| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2011 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Tangent with given gradient |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring standard technique to find dy/dx, then solving simultaneous equations (the curve equation plus dy/dx = -1). While it involves multiple steps, each step uses routine A-level methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(2y\frac{dy}{dx}\) as derivative of \(y^2\), or equivalent | B1 | |
| Equate derivative of LHS to zero and solve for \(\frac{dy}{dx}\) | M1 | |
| Obtain given answer correctly | A1 | [3] |
| (ii) Equate gradient expression to \(-1\) and rearrange | M1 | |
| Obtain \(y = 2x\) | A1 | |
| Substitute into original equation to obtain an equation in \(x^2\) (or \(y^2\)) | M1 | |
| Obtain \(2x^2 - 3x - 2 = 0\) (or \(y^2 - 3y - 4 = 0\)) | A1 | |
| Correct method to solve their quadratic equation | M1 | |
| State answers \((-\frac{1}{2}, -1)\) and \((2, 4)\) | A1 | [6] |
**(i)** State $2y\frac{dy}{dx}$ as derivative of $y^2$, or equivalent | B1 |
Equate derivative of LHS to zero and solve for $\frac{dy}{dx}$ | M1 |
Obtain given answer correctly | A1 | [3]
**(ii)** Equate gradient expression to $-1$ and rearrange | M1 |
Obtain $y = 2x$ | A1 |
Substitute into original equation to obtain an equation in $x^2$ (or $y^2$) | M1 |
Obtain $2x^2 - 3x - 2 = 0$ (or $y^2 - 3y - 4 = 0$) | A1 |
Correct method to solve their quadratic equation | M1 |
State answers $(-\frac{1}{2}, -1)$ and $(2, 4)$ | A1 | [6]8 The equation of a curve is $2 x ^ { 2 } - 3 x - 3 y + y ^ { 2 } = 6$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x - 3 } { 3 - 2 y }$.\\
(ii) Find the coordinates of the two points on the curve at which the gradient is - 1 .
\hfill \mbox{\textit{CAIE P2 2011 Q8 [9]}}