| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find vertical tangent points |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question with standard techniques. Part (i) is routine application of implicit differentiation, part (ii) requires setting the numerator to zero (impossible since y≠0 on the curve), and part (iii) involves setting the denominator to zero and solving simultaneously with the original equation. All steps are mechanical with no novel insight required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Obtain \(3y^2 \frac{dy}{dx}\) as derivative of \(y^3\) | B1 | |
| Obtain \(4y + 4x \frac{dy}{dx}\) as derivative of \(4xy\) | B1 | |
| Equate derivative of left-hand side to zero and solve for \(\frac{dy}{dx}\), must be from implicit differentiation | M1 | |
| Confirm given answer \(\frac{dy}{dx} = -\frac{4y}{3y^2 + 4x}\) correctly | A1 | [4] |
| (ii) State or imply \(y = 0\) | B1 | |
| Substitute in equation of curve and show contradiction | B1 | [2] |
| (iii) State or imply \(3y^2 + 4x = 0\) | B1 | |
| Eliminate one variable from equation of curve using \(3y^2 + 4x = 0\) | M1 | |
| Obtain \(y = -2\) | A1 | |
| Obtain \(x = -3\) | A1 | [4] |
(i) Obtain $3y^2 \frac{dy}{dx}$ as derivative of $y^3$ | B1 |
Obtain $4y + 4x \frac{dy}{dx}$ as derivative of $4xy$ | B1 |
Equate derivative of left-hand side to zero and solve for $\frac{dy}{dx}$, must be from implicit differentiation | M1 |
Confirm given answer $\frac{dy}{dx} = -\frac{4y}{3y^2 + 4x}$ correctly | A1 | [4]
(ii) State or imply $y = 0$ | B1 |
Substitute in equation of curve and show contradiction | B1 | [2]
(iii) State or imply $3y^2 + 4x = 0$ | B1 |
Eliminate one variable from equation of curve using $3y^2 + 4x = 0$ | M1 |
Obtain $y = -2$ | A1 |
Obtain $x = -3$ | A1 | [4]7 The equation of a curve is
$$y ^ { 3 } + 4 x y = 16$$
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 4 y } { 3 y ^ { 2 } + 4 x }$.\\
(ii) Show that the curve has no stationary points.\\
(iii) Find the coordinates of the point on the curve where the tangent is parallel to the $y$-axis.
\hfill \mbox{\textit{CAIE P2 2015 Q7 [10]}}