A2 October 2021 Paper 2 Q5
5. The curve \(C\) has equation
\[y = \arccos\left(\frac{1}{2}x\right) \qquad -2 \leqslant x \leqslant 2\]The normal to \(C\), at the point where \(x = 1\), crosses the \(x\)-axis at the point \(A\) and crosses the \(y\)-axis at the point \(B\).
Given that \(O\) is the origin,
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-\lambda}{\sqrt{1 - \beta x^2}}\) where \(\lambda \gt 0\) and \(\beta \gt 0\) and \(\beta \neq 1\) Alternatively \(2\cos y = x \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}y} = \alpha\sin y \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\alpha\sin y}\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-\frac{1}{2}}{\sqrt{1 - \frac{1}{4}x^2}}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-1}{2\sqrt{1 - \frac{1}{4}x^2}}\) o.e. or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{2\sin y}\) | A1 | 1.1b |
| States that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} \neq 0\) therefore \(C\) has no stationary points. Tries to solve \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) and ends up with a contradiction e.g. \(-1 = 0\) therefore \(C\) has no stationary points. As \(\operatorname{cosec} y \gt 1\) therefore \(C\) has no stationary points. | A1 | 2.4 |
| (3) |
Notes
(a)
M1: Finds the correct form for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
A1: Correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
A1: States or shows that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} \neq 0\) and draws the required conclusion. This mark can be scored as long as the M mark has been awarded.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-1}{2\sqrt{1 - \frac{1}{4} \times 1^2}} = \left\{-\dfrac{1}{\sqrt{3}}\right\}\) | M1 | 1.1b |
| Normal gradient \(= -\dfrac{1}{m}\) and \(y - \dfrac{\pi}{3} = m_n(x - 1)\) Alternatively \(\dfrac{\pi}{3} = m_n(1) + c \Rightarrow c = \ldots\left\{\dfrac{\pi}{3} - \sqrt{3}\right\}\) and then \(y = m_n x + c\) | M1 | 1.1b |
| \(y = 0 \Rightarrow 0 - \dfrac{\pi}{3} = \sqrt{3}(x_A - 1) \Rightarrow x_A = \ldots\left\{1 - \dfrac{\pi}{3\sqrt{3}} \text{ or } 1 - \dfrac{\pi\sqrt{3}}{9}\right\}\) and \(x = 0 \Rightarrow y_B - \dfrac{\pi}{3} = \sqrt{3}(0 - 1) \Rightarrow y_B = \ldots\left\{\dfrac{\pi}{3} - \sqrt{3}\right\}\) | M1 | 3.1a |
| Area \(= \dfrac{1}{2} \times x_A \times -y_B = \dfrac{1}{2}\left(1 - \dfrac{\pi}{3\sqrt{3}}\right)\left(\sqrt{3} - \dfrac{\pi}{3}\right)\) | M1 | 1.1b |
| Area \(\dfrac{1}{54}\left(27\sqrt{3} - 18\pi + \sqrt{3}\pi^2\right)\) \(\quad(p = 27,\ q = -18,\ r = 1)\) | A1 | 2.1 |
| (5) | ||
| (8 marks) |
Notes
(b)
M1: Substitutes \(x = 1\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
M1: Finds the normal gradient and finds the equation of the normal using \(y - \dfrac{\pi}{3} = m_n(x - 1)\)
M1: Finds where their normal cuts the \(x\)-axis and the \(y\)-axis.
M1: Finds the area of the triangle \(OAB = \dfrac{1}{2} \times x_A \times -y_B\).
A1: Correct area
Special case: If finds the tangent to the curve, the \(x\) and \(y\) intercepts and the area of the triangle max score M1 M0 M1 M0 A0
Note common error
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-1}{\sqrt{1 - \frac{1}{4}x^2}}\) In part (b) this leads to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-2}{\sqrt{3}}\) leading to normal gradient \(\dfrac{\sqrt{3}}{2}\) and
\(y = \dfrac{\sqrt{3}}{2}x - \dfrac{\sqrt{3}}{2} + \dfrac{\pi}{3}\) and \(\left(0, \dfrac{\pi}{3} - \dfrac{\sqrt{3}}{2}\right)\) and \(\left(1 - \dfrac{2\pi}{3\sqrt{3}}, 0\right)\) therefore area \(= \dfrac{1}{2}\left(\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{2}\right)\left(\dfrac{2\pi}{3\sqrt{3}} - 1\right)\)
This can score M1 M1 M1 M1 A0