Question 2 - A-Level Maths

Edexcel C34 2014 June — Question 2 6 marks

Exam Board	Edexcel
Module	C34 (Core Mathematics 3 & 4)
Year	2014
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Implicit equations and differentiation
Type	Find tangent equation at point
Difficulty	Standard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate both sides with respect to x, substitute the given point to find dy/dx, then write the tangent equation. While implicit differentiation is a step above basic differentiation, this is a standard textbook exercise with clear steps and no conceptual surprises, making it slightly easier than average.
Spec	1.07s Parametric and implicit differentiation

2. A curve $C$ has the equation $$x ^ { 3 } - 3 x y - x + y ^ { 3 } - 11 = 0$$ Find an equation of the tangent to $C$ at the point $( 2 , - 1 )$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.

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Question 2:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$3x^2 - \left(3y + 3x\frac{dy}{dx}\right) - 1 + 3y^2\frac{dy}{dx} = 0$	M1	Differentiates implicitly to include either $\pm ky^2\frac{dy}{dx}$ or $\pm 3x\frac{dy}{dx}$
A1	$x^3 \to 3x^2$ and $-x + y^3 - 11 \to -1 + 3y^2\frac{dy}{dx}$; $= 0$ needed here or implied
M1	Product rule applied: $-3xy \to -\left(3y + 3x\frac{dy}{dx}\right)$ or $\pm 3y \pm 3x\frac{dy}{dx}$
$\left\{\frac{dy}{dx} = \frac{3x^2 - 3y - 1}{3x - 3y^2}\right\}$ (not necessarily required)
At $(2,-1)$: $m(\mathbf{T}) = \frac{dy}{dx} = \frac{3(2)^2 - 3(-1) - 1}{3(2) - 3(-1)^2} = \frac{14}{3}$	M1	Correct method to collect two (not three) $dy/dx$ terms and evaluate gradient at $x=2$, $y=-1$
$\mathbf{T}: y - -1 = \frac{14}{3}(x - 2)$	dM1	Dependent on all previous M marks; uses line equation with their $\frac{14}{3}$; may use $y = \frac{14}{3}x + c$ and substitute $x=2$, $y=-1$
$\mathbf{T}: 14x - 3y - 31 = 0$ or equivalent	A1	Any positive or negative whole number multiple acceptable; must have $= 0$

# Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $3x^2 - \left(3y + 3x\frac{dy}{dx}\right) - 1 + 3y^2\frac{dy}{dx} = 0$ | M1 | Differentiates implicitly to include either $\pm ky^2\frac{dy}{dx}$ **or** $\pm 3x\frac{dy}{dx}$ |
| | A1 | $x^3 \to 3x^2$ **and** $-x + y^3 - 11 \to -1 + 3y^2\frac{dy}{dx}$; $= 0$ needed here or implied |
| | M1 | Product rule applied: $-3xy \to -\left(3y + 3x\frac{dy}{dx}\right)$ or $\pm 3y \pm 3x\frac{dy}{dx}$ |
| $\left\{\frac{dy}{dx} = \frac{3x^2 - 3y - 1}{3x - 3y^2}\right\}$ (not necessarily required) | | |
| At $(2,-1)$: $m(\mathbf{T}) = \frac{dy}{dx} = \frac{3(2)^2 - 3(-1) - 1}{3(2) - 3(-1)^2} = \frac{14}{3}$ | M1 | Correct method to collect **two (not three)** $dy/dx$ terms and evaluate gradient at $x=2$, $y=-1$ |
| $\mathbf{T}: y - -1 = \frac{14}{3}(x - 2)$ | dM1 | Dependent on all previous M marks; uses line equation with their $\frac{14}{3}$; may use $y = \frac{14}{3}x + c$ and substitute $x=2$, $y=-1$ |
| $\mathbf{T}: 14x - 3y - 31 = 0$ or equivalent | A1 | Any positive or negative whole number multiple acceptable; must have $= 0$ |

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2. A curve $C$ has the equation

$$x ^ { 3 } - 3 x y - x + y ^ { 3 } - 11 = 0$$

Find an equation of the tangent to $C$ at the point $( 2 , - 1 )$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\

\hfill \mbox{\textit{Edexcel C34 2014 Q2 [6]}}

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