# Question 1:

## Part (a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a\tan y = x \Rightarrow a\sec^2 y \frac{dy}{dx} = 1$ | M1 | Differentiating with respect to $x$ or $y$; $\frac{dx}{dy} = a\sec^2 y$ |
| $\Rightarrow \frac{dy}{dx} = \frac{1}{a\sec^2 y}$ | A1 | For $\frac{dy}{dx}$; or $a\frac{dy}{dx} = \frac{1}{\sec^2 y}$ |
| $\Rightarrow \frac{dy}{dx} = \frac{1}{a\left(1+\frac{x^2}{a^2}\right)}$ | | |
| $\Rightarrow \frac{dy}{dx} = \frac{a}{a^2+x^2}$ | A1(ag) | Completion www with sufficient detail |
| **[3]** | | |

## Part (a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 - 4x + 8 = (x-2)^2 + 4$ | B1 | |
| $\int_0^4 \frac{1}{x^2-4x+8}dx = \frac{1}{2}\left[\arctan\frac{x-2}{2}\right]_0^4$ | M1 | Integral of form $a$ arctan $bu$ or any appropriate substitution |
| | A1 | Correct integral with consistent limits; $\frac{1}{2}\left[\arctan\frac{u}{2}\right]_{-2}^{2}$ |
| $= \frac{1}{2}(\arctan(1) - \arctan(-1))$ | | |
| $= \frac{\pi}{4}$ | A1 | Evaluated in terms of $\pi$ |
| **[4]** | | |

## Part (a)(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int 1 \times \arctan x\, dx$ | M1 | Using parts with $u = \arctan x$ and $v' = 1$; allow one other error |
| $= x\arctan x - \int\frac{x}{1+x^2}dx$ | A1 | |
| | M1 | $\int\frac{x}{1+x^2}dx = a\ln(1+x^2)$ |
| $= x\arctan x - \frac{1}{2}\ln(1+x^2) + c$ | A1 | $a = \frac{1}{2}$. Condone omitted $c$ |
| **[4]** | | |

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# Question 1(b):

## Part (b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = 2\cos\theta \Rightarrow r^2 = 2r\cos\theta$ | M1 | Using $r^2 = x^2+y^2$ and $x = r\cos\theta$ |
| $\Rightarrow x^2 + y^2 = 2x$ | A1 | A correct cartesian equation in any form |
| $\Rightarrow (x-1)^2 + y^2 = 1$ | A1(ag) | Explaining that the curve is a circle; e.g. writing as $(x-\alpha)^2+(y-\beta)^2=r^2$ |
| **OR** $x = r\cos\theta \Rightarrow x = 2\cos^2\theta$ | | |
| $y = r\sin\theta \Rightarrow y = 2\cos\theta\sin\theta = \sin 2\theta$ | M1 | Using $x = r\cos\theta$, $y = r\sin\theta$ and linking $x$ in terms of $\cos 2\theta$ |
| $\cos 2\theta = 2\cos^2\theta - 1 \Rightarrow x = \cos 2\theta + 1$ | A1 | |
| $\Rightarrow (x-1)^2 + y^2 = 1$ | A1(ag) | Explaining that the curve is a circle; e.g. writing as $(x-\alpha)^2+(y-\beta)^2=r^2$ |
| Centre $(1, 0)$ | B1 | Independent |
| Radius $1$ | B1 | Independent |
| **[5]** | | |

## Part (b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 + (y-2)^2 = 4 \Rightarrow x^2 + y^2 = 4y$ | | |
| $\Rightarrow r^2 = 4r\sin\theta$ | M1 | Using $r^2 = x^2+y^2$ and $y = r\sin\theta$ |
| $\Rightarrow r = 4\sin\theta$ | A1 | For answer alone www: B1 for $r = k\sin\theta$, B1 for $k=4$ |
| **[2]** | | |

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