| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Region bounded by curve and tangent lines |
| Difficulty | Challenging +1.8 This is a challenging Further Maths polar coordinates problem requiring: (a) integration of a non-standard polar curve over two separate intervals, (b) finding tangent lines to determine rectangle dimensions, and (c) calculating shaded areas. The curve r = 3(cos 2θ)^(1/2) is more complex than typical polar examples, and part (b) requires geometric insight to relate the rectangle to the curve. However, the structure is guided by the diagram and follows standard polar area techniques, making it demanding but not requiring exceptional creativity. |
| Spec | 4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(A = (4\times)\int_0^{\frac{\pi}{4}} \frac{9}{2}\cos 2\theta\, d\theta\) | M1A1 | M1 for \(\frac{1}{2}\int r^2\,d\theta = \frac{1}{2}\int\alpha\cos2\theta\,d\theta\) with \(\alpha=3\) or \(9\); A1 for correct limits \((0,\frac{\pi}{4})\) with multiple 4 |
| \(= 18\left[\frac{\sin 2\theta}{2}\right]_0^{\frac{\pi}{4}}\) | M1 | \(\cos2\theta \to \pm\frac{1}{2}\sin2\theta\) |
| \(9\left[\sin\frac{\pi}{2} - 0\right] = 9\) | A1 | Using limits and factor 4 or 2 to obtain 9 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(r\sin\theta = 3(\cos2\theta)^{\frac{1}{2}}\sin\theta\) | M1 | For \(r\sin\theta = 3(\cos2\theta)^{\frac{1}{2}}\sin\theta\) or \(r^2\sin^2\theta = 9\cos2\theta\sin^2\theta\) |
| \(\frac{d}{d\theta}(r\sin\theta) = \left\{-3\times\frac{1}{2}(\cos2\theta)^{-\frac{1}{2}}\times 2\sin2\theta\sin\theta + 3(\cos2\theta)^{\frac{1}{2}}\cos\theta\right\}\) | M1depA1 | Differentiating rhs wrt \(\theta\); product and chain rule must be used |
| At max/min: \(\frac{-3\sin2\theta\sin\theta}{(\cos2\theta)^{\frac{1}{2}}} + 3(\cos2\theta)^{\frac{1}{2}}\cos\theta = 0\) | M1 | Equating expression to 0 |
| \(\sin2\theta\sin\theta = \cos2\theta\cos\theta\) | ||
| \(2\sin^2\theta\cos\theta = (1-2\sin^2\theta)\cos\theta\) | ||
| \(\cos\theta(1-4\sin^2\theta)=0\) | ||
| \(\sin\theta = \pm\frac{1}{2},\quad \theta = \pm\frac{\pi}{6}\) | M1A1 | M1dep for solving to \(\sin k\theta=\ldots\) or \(\cos k\theta=\ldots\) using correct trig formulae; ignore extra answers |
| \(r\sin\frac{\pi}{6} = 3\left(\cos\frac{\pi}{3}\right)^{\frac{1}{2}}\times\frac{1}{2} = \frac{3\sqrt{2}}{4}\) | B1 | Length of \(\frac{1}{2}PS = \frac{3\sqrt{2}}{4}\); may not be shown explicitly |
| Shaded area \(= 6\times\frac{3\sqrt{2}}{2} - 9 = 9\sqrt{2}-9\) | M1, A1 | M1 for \(PS\times 6 - \) their (a); A1 for \(9\sqrt{2}-9\) (awrt 3.73) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(3(\cos 2\theta)^{\frac{1}{2}}\sin\theta\) | M1 | First M mark |
| \(3(\cos 2\theta)^{\frac{1}{2}}\cos\theta - 3\left(\frac{1}{2}\right)(\cos 2\theta)^{-\frac{1}{2}}(2)\sin 2\theta\sin\theta = 0\) | M1 (dep) A1 | Second (dependent) M mark for differentiating using the product rule; A1 for correct derivative; M1 for setting derivative equal to 0 |
| \(3(\cos 2\theta)^{\frac{1}{2}}\cos\theta - 3(\cos 2\theta)^{-\frac{1}{2}}\sin 2\theta\sin\theta = 0\) | ||
