# Question 4:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\theta = -\pi/2$: $O(0,0)$ | B1 | Origin or O, condone omission of $(0,0)$ or O |
| $\theta = 0$: $P(2,0)$ | B1 | Or, say at P $x=2$, $y=0$, need P stated |
| $\theta = \pi/2$: $O(0,0)$ | B1 [3] | Origin or O, condone omission of $(0,0)$ or O |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2\cos 2\theta}{-2\sin\theta} = -\frac{\cos 2\theta}{\sin\theta}$ | M1, A1 | their $\frac{dy/d\theta}{dx/d\theta}$; any equivalent form www (not from $-2\cos 2\theta / 2\sin\theta$) |
| When $\theta = \pi/2$: $\frac{dy}{dx} = -\cos\pi/\sin(\pi/2) = 1$ | M1 | subst $\theta = \pi/2$ in their equation |
| When $\theta = -\pi/2$: $\frac{dy}{dx} = -\cos(-\pi)/\sin(-\pi/2) = -1$ | A1 | Obtaining $\frac{dy}{dx} = 1$ and $\frac{dy}{dx} = -1$ shown (or explaining using symmetry of curve) www |
| Either $1\times -1 = -1$ so perpendicular. Or gradient tangent $=1 \Rightarrow$ meets axis at $45°$, similarly gradient $=-1 \Rightarrow$ meets axis at $45°$ oe | A1 [5] | justification that tangents are perpendicular www dependent on previous A1 |

## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| At Q, $\sin 2\theta = 1 \Rightarrow 2\theta = \pi/2$, $\theta = \pi/4$ | M1 | or, using the derivative, $\cos 2\theta = 0$ soi or their $\frac{dy}{dx} = 0$ to find $\theta$. If the only error is in the sign or the coeff of the derivative in (ii), allow full marks in this part (condone $\theta = 45°$) |
| coordinates of Q are $(2\cos\pi/4,\ \sin\pi/2) = (\sqrt{2}, 1)$ | A1 A1 [3] | www (exact only) accept $2/\sqrt{2}$ |

## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin^2\theta = (1-\cos^2\theta) = 1 - \frac{1}{4}x^2$ | B1 | oe, eg may be $x^2 = \ldots$ |
| $y = \sin 2\theta = 2\sin\theta\cos\theta$ | M1 | **Use** of $\sin 2\theta = 2\sin\theta\cos\theta$ |
| $= (\pm)\ x\sqrt{1-\frac{1}{4}x^2}$ | A1 | subst for $x$ **or** $y^2 = 4\sin^2\theta\cos^2\theta$ (squaring); either order oe |
| $\Rightarrow y^2 = x^2(1-\frac{1}{4}x^2)$ | A1 [4] | squaring **or** subst for $x$; either order oe. **AG** |

## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = \int_0^2 \pi x^2\left(1-\frac{1}{4}x^2\right)dx$ | M1 | integral including correct limits but ft their '2' from (i); condone omission of $dx$ if intention clear |
| $= \int_0^2\left(\pi x^2 - \frac{1}{4}\pi x^4\right)dx$ | | |
| $= \pi\left[\frac{1}{3}x^3 - \frac{1}{20}x^5\right]_0^2$ | B1 | $\left[\frac{1}{3}x^3 - \frac{1}{20}x^5\right]$ ie allow if no $\pi$ and/or incorrect/no limits |
| $= \pi\left[\frac{8}{3} - \frac{32}{20}\right]$ | A1 | substituting limits into correct expression (including $\pi$) ft their '2' |
| $= \frac{16\pi}{15}$ | A1 [4] | cao oe, 3.35 or better (any multiple of $\pi$ must round to 3.35 or better) |

## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AA'} = \begin{pmatrix}2\\4\\1\end{pmatrix} - \begin{pmatrix}1\\2\\4\end{pmatrix} = \begin{pmatrix}1\\2\\-3\end{pmatrix}$ | B1 | finding $\overrightarrow{AA'}$ or $\overrightarrow{A'A}$ by subtraction, **subtraction must be seen**. B0 if $\overrightarrow{AA'}$, $\overrightarrow{A'A}$ confused. Assume they have found $\overrightarrow{AA'}$ if no label |
| This vector is normal to $x + 2y - 3z = 0$ | B1 | reference to normal or **n**, or perpendicular to $x+2y-3z=0$, or statement that vector matches coefficients of plane and is therefore perpendicular, or showing $AA'$ is perpendicular to **two** vectors in the plane |
| M is $\left(1\frac{1}{2},\ 3,\ 2\frac{1}{2}\right)$ | M1 | for finding M correctly (can be implied by two correct coordinates) |
| $x + 2y - 3z = 1\frac{1}{2} + 6 - 7\frac{1}{2} = 0 \Rightarrow$ M lies in plane | A1 [4] | showing numerical subst of M in plane $= 0$ |

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