OCR MEI Paper 2 2024 June — Question 13 9 marks

Exam BoardOCR MEI
ModulePaper 2 (Paper 2)
Year2024
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind stationary points
DifficultyStandard +0.8 This requires implicit differentiation to find dy/dx, setting it to zero, then solving the resulting system of equations (the original curve equation plus dy/dx = 0) simultaneously. While implicit differentiation itself is standard, solving the coupled system algebraically to find exact coordinates of turning points requires careful manipulation and is more demanding than routine differentiation questions.
Spec1.07s Parametric and implicit differentiation

13 Determine the coordinates of the turning points on the curve with equation $$y ^ { 2 } + x y + x ^ { 2 } - x = 1 .$$

Question 13:
AnswerMarks Guidance
\(2y\dfrac{dy}{dx}\)B1 AO 1.1
\(y + x\dfrac{dy}{dx}\)B1 AO 3.1a
(their previous terms) \(+ 2x - 1 = 0\)B1 AO 1.1
their \(y + 2x - 1 = 0\)M1 AO 2.1
\((1-2x)^2 + x(1-2x) + x^2 - x = 1\) or \(y^2 + \dfrac{(1-y)y}{2} + \dfrac{(1-y)^2}{4} - \dfrac{1-y}{2} = 1\)M1 AO 3.1a
\(3x^2 - 4x [= 0]\) or \(3y^2 + 2y - 5 [= 0]\)A1 AO 1.1
\(x = 0,\ x = \dfrac{4}{3}\)M1 AO 1.1
\(y = 1,\ y = -\dfrac{5}{3}\)M1 AO 1.1
\((0, 1)\) and \(\left(\dfrac{4}{3}, -\dfrac{5}{3}\right)\) or \(x = 0,\ y = 1\) and \(x = \dfrac{4}{3},\ y = -\dfrac{5}{3}\)A1 AO 3.2a
## Question 13:

$2y\dfrac{dy}{dx}$ | B1 | AO 1.1 | chain rule

$y + x\dfrac{dy}{dx}$ | B1 | AO 3.1a | product rule

(their previous terms) $+ 2x - 1 = 0$ | B1 | AO 1.1 | may award if "$= 0$" seen later, but not if RHS is $\dfrac{dy}{dx}$

their $y + 2x - 1 = 0$ | M1 | AO 2.1 | substitution of $\dfrac{dy}{dx} = 0$; dependent on award of at least one B mark; **NB** $\dfrac{dy}{dx} = \dfrac{1-2x-y}{2y+x}$

$(1-2x)^2 + x(1-2x) + x^2 - x = 1$ **or** $y^2 + \dfrac{(1-y)y}{2} + \dfrac{(1-y)^2}{4} - \dfrac{1-y}{2} = 1$ | M1 | AO 3.1a | elimination of $x$ or $y$ using expression or value obtained from use of $\dfrac{dy}{dx} = 0$; dependent on award of at least one B mark

$3x^2 - 4x [= 0]$ **or** $3y^2 + 2y - 5 [= 0]$ | A1 | AO 1.1 |

$x = 0,\ x = \dfrac{4}{3}$ | M1 | AO 1.1 | values of $x$ **or** $y$ found from their quadratic

$y = 1,\ y = -\dfrac{5}{3}$ | M1 | AO 1.1 | 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 $\dfrac{1}{3}$ from substitution into original equation

$(0, 1)$ and $\left(\dfrac{4}{3}, -\dfrac{5}{3}\right)$ **or** $x = 0,\ y = 1$ and $x = \dfrac{4}{3},\ y = -\dfrac{5}{3}$ | A1 | AO 3.2a | **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
13 Determine the coordinates of the turning points on the curve with equation

$$y ^ { 2 } + x y + x ^ { 2 } - x = 1 .$$

\hfill \mbox{\textit{OCR MEI Paper 2 2024 Q13 [9]}}