| \(3(\cos 2\theta)^{\frac{1}{2}} - 6(\cos 2\theta)^{-\frac{1}{2}}\sin^2\theta = 0\) | Use of \(\sin 2\theta = 2\sin\theta\cos\theta\), division by \(3\cos\theta\) and multiplication by \((\cos 2\theta)^{\frac{1}{2}}\) simplify but do not provide specific M marks | |
| \(\cos 2\theta - 2\sin^2\theta = 0\) | ||
| \((1 - 2\sin^2\theta) - 2\sin^2\theta = 0\) | M1 | Use of \(\cos 2\theta = 1 - 2\sin^2\theta\) gives next M mark provided a value of \(\sin\theta\) or alt is reached with no errors seen |
| \(4\sin^2\theta = 1\) | ||
| \(\sin\theta = \pm\frac{1}{2}\) | M1 | Value of \(\sin\theta\) reached with use of \(\cos 2\theta = \ldots\) and no method errors seen (arithmetic slips condoned); gives final M mark |
| \(\theta = \dfrac{\pi}{6}\) | A1 | Second accuracy mark given here |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\sin\theta\) | M1 | First M mark; use of \(\cos 2\theta = \cos^2\theta - \sin^2\theta\) gives 4th M mark provided value of \(\sin\theta\) or alt reached with no errors seen after differentiation |
| \(3\left(\frac{1}{2}\right)(\cos^2\theta - \sin^2\theta)^{-\frac{1}{2}}(-2\cos\theta\sin\theta - 2\sin\theta\cos\theta)\sin\theta + 3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\cos\theta = 0\) | M1 (dep) A1 | Second (dependent) M mark for product rule; A1 for correct derivative; M1 for setting equal to 0 |
| \(-6(\cos^2\theta - \sin^2\theta)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\cos\theta = 0\) | ||
| \(-6\sin^2\theta + 3(\cos^2\theta - \sin^2\theta) = 0\) | M1 | Multiplication by \((\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\), division by \(3\cos\theta\) and use of \(\cos^2\theta = 1 - \sin^2\theta\) simplify but do not provide specific M marks |
| \(4\sin^2\theta = 1\) | ||
| \(\sin\theta = \pm\frac{1}{2}\) | M1 | Value of \(\sin\theta\) reached with use of \(\cos 2\theta = \ldots\) and no method errors seen (arithmetic slips condoned) |
| \(\theta = \dfrac{\pi}{6}\) | A1 | Second accuracy mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(3(2\cos^2\theta - 1)^{\frac{1}{2}}\sin\theta\) | M1 | First M mark; use of \(\cos 2\theta = 2\cos^2\theta - 1\) gives 4th M mark provided value of \(\sin\theta\) or alt reached with no errors seen after differentiation |
| \(3\left(\frac{1}{2}\right)(2\cos^2\theta - 1)^{-\frac{1}{2}}(-4\cos\theta\sin\theta)\sin\theta + 3(2\cos^2\theta - 1)^{\frac{1}{2}}\cos\theta = 0\) | M1 (dep) A1 | Second (dep) M mark for product rule; A1 for correct derivative; M1 for setting equal to 0 |
| \(-6(2\cos^2\theta - 1)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(2\cos^2\theta - 1)^{\frac{1}{2}}\cos\theta = 0\) | ||
| \(-6\sin^2\theta + 3(2\cos^2\theta - 1) = 0\) | M1 | Multiplication by \((\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\), division by \(3\cos\theta\) and use of \(\sin^2\theta = 1 - \cos^2\theta\) or vice versa simplify but do not provide specific M marks |
| \(4\sin^2\theta = 1\) or \(4\cos^2\theta = 3\) | ||
| \(\sin\theta = \pm\frac{1}{2}\) or \(\cos\theta = \pm\frac{\sqrt{3}}{2}\) | M1 | Value of \(\sin\theta\) or \(\cos\theta\) reached with use of \(\cos 2\theta = \ldots\) and no method errors seen (arithmetic slips condoned); gives final M mark |
