## Question 9(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin t \sin 2t = 0$ oe seen | M1 | **NB** $t = 0,\ \frac{1}{2}\pi,\ \pi$ |
| $(0,0)$, $(1,0)$ and $(2,0)$ or $x=0,\ x=1,\ x=2$ cao | A2 | **A1** for two of three correct; deduct 1 mark if all three correct plus extra values; if **A0**, allow **SC1** for $t = 0,\ \frac{1}{2}\pi,\ \pi$; if unsupported, full marks for all three values correct |

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## Question 9(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[\frac{dy}{dt}\right] = 2\sin t\cos 2t + \cos t\sin 2t$ | B1 | or $4\sin t\cos^2 t - 2\sin^3 t$ |
| $\frac{(2\sin t\cos 2t + \cos t\sin 2t)}{\sin t}$ or $\frac{(4\sin t\cos^2 t - 2\sin^3 t)}{\sin t}$ | M1 | Allow sign errors and/or one incorrect coefficient |
| Substitution of $\sin 2t = 2\sin t\cos t$ in their expression | M1 | May be seen before differentiation |
| $\frac{(2\sin t\cos 2t + \cos t\sin 2t)}{\sin t}$ and completion to $2\cos 2t + 2\cos^2 t$ www **NB AG** | A1 | At least one correct intermediate step needed |
| e.g. $2(2\cos^2 t - 1) + 2\cos^2 t = 0$ or $2\cos 2t + 2 \times \frac{1}{2}(1 + \cos 2t) = 0$ | M1 | Use of double angle formula to obtain quadratic equation in e.g. $\cos t$ or linear equation in $\cos 2t$; may be seen before differentiation; mark intent: allow sign error, bracket error, omission of one coefficient |
| $\left(1 + \frac{1}{\sqrt{3}},\ \frac{-4}{3\sqrt{3}}\right)$ oe isw | A1 | e.g. $\left(\frac{\sqrt{3}+3}{3},\ \frac{-4\sqrt{3}}{9}\right)$ |
| $\left(1 - \frac{1}{\sqrt{3}},\ \frac{4}{3\sqrt{3}}\right)$ oe isw | A1 | If **A0, A0**, allow **A1** for both $x$ values correct |

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## Question 9(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 2(1-\cos^2 t)\cos t$ oe; may be implicit equation, may be implied by partial substitution for $\cos t$, e.g. $(1-x)^2 + \frac{y}{2\cos t} = 1$ | M1 | or $\frac{dy}{dx} = 6\cos^2 t - 2$; use of double angle formula (and Pythagoras) to obtain expression for $y$ and $\frac{dy}{dx}$ in terms of $\cos t$ only |
| $y = 2(1-(1-x)^2)(1-x)$ | M1 | or $\frac{dy}{dx} = 6(1-x)^2 - 2$; substitution of $\cos t = \pm 1 \pm x$ to obtain expression in terms of $x$ only; allow sign errors, bracket errors or minor slips in arithmetic |
| $y = 2x^3 - 6x^2 + 4x$ or $y = 2x(x^2-3x+2)$ or $y = 2x(x-1)(x-2)$ oe cao | A1 | Integration and substitution of e.g. $(0,0)$ to obtain correct answer; must see $y =$ at some stage for **A1** |

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## Question 9(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Cubic with two turning points and of correct orientation through $(0,0)$ | M1 | |
| $x$-intercepts correct and only for $0 \leq x \leq 2$ | A1 | |

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