## Question 8(a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = 2$ | B1 | Correct value; condone $Q=2$ |

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## Question 8(b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $(2.5,\ 1.5)$ | B1 | Allow stated separately as $x=2.5,\ y=1.5$ |

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## Question 8(c):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}}$ | M1 | Attempts to differentiate both parameters and uses $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ to achieve $\frac{1\pm\frac{1}{t^2}}{1\pm\frac{1}{t^2}}$ or equivalent via quotient rule |
| $t=2 \Rightarrow \frac{dy}{dx} = \frac{5}{3} \Rightarrow y-1.5 = -\frac{3}{5}(x-2.5)$ | dM1 | Uses $t=2$ in $\frac{dy}{dx}$ and uses negative reciprocal gradient with point $P$ to form normal; dependent on previous M |
| $3x + 5y = 15$ | A1* | Obtains printed answer with no errors and sufficient working shown |

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## Question 8(d):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $l$ crosses $x$-axis at $x=5$ | B1 | Correct $x$-intercept for $l$; may appear on figure or implied by cone volume calculation |
| $\frac{1}{3}\pi \times 1.5^2(5-2.5) = \frac{15}{8}\pi$ | M1 | Fully correct method for cone volume $\frac{1}{3}\pi \times ``1.5"^2(``5"-``2.5")$ |
| $V = \pi\int y^2\,dx = \pi\int y^2\frac{dx}{dt}\,dt = \pi\int\left(t-\frac{1}{t}\right)^2\left(1-\frac{1}{t^2}\right)dt$ | M1 | Correct strategy for volume under curve; condone sign slip on $\frac{dx}{dt}$; $\pi$ may be added later |
| $\pi\int\left(t^2-2+\frac{1}{t^2}\right)\left(1-\frac{1}{t^2}\right)dt = \pi\int\left(t^2+\frac{3}{t^2}-\frac{1}{t^4}-3\right)dt = \ldots$ | M1 | Attempt to expand $\left(t-\frac{1}{t}\right)^2\left(1\pm\frac{k}{t^2}\right)$ and integrate term by term; look for polynomial in $t$ with positive and negative indices |
| $\pi\left[\frac{t^3}{3} - \frac{3}{t} + \frac{1}{3t^3} - 3t\right]$ | A1 | Correct integration; does not need to be simplified |
| $\pi\left[\frac{t^3}{3}-\frac{3}{t}+\frac{1}{3t^3}-3t\right]_1^2 = \pi\left(\left(\frac{8}{3}-\frac{3}{2}+\frac{1}{24}-6\right)-\left(\frac{1}{3}-3+\frac{1}{3}-3\right)\right)$ | M1 | Applies $t$ limits 1 and 2 to an attempt at the integral |
| $V = \frac{13}{24}\pi + \frac{15}{8}\pi = \frac{29}{12}\pi$ | A1 | $\frac{29}{12}\pi$ or exact equivalent |

## Question (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = t + \frac{1}{t},\ y = t - \frac{1}{t} \Rightarrow x^2 - y^2 = 4 \Rightarrow 2x - 2y\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ **or** $y^2 = x^2 - 4 \Rightarrow y = \left(x^2 - 4\right)^{\frac{1}{2}} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = x\left(x^2 - 4\right)^{-\frac{1}{2}}$ | M1 | Attempts to differentiate a Cartesian equation of the form $x^2 - y^2 = k$ to obtain either $\alpha x + \alpha y\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ or $\frac{\mathrm{d}y}{\mathrm{d}x} = kx\left(x^2-4\right)^{-\frac{1}{2}}$ |
| $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{y} = \frac{t + \frac{1}{t}}{t - \frac{1}{t}} = \frac{2.5}{1.5}$ or $\frac{\mathrm{d}y}{\mathrm{d}x} = 2.5\left(2.5^2 - 4\right)^{-\frac{1}{2}}$ $\Rightarrow y - 1.5 = -\frac{3}{5}(x - 2.5)$ | M1 | Uses $t = 2$ in their $\frac{\mathrm{d}y}{\mathrm{d}x}$ and uses the negative reciprocal gradient with their point $P$. If using explicit differentiation, must attempt $x$ using $t=2$ first |
| $3x + 5y = 15$ | A1* | Obtains the printed answer with no errors and sufficient working shown |
| | **(3)** | |

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## Question (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $l$ crosses $x$-axis at $x = 5$ | B1 | |
| $\frac{1}{3}\pi \times 1.5^2(5 - 2.5)\left(= \frac{15}{8}\pi\right)$ | M1 | Fully correct method for the cone volume |
| $V = \pi\int y^2\,\mathrm{d}x = \pi\int\left(x^2 - 4\right)\mathrm{d}x$ | M1 | Correct strategy for the other volume. Condone $V = \pi\int y^2\,\mathrm{d}x = \pi\int\left(x^2 - k\right)\mathrm{d}x$ |
| $\pi\int\left(x^2 - 4\right)\mathrm{d}x = \pi\left[\frac{x^3}{3} - 4x\right]$ | M1A1 | M1: Attempt to integrate $\int\left(x^2 - k\right)\mathrm{d}x$; A1: Correct integration of $\int\left(x^2 - 4\right)\mathrm{d}x$ |
| $\pi\left[\frac{x^3}{3} - 4x\right]_{2}^{2.5}$ | M1 | Applies the limits "2" and their $x$ from $t = 2$ to an attempted integral of $(\pi)\int\left(x^2 - k\right)\mathrm{d}x$ |
| $V = \frac{13}{24}\pi + \frac{15}{8}\pi = \frac{29}{12}\pi$ | A1 | Cao |
| | **(7)** | |