# Question 8:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| When $y=1$: $1=1-2\cos t \Rightarrow t=-\frac{\pi}{2}, \frac{\pi}{2}$ | M1 | Sets $y=1$ to find $t$ and uses their $t$ to find $x$ |
| $k \text{ (or }x)=\frac{\pi}{2}-4\sin\left(\frac{\pi}{2}\right)$ or $x=-\frac{\pi}{2}-4\sin\left(-\frac{\pi}{2}\right)$ | | |
| When $t=-\frac{\pi}{2}$, $k>0$, so $k=4-\frac{\pi}{2}$ or $\frac{8-\pi}{2}$ | A1 | $x$ or $k=4-\frac{\pi}{2}$ |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt}=1-4\cos t$ | B1 | At least one of $\frac{dx}{dt}$ or $\frac{dy}{dt}$ correct |
| $\frac{dy}{dt}=2\sin t$ | B1 | Both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ correct |
| $\frac{dy}{dx}=\frac{2\sin t}{1-4\cos t}$ | M1 | Applies $\frac{dy}{dt}$ divided by $\frac{dx}{dt}$ and substitutes $t$ into $\frac{dy}{dx}$ |
| At $t=-\frac{\pi}{2}$: $\frac{dy}{dx}=\frac{2\sin(-\frac{\pi}{2})}{1-4\cos(-\frac{\pi}{2})}=-2$ | A1 cao cso | Correct value for $\frac{dy}{dx}$ of $-2$ |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2\sin t}{1-4\cos t}=-\frac{1}{2}$ | M1 | Sets $\frac{dy}{dx}=-\frac{1}{2}$ |
| $4\sin t - 4\cos t = -1$ | A1 | See notes |
| $4\sqrt{2}\sin\left(t-\frac{\pi}{4}\right)=-1$ or $-4\sqrt{2}\cos\left(t+\frac{\pi}{4}\right)=-1$ | M1; A1 | See notes |
| $t=\sin^{-1}\left(\frac{-1}{4\sqrt{2}}\right)+\frac{\pi}{4}$ or $t=\cos^{-1}\left(\frac{1}{4\sqrt{2}}\right)-\frac{\pi}{4}$ | dM1 | See notes |
| $t=0.6076875626... = 0.6077$ (4 dp) | A1 | Anything that rounds to 0.6077 |

# Question 8:

## Part (c) – Alternative Method 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2\sin t}{1-4\cos t} = -\frac{1}{2}$ | M1 | Sets their $\frac{dy}{dx} = -\frac{1}{2}$ |
| e.g. $\left(\frac{2\sin t}{1-4\cos t}\right)^2 = \frac{1}{4}$ or $(4\sin t)^2 = (4\cos t - 1)^2$ or $(4\sin t + 1)^2 = (4\cos t)^2$ etc. | A1 | Squaring to give a correct equation. This mark can be implied by a "squared" correct equation. |
| Squares equation, applies $\sin^2 t + \cos^2 t = 1$, achieves **three term quadratic** of form $\pm a\cos^2 t \pm b\cos t \pm c = 0$ or $\pm a\sin^2 t \pm b\sin t \pm c = 0$ where $a\neq 0, b\neq 0, c\neq 0$ | M1 | |
| Either $32\cos^2 t - 8\cos t - 15 = 0$ or $32\sin^2 t + 8\sin t - 15 = 0$ | A1 | For a correct **three term quadratic equation** |
| Either $\cos t = \frac{8 \pm \sqrt{1984}}{64} = \frac{1+\sqrt{31}}{8} \Rightarrow t = \cos^{-1}(\ldots)$ or $\sin t = \frac{-8 \pm \sqrt{1984}}{64} = \frac{-1\pm\sqrt{31}}{8} \Rightarrow t = \sin^{-1}(\ldots)$ | dM1 | **Dependent on 2nd M1.** Uses correct algebraic processes to give $t = \ldots$ |
| $t = 0.6076875626\ldots = 0.6077$ (4 dp) | A1 | Anything that rounds to $0.6077$ |

