# Question 1:

## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Fully correct curve (cardioid shape) | G2 | A fully correct curve; Give G1 for one error, e.g. incorrect form at O, lack of clear symmetry, sharp point at RH extremity |

## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Area} = \frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2\,d\theta$ | M1 | Integral expression involving $(1+\cos\theta)^2$ |
| $= \frac{1}{2}\int_0^{2\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta$ | A1 | Correct expanded integral expression, incl. limits. Limits may be implied by later work. Penalise missing $\frac{1}{2}$ here (max. 4/6) |
| $= \frac{1}{2}\int_0^{2\pi}\left(\frac{3}{2}+2\cos\theta+\frac{1}{2}\cos 2\theta\right)d\theta$ | M1 | Using $\cos^2\theta = \frac{1}{2}+\frac{1}{2}\cos 2\theta$. Allow sign or factor errors |
| $= \frac{1}{2}\left[\frac{3}{2}\theta+2\sin\theta+\frac{1}{4}\sin 2\theta\right]_0^{2\pi}$ | A2 | Correct result of integration. Give A1 for one error in this expression |
| $= \frac{3}{2}\pi$ | A1 | Dependent on previous A2 |

## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin x = x - \frac{1}{6}x^3\ldots$ and $\cos x = 1 - \frac{1}{2}x^2\ldots$ | B1 | Both series correct as far as second term. Ignore higher-order terms. Allow denominators left as $2!$, $3!$ |
| $\tan x \approx \left(x - \frac{1}{6}x^3\right)\left(1-\frac{1}{2}x^2\right)^{-1}$ | M1 | Using $\tan x = \frac{\sin x}{\cos x}$. Allow even if no further progress but must be used, not just stated |
| $= \left(x-\frac{1}{6}x^3\right)\left(1+\frac{1}{2}x^2+\ldots\right)$ | M1 | Using binomial expansion. Dependent on first M1. If methods mixed, mark to benefit of candidate |
| $= x + \frac{1}{2}x^3 - \frac{1}{6}x^3 + \ldots$ | M1 | Expanding brackets. Dependent on previous M1 |
| $= x + \frac{1}{3}x^3\ldots$ | A1A1 | $a=1$, $b=\frac{1}{3}$ correctly obtained. Dependent on both M1s. Deduct 1 for each additional term (*) |
| **OR** $\frac{x-\frac{1}{6}x^3}{1-\frac{1}{2}x^2} = ax+bx^3$ | M1 | Using $\tan x = \frac{\sin x}{\cos x}$ |
| $\Rightarrow x - \frac{1}{6}x^3 = \left(1-\frac{1}{2}x^2\right)(ax+bx^3) = ax+(b-\frac{1}{2}a)x^3+\ldots$ | M1 | Attempting to compare coefficients |
| $\Rightarrow a=1$ | A1 | Correctly obtained. As (*) |
| $b - \frac{1}{2}a = -\frac{1}{6}$ | M1 | Obtaining $b$ |
| $\Rightarrow b = \frac{1}{3}$ | A1 | Correctly obtained. As (*) |
| **OR** $f(x)=\tan x \Rightarrow f'(x)=\sec^2 x$; $f''(x)=2\sec^2 x\tan x$; $f'''(x)=4\sec^2 x\tan^2 x+2\sec^4 x$; $f(0)=0,\,f'(0)=1,\,f''(0)=0,\,f'''(0)=2$ | M1, M1 | Attempting first two derivatives; Attempting third derivative. Using the product rule |
| $f(x)=f(0)+xf'(0)+\frac{x^2 f''(0)}{2!}+\ldots$ | M1 | Applying Maclaurin series. Dependent on first M1 |
| $= x+\frac{1}{3}x^3\ldots$ | A1A1 | Correctly obtained. As (*) |

## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 \frac{1}{\sqrt{1-\frac{1}{4}x^2}}\,dx = \int_0^1 \frac{2}{\sqrt{4-x^2}}\,dx = \left[2\arcsin\frac{x}{2}\right]_0^1$ | M1 | arcsin alone, or any sine substitution |
| | A1 | $2$ and $\frac{x}{2}$ |
| $= 2\left(\frac{\pi}{6}-0\right)$ | M1 | Using limits. Dependent on first M1. No need to see explicit use of $x=0$. Limits wrong way round M0 |
| $= \frac{\pi}{3}$ | A1 | Evaluated in terms of $\pi$ |

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