## Question 5:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Choosing and applying the quotient rule, product rule or implicit differentiation to the given function | M1 | For implicit differentiation look for $(\pm\cdots\sin\theta \pm\cdots\cos\theta)y + \frac{dy}{d\theta}(4\cos\theta + 4\sin\theta) = \cdots\sin\theta$ |
| $\frac{dy}{d\theta} = \frac{(4\cos\theta+4\sin\theta)(-5\sin\theta)-(5\cos\theta)(-4\sin\theta+4\cos\theta)}{(4\cos\theta+4\sin\theta)^2}$ | A1 | Correct expression for $\frac{dy}{d\theta}$ |
| Expands and uses $\sin^2\theta + \cos^2\theta = 1$ at least once in numerator or denominator, OR uses $2\sin\theta\cos\theta = \sin 2\theta$ in $\frac{dy}{d\theta} = \frac{\cdots}{\cdots C\sin\theta\cos\theta}$ | M1 | |
| Expands and uses $\sin^2\theta + \cos^2\theta = 1$ in numerator AND denominator, AND uses $2\sin\theta\cos\theta = \sin 2\theta$ in denominator to reach $\frac{dy}{d\theta} = \frac{P}{Q + R\sin 2\theta}$ | M1 | |
| Fully correct proof with $A = -\frac{5}{4}$ giving $\frac{dy}{d\theta} = -\frac{5}{4(1+\sin 2\theta)}$ | A1 | |

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## Question 6a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correctly deduces gradient of $AQ$ is $\frac{-1}{-4} = \frac{1}{4}$ (perpendicular to $l_1$) | M1 | By circle theorems, radius $AQ \perp$ tangent $l_1$ |
| Finding equation of line with gradient $\frac{1}{4}$ through $(-3, 1)$: $y - 1 = \frac{1}{4}(x+3)$ | M1 | |
| $4y - x = 7$ | A1 | Must be in required form $ax + by = c$ |

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## Question 6b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt at solving $4y - x = 7$ and $y = -4x + 6$ simultaneously to find coordinates of $Q$ | M1 | |
| $Q = (1, 2)$ | A1 | |
| Attempt to use Pythagoras to find radius or $r^2$ using $Q(1,2)$ and $A(-3,1)$: $r = \sqrt{(1-(-3))^2+(2-1)^2} = \sqrt{17}$ | M1 | |
| $(x+3)^2 + (y-1)^2 = 17$ | A1 | |

## Question 6c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $y = -4x + k$ into circle equation: $(x+3)^2 + (-4x+k-1)^2 = 17$ | M1 | For solving $l_2$ with their equation of the circle |
| $17x^2 + (14-8k)x + k^2 - 2k - 7 = 0$ | | |
| Uses $b^2 - 4ac = 0$: $(14-8k)^2 - 4(17)(k^2-2k-7) = 0$ | M1 | Uses $b^2 - 4ac = 0$ and finds a value of $k$ |
| $-4k^2 - 88k + 672 = 0 \Rightarrow k = 6$ or $k = -28$ | | |
| $k = -28$ (since $k \neq 6$) | A1 | |
| **OR:** $\frac{x+1}{2} = -3 \Rightarrow x = -7$, $\frac{y+2}{2} = 1 \Rightarrow y = 0$ | M1 | Finding end point of diameter using mid-point formula |
| Substitute $(-7, 0)$ into $y = -4x + k$ | M1 | Substituting $(-7,0)$ into $y = -4x+k$ and proceeds to $k=\ldots$ |
| $k = -28$ | A1 | |

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## Question 7a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{6}{(7k-2x)}\,dx = A\ln(7k-2x)$ | M1 | |
| $\int \frac{6}{(7k-2x)}\,dx = -3\ln(7k-2x)$ | A1 | |
| Substituting $3k$ and $2k$ into $A\ln(7k-2x)$ and subtracting | M1 | Substituting $3k$ and $2k$ and subtracting either way around |
| $[-3\ln(7k-2x)]_{2k}^{3k} = -3(\ln(k) - \ln(3k)) = -3\ln\!\left(\frac{1}{3}\right)$ | A1 | Uses correct ln work and notation to show $I = -3\ln\!\left(\frac{1}{3}\right)$ or $\ln 27$, i.e. independent of $k$ |

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## Question 7b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{2}{3(2x-k)^2}\,dx = \frac{B}{(2x-k)}$ | M1 | |
| Substituting $2k$ and $k$ into $\frac{B}{(2x-k)}$ and subtracting | M1 | Substituting $2k$ and $k$ into their $\frac{B}{(2x-k)}$ and subtracting |
| $= \frac{-1}{3(3k)} - \frac{-1}{3k} = \frac{2}{9k}$, which is of the form $\frac{C}{k}$ with $C = \frac{2}{9}$ | A1 | Shows it is inversely proportional to $k$ |

