A2 June 2023 Paper 2 Q4
4 In this question you must show detailed reasoning.
The region \(R\) is bounded by the curve with equation \(y = \dfrac{1}{\sqrt{3x^2 - 3x + 1}}\), the \(x\)-axis and the lines with equations \(x = \dfrac{1}{2}\) and \(x = 1\) (see diagram). The units of the axes are cm.
A pendant is to be made out of a precious metal. The shape of the pendant is modelled as the shape formed when \(R\) is rotated by \(2\pi\) radians about the \(x\)-axis.
Find the exact value of the volume of precious metal required to make the pendant, according to the model. [4]
| Scheme | Marks | AO |
|---|---|---|
| DR \(V = \pi\displaystyle\int_{\frac{1}{2}}^{1}\left(\left(3x^2 - 3x + 1\right)^{-\frac{1}{2}}\right)^2\mathrm{d}x\) | M1 | 3.3 |
| \(V = \pi\displaystyle\int\frac{1}{3x^2 - 3x + 1}\,\mathrm{d}x = \frac{1}{3}\pi\int\frac{1}{x^2 - x + \frac{1}{3}}\,\mathrm{d}x\) \(= \dfrac{1}{3}\pi\displaystyle\int\frac{1}{\left(x - \frac{1}{2}\right)^2 - \frac{1}{4} + \frac{1}{3}}\,\mathrm{d}x = \frac{1}{3}\pi\int\frac{1}{\left(x - \frac{1}{2}\right)^2 + \frac{1}{12}}\,\mathrm{d}x\) | M1 | 2.2a |
| \(\displaystyle\int\frac{1}{\left(x - \frac{1}{2}\right)^2 + \frac{1}{12}}\,\mathrm{d}x = \left[\frac{1}{\sqrt{\frac{1}{12}}}\tan^{-1}\left(\frac{x - \frac{1}{2}}{\sqrt{\frac{1}{12}}}\right)\right]\) | A1 | 1.1 |
| \(= \dfrac{2}{\sqrt{3}}\pi\left[\tan^{-1}\left(\sqrt{3}\right) - \tan^{-1}(0)\right]\) \(= \dfrac{2}{\sqrt{3}}\pi\dfrac{\pi}{3} = \dfrac{2\sqrt{3}}{9}\pi^2\) so \(\dfrac{2\sqrt{3}}{9}\pi^2\ \text{cm}^3\) | A1 | 3.4 |
| [4] |
Notes
M1: (1st) Using \(V = \pi\int_a^b y^2\,\mathrm{d}x\) with limits. Accept squared out expression. Condone omission of \(\mathrm{d}x\)
M1: (2nd) Expressing the integral in completed square form, or \(\pi\displaystyle\int\frac{1}{3\left(x - \frac{1}{2}\right)^2 + \frac{1}{4}}\,\mathrm{d}x\) or \(\pi\displaystyle\int\frac{1}{\left(\sqrt{3}x - \frac{\sqrt{3}}{2}\right)^2 + \frac{1}{4}}\,\mathrm{d}x\)
A1: (1st) \(= \left[\frac{2}{\sqrt{3}}\pi\tan^{-1}\left(\sqrt{3}(2x - 1)\right)\right]\); may be equivalent based on their form and their substitution. Could see a substitution, e.g. \(u = \sqrt{3}\left(x - \frac{1}{2}\right)\) with \(\mathrm{d}u = \sqrt{3}\,\mathrm{d}x\)
A1: (2nd) oe. Do not penalise missing units