| \(\theta = \dfrac{\pi}{6}\) | A1 | Second accuracy mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(3(1 - 2\sin^2\theta)^{\frac{1}{2}}\sin\theta\) | M1 | First M mark; use of \(\cos 2\theta = 2\cos^2\theta - 1\) gives 4th M mark provided value of \(\sin\theta\) or alt reached with no errors seen after differentiation |
| \(3\left(\frac{1}{2}\right)(1 - 2\sin^2\theta)^{-\frac{1}{2}}(-4\cos\theta\sin\theta)\sin\theta + 3(1 - 2\sin^2\theta)^{\frac{1}{2}}\cos\theta = 0\) | M1 (dep) A1 | Second (dependent) M mark for product rule; A1 for correct derivative; M1 for setting equal to 0 |
| \(-6(1 - 2\sin^2\theta)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(1 - 2\sin^2\theta)^{\frac{1}{2}}\cos\theta = 0\) | ||
| \(-6\sin^2\theta + 3(1 - 2\cos^2\theta) = 0\) | M1 | Multiplication by \((\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\), division by \(3\cos\theta\) and use of \(\sin^2\theta = 1 - \cos^2\theta\) or vice versa simplify but do not provide specific M marks |
| \(4\sin^2\theta = 1\) or \(4\cos^2\theta = 3\) | ||
| \(\sin\theta = \pm\frac{1}{2}\) or \(\cos\theta = \pm\frac{\sqrt{3}}{2}\) | M1 | Value of \(\sin\theta\) or \(\cos\theta\) reached with use of \(\cos 2\theta = \ldots\) and no method errors seen (arithmetic slips condoned) |
| \(\theta = \dfrac{\pi}{6}\) | A1 | Second A mark given here |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(9\cos 2\theta\sin^2\theta\) | M1 | First M mark even if they have a slip on the 9 and use 3, but must be \(\sin^2\theta\) |
| \(-9(2)\sin 2\theta\sin^2\theta + 9(2)\cos 2\theta\sin\theta\cos\theta = 0\) | M1 (dep) A1 | Second (dependent) M mark for product rule; A1 for correct derivative; M1 for setting equal to 0 |
| \(-2\sin^2\theta + \cos 2\theta = 0\) | M1 | Division by \(9\sin 2\theta\) or \(18\sin\theta\) and use of \(\sin 2\theta = 2\sin\theta\cos\theta\) followed by division by \(\cos\theta\) simplifies but does not provide specific M marks |
| or \(-\sin 2\theta\sin\theta + \cos 2\theta\cos\theta = 0\) leading to \(-2\sin^2\theta + \cos 2\theta = 0\) or \(\cos 3\theta = 1\) (compound angle formula) | ||
| \(-2\sin^2\theta + 1 - 2\sin^2\theta = 0\) | M1 | Use of \(\cos 2\theta = 1 - 2\sin^2\theta\) gives next M mark provided value of \(\sin\theta\) or alt reached with no errors seen |
| \(4\sin^2\theta = 1\) | ||
| \(\sin\theta = \pm\frac{1}{2}\) or \(3\theta = 2\pi\) (from \(\cos 3\theta = 1\)) | M1 | Value of \(\sin\theta\) or alt reached with use of \(\cos 2\theta = \ldots\) and no method errors seen (arithmetic slips condoned); gives final M mark |
| \(\theta = \dfrac{\pi}{6}\) | A1 | Second accuracy mark |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of \(9(\cos^2\theta - \sin^2\theta)\sin^2\theta\) | M1 | First M mark even if slip on the 9, but must be \(\sin^2\theta\) |
| Correct differentiation using product rule: \(9(-2\cos\theta\sin\theta - 2\sin\theta\cos\theta)\sin^2\theta + 9(\cos^2\theta - \sin^2\theta)2\sin\theta\) | dM1 | Second (dependent) M mark for differentiating using the product rule |
| Correct derivative obtained and set equal to 0 | A1, M1 | A1 for correct derivative; M1 for setting derivative equal to 0 |
