June 2024 Paper 2 Q13
13 Determine the coordinates of the turning points on the curve with equation
\(y^2 + xy + x^2 - x = 1\). [9]
| Scheme | Marks | AO |
|---|---|---|
| \(2y\frac{\mathrm{d}y}{\mathrm{d}x}\) | B1 | 1.1 |
| \(y + x\frac{\mathrm{d}y}{\mathrm{d}x}\) | B1 | 3.1a |
| (their previous terms) \(+ 2x - 1 = 0\) | B1 | 1.1 |
| their \(y + 2x - 1 = 0\) | M1 | 2.1 |
| \((1-2x)^2 + x(1-2x) + x^2 - x = 1\) or \(y^2 + \frac{(1-y)y}{2} + \frac{(1-y)^2}{4} - \frac{1-y}{2} = 1\) | M1 | 3.1a |
| \(3x^2 - 4x\,[= 0]\) or \(3y^2 + 2y - 5\,[= 0]\) | A1 | 1.1 |
| \(x = 0, x = \frac{4}{3}\) | M1 | 1.1 |
| \(y = 1, y = -\frac{5}{3}\) | M1 | 1.1 |
| \((0, 1)\) and \(\left(\frac{4}{3}, -\frac{5}{3}\right)\) or \(x = 0, y = 1\) and \(x = \frac{4}{3}, y = -\frac{5}{3}\) | A1 | 3.2a |
| [9] |
Notes
B1: chain rule
B1: product rule
B1: may award if “= 0” seen later, but not if RHS is \(\frac{\mathrm{d}y}{\mathrm{d}x}\)
M1: substitution of \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\); may follow (incorrect) rearrangement; dependent on award of at least one B mark
NB \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1-2x-y}{2y+x}\)
M1: elimination of \(x\) or \(y\) using expression or value obtained from use of \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\); dependent on award of at least one B mark
M1: values of \(x\) or \(y\) found from their quadratic
M1: values of \(y\) or \(x\) found from substitution of both \(x\) or both \(y\) values; must see substitution unless values correct
NB may see extra points \(y = -1\) or \(\frac{1}{3}\) from substitution into original equation
A1: A0 if extra points in final answer; dependent on fully correct working throughout;
if M0M0 allow SCB1 for 1 correct pair of coordinates and no others
Alternatively
| Scheme | Marks | AO |
|---|---|---|
| \(2x\frac{\mathrm{d}x}{\mathrm{d}y} - \frac{\mathrm{d}x}{\mathrm{d}y}\) | B1 | |
| \(x + y\frac{\mathrm{d}x}{\mathrm{d}y}\) | B1 | |
| (their previous terms) \(+ 2y = 0\) | B1 | |
| their \(1 - 2x - y = 0\) | M1 | |
| \((1-2x)^2 + x(1-2x) + x^2 - x = 1\) or \(y^2 + \frac{(1-y)y}{2} + \frac{(1-y)^2}{4} - \frac{1-y}{2} = 1\) | M1 | |
| \(3x^2 - 4x\,[= 0]\) or \(3y^2 + 2y - 5\,[= 0]\) | A1 | |
| \(x = 0, x = \frac{4}{3}\) | M1 | |
| \(y = 1, y = -\frac{5}{3}\) | M1 | |
| \((0, 1)\) and \(\left(\frac{4}{3}, -\frac{5}{3}\right)\) or \(x = 0, y = 1\) and \(x = \frac{4}{3}, y = -\frac{5}{3}\) | A1 | |
| 9 |
B1: chain rule
B1: product rule
B1: may award if “= 0” seen later, but not if RHS is \(\frac{\mathrm{d}y}{\mathrm{d}x}\)
M1: from setting denominator of \(\frac{\mathrm{d}x}{\mathrm{d}y} = \frac{2y+x}{1-2x-y}\) equal to 0 or rearranging to find \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1-2x-y}{2y+x}\) and setting equal to 0; dependent on award of at least one B mark
M1: elimination of \(x\) or \(y\) using expression or value obtained from use of \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\); dependent on award of at least one B mark
M1: values of \(x\) or \(y\) found from their quadratic
M1: values of \(y\) or \(x\) found from substitution of both \(x\) or both \(y\) values; must see substitution unless values correct
NB may see extra points \(y = -1\) or \(\frac{1}{3}\) from substitution into original equation
A1: A0 if extra points in final answer; dependent on fully correct working throughout
if M0M0 allow SCB1 for 1 correct pair of coordinates and no others