A2 June 2022 Q2

EdexcelCurrent spec7 markst-Formulae

2. During 2029, the number of hours of daylight per day in London, \(H\), is modelled by the equation

\[H = 0.3\sin\left(\frac{x}{60}\right) - 4\cos\left(\frac{x}{60}\right) + 11.5 \qquad 0 \leqslant x \lt 365\]

where \(x\) is the number of days after 1st January 2029 and the angle is in radians.

(a) Show that, according to the model, the number of hours of daylight in London on the 31st January 2029 will be 8.13 to 3 significant figures. (1)

(b) Use the substitution \(t = \tan\left(\dfrac{x}{120}\right)\) to show that \(H\) can be written as\[H = \frac{at^2 + bt + c}{1 + t^2}\]where \(a\), \(b\) and \(c\) are constants to be determined. (2)

(c) Hence determine, according to the model, the date of the first day of 2029 when there will be at least 12 hours of daylight in London. (4)

Mark scheme (a)

Scheme	Marks	AO
\((H =)\,0.3\sin\left(\dfrac{30}{60}\right) - 4\cos\left(\dfrac{30}{60}\right) + 11.5 = 8.13\) {hours}*	B1*	3.4
	(1)

Notes

B1*: Uses \(x = 30\) to show that \(H = 8.13\). Accept \(x = 30\) seen substituted followed by 8.13, or \(x = 30\) identified followed by 8.133… before rounding to 8.13.

Mark scheme (b)

Scheme	Marks	AO
Substitutes \(\sin\left(\dfrac{x}{60}\right) = \dfrac{2t}{1 + t^2}\) and \(\cos\left(\dfrac{x}{60}\right) = \dfrac{1 - t^2}{1 + t^2}\) into \(H\) \((H =)\,0.3\left(\dfrac{2t}{1 + t^2}\right) - 4\left(\dfrac{1 - t^2}{1 + t^2}\right) + 11.5\)	M1	1.1b
\((H =)\,\dfrac{0.6t - 4 + 4t^2 + 11.5\left(1 + t^2\right)}{1 + t^2} = \dfrac{15.5t^2 + 0.6t + 7.5}{1 + t^2}\)	A1	2.1
	(2)

Notes

M1: Uses the correct \(t\)-formulae \(\sin\left(\dfrac{x}{60}\right) = \dfrac{2t}{1 + t^2}\) and \(\cos\left(\dfrac{x}{60}\right) = \dfrac{1 - t^2}{1 + t^2}\), attempts to substitute into \(H\).

A1: Fully correct method, expresses as a single fraction with a denominator of \(1 + t^2\) to achieve \(H = \dfrac{15.5t^2 + 0.6t + 7.5}{1 + t^2}\) (oe with fractions or accept values for \(a\), \(b\) and \(c\) stated).

Mark scheme (c)

Scheme	Marks	AO
\(H = \dfrac{15.5t^2 + 0.6t + 7.5}{1 + t^2} = 12 \Rightarrow 3.5t^2 + 0.6t - 4.5 = 0\)	M1	3.4
\(\Rightarrow t = \dfrac{-0.6 \pm \sqrt{0.6^2 - 4(3.5)(-4.5)}}{7}\) \(= \ldots\ (1.051\ldots, -1.222\ldots) \Rightarrow x = 120\tan^{-1}(\text{“}1.051\ldots\text{”}) = \ldots\ (97.254\ldots)\)	dM1	3.1b
\(x =\) awrt 97	A1	1.1b
8^th or 9^th April	A1	3.2a
	(4)
	(7 marks)

Notes

M1: Sets \(H = 12\) (or any inequality in between) and rearranges to form a quadratic equation for \(t\).

dM1: Dependent on the previous method mark. Solves the quadratic by any means (accept one correct answer for their quadratic if no method shown) and uses this to find a value for \(x\).

A1: Correct value for \(x =\) awrt 97 or accept 98 following a correct value for \(t\).

A1: Correct day of the year. Accept 8^th or 9^th April following awrt 97 from a correct method.

Note: Question says hence, so answers by graphical methods or trial and improvement are not acceptable for full credit. They can score a SC M0dM0A0B1 for achieving a correct date.