# Question 14:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\cos 2t = 1 - 2\sin^2 t \Rightarrow \frac{y-4}{2} = 1 - 2\left(\frac{x-3}{2}\right)^2$ | M1 | Uses $\cos 2t = 1 - 2\sin^2 t$ to eliminate $t$ |
| $y - 4 = 2 - 4 \times \frac{(x-3)^2}{4} \Rightarrow y = 6-(x-3)^2$ | A1* | Correct given answer |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cap$ shaped parabola | M1 | |
| Fully correct with 'ends' at $(1,2)$ and $(5,2)$ | A1 | |
| Suitable reason: e.g. states $x = 3 + 2\sin t$, $1 \leqslant x \leqslant 5$ | B1 | |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Either finds lower value $k = 7$ or deduces $k < \frac{37}{4}$ | B1 | |
| Finds where $x + y = k$ meets $y = 6-(x-3)^2$: $k - x = 6-(x-3)^2$ and proceeds to 3TQ in $x$ or $y$ | M1 | |
| Correct 3TQ in $x$: $x^2 - 7x + (k+3) = 0$ or in $y$: $y^2 + (7-2k)y + (k^2-6k+3) = 0$ | A1 | |
| Uses $b^2 - 4ac = 0$: $49 - 4(k+3) = 0 \Rightarrow k = \frac{37}{4}$ or $(7-2k)^2 - 4(k^2-6k+3) = 0 \Rightarrow k = \frac{37}{4}$ | M1 | |
| Range: $\left\{k : 7 \leqslant k < \frac{37}{4}\right\}$ | A1 | |

# Mark Scheme Extraction

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Proceeds to $y = 6-(x-3)^2$ without any errors | A1* | Allow proof starting with $y=6-(x-3)^2$ substituting parametric coordinates. M1 scored for correct $\cos 2t = 1-2\sin^2 t$; A1 only when both sides seen to be same AND comment made |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sketching a $\cap$ parabola with maximum in quadrant one | M1 | Does not need to be symmetrical |
| Sketching a $\cap$ parabola with maximum in quadrant one with end coordinates $(1,2)$ and $(5,2)$ | A1 | — |
| Any suitable explanation why $C$ does not include all points of $y=6-(x-3)^2$, $x\in\mathbb{R}$, referencing **limits on sin or cos** with **link to restriction on $x$ or $y$** | B1 | e.g. "As $-1\leqslant\sin t\leqslant 1$ then $1\leqslant x\leqslant 5$"; "As $\sin t\leqslant 1$ then $x\leqslant 5$"; "As $-1\leqslant\cos(2t)\leqslant 1$ then $2\leqslant y\leqslant 6$". Withhold if domain stated as $2\leqslant x\leqslant 5$. Do not allow statement on top limit of $y$ only |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces lower value $k=7$, e.g. substituting $(5,2)$: $x+y=k\Rightarrow k=7$, or $x=5$ into $x^2-7x+(k+3)=0\Rightarrow 25-35+k+3=0\Rightarrow k=7$ | B1 | May be awarded from later work |
| Deduces $k<\dfrac{37}{4}$ | B1 | May be awarded from later work |
| Attempt at upper value for $k$: finds where $x+y=k$ meets $y=6-(x-3)^2$ using appropriate method, e.g. sets $k-x=6-(x-3)^2$ and proceeds to a 3TQ | M1 | — |
| Correct 3TQ: $x^2-7x+(k+3)=0$ | A1 | The $=0$ may be implied by subsequent work |
| Uses discriminant condition: $b^2=4ac$ or $b^2\ldots 4ac$ leading to critical value for $k$, e.g. $49-4\times1\times(k+3)=0\Rightarrow k=\dfrac{37}{4}$ | M1 | Accept use of $b^2=4ac$ or $b^2\ldots 4ac$ where $\ldots$ is any inequality |
| Range: $k=\left\{k: 7\leqslant k<\dfrac{37}{4}\right\}$, accept $k\in\left[7,\dfrac{37}{4}\right)$ or exact equivalent | A1 | — |

**ALT for upper value:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets gradient of $y=6-(x-3)^2$ equal to $-1$ | M1 | Finding where $x+y=k$ meets curve using tangency condition |
| $-2x+6=-1\Rightarrow x=3.5$ | A1 | — |
| Finds point of intersection and uses to find upper $k$: $y=6-(3.5-3)^2=5.75$, hence $k=3.5+5.75=9.25$ | M1 | — |
| Range: $k=\{k:7\leqslant k<9.25\}$ | A1 | — |