| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2010 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Show specific gradient value |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate implicitly to find dy/dx, substitute a point to verify zero gradient, then find another point and its tangent equation. While it involves multiple steps, each is routine application of standard techniques with no conceptual challenges beyond basic implicit differentiation. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| State \(2y\frac{dy}{dx}\) as derivative of \(y^2\), or equivalent | B1 | |
| State \(2y + 2x\frac{dy}{dx}\) as derivative of \(2xy\), or equivalent | B1 | |
| Substitute \(x = -2\) and \(y = 2\) and evaluate \(\frac{dy}{dx}\) | M1 | |
| Obtain zero correctly and make correct conclusion | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Substitute \(x = -2\) into given equation and solve | M1 | |
| Obtain \(y = -6\) correctly | A1 | |
| Obtain \(\frac{dy}{dx} = 2\) correctly | B1 | |
| Form the equation of the tangent at \((-2, -6)\) | M1 | |
| Obtain answer \(y = 2x - 2\) | A1 | [5] |
**(i)**
| State $2y\frac{dy}{dx}$ as derivative of $y^2$, or equivalent | B1 |
| State $2y + 2x\frac{dy}{dx}$ as derivative of $2xy$, or equivalent | B1 |
| Substitute $x = -2$ and $y = 2$ and evaluate $\frac{dy}{dx}$ | M1 |
| Obtain zero correctly and make correct conclusion | A1 | [4] |
**(ii)**
| Substitute $x = -2$ into given equation and solve | M1 |
| Obtain $y = -6$ correctly | A1 |
| Obtain $\frac{dy}{dx} = 2$ correctly | B1 |
| Form the equation of the tangent at $(-2, -6)$ | M1 |
| Obtain answer $y = 2x - 2$ | A1 | [5] |8 The equation of a curve is
$$x ^ { 2 } + 2 x y - y ^ { 2 } + 8 = 0$$
(i) Show that the tangent to the curve at the point $( - 2,2 )$ is parallel to the $x$-axis.\\
(ii) Find the equation of the tangent to the curve at the other point on the curve for which $x = - 2$, giving your answer in the form $y = m x + c$.
\hfill \mbox{\textit{CAIE P2 2010 Q8 [9]}}