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## Question 8a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $D(90) = 12 + 9\sin\!\left(\frac{360(90)}{365} - 63.435\right) = 15.851$ hours | B1 | Finding $D(90) = 15.85\ldots$ hours |

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## Question 8b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $17 = 12 + 9\sin\!\left(\frac{360t}{365} - 63.435\right)$ proceeding to $\sin\!\left(\frac{360t}{365} - 63.435\right)^{\circ} = k$, $|k| \leq 1$ | M1 | |
| $\sin\!\left(\frac{360t}{365} - 63.435\right)^{\circ} = \frac{5}{9}$; correct order to find one correct value of $t$ | A1 | |
| Using correct order to find a second value of $t$ | M1 | |
| $t = 98.5 \approx 99$ days and $t = 212.598 \approx 213$ days | A1 | |

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## Question 9a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d}{dx}(x^2 + y^3 - 10x - 12y - 5) = 2x + 3y^2\frac{dy}{dx} - 10 - 12\frac{dy}{dx} = 0$ | M1 | Either $y^3 \to Ay^2\frac{dy}{dx}$ or $12y \to 12\frac{dy}{dx}$ |
| $2x + 3y^2\frac{dy}{dx} - 10 - 12\frac{dy}{dx} = 0$ | A1 | Fully correct derivative |
| $\frac{dy}{dx}(3y^2 - 12) = 10 - 2x$ | M1 | Valid attempt making $\frac{dy}{dx}$ the subject with two terms in $\frac{dy}{dx}$ from $x^2$ and $2y$ |
| $\frac{dy}{dx} = \frac{10-2x}{3y^2-12}$ | A1 | With no errors |

## Question 9b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $3y^2 - 12 = 0$ and proceeds to find $y$ | M1 | |
| $y = 2$ | A1 | From diagram, $y$ is positive at $Q$ |
| Substitutes $y = 2$ in $x^2 + y^3 - 10x - 12y - 5 = 0$ giving $x^2 - 10x - 21 = 0$ | M1 | Creates 3-term quadratic and solves |
| $Q(5 + \sqrt{46}, 2)$ | A1 | $x = 5 \pm \sqrt{46}$, positive value taken from diagram |

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## Question 10a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{30}\cos\frac{\pi}{6}t \; dt = A\sin\frac{\pi}{6}t + c$ | M1 | |
| $\frac{1}{5\pi}\sin\frac{\pi}{6}t + c$ | A1 | Correct integration |

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## Question 10b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Separates variables: $\int \frac{1}{X} \, dX = \int \frac{1}{30}\cos\frac{\pi}{6}t \; dt$ | M1 | Separation of variables |
| $\ln X = \frac{1}{5\pi}\sin\frac{\pi}{6}t + c$ | A1 | Integrates both sides |
| Rearranges to $X = ke^{\frac{1}{5\pi}\sin(\frac{\pi}{6}t)}$ | M1 | |
| Substitutes $t=0, X=6$ | M1 | |
| $k = 6$ | A1 | |

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## Question 10c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Maximum of $\frac{1}{5\pi}\sin\frac{\pi}{6}t = \frac{1}{5\pi}$, so $X = 6e^{\frac{1}{5\pi}} = 6.3944$ | B1 | Shows maximum height is 6.39 m |

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## Question 10d:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin\left(\frac{\pi}{6}t\right) = 1$, so $\frac{\pi}{6}t = \frac{\pi}{2} + 2k\pi, \; k \in \mathbb{Z}$; identifies second maximum when $\frac{\pi}{6}t = \frac{5\pi}{2}$ | M1 | |
| $t = 15$ | A1 | |

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## Question 11a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Takes out factor $4^{-\frac{1}{2}} = \frac{1}{2}$: $(4-x)^{-\frac{1}{2}} = \frac{1}{2}\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}}$ | B1 | |
| Binomial expansion with $n = -\frac{1}{2}$ and term $-\frac{x}{4}$ | M1 | Condone sign slips |
| Any unsimplified correct expansion | M1 | Bracketing must be correct |
| $\frac{1}{2} + \frac{x}{16} + \frac{3x^2}{256} + \cdots$ | A1 | Correct solution only |

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## Question 11c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to multiply $(1+2x)^{\frac{1}{2}} \times (4-x)^{-\frac{1}{2}}$ | B1 | |
| $(1+2x)^{\frac{1}{2}} = 1 + x - \frac{1}{2}x^2 + \cdots$; multiplies two expansions to quadratic | M1 | |
| $f(x) = \frac{1}{2} + \frac{9x}{16} - \frac{45x^2}{256} + \cdots$, hence $A = \frac{45}{256}$ | A1 | |