Question 7 - A-Level Maths

CAIE P2 2014 June — Question 7 10 marks

Exam Board	CAIE
Module	P2 (Pure Mathematics 2)
Year	2014
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Implicit equations and differentiation
Type	Show no stationary points exist
Difficulty	Standard +0.3 This is a straightforward implicit differentiation question requiring standard techniques: (i) differentiate implicitly, substitute a point to find gradient, then write tangent equation; (ii) set dy/dx = 0 and show no real solutions exist. Both parts follow routine procedures with no novel insight required, making it slightly easier than average.
Spec	1.07m Tangents and normals: gradient and equations 1.07n Stationary points: find maxima, minima using derivatives 1.07s Parametric and implicit differentiation

7 The equation of a curve is $$2 x ^ { 2 } + 3 x y + y ^ { 2 } = 3$$

Find the equation of the tangent to the curve at the point $( 2 , - 1 )$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
Show that the curve has no stationary points.

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Answer	Marks	Guidance
(i) Obtain $3y + 3x\frac{d\Phi}{dx}$ as derivative of $3xy$	B1
Obtain $2y\frac{d\Phi}{dx}$ as derivative of $y^2$	B1
State $4x + 3y + 3x\frac{d\Phi}{dx} + 2y\frac{d\Phi}{dx} = 0$	B1
Substitute $2$ and $-1$ to find gradient of curve (dependent on at least one B1)	M1
Form equation of tangent through $(2, -1)$ with numerical gradient (dependent on previous M1)	DM1
Obtain $5x + 4y - 6 = 0$ or equivalent of required form	A1	[6]
(ii) Use $\frac{d\Phi}{dx} = 0$ to find relation between $x$ and $y$ (dependent on at least one B1 from part (i))	M1
Obtain $4x + 3y = 0$ or equivalent	A1
Substitute for $x$ or $y$ in equation of curve	M1
Obtain $-\frac{1}{8}y^2 = 3$ or $-\frac{3}{8}x^2 = 3$ or equivalent and conclude appropriately	A1	[4]

**(i)** Obtain $3y + 3x\frac{d\Phi}{dx}$ as derivative of $3xy$ | B1 |
Obtain $2y\frac{d\Phi}{dx}$ as derivative of $y^2$ | B1 |
State $4x + 3y + 3x\frac{d\Phi}{dx} + 2y\frac{d\Phi}{dx} = 0$ | B1 |
Substitute $2$ and $-1$ to find gradient of curve (dependent on at least one B1) | M1 |
Form equation of tangent through $(2, -1)$ with numerical gradient (dependent on previous M1) | DM1 |
Obtain $5x + 4y - 6 = 0$ or equivalent of required form | A1 | [6]

**(ii)** Use $\frac{d\Phi}{dx} = 0$ to find relation between $x$ and $y$ (dependent on at least one B1 from part (i)) | M1 |
Obtain $4x + 3y = 0$ or equivalent | A1 |
Substitute for $x$ or $y$ in equation of curve | M1 |
Obtain $-\frac{1}{8}y^2 = 3$ or $-\frac{3}{8}x^2 = 3$ or equivalent and conclude appropriately | A1 | [4]

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7 The equation of a curve is

$$2 x ^ { 2 } + 3 x y + y ^ { 2 } = 3$$

(i) Find the equation of the tangent to the curve at the point $( 2 , - 1 )$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\
(ii) Show that the curve has no stationary points.

\hfill \mbox{\textit{CAIE P2 2014 Q7 [10]}}

This paper (7 questions)

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Q1 4 Q2 6 Q3 6 Q4 8 Q5 8 Q6 8 Q7 10

Answer	Marks	Guidance
(i) Obtain \(3y + 3x\frac{d\Phi}{dx}\) as derivative of \(3xy\)	B1
Obtain \(2y\frac{d\Phi}{dx}\) as derivative of \(y^2\)	B1
State \(4x + 3y + 3x\frac{d\Phi}{dx} + 2y\frac{d\Phi}{dx} = 0\)	B1
Substitute \(2\) and \(-1\) to find gradient of curve (dependent on at least one B1)	M1
Form equation of tangent through \((2, -1)\) with numerical gradient (dependent on previous M1)	DM1
Obtain \(5x + 4y - 6 = 0\) or equivalent of required form	A1	[6]
(ii) Use \(\frac{d\Phi}{dx} = 0\) to find relation between \(x\) and \(y\) (dependent on at least one B1 from part (i))	M1
Obtain \(4x + 3y = 0\) or equivalent	A1
Substitute for \(x\) or \(y\) in equation of curve	M1
Obtain \(-\frac{1}{8}y^2 = 3\) or \(-\frac{3}{8}x^2 = 3\) or equivalent and conclude appropriately	A1	[4]