June 2023 Paper 3 Q5
5 In this question you must show detailed reasoning.
This question is about the curve \(y = x^3 - 5x^2 + 6x\).
(a) Find the equation of the tangent, T, to the curve at the point \((0, 0)\). [3]
(b) Find the equation of the normal, N, to the curve at the point \((1, 2)\). [3]
(c) Find the coordinates of the point of intersection of T and N. [2]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 10x + 6\) | M1 | 1.1 |
| When \(x = 0\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6\) | M1 | 1.1 |
| Tangent goes through origin so equation is \(y = 6x\) cao | A1 | 1.1 |
| [3] |
Notes
M1: At least two terms correct
M1: FT their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
A1: Need reasoning eg \(y - 0 = 6(x - 0)\) or use of \(y = mx + c\)
\(y = 6x\) implies previous M mark
| Scheme | Marks | AO |
|---|---|---|
| DR When \(x = 1\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -1\) | M1 | 1.1 |
| Gradient of normal is 1 | M1 | 1.1 |
| [\((y-2) = (x-1)\) so] \(y = x + 1\) | A1 | 1.1 |
| [3] |
Notes
M1: FT their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) from (a)
M1: FT negative reciprocal of their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
A1: oe but constant terms should be collected
| Scheme | Marks | AO |
|---|---|---|
| DR \(6x = x + 1\) | M1 | 1.1 |
| \(x = \dfrac{1}{5},\ y = \dfrac{6}{5}\) oe | A1 | 1.1 |
| [2] |
Notes
M1: Eliminate a variable
FT their equations from (a) and (b) for M1 only