| \(18\cos^3\theta\sin\theta - 54\sin^3\theta\cos\theta = 0\) leading to \(\cos^2\theta - 3\sin^2\theta = 0\) | — | Division by \(18\cos\theta\sin\theta\); use of \(\sin^2\theta = 1 - \cos^2\theta\) simplifies but does not give M marks |
| \(1 - 4\sin^2\theta = 0\) or \(4\cos^2\theta - 3 = 0\) | dM1 | Use of \(\cos 2\theta = 2\cos^2\theta - 1\) gives 4th M mark provided value of \(\sin\theta\) or alt reached with no errors after differentiation |
| \(\sin\theta = \pm\dfrac{1}{2}\) or \(\cos\theta = \pm\dfrac{\sqrt{3}}{2}\) | M1 | Value of \(\sin\theta\) or \(\cos\theta\) reached with use of \(\cos 2\theta = \ldots\) and no method errors seen (arithmetic slips condoned) |
| \(\theta = \dfrac{\pi}{6}\) | A1 | Second accuracy mark |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of \(9(2\cos^2\theta - 1)\sin^2\theta\) | M1 | Must be \(\sin^2\theta\) |
| \(9(-4\cos\theta\sin\theta)\sin^2\theta + 9(2\cos^2\theta - 1)2\sin\theta\cos\theta = 0\) | dM1 | Product rule differentiation |
| Correct derivative set equal to 0 | A1, M1 | |
| \(-2\sin^2\theta + 2\cos^2\theta - 1 = 0\), giving \(2\cos 2\theta = 1\) or \(1 - 4\sin^2\theta = 0\) or \(4\cos^2\theta - 3 = 0\) | dM1 | Can also use \(\cos^2\theta - \sin^2\theta = \cos 2\theta\) |
| \(\sin\theta = \pm\dfrac{1}{2}\) or \(\cos\theta = \pm\dfrac{\sqrt{3}}{2}\) or \(\cos 2\theta = \dfrac{1}{2}\) | M1 | |
| \(\theta = \dfrac{\pi}{6}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of \(9(1 - 2\sin^2\theta)\sin^2\theta\) | M1 | Must be \(\sin^2\theta\) |
| Derivative: \(18\sin\theta\cos\theta - 72\cos\theta\sin^3\theta\) | dM1 | |
| Set to 0: \(1 - 4\sin^2\theta = 0\) | M1 | Division by \(18\cos\theta\sin\theta\) |
| \(\sin\theta = \pm\dfrac{1}{2}\) or \(\cos\theta = \pm\dfrac{\sqrt{3}}{2}\) or \(\cos 2\theta = \dfrac{1}{2}\) | M1 | Use of \(\cos 2\theta = \ldots\) with no method errors |
| \(\theta = \dfrac{\pi}{6}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of \(3(\cos 2\theta)^{\frac{1}{2}}\sin\theta\) | M1 | |
| \(3\left(\dfrac{1}{2}\right)(\cos 2\theta)^{-\frac{1}{2}}(-2\sin 2\theta)\sin\theta + 3(\cos 2\theta)^{\frac{1}{2}}\cos\theta = 0\) | M1A1 | Correct differentiation using product and chain rule |
| Setting derivative equal to zero | M1 | |
| Multiply by \((\cos 2\theta)^{\frac{1}{2}}\), divide by 3: \(\cos 2\theta\cos\theta - \sin 2\theta\sin\theta = 0\) | dM1 | Use correct trig formulae to reduce to \(\cos k\theta = \ldots\) |
| \(\cos 3\theta = 0\), so \(3\theta = \dfrac{\pi}{2}\), giving \(\theta = \dfrac{\pi}{6}\) | A1 | A mark requires \(\cos\theta = \ldots\) or \(\theta = \dfrac{\pi}{6}\) |
## Question 8:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $A = (4\times)\int_0^{\frac{\pi}{4}} \frac{9}{2}\cos 2\theta\, d\theta$ | M1A1 | M1 for $\frac{1}{2}\int r^2\,d\theta = \frac{1}{2}\int\alpha\cos2\theta\,d\theta$ with $\alpha=3$ or $9$; A1 for correct limits $(0,\frac{\pi}{4})$ with multiple 4 |
| $= 18\left[\frac{\sin 2\theta}{2}\right]_0^{\frac{\pi}{4}}$ | M1 | $\cos2\theta \to \pm\frac{1}{2}\sin2\theta$ |
| $9\left[\sin\frac{\pi}{2} - 0\right] = 9$ | A1 | Using limits and factor 4 or 2 to obtain 9 |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $r\sin\theta = 3(\cos2\theta)^{\frac{1}{2}}\sin\theta$ | M1 | For $r\sin\theta = 3(\cos2\theta)^{\frac{1}{2}}\sin\theta$ or $r^2\sin^2\theta = 9\cos2\theta\sin^2\theta$ |