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## Part (c) – Alternative Method 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2\sin t}{1-4\cos t} = -\frac{1}{2}$ | M1 | Sets their $\frac{dy}{dx} = -\frac{1}{2}$ |
| e.g. $(4\sin t - 4\cos t)^2 = (-1)^2$ | A1 | Squaring to give a correct equation. Can be implied by a correct equation. Note: can also give 1st A1 for $4\sin t - 4\cos t = -1$ as in main scheme |
| So $16\sin^2 t - 32\sin t\cos t + 16\cos^2 t = 1$; applies both $\sin^2 t + \cos^2 t = 1$ and $\sin 2t = 2\sin t\cos t$, achieves form $\pm a \pm b\sin 2t = \pm c$ | M1 | |
| $16 - 16\sin 2t = 1$ or equivalent | A1 | |
| $\left\{\sin 2t = \frac{15}{16}\right\} \Rightarrow t = \frac{\sin^{-1}(\ldots)}{2}$ | dM1 | **Dependent on 2nd M1.** Uses correct algebraic processes to give $t = \ldots$ |
| $t = 0.6076875626\ldots = 0.6077$ (4 dp) | A1 | Anything that rounds to $0.6077$ |

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## Question 8 Notes:

### Part (a):
| Note | Mark | Guidance |
|---|---|---|
| Sets $y=1$ to find $t$, uses $t$ to find $x$ | M1 | Can be implied by $x$ or $k = 4 - \frac{\pi}{2}$ or $2.429\ldots$ or $\frac{\pi}{2}-4$ or $-2.429\ldots$ |
| $x$ or $k = 4-\frac{\pi}{2}$ or $\frac{8-\pi}{2}$ | A1 | Decimal answer of $2.429\ldots$ without exact answer is A0 |

### Part (b):
| Note | Mark | Guidance |
|---|---|---|
| At least one of $\frac{dx}{dt}$ or $\frac{dy}{dt}$ correct | B1 | Can be implied from working |
| Both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ correct | B1 | Can be implied from working |
| Applies $\frac{dy}{dt}$ divided by $\frac{dx}{dt}$ **and** attempts to substitute $t$ into $\frac{dy}{dx}$ | M1 | i.e. $\frac{dy}{dx} = \frac{2\sin t}{1-4\cos t}$ followed by $-2$ (from $t=-\frac{\pi}{2}$) or $2$ (from $t=\frac{\pi}{2}$) |
| Using $t = -\frac{\pi}{2}$ (not $t=\frac{3\pi}{2}$) to find correct $\frac{dy}{dx}$ of $-2$ **by correct solution only** | A1 | |

### Part (c):
| Note | Mark | Guidance |
|---|---|---|
| If incorrect $\frac{dy}{dx}$ used in (c), accuracy marks not obtainable | NOTE | |
| Sets $\frac{dy}{dx} = -\frac{1}{2}$ | 1st M1 | |
| Rearranges to correct equation with $\sin t$ and $\cos t$ **on the same side**, e.g. $4\sin t - 4\cos t = -1$ or $4\cos t - 4\sin t = 1$ or $\sin t - \cos t = -\frac{1}{4}$ etc. | 1st A1 | |
| Rewrites $\pm\lambda\sin t \pm \mu\cos t$ in form $R\cos(t\pm\alpha)$ or $R\sin(t\pm\alpha)$, where $R\neq 1$ or $0$, $\alpha\neq 0$ | 2nd M1 | |
| Correct equation, e.g. $4\sqrt{2}\sin\left(t-\frac{\pi}{4}\right)=-1$ or $-4\sqrt{2}\cos\left(t+\frac{\pi}{4}\right)=-1$ or $\sqrt{2}\sin\left(t-\frac{\pi}{4}\right)=-\frac{1}{4}$ or $\sqrt{2}\cos\left(t+\frac{\pi}{4}\right)=\frac{1}{4}$ etc. | 2nd A1 | |
| **Dependent on 2nd M1.** Uses correct algebraic processes to give $t=\ldots$ | 3rd M1 | |
| Anything that rounds to $0.6077$ | 4th A1 | Do **not** give if extra solutions in range $-\frac{2\pi}{3}\leqslant t \leqslant \frac{2\pi}{3}$; **do** give if extra solutions outside this range |