AS October 2020 Q3

EdexcelAS paperCurrent spec11 markst-Formulae

(i) Use the substitution \(t = \tan\left(\dfrac{x}{2}\right)\) to prove that\[\cot x + \tan\left(\frac{x}{2}\right) = \operatorname{cosec} x \qquad x \neq n\pi,\ n \in \mathbb{Z}\] (2)

(ii)

Figure 1: wind turbine on horizontal ground, with the height H metres marked from the ground up to the tip of one blade — Figure 1

An engineer models the vertical height above the ground of the tip of one blade of a wind turbine, shown in Figure 1. The ground is assumed to be horizontal.

The vertical height of the tip of the blade above the ground, \(H\) metres, at time \(x\) seconds after the wind turbine has reached its constant operating speed, is modelled by the equation

\[H = 90 - 30\cos(120x)^\circ - 40\sin(120x)^\circ \qquad (\text{I})\]

(a) Show that \(H = 60\) when \(x = 0\) (1)

Using the substitution \(t = \tan(60x)^\circ\)

(b) show that equation \((\text{I})\) can be rewritten as\[H = \frac{120t^2 - 80t + 60}{1 + t^2}\] (3)

(c) Hence find, according to the model, the value of \(x\) when the tip of the blade is 100 m above the ground for the first time after the wind turbine has reached its constant operating speed. (5)

Mark scheme (i)

Scheme	Marks	AO
\(\text{lhs} = \cot x + \tan\left(\dfrac{x}{2}\right) = \dfrac{1 - t^2}{2t} + t\)	M1	1.1a
\(\dfrac{1 - t^2}{2t} + t = \dfrac{1 + t^2}{2t}\left(= \dfrac{1}{\sin x}\right) = \operatorname{cosec} x\ *\)	A1*	2.1
	(2)

Notes

M1: Selects the correct expression for \(\cot x\) in terms of \(t\) and substitutes this and \(t\) into the lhs

A1*: Fully correct proof. Allow correct work leading to \(\dfrac{1 + t^2}{2t} = \operatorname{cosec} x\)

Scheme	Marks	AO
\(x = 0 \Rightarrow H = 90 - 30\cos(0) - 40\sin(0) = 90 - 30 = 60\)	B1	1.1b
	(1)

Mark scheme (ii)(b)

Scheme	Marks	AO
\(H = 90 - 30\cos 120x - 40\sin 120x = 90 - 30\left(\dfrac{1 - t^2}{1 + t^2}\right) - 40\left(\dfrac{2t}{1 + t^2}\right)\)	M1	1.1b
\(= \dfrac{90 + 90t^2 - 30 + 30t^2 - 80t}{1 + t^2}\)	M1	1.1b
\(= \dfrac{120t^2 - 80t + 60}{1 + t^2}\ *\)	A1*	2.1
	(3)

Notes

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

M1: Correct method to obtain a common denominator

A1*: Collects terms and simplifies to obtain the printed answer with no errors

Mark scheme (ii)(c)

Scheme	Marks	AO
\(\dfrac{120t^2 - 80t + 60}{1 + t^2} = 100 \Rightarrow 120t^2 - 80t + 60 = 100 + 100t^2\)	M1	3.4
\(20t^2 - 80t - 40 = 0\)	A1	1.1b
\(t = \dfrac{4 \pm \sqrt{16 + 8}}{2} \Rightarrow 60x = \tan^{-1}\left(2 + \sqrt{6}\right)\) or \(60x = \tan^{-1}\left(2 - \sqrt{6}\right)\)	M1	3.4
\(60x = \tan^{-1}\left(2 + \sqrt{6}\right) = 77.33\ldots \Rightarrow x = \ldots\)	dM1	3.1b
\(x = 1.29\)	A1	3.2a
	(5)
	(11 marks)

Notes

M1: Uses \(H = 100\) with the model and multiplies up to obtain a quadratic equation in \(t\)

A1: Correct 3TQ

M1: Solves their 3TQ in \(t\) and proceeds to obtain values of \(60x\) as suggested by the model

M1: A fully correct strategy to identify the required value of \(x\) from the positive root of the quadratic equation in \(t\)

A1: awrt 1.29

Attempts in radians can score all but the final mark in (c). (Gives \(60x = 1.3\ldots\) etc.)