## Question 11 (EITHER):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x^2+y^2)^2 = a^2(x^2-y^2) \Rightarrow r^2 = a^2\left(\frac{x^2}{r^2}-\frac{y^2}{r^2}\right) = a^2(\cos^2\theta - \sin^2\theta) = a^2\cos 2\theta$ | M1, A1 | Uses $x^2+y^2=r^2$, $x=r\cos\theta$, $y=r\sin\theta$; 2 marks |
| Sketches $C$ | B2,1,0 | One mark for each loop or half of whole curve; 2 marks |
| Area $= \frac{1}{2}\int_{-\pi/4}^{\pi/4} a^2\cos 2\theta\, d\theta \left(= \int_0^{\pi/4} a^2\cos 2\theta\, d\theta\right)$ | M1 | Uses sector area formula |
| $= a^2\left[\frac{\sin 2\theta}{2}\right]_0^{\pi/4} = \frac{a^2}{2}$ | A1A1 | S.C. omission of $\frac{1}{2}$ factor but correct integration gets B1; 3 marks |
| $2(x^2+y^2)(2x+2yy') = a^2(2x-2yy')$ | B1B1 | Differentiates |
| $y'=0 \Rightarrow 2x(x^2+y^2) = a^2 x$ | M1 | Puts $y'=0$ |
| $\Rightarrow 2r^2 = a^2 \Rightarrow r = \frac{a}{\sqrt{2}}$ $(r\geq 0)$ | A1 | Obtains coordinates |
| $\Rightarrow \cos 2\theta = \frac{1}{2}$ | M1 | |
| $\Rightarrow \theta = \pm\frac{\pi}{6}, \pm\frac{5\pi}{6}$ | A1A1 | 7 marks total |
| i.e. $\left(\frac{a}{\sqrt{2}}, \pm\frac{\pi}{6}\right)$ and $\left(\frac{a}{\sqrt{2}}, \pm\frac{5\pi}{6}\right)$ | | |

**Alternative last 7 marks:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx}=0 \Rightarrow \frac{dy}{d\theta}=0 \Rightarrow r\cos\theta + \frac{dr}{d\theta}\sin\theta = 0$ | M1A1 | Obtains condition for tangent parallel to initial line |
| $2r\frac{dr}{d\theta} = -2a^2\sin 2\theta \Rightarrow \frac{dr}{d\theta} = -\frac{a^2}{r}\sin 2\theta$ | A1 | Differentiates equation of $C$ |
| $\therefore r\cos\theta - \frac{a^2}{r}\sin 2\theta\sin\theta = 0$ | M1 | Forms equation |
| $\therefore a^2\cos 2\theta\cos\theta = a^2\sin 2\theta\sin\theta \therefore \frac{1}{\tan 2\theta} = \tan\theta$ | | |
| $\therefore \frac{1-t^2}{2t} = t \Rightarrow 2t^2 = 1-t^2 \Rightarrow 3t^2=1$ | M1 | |
| $\therefore t = \pm\frac{1}{\sqrt{3}}$ | | Solves for $\tan\theta$ |
| $\therefore \theta = \pm\frac{\pi}{6}, \pm\frac{5\pi}{6}$ | A1 | |
| $\left(\frac{a}{\sqrt{2}}, \pm\frac{\pi}{6}\right)$, $\left(\frac{a}{\sqrt{2}}, \pm\frac{5\pi}{6}\right)$ | A1 | Writes coordinates of points; award final A1 if $r$ found but result not written out |

# Question 11:

**OR method (substitution $y = xz$)**

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**Differentiates once:**

$\frac{dy}{dx} = z + x\frac{dz}{dx}$ | B1 | First derivative correct

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**Differentiates again:**

$\frac{d^2y}{dx^2} = 2\frac{dz}{dx} + x\frac{d^2z}{dx^2}$ | B1 | Second derivative correct

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**Substitutes:**

$\frac{d^2z}{dx^2} + \left(\frac{2}{x} + 6 - \frac{2}{x}\right)\frac{dz}{dx} + \left(\frac{6z}{x} - \frac{2z}{x^2} + 9z - \frac{6z}{x} + \frac{2z}{x^2}\right) = 169\sin 2x$ | M1 | Substitutes both derivatives into ODE and simplifies

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**Obtains result:**

$\Rightarrow \frac{d2z}{dx^2} + 6\frac{dz}{dx} + 9z = 169\sin 2x \quad \text{(AG)}$ | A1 | Correct reduced equation (Mark similarly if substitution rearranged to $z = \frac{y}{x}$) | **Part Mark: 4**

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**Finds CF:**

$m^2 + 6m + 9 = (m+3)^2 = 0 \Rightarrow m = -3$ | M1 |
CF: $Ae^{-3x} + Bxe^{-3x}$ | A1 |

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**Finds PI:**

PI: $y = p\sin 2x + q\cos 2x$ | M1 | Correct form of trial PI
$y' = 2p\cos 2x - 2q\sin 2x$
$y'' = -4p\sin 2x - 4q\cos 2x$

$5p - 12q = 169$
$12p + 5q = 0$ | M1 | Substitutes and equates coefficients

$\Rightarrow p = 5$ and $q = -12$ | A1 |

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**Finds GS:**

$z = Ae^{-3x} + Bxe^{-3x} + 5\sin 2x - 12\cos 2x$ | A1 |

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**Evaluates coefficients from initial conditions:**

$-10 = A - 12 \Rightarrow A = 2$ | B1 |
$z' = -6e^{-3x} + Be^{-3x} - 3Bxe^{-3x} + 10\cos 2x + 24\sin 2x$
$5 = -6 + B + 10 \Rightarrow B = 1$ | M1, A1 |

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**Finds particular solution:**

$z = 2e^{-3x} + xe^{-3x} + 5\sin 2x - 12\cos 2x$

$\therefore y = 2xe^{-3x} + x^2e^{-3x} + 5x\sin 2x - 12x\cos 2x$ | A1 | **Part Mark: 10** | **Total: [14]**