OCR MEI Paper 2 2018 June — Question 15 9 marks

Exam BoardOCR MEI
ModulePaper 2 (Paper 2)
Year2018
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind dy/dx at a point
DifficultyStandard +0.8 This question requires implicit differentiation with a product rule and a fractional power, then solving a cubic equation to find x-values where y=1, and finally substituting to find gradients at multiple points. The combination of algebraic manipulation, implicit differentiation technique, and solving the resulting cubic makes this moderately harder than a standard implicit differentiation question.
Spec1.07s Parametric and implicit differentiation

15 You must show detailed reasoning in this question. The equation of a curve is $$y ^ { 3 } - x y + 4 \sqrt { x } = 4 .$$ Find the gradient of the curve at each of the points where \(y = 1\).

Question 15:
AnswerMarks Guidance
substitution of \(y = 1\)M1 AO 1.1a
\(x - 4\sqrt{x} + 3 = 0\) or \(4\sqrt{x} = x + 3\); \(x = 1\) or \(9\)A1 AO 2.1
\(3y^2\frac{dy}{dx}\)A1 AO 1.1
\(-x\times\frac{dy}{dx} - y\) or \(x\times\frac{dy}{dx} + y\)B1 AO 3.1a
\(3y^2\frac{dy}{dx} - x\frac{dy}{dx} - y + \frac{2}{\sqrt{x}} [= 0]\); substitution of \(y=1\) and their \(x=1\) or their \(x=9\)M1 AO 2.1
\(m = -\frac{1}{2}\) [at \((1,1)\)]A1 AO 1.1
\(m = -\frac{1}{18}\) [at \((1,9)\)]M1, A1, A1 [9] AO 1.1
## Question 15:

substitution of $y = 1$ | M1 | AO 1.1a |

$x - 4\sqrt{x} + 3 = 0$ or $4\sqrt{x} = x + 3$; $x = 1$ or $9$ | A1 | AO 2.1 |

$3y^2\frac{dy}{dx}$ | A1 | AO 1.1 |

$-x\times\frac{dy}{dx} - y$ or $x\times\frac{dy}{dx} + y$ | B1 | AO 3.1a |

$3y^2\frac{dy}{dx} - x\frac{dy}{dx} - y + \frac{2}{\sqrt{x}} [= 0]$; substitution of $y=1$ and their $x=1$ or their $x=9$ | M1 | AO 2.1 | allow one sign error

$m = -\frac{1}{2}$ [at $(1,1)$] | A1 | AO 1.1 | dependent on at least two terms correct on LHS following differentiation; allow following wrong rearrangement after differentiating

$m = -\frac{1}{18}$ [at $(1,9)$] | M1, A1, A1 [9] | AO 1.1 | allow $-0.05555$...to 2 sf or better

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15 You must show detailed reasoning in this question.

The equation of a curve is

$$y ^ { 3 } - x y + 4 \sqrt { x } = 4 .$$

Find the gradient of the curve at each of the points where $y = 1$.

\hfill \mbox{\textit{OCR MEI Paper 2 2018 Q15 [9]}}