| Question | Answer | Marks | AO | Guidance |
| 7 | (a) | | \(\begin{aligned} | x = r \cos \theta , y = r \sin \theta \Rightarrow ( r \cos \theta ) ^ { 3 } + ( r \sin \theta ) ^ { 3 } = 2 r \cos \theta \cdot r \sin \theta |
| \Rightarrow r \left( \cos ^ { 3 } \theta + \sin ^ { 3 } \theta \right) = 2 \cos \theta \sin \theta |
| \Rightarrow r = \frac { 2 \cos \theta \sin \theta } { \cos ^ { 3 } \theta + \sin ^ { 3 } \theta } \end{aligned}\) | | | | Substitution | | May see "or \(r = 0\) " but not required. |
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| | | | [2] | | |
| (b) | | \(\begin{aligned} f \left( \frac { 1 } { 2 } \pi - \theta \right) | = \frac { 2 \cos \left( \frac { 1 } { 2 } \pi - \theta \right) \sin \left( \frac { 1 } { 2 } \pi - \theta \right) } { \cos ^ { 3 } \left( \frac { 1 } { 2 } \pi - \theta \right) + \sin ^ { 3 } \left( \frac { 1 } { 2 } \pi - \theta \right) } |
| = \frac { 2 \sin \theta \cos \theta } { \sin ^ { 3 } \theta + \cos ^ { 3 } \theta } \end{aligned}\) | | | Correct substitution into their \(\mathrm { f } ( \theta )\) |
| | | | [2] | | |
| (c) | | So the line of symmetry is \(\theta = \frac { \pi } { 4 }\) | B1 | 2.2a | | Allow \(y = x\). | | Must have \(\theta =\) |
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| | | | [1] | | |
| (d) | | \(r = \mathrm { f } \left( \frac { 1 } { 4 } \pi \right) = \sqrt { 2 }\) | B1 | 1.1 | BC |
| | | | [1] | | |
| (e) | | | \(r = 0\) when \(\theta = 0\). | | \(r = 0\) also when \(\theta = \frac { \pi } { 2 }\) | | In range \(0 < \theta < \frac { \pi } { 2 } , r > 0\) and is continuous | | So there is a loop |
| B1 | 3.1a | | For both, ignore extras. | | Conclusion - both statements for \(r\) need to be mentioned |
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| | | | [2] | | |