October 2021 Paper 3 Q2
2
The diagram shows triangle \(ABC\) in which angle \(A\) is \(60^\circ\) and the lengths of \(AB\) and \(AC\) are \((4 + h)\,\mathrm{cm}\) and \((4 - h)\,\mathrm{cm}\) respectively.
| Scheme | Marks | AO |
|---|---|---|
| \(\left(p^2 =\right)(4 + h)^2 + (4 - h)^2 - 2(4 + h)(4 - h)\cos 60\) \(= \left(16 + 8h + h^2\right) + \left(16 - 8h + h^2\right) - \left(16 - h^2\right)\) | M1 | 1.1 |
| \(p^2 = 16 + 3h^2\) | A1 | 2.2a |
| [2] |
Notes
M1: Correct application of cosine rule
A1: AG – at least one line of intermediate working (must have \(p^2 = \ldots\))
Any errors or missing brackets then A0
| Scheme | Marks | AO |
|---|---|---|
| \(\left(16 + 3h^2\right)^{\frac{1}{2}} = 4(1 + \ldots)^{\frac{1}{2}}\) | B1 | 1.1 |
| \(\left(1 + kh^2\right)^{\frac{1}{2}} = 1 + \frac{1}{2}kh^2 + \ldots\) | M1 | 1.1 |
| \(\ldots + \dfrac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}\left(kh^2\right)^2\) | A1ft | 1.1 |
| \((p =)\,4 + \frac{3}{8}h^2 - \frac{9}{512}h^4 + \ldots\) | A1 | 1.1 |
| [4] |
Notes
B1: For reference: \(4\left(1 + \frac{3}{16}h^2\right)^{\frac{1}{2}}\)
or for \(16^{\frac{1}{2}}(1 + \ldots)^{\frac{1}{2}}\)
M1: Correct first two terms for their \(k\)
\(k \neq 1\)
A1ft: Correct third term following through their \(k\)
A1: \(\lambda = \frac{3}{8}, \mu = -\frac{9}{512}\) (oe)
SC if candidates assume that \(p = 4 + \lambda h^2 + \mu h^4\) and then substitute into \(p^2 = 16 + 3h^2\) to find \(\lambda\) and \(\mu\) then B1 for correct \(\lambda\) and B1 for correct \(\mu\) (so 2/4 max.)