Higher June 2024 Paper 1 Q25
25 Show that the value of \(\quad 6\sin 30^\circ + 2\cos 30^\circ \times 4\tan 30^\circ \quad\) is an integer. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: substitutes values | ||
| \((\sin 30^\circ =)\ \dfrac{1}{2}\) or \(6\sin 30^\circ = 3\) or \((\cos 30^\circ =)\ \dfrac{\sqrt{3}}{2}\) or \(2\cos 30^\circ = \sqrt{3}\) or \((\tan 30^\circ =)\ \dfrac{1}{\sqrt{3}}\) or \(\dfrac{\sqrt{3}}{3}\) or \(4\tan 30^\circ = \dfrac{4}{\sqrt{3}}\) or \(\dfrac{4\sqrt{3}}{3}\) | M1 | may be seen beside the given expression or in a table |
| \(6\left(\dfrac{1}{2}\right)\) and \(2\left(\dfrac{\sqrt{3}}{2}\right)\) and \(4\left(\dfrac{1}{\sqrt{3}}\right)\) or \(6\left(\dfrac{1}{2}\right)\) and \(2\left(\dfrac{\sqrt{3}}{2}\right)\) and \(4\left(\dfrac{\sqrt{3}}{3}\right)\) or \(\dfrac{6}{2}\) and \(\dfrac{2\sqrt{3}}{2}\) and \(\dfrac{4\sqrt{3}}{3}\) | M1dep | oe |
| Processing at least as far as \(\dfrac{6}{2} + \dfrac{8\sqrt{3}}{2\sqrt{3}}\) or \(\dfrac{6}{2} + \dfrac{8\sqrt{3}\sqrt{3}}{6}\) or \(\dfrac{6}{2} + \dfrac{24}{6}\) | M1dep | oe |
| 7 from correct working | A1 | SC2 \(4 + 4\sqrt{3}\) oe |
| Alternative method 2: uses a trig identity | ||
| \(6\sin 30^\circ + 2\cos 30^\circ \times 4\dfrac{\sin 30^\circ}{\cos 30^\circ}\) | M1 | oe |
| \(6\sin 30^\circ + 8\sin 30^\circ\) or \(14\sin 30^\circ\) | M1dep | oe |
| \(14 \times \dfrac{1}{2}\) | M1dep | oe |
| 7 from correct working | A1 | SC2 \(4 + 4\sqrt{3}\) oe |
Additional guidance
| Alt 2 is not on this specification, but may be seen if other qualifications have been studied, eg AQA Certificate – Level 2 Further Maths | |
| Incorrect order of operations gives \(4 + 4\sqrt{3}\) oe | SC2 |
| Allow \(\sqrt{1}\) for 1 throughout |