### 1(a)(i)
Correct general shape (not multiple-valued, not straight, negative gradient throughout) relative to axes. Reasonably vertical at ends. Correct domain (labelled at −1 and 1). Correct range (labelled at π). Correct y-intercept (labelled at π/2) | B1, B1 | SC B1B0 for a fully correct curve in [−1,1] × [0,π] but multiple-valued

### 1(a)(ii)
$\cos y = x \Rightarrow -\sin y \frac{dy}{dx} = 1$

$\Rightarrow \frac{dy}{dx} = \frac{-1}{\sin y}$

$\sin^2 y + \cos^2 y = 1 \Rightarrow \sin y = (\pm)\sqrt{1-x^2}$

$\Rightarrow \frac{dy}{dx} = \pm\frac{1}{\sqrt{1-x^2}}$ or $\frac{-1}{\sqrt{1-x^2}}$

Taking − sign because gradient is negative | M1, A1(ag), B1 | Differentiating w.r.t. x or y | $\frac{dx}{dy} = -\sin y$; Completion www with intermediate step dependent on B1 below; Validly rejecting + sign. Dependent on A1 above | Or $0 \le y \le \pi \Rightarrow \sin y \ge 0 \Rightarrow \frac{dy}{dx} \le 0$ Or f(x) is decreasing

### 1(a)(iii)
$f(x) = \arccos x$

$\Rightarrow f'(x) = -(1-x^2)^{-\frac{1}{2}}$

$\Rightarrow f''(x) = \frac{1}{2}(1-x^2)^{-\frac{3}{2}} \times -2x = -x(1-x^2)^{-\frac{3}{2}}$

$\Rightarrow f'''(x) = -(1-x^2)^{-\frac{3}{2}} - x \times -\frac{3}{2}(1-x^2)^{-\frac{5}{2}} \times -2x$

$= -(1-x^2)^{-\frac{3}{2}} - 3x^2(1-x^2)^{-\frac{5}{2}}$

$\Rightarrow f(0) = \frac{\pi}{2}$

$f'(0) = -1, f''(0) = 0, f'''(0) = -1$

$\Rightarrow f(x) = \frac{\pi}{2} - x - \frac{x^3}{6} + ...$ | M1, A1, M1, A1, B1, B1B1 | Derivative in the form $kx(1-x^2)^{-\frac{3}{2}}$ o.e. | Any correct form www; Differentiating $f''(x)$ using product or quotient and chain rules. Dep. on 1st M1; Any correct form www; As first term of expansion; −x www., $\frac{x^3}{6}$ www | Allow a clear explanation that only the first term contributes to McLaurin expansion for 7/7; Incorrect simplification above loses the last B1

### 1(b)(i)
$r = \theta + \sin \theta$

$\Rightarrow \frac{dr}{d\theta} = 1 + \cos \theta$

$\cos \theta \ge -1 \Rightarrow \frac{dr}{d\theta} \ge 0$, so r increases as θ increases | B1, B1 | Do not condone > | $\frac{dr}{d\theta} \ge 0$ stated. Dependent on first B1 | No wrong statements

One complete revolution with r(0) = 0 and r(2π) ≥ r(3π/2) ≥ r(π) ≥ r(π/2) > 0. Correct general shape with two complete revolutions | B1, B1 | Independent. Condone r(0) > 0 for B0B1

### 1(b)(ii)
$\text{Area} = \frac{1}{2}\int_0^a r^2 d\theta = \frac{1}{2}\int_0^a (\theta + \sin \theta)^2 d\theta$

For small θ, $\sin \theta \approx \theta \Rightarrow r \approx 2\theta$

$\text{Area} \approx \frac{1}{2}\int_0^a (2\theta)^2 d\theta = \frac{1}{2}\left[\frac{4}{3}\theta^3\right]_0^a = \frac{2}{3}a^3$ | M1, M1, A1 | Forming an integral expression in θ for the required area | Using $\sin \theta \approx \theta$ and a complete method for integrating their expression | Condone only omitted limits or $\frac{1}{2}$; Dependent on first M1

---