| $\frac{d}{d\theta}(r\sin\theta) = \left\{-3\times\frac{1}{2}(\cos2\theta)^{-\frac{1}{2}}\times 2\sin2\theta\sin\theta + 3(\cos2\theta)^{\frac{1}{2}}\cos\theta\right\}$ | M1depA1 | Differentiating rhs wrt $\theta$; product and chain rule must be used |
| At max/min: $\frac{-3\sin2\theta\sin\theta}{(\cos2\theta)^{\frac{1}{2}}} + 3(\cos2\theta)^{\frac{1}{2}}\cos\theta = 0$ | M1 | Equating expression to 0 |
| $\sin2\theta\sin\theta = \cos2\theta\cos\theta$ | | |
| $2\sin^2\theta\cos\theta = (1-2\sin^2\theta)\cos\theta$ | | |
| $\cos\theta(1-4\sin^2\theta)=0$ | | |
| $\sin\theta = \pm\frac{1}{2},\quad \theta = \pm\frac{\pi}{6}$ | M1A1 | M1dep for solving to $\sin k\theta=\ldots$ or $\cos k\theta=\ldots$ using correct trig formulae; ignore extra answers |
| $r\sin\frac{\pi}{6} = 3\left(\cos\frac{\pi}{3}\right)^{\frac{1}{2}}\times\frac{1}{2} = \frac{3\sqrt{2}}{4}$ | B1 | Length of $\frac{1}{2}PS = \frac{3\sqrt{2}}{4}$; may not be shown explicitly |
| Shaded area $= 6\times\frac{3\sqrt{2}}{2} - 9 = 9\sqrt{2}-9$ | M1, A1 | M1 for $PS\times 6 - $ their (a); A1 for $9\sqrt{2}-9$ (awrt 3.73) |
# Question (Polar Curves - Finding Tangent Direction)
## Option 1 – using $r\sin\theta$ with/without manipulation of $\cos 2\theta$ before differentiation
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $3(\cos 2\theta)^{\frac{1}{2}}\sin\theta$ | M1 | First M mark |
| $3(\cos 2\theta)^{\frac{1}{2}}\cos\theta - 3\left(\frac{1}{2}\right)(\cos 2\theta)^{-\frac{1}{2}}(2)\sin 2\theta\sin\theta = 0$ | M1 (dep) A1 | Second (dependent) M mark for differentiating using the product rule; A1 for correct derivative; M1 for setting derivative equal to 0 |
| $3(\cos 2\theta)^{\frac{1}{2}}\cos\theta - 3(\cos 2\theta)^{-\frac{1}{2}}\sin 2\theta\sin\theta = 0$ | | |
| $3(\cos 2\theta)^{\frac{1}{2}} - 6(\cos 2\theta)^{-\frac{1}{2}}\sin^2\theta = 0$ | | Use of $\sin 2\theta = 2\sin\theta\cos\theta$, division by $3\cos\theta$ and multiplication by $(\cos 2\theta)^{\frac{1}{2}}$ simplify but do not provide specific M marks |
| $\cos 2\theta - 2\sin^2\theta = 0$ | | |
| $(1 - 2\sin^2\theta) - 2\sin^2\theta = 0$ | M1 | Use of $\cos 2\theta = 1 - 2\sin^2\theta$ gives next M mark provided a value of $\sin\theta$ or alt is reached with no errors seen |
| $4\sin^2\theta = 1$ | | |
| $\sin\theta = \pm\frac{1}{2}$ | M1 | Value of $\sin\theta$ reached with use of $\cos 2\theta = \ldots$ and no method errors seen (arithmetic slips condoned); gives final M mark |
| $\theta = \dfrac{\pi}{6}$ | A1 | Second accuracy mark given here |
---
## Option 1 (continued) – using $3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\sin\theta$
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\sin\theta$ | M1 | First M mark; use of $\cos 2\theta = \cos^2\theta - \sin^2\theta$ gives 4th M mark provided value of $\sin\theta$ or alt reached with no errors seen after differentiation |
| $3\left(\frac{1}{2}\right)(\cos^2\theta - \sin^2\theta)^{-\frac{1}{2}}(-2\cos\theta\sin\theta - 2\sin\theta\cos\theta)\sin\theta + 3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\cos\theta = 0$ | M1 (dep) A1 | Second (dependent) M mark for product rule; A1 for correct derivative; M1 for setting equal to 0 |
