Higher January 2023 Paper 2R Q24
24 The diagram shows a solid cone and a solid sphere.
Diagram NOT accurately drawn
The cone has base radius \(r\), slant height \(l\) and perpendicular height \(h\)
The sphere has radius \(r\)
The base radius of the cone is equal to the radius of the sphere.
Given that
\(k \times\) volume of the cone = volume of the sphere
show that the total surface area of the cone can be written in the form
\(\pi r^2\left(\dfrac{k + \sqrt{k^2 + a}}{k}\right)\)
where \(a\) is a constant to be found.
(6)
| Scheme | Marks |
|---|---|
eg \(k \times \dfrac{1}{\cancel{3}}\cancel{\pi}r^2h = \dfrac{4}{\cancel{3}}\cancel{\pi}r^3\) or \(k \times \dfrac{1}{\cancel{3}}\pi\cancel{r^2}h = \dfrac{4}{\cancel{3}}\pi r^{\cancel{3}}\) or \(k \times \dfrac{1}{3}\cancel{\pi}\cancel{r^2}h = \dfrac{4}{3}\cancel{\pi}r^{\cancel{3}}\) or \(k \times h = 4r\) | M1 |
| \(h = \dfrac{4r}{k}\) | M1 |
| eg \(l^2 = r^2 + \left(\dfrac{4r}{k}\right)^2\) or \(l = \sqrt{r^2 + \left(\dfrac{4r}{k}\right)^2}\) | M1 |
eg \(l = r\sqrt{1 + \dfrac{16}{k^2}}\) or \(l = r\sqrt{\dfrac{k^2 + 16}{k^2}}\) or \(l = r\dfrac{\sqrt{k^2 + 16}}{k}\) | M1 |
| eg \(\pi r^2\left(\sqrt{1 + \dfrac{16}{k^2}} + 1\right)\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\pi r^2\left(\dfrac{k + \sqrt{k^2 + 16}}{k}\right)\) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for setting up an equation with volumes and some simplification (minimum of 2 terms simplified)
M1: for finding \(h\) in terms of \(r\) and \(k\) in its simplest form (may be seen at a later stage)
M1: for correct substitution into Pythagoras’ theorem (accept substitution of \(h = \dfrac{4\pi r}{\pi k}\))
M1: for rearranging and removing the \(r\) from the square root (may be seen at a later stage)
M1: for a correct expression for surface area in terms of \(r\) and \(k\) with \(\pi r^2\) removed as a factor