denominator,
\(A + \dfrac{B(1 - t^2)}{1 + t^2} + \dfrac{Ct}{1 + t^2} = \dfrac{A(1 + t^2) + B(1 - t^2) + Ct}{(1 + t^2)}\)

A1: Correct answer in the correct form

(Corrected from the printed mark scheme: the second line is printed as \(\dfrac{120(1 + t^2) - 75(1 - t^2)3 - 54t}{10(1 + t^2)}\), with a stray 3.)

Mark scheme (c)

Scheme	Marks	AO
\(\dfrac{195t^2 - 54t + 45}{10(1 + t^2)} = 10 \Rightarrow 195t^2 - 54t + 45 = 100 + 100t^2\)	M1	3.4
\(95t^2 - 54t - 55 = 0\)	A1	1.1b
\(t = 1.09\ldots, \Rightarrow \dfrac{x}{2} = \tan^{-1} 1.09\ldots = \ldots\{47.4\ldots \text{ or } 0.828\ldots\}\) or \(-0.528\ldots \Rightarrow \dfrac{x}{2} = \tan^{-1}(-0.528) = \ldots\{-27.83\ldots \text{ or } -0.485\ldots\}\) Alternative: Leading to a value for x \(t = 1.09\ldots \Rightarrow \sin x = \dfrac{2(1.09)}{1 + (1.09)^2}\) or \(\cos x = \dfrac{1 - (1.09)^2}{1 + (1.09)^2}\) Or \(t = -0.528\ldots \Rightarrow \sin x = \dfrac{2(-0.528)}{1 + (-0.528)^2}\) or \(\cos x = \dfrac{1 - (-0.528)^2}{1 + (-0.528)^2}\)	dM1	3.4
\(\dfrac{x}{2} = \tan^{-1}(-0.528) + 180 = 152.1\ldots \Rightarrow x = \ldots\) \(x = 360 + \sin^{-1}\left(\dfrac{2(-0.528)}{1 + (-0.528)^2}\right)\) \(x = 360 - \cos^{-1}\left(\dfrac{1 - (-0.528)^2}{1 + (-0.528)^2}\right)\)	ddM1	3.1b
304 days	A1	3.2a
	(5)
	(9 marks)

Notes

M1: Uses \(S = 10\) with the model and multiplies up to obtain a quadratic equation in \(t\)

A1: Correct 3TQ

dM1: Solves their 3TQ in \(t\) and proceeds to obtain at least one value of \(\dfrac{x}{2}\) as suggested by the model. Alternative uses \(t\) formulae for \(\sin x\) or \(\cos x\) to find a value for \(x\), degrees or radians

ddM1: A fully correct strategy to find the required value of \(x\) from the negative root of the quadratic equation in \(t\). They must be using degrees for this mark. This mark may be award even if more than one value found e.g \(x = 95.2,\ 304.3\)

A1: Correct number of days, must be clear that this is their answer