CAIE P2 2008 June — Question 7 9 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2008
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind stationary points
DifficultyStandard +0.3 This is a straightforward implicit differentiation question requiring standard technique to find dy/dx, then solving 2y-x=0 with the original equation to find stationary points. While it involves multiple steps, each is routine for P2 level with no novel insight required, making it slightly easier than average.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation

7 The equation of a curve is $$x ^ { 2 } + y ^ { 2 } - 4 x y + 3 = 0$$
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y - x } { y - 2 x }\).
  2. Find the coordinates of each of the points on the curve where the tangent is parallel to the \(x\)-axis.

AnswerMarks
(i) State \(2y\frac{dy}{dx}\) as derivative of \(y^2\), or equivalentB1
State \(4y + 4x\frac{dy}{dx}\) as derivative of \(4xy\), or equivalentB1
Equate derivative of LHS to zero and solve for \(\frac{dy}{dx}\)M1
Obtain given answer correctly [The M1 is dependent on at least one of the B marks being obtained.]A1
[4]
(ii) State or imply that the coordinates satisfy \(2y - x = 0\)B1
Obtain an equation in \(x^2\) (or \(y^2\))M1
Solve and obtain \(x^2 = 4\) (or \(y^2 = 1\))A1
State answer \((2, 1)\)A1
State answer \((-2, -1)\)A1
[5]
**(i)** State $2y\frac{dy}{dx}$ as derivative of $y^2$, or equivalent | B1 |
State $4y + 4x\frac{dy}{dx}$ as derivative of $4xy$, or equivalent | B1 |
Equate derivative of LHS to zero and solve for $\frac{dy}{dx}$ | M1 |
Obtain given answer correctly [The M1 is dependent on at least one of the B marks being obtained.] | A1 |
| [4] |

**(ii)** State or imply that the coordinates satisfy $2y - x = 0$ | B1 |
Obtain an equation in $x^2$ (or $y^2$) | M1 |
Solve and obtain $x^2 = 4$ (or $y^2 = 1$) | A1 |
State answer $(2, 1)$ | A1 |
State answer $(-2, -1)$ | A1 |
| [5] |
7 The equation of a curve is

$$x ^ { 2 } + y ^ { 2 } - 4 x y + 3 = 0$$

(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y - x } { y - 2 x }$.\\
(ii) Find the coordinates of each of the points on the curve where the tangent is parallel to the $x$-axis.

\hfill \mbox{\textit{CAIE P2 2008 Q7 [9]}}