| $-6(\cos^2\theta - \sin^2\theta)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}\cos\theta = 0$ | | |
| $-6\sin^2\theta + 3(\cos^2\theta - \sin^2\theta) = 0$ | M1 | Multiplication by $(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}$, division by $3\cos\theta$ and use of $\cos^2\theta = 1 - \sin^2\theta$ simplify but do not provide specific M marks |
| $4\sin^2\theta = 1$ | | |
| $\sin\theta = \pm\frac{1}{2}$ | M1 | Value of $\sin\theta$ reached with use of $\cos 2\theta = \ldots$ and no method errors seen (arithmetic slips condoned) |
| $\theta = \dfrac{\pi}{6}$ | A1 | Second accuracy mark |
---
## Option 1 (continued) – using $3(2\cos^2\theta - 1)^{\frac{1}{2}}\sin\theta$
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $3(2\cos^2\theta - 1)^{\frac{1}{2}}\sin\theta$ | M1 | First M mark; use of $\cos 2\theta = 2\cos^2\theta - 1$ gives 4th M mark provided value of $\sin\theta$ or alt reached with no errors seen after differentiation |
| $3\left(\frac{1}{2}\right)(2\cos^2\theta - 1)^{-\frac{1}{2}}(-4\cos\theta\sin\theta)\sin\theta + 3(2\cos^2\theta - 1)^{\frac{1}{2}}\cos\theta = 0$ | M1 (dep) A1 | Second (dep) M mark for product rule; A1 for correct derivative; M1 for setting equal to 0 |
| $-6(2\cos^2\theta - 1)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(2\cos^2\theta - 1)^{\frac{1}{2}}\cos\theta = 0$ | | |
| $-6\sin^2\theta + 3(2\cos^2\theta - 1) = 0$ | M1 | Multiplication by $(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}$, division by $3\cos\theta$ and use of $\sin^2\theta = 1 - \cos^2\theta$ or vice versa simplify but do not provide specific M marks |
| $4\sin^2\theta = 1$ or $4\cos^2\theta = 3$ | | |
| $\sin\theta = \pm\frac{1}{2}$ or $\cos\theta = \pm\frac{\sqrt{3}}{2}$ | M1 | Value of $\sin\theta$ or $\cos\theta$ reached with use of $\cos 2\theta = \ldots$ and no method errors seen (arithmetic slips condoned); gives final M mark |
| $\theta = \dfrac{\pi}{6}$ | A1 | Second accuracy mark |
---
## Option 1 (continued) – using $3(1 - 2\sin^2\theta)^{\frac{1}{2}}\sin\theta$
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $3(1 - 2\sin^2\theta)^{\frac{1}{2}}\sin\theta$ | M1 | First M mark; use of $\cos 2\theta = 2\cos^2\theta - 1$ gives 4th M mark provided value of $\sin\theta$ or alt reached with no errors seen after differentiation |
| $3\left(\frac{1}{2}\right)(1 - 2\sin^2\theta)^{-\frac{1}{2}}(-4\cos\theta\sin\theta)\sin\theta + 3(1 - 2\sin^2\theta)^{\frac{1}{2}}\cos\theta = 0$ | M1 (dep) A1 | Second (dependent) M mark for product rule; A1 for correct derivative; M1 for setting equal to 0 |
| $-6(1 - 2\sin^2\theta)^{-\frac{1}{2}}\cos\theta\sin^2\theta + 3(1 - 2\sin^2\theta)^{\frac{1}{2}}\cos\theta = 0$ | | |
| $-6\sin^2\theta + 3(1 - 2\cos^2\theta) = 0$ | M1 | Multiplication by $(\cos^2\theta - \sin^2\theta)^{\frac{1}{2}}$, division by $3\cos\theta$ and use of $\sin^2\theta = 1 - \cos^2\theta$ or vice versa simplify but do not provide specific M marks |
| $4\sin^2\theta = 1$ or $4\cos^2\theta = 3$ | | |
| $\sin\theta = \pm\frac{1}{2}$ or $\cos\theta = \pm\frac{\sqrt{3}}{2}$ | M1 | Value of $\sin\theta$ or $\cos\theta$ reached with use of $\cos 2\theta = \ldots$ and no method errors seen (arithmetic slips condoned) |
