Higher June 2019 Paper 2 Q18
18. A cone has a radius \(x\) cm, a perpendicular height \((x + 2)\) cm and a slant height 16·4 cm.
Diagram not drawn to scale
Show that \(x\) satisfies the equation \(x^2 + 2x - 132{\cdot}48 = 0\) and calculate the volume of the cone.
You must show all your working. [7]
| Scheme | Marks | Comments |
|---|---|---|
| Showing \(x^2 + 2x - 132.48 = 0\) or \(2x^2 + 4x - 264.96 = 0\) | B2 | B1 for \(x^2 + (x + 2)^2 = 16.4^2\) |
| \(x = \dfrac{-4 \pm \sqrt{(4^2 - 4 \times 2 \times -264.96)}}{2 \times 2}\) or \(x = \dfrac{-2 \pm \sqrt{(2^2 - 4 \times 1 \times -132.48)}}{2 \times 1}\) or \((x + 1)^2 - 133.48 = 0\) | M1 | FT for equivalent level of difficulty Allow 1 slip in substitution, not use of incorrect formula |
| \(x = \dfrac{-4 \pm \sqrt{2135.68}}{4}\) or \(x = \dfrac{-2 \pm \sqrt{533.92}}{2}\) or \(x + 1 = \sqrt{133.48}\) | A1 | Either negative \(x\)-value not given or ignored in further working |
| \(x = 10.55\ldots\)(cm) or 10.6(cm) | A1 | Candidate must not show working with negative \(x\)-value |
| (Volume =) \(\frac13 \times \pi \times 10.55\ldots^2 \times 12.55\ldots\) | M1 | FT provided at least 2 marks previously awarded |
| Answer in the range 1462 (cm3) to 1483 (cm3) | A1 | Must be from correct working FT for an answer in a similar range, not allowing truncation of ‘their \(x\)’ to a whole number or 1 d.p. |
| (7) |