June 2024 Paper 3 Q4

Notes

M1: Use of both \(\cot x \equiv \dfrac{\cos x}{\sin x}\) and \(\mathrm{cosec}\,x \equiv \dfrac{1}{\sin x}\)
Condone use of \(s\) and \(c\) throughout but final answer must be in terms of sine – allow e.g. \(\theta\) for \(x\) for M marks but must be \(x\) for the A mark

M1: Correct use of \(\sin^2 x + \cos^2 x \equiv 1\) to obtain an equation in \(\sin x\) only
Not dependent on the first M mark

A1: AG (so must be equal to zero) – sufficient working must be shown – any errors seen is A0
A0 if an angle is missing from any trig. expression used in their working

Alternative method

M1: Correct use of \(1 + \cot^2 x \equiv \mathrm{cosec}^2 x\)

M1: Replacing \(\mathrm{cosec}\,x\) with \(\dfrac{1}{\sin x}\) to obtain an equation in \(\sin x\) only
Not dependent on first M mark
Note: \(2\,\mathrm{cosec}^2 x - 9\,\mathrm{cosec}\,x - 5 = 0 \Rightarrow 2 - 9\sin x - 5\sin^2 x = 0\) with no intermediate working is M0 unless explicit mention is made of multiplying through by \(\sin^2 x\) or explicit mention is made of dividing by \(\mathrm{cosec}^2 x\)

A1: AG (so must be equal to zero) – sufficient working must be shown – any errors seen is A0
A0 if an angle is missing from any trig. expression used in their working

Notes

(b)(i)

M1: SEE APPENDIX for awarding this mark (solving 3TQ expressions)
condone using \(x\) for \(\theta\) or \(2\theta\) for the M mark – condone for M1 only \((5\sin\theta - 1)(\sin\theta + 2)\)

B1: Correctly stating that \(\sin 2\theta = 0.2\) and that \(\sin 2\theta\) cannot equal \(-2\) (must explicitly reject the \(-2\) (but no rationale required) - this mark is not implied by correct values for \(\theta\) (as DR required)
Must be solving \(5\sin^2 2\theta + 9\sin 2\theta - 2 = 0\) for the B marks
condone \(\sin 2x = 0.2\)

B1: awrt 0.101 (0.1006789…) www

B1: awrt 1.470 (1.4701173…) www

Ignore additional solutions outside of the range \(0 \lt \theta \lt \pi\), but if any other solutions inside the range, award at most one of the two final B marks for one correct value
SC B1 for awrt 0.10 and awrt 1.47 only if 3 dp or better) not seen
SC B1 for awrt 5.77 and awrt 84.2 only (working in degrees)

(b)(ii)

M1: Use of the small angle approximation \(\sin 2\theta \approx 2\theta\) twice in the given answer from (a) to obtain a three-term quadratic in \(\theta\) (allow un-simplified)
Award M1 only for \(5\theta^2 + 9\theta - 2\,[= 0]\) (so for using \(\theta\) instead of \(2\theta\)) – allow e.g. \(x\) for \(\theta\)

A1: State 0.10 (or better e.g. 0.100…) as a decimal following a correct quadratic in \(\theta\) seen (no method required for solving the quadratic) and comment that this is accurate to 2 dp (as a minimum must mention ‘2 dp’ with the value of 0.10(000…) appearing in this part and 0.10 or 0.101 or 0.100(6789…) appearing in part (b)(i))
This mark is dependent on an awrt 0.10 seen in part (b)(i) or a correct sign change test (see below)
Ignore any consideration of other root(s)

Alternative for M mark

M1: Re-writing at least one of their equations \(\sin 2\theta = k\) with \(-1 \lt k \lt 1\) (from part (a)) as \(2\theta = k\)

Alternative for A mark

A1: Correct values to at least 1 dp (rot) with explanation (‘change of sign’ either stated or comparing values with zero) and correct conclusion (as a minimum must mention ‘2 dp’)

Alternative for (b)(ii) (Appendix)

M1: Use of small angle approximations \(\sin 2\theta \approx 2\theta\) and \(\tan 2\theta \approx 2\theta\) in the given equation from (a) to obtain what would be equivalent to a three-term quadratic in \(\theta\) (allow un-simplified). Allow M1 only for \(\dfrac{2}{\theta^2} - \dfrac{9}{\theta} - 3\,[= 0]\) (so for using \(\theta\) instead of \(2\theta\)) – allow e.g. \(x\) for \(\theta\)

A1: State 0.10 (or better e.g. 0.1039…) as a decimal following a correct quadratic in \(\theta\) (no method required for solving quadratic) and comment that this is accurate to 2 dp (as a minimum must mention ‘2 dp’ with the value of 0.10(39…) appearing in this part and 0.10 or 0.101 or 0.100(6789…) appearing in part (b)(i))

Appendix: rules for solving quadratics in questions 2 and 4(b)(i) only

In questions 2 and 4(b)(i) candidates are required to solve 3 term quadratics (3TQ) using DR – therefore we must see a correct, complete method for solving these quadratics – the correct answers do not imply the corresponding M mark, for example in question 2, \(9x^2 - 38x + 8 = 0 \Rightarrow x = 4\) or \(x = \frac{2}{9}\) is M0

Rules for factorising:
\(at^2 + bt + c \Rightarrow (mt + n)(pt + q)\) where \(a = mp\) and one of \(mq + np = b\) or \(c = nq\) (so when expanding their factorised expression it must give the correct quadratic term and one other term correct of the preceding 3TQ expression/equation)
e.g. in question 2 (and similarly for question 4(b)(i)):
\(9x^2 - 38x + 8 = \left(x - \frac{2}{9}\right)(x - 4)\) is M0 (but the following B1 for the correct c.v. of \(\frac{2}{9}\) and 4 in qu. 2 can still be awarded as they follow from these two factors)
\(9x^2 - 38x + 8 = (3x + 8)(3x + 1)\) is M1 (when expanded the \(x^2\) and constant terms are correct)
Allow correct part factorisation for their 3TQ expression e.g. if correct 3TQ then in question 2 the expression \(9x(x - 4) - 2(x - 4)\) scores M1

Rules for the formula:
Must apply the correct formula for their three-term quadratic (no errors even if correct formula is stated) – note that stating the formula (in terms of \(a\), \(b\) and \(c\)) followed immediately by the corresponding roots is M0 – we must see the formula being applied e.g. \(9x^2 - 38x + 8 = 0 \Rightarrow x = \dfrac{38 \pm \sqrt{38^2 - 4(9)(8)}}{2(9)}\).
Minimal acceptable working would be \(x = \dfrac{38 \pm \sqrt{1156}}{18}\) (so must explicitly see the discriminant) for M1

Rules for completing the square – using \(9x^2 - 38x + 8 = 0\) as an example:
The M1 is not awarded until correctly getting to the stage of \(x - \frac{19}{9} = \pm\sqrt{\frac{289}{81}}\) (must include \(\pm\) so implying two roots) with no errors (so consistent with applying the formula correctly)