| $\theta = \dfrac{\pi}{6}$ | A1 | Second A mark given here |
---
## Option 2 – using $r^2\sin^2\theta$ with/without manipulation of $\cos 2\theta$ before differentiation
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $9\cos 2\theta\sin^2\theta$ | M1 | First M mark even if they have a slip on the 9 and use 3, but must be $\sin^2\theta$ |
| $-9(2)\sin 2\theta\sin^2\theta + 9(2)\cos 2\theta\sin\theta\cos\theta = 0$ | M1 (dep) A1 | Second (dependent) M mark for product rule; A1 for correct derivative; M1 for setting equal to 0 |
| $-2\sin^2\theta + \cos 2\theta = 0$ | M1 | Division by $9\sin 2\theta$ or $18\sin\theta$ and use of $\sin 2\theta = 2\sin\theta\cos\theta$ followed by division by $\cos\theta$ simplifies but does not provide specific M marks |
| or $-\sin 2\theta\sin\theta + \cos 2\theta\cos\theta = 0$ leading to $-2\sin^2\theta + \cos 2\theta = 0$ or $\cos 3\theta = 1$ (compound angle formula) | | |
| $-2\sin^2\theta + 1 - 2\sin^2\theta = 0$ | M1 | Use of $\cos 2\theta = 1 - 2\sin^2\theta$ gives next M mark provided value of $\sin\theta$ or alt reached with no errors seen |
| $4\sin^2\theta = 1$ | | |
| $\sin\theta = \pm\frac{1}{2}$ or $3\theta = 2\pi$ (from $\cos 3\theta = 1$) | M1 | Value of $\sin\theta$ or alt reached with use of $\cos 2\theta = \ldots$ and no method errors seen (arithmetic slips condoned); gives final M mark |
| $\theta = \dfrac{\pi}{6}$ | A1 | Second accuracy mark |
# Mark Scheme Extraction
## Question (Differentiation of trigonometric expression):
### Method 1: Using $9(\cos^2\theta - \sin^2\theta)\sin^2\theta$
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $9(\cos^2\theta - \sin^2\theta)\sin^2\theta$ | M1 | First M mark even if slip on the 9, but must be $\sin^2\theta$ |
| Correct differentiation using product rule: $9(-2\cos\theta\sin\theta - 2\sin\theta\cos\theta)\sin^2\theta + 9(\cos^2\theta - \sin^2\theta)2\sin\theta$ | dM1 | Second (dependent) M mark for differentiating using the product rule |
| Correct derivative obtained and set equal to 0 | A1, M1 | A1 for correct derivative; M1 for setting derivative equal to 0 |
| $18\cos^3\theta\sin\theta - 54\sin^3\theta\cos\theta = 0$ leading to $\cos^2\theta - 3\sin^2\theta = 0$ | — | Division by $18\cos\theta\sin\theta$; use of $\sin^2\theta = 1 - \cos^2\theta$ simplifies but does not give M marks |
| $1 - 4\sin^2\theta = 0$ or $4\cos^2\theta - 3 = 0$ | dM1 | Use of $\cos 2\theta = 2\cos^2\theta - 1$ gives 4th M mark provided value of $\sin\theta$ or alt reached with no errors after differentiation |
| $\sin\theta = \pm\dfrac{1}{2}$ or $\cos\theta = \pm\dfrac{\sqrt{3}}{2}$ | M1 | Value of $\sin\theta$ or $\cos\theta$ reached with use of $\cos 2\theta = \ldots$ and no method errors seen (arithmetic slips condoned) |
| $\theta = \dfrac{\pi}{6}$ | A1 | Second accuracy mark |
---
### Method 2: Using $9(2\cos^2\theta - 1)\sin^2\theta$
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $9(2\cos^2\theta - 1)\sin^2\theta$ | M1 | Must be $\sin^2\theta$ |
| $9(-4\cos\theta\sin\theta)\sin^2\theta + 9(2\cos^2\theta - 1)2\sin\theta\cos\theta = 0$ | dM1 | Product rule differentiation |
