AS June 2018 Q1

EdexcelAS paperCurrent spec7 markst-Formulae

(a) Use the substitution \(t = \tan\left(\dfrac{x}{2}\right)\) to show that the equation\[5\sin x + 12\cos x = 2\]can be written in the form\[7t^2 - 5t - 5 = 0\] (3)

(b) Hence solve, for \(-180^\circ \lt x \lt 180^\circ\), the equation\[5\sin x + 12\cos x = 2\]giving your answers to one decimal place. (4)

Mark scheme (a)

Scheme	Marks	AO
\(t = \tan\left(\dfrac{x}{2}\right),\ 5\sin x + 12\cos x = 2 \Rightarrow 7t^2 - 5t - 5 = 0\)
\(\{5\sin x + 12\cos x =\}\ 5\left(\dfrac{2t}{1 + t^2}\right) + 12\left(\dfrac{1 - t^2}{1 + t^2}\right)\)	M1	1.1b
\(5\left(\dfrac{2t}{1 + t^2}\right) + 12\left(\dfrac{1 - t^2}{1 + t^2}\right) = 2 \Rightarrow 5(2t) + 12(1 - t^2) = 2(1 + t^2)\)	M1	1.1b
\(7t^2 - 5t - 5 = 0\ *\)	A1*	2.1
	(3)

Notes

M1: Uses at least one of \(\sin x = \dfrac{2t}{1 + t^2}\) or \(\cos x = \dfrac{1 - t^2}{1 + t^2}\) to express \(5\sin x + 12\cos x\) in terms of \(t\) only

M1: Uses both correct formula \(\sin x = \dfrac{2t}{1 + t^2}\) and \(\cos x = \dfrac{1 - t^2}{1 + t^2}\) in \(5\sin x + 12\cos x\), equates their expression to 2 and eliminates the fractions

A1*: Collects terms to one side and simplifies to obtain the printed answer

Mark scheme (b)

Scheme	Marks	AO
\(t = \dfrac{5 \pm \sqrt{5^2 - 4(7)(-5)}}{2(7)}\ \left\{= \dfrac{5 \pm \sqrt{165}}{14} = 1.2746\ldots, -0.5603\ldots\right\}\)	M1	1.1a
\(\dfrac{x}{2} = \arctan\left(\dfrac{5 + \sqrt{165}}{14}\right)\) or \(\dfrac{x}{2} = \arctan\left(\dfrac{5 - \sqrt{165}}{14}\right)\) and \(\Rightarrow x = \ldots\)	M1	3.1a
\(x = \text{awrt } 104^\circ\) or \(x = \) any answer in the range \([-58.6^\circ, -58^\circ]\)	A1	1.1b
\(x = 103.8^\circ\) and \(x = -58.5^\circ\)	A1	1.1b
	(4)
	(7 marks)

Notes

M1: Selects a correct process (e.g. using the quadratic formula, completing the square or calculator approach) to solve \(7t^2 - 5t - 5 = 0\)

Note: Allow 1^st M1 for at least one of awrt 1.3 or awrt \(-0.6\) or for a correct exact value of \(t\)

Note: Do not allow an attempt at factorisation of \(7t^2 - 5t - 5\) for the 1^st M1

M1: Adopts a correct applied strategy of taking \(\arctan(\text{their found } t)\) and multiplying the result by 2 to obtain at least one value for \(x\) within the range \(-180^\circ \lt x \lt 180^\circ\) (or in radians \(-\pi \lt x \lt \pi\))

A1: See scheme

A1: For both 103.8 and \(-58.5\)

Note: Give final A0 for extra solutions given within the range \(-180^\circ \lt x \lt 180^\circ\)

Note: Ignore extra solutions outside the range \(-180^\circ \lt x \lt 180^\circ\) for the final A mark

Note: In degrees, \(\dfrac{x}{2} = \{51.88\ldots,\ -128.11\ldots,\ -29.26\ldots,\ 150.73\ldots\}\)

Note: Working in radians gives \(\dfrac{x}{2} = \{0.905\ldots, -0.510\ldots\} \Rightarrow x = \{1.81\ldots, -1.02\ldots\}\)

Note: Give 2^nd M0 for \(\dfrac{x}{2} = \{51.88\ldots, -29.26\ldots\} \Rightarrow x = \{25.9, -14.6\}\)