| Correct derivative set equal to 0 | A1, M1 | |
| $-2\sin^2\theta + 2\cos^2\theta - 1 = 0$, giving $2\cos 2\theta = 1$ or $1 - 4\sin^2\theta = 0$ or $4\cos^2\theta - 3 = 0$ | dM1 | Can also use $\cos^2\theta - \sin^2\theta = \cos 2\theta$ |
| $\sin\theta = \pm\dfrac{1}{2}$ or $\cos\theta = \pm\dfrac{\sqrt{3}}{2}$ or $\cos 2\theta = \dfrac{1}{2}$ | M1 | |
| $\theta = \dfrac{\pi}{6}$ | A1 | |
---
### Method 3: Using $9(1 - 2\sin^2\theta)\sin^2\theta$
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $9(1 - 2\sin^2\theta)\sin^2\theta$ | M1 | Must be $\sin^2\theta$ |
| Derivative: $18\sin\theta\cos\theta - 72\cos\theta\sin^3\theta$ | dM1 | |
| Set to 0: $1 - 4\sin^2\theta = 0$ | M1 | Division by $18\cos\theta\sin\theta$ |
| $\sin\theta = \pm\dfrac{1}{2}$ or $\cos\theta = \pm\dfrac{\sqrt{3}}{2}$ or $\cos 2\theta = \dfrac{1}{2}$ | M1 | Use of $\cos 2\theta = \ldots$ with no method errors |
| $\theta = \dfrac{\pi}{6}$ | A1 | |
---
### Method 4: Using factor formulae after differentiating $3(\cos 2\theta)^{\frac{1}{2}}\sin\theta$
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $3(\cos 2\theta)^{\frac{1}{2}}\sin\theta$ | M1 | |
| $3\left(\dfrac{1}{2}\right)(\cos 2\theta)^{-\frac{1}{2}}(-2\sin 2\theta)\sin\theta + 3(\cos 2\theta)^{\frac{1}{2}}\cos\theta = 0$ | M1A1 | Correct differentiation using product and chain rule |
| Setting derivative equal to zero | M1 | |
| Multiply by $(\cos 2\theta)^{\frac{1}{2}}$, divide by 3: $\cos 2\theta\cos\theta - \sin 2\theta\sin\theta = 0$ | dM1 | Use correct trig formulae to reduce to $\cos k\theta = \ldots$ |
| $\cos 3\theta = 0$, so $3\theta = \dfrac{\pi}{2}$, giving $\theta = \dfrac{\pi}{6}$ | A1 | A mark requires $\cos\theta = \ldots$ or $\theta = \dfrac{\pi}{6}$ |
The image provided appears to be just the back cover/colophon page of an Edexcel mark scheme document (Summer 2013, Order Code UA035974). It contains only publication/contact information for Edexcel Publications and logos for Ofqual, Welsh Assembly Government, and CEA.
**There is no mark scheme content visible on this page.** No questions, answers, mark allocations, or guidance notes are present to extract.
If you have additional pages from this mark scheme document, please share those and I'll be happy to extract and format the marking content for you.8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6d1485a6-e52b-4492-8d3b-eadca26962db-14_360_1109_237_566}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a closed curve $C$ with equation
$$r = 3 ( \cos 2 \theta ) ^ { \frac { 1 } { 2 } } , \quad \text { where } - \frac { \pi } { 4 } < \theta \leqslant \frac { \pi } { 4 } , \frac { 3 \pi } { 4 } < \theta \leqslant \frac { 5 \pi } { 4 }$$
The lines $P Q , S R , P S$ and $Q R$ are tangents to $C$, where $P Q$ and $S R$ are parallel to the initial line and $P S$ and $Q R$ are perpendicular to the initial line. The point $O$ is the pole.
\begin{enumerate}[label=(\alph*)]
\item Find the total area enclosed by the curve $C$, shown unshaded inside the rectangle in Figure 1.
\item Find the total area of the region bounded by the curve $C$ and the four tangents, shown shaded in Figure 1.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2013 Q8 [13]}}