| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Moderate -0.3 This is a straightforward implicit differentiation question requiring product rule and logarithm differentiation, followed by finding a tangent equation at a given point. The algebra is clean (y=1 simplifies the logarithm), and both parts follow standard procedures with no conceptual surprises, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) EITHER: State or imply \(\frac{1}{y}\frac{dy}{dx}\) as derivative of \(\ln y\) | B1 | |
| State correct derivative of LHS, e.g. \(\ln y + \frac{x}{y}\frac{dy}{dx}\) | B1 | |
| Differentiate RHS and obtain an expression for \(\frac{dy}{dx}\) | M1 | |
| Obtain given answer | A1 | |
| OR 1: State \(\ln y = \frac{2x + 1}{x}\), or equivalent, and differentiate both sides | M1 | |
| State correct derivative of LHS, e.g. \(\frac{1}{y}\frac{dy}{dx}\) | B1 | |
| State correct derivative of RHS, e.g. \(-1/x^2\) | B1 | |
| Rearrange and obtain given answer | A1 | |
| OR 2: State \(y = \exp(2 + 1/x)\), or equivalent, and attempt differentiation by chain rule | M1 | |
| State correct derivative of RHS, e.g. \(-\exp(2 + 1/x) / x^2\) | B1 + B1 | |
| Obtain given answer | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) State or imply \(x = -\frac{1}{2}\) when \(y = 1\) | B1 | |
| Substitute and obtain gradient of \(-4\) | B1√ | |
| Correctly form equation of tangent | M1 | |
| Obtain final answer \(y + 4x + 1 = 0\), or equivalent | A1 | [4] |
**(i)** **EITHER:** State or imply $\frac{1}{y}\frac{dy}{dx}$ as derivative of $\ln y$ | B1 |
State correct derivative of LHS, e.g. $\ln y + \frac{x}{y}\frac{dy}{dx}$ | B1 |
Differentiate RHS and obtain an expression for $\frac{dy}{dx}$ | M1 |
Obtain given answer | A1 |
**OR 1:** State $\ln y = \frac{2x + 1}{x}$, or equivalent, and differentiate both sides | M1 |
State correct derivative of LHS, e.g. $\frac{1}{y}\frac{dy}{dx}$ | B1 |
State correct derivative of RHS, e.g. $-1/x^2$ | B1 |
Rearrange and obtain given answer | A1 |
**OR 2:** State $y = \exp(2 + 1/x)$, or equivalent, and attempt differentiation by chain rule | M1 |
State correct derivative of RHS, e.g. $-\exp(2 + 1/x) / x^2$ | B1 + B1 |
Obtain given answer | A1 | [4]
[The B marks are for the exponential term and its multiplier.]
**(ii)** State or imply $x = -\frac{1}{2}$ when $y = 1$ | B1 |
Substitute and obtain gradient of $-4$ | B1√ |
Correctly form equation of tangent | M1 |
Obtain final answer $y + 4x + 1 = 0$, or equivalent | A1 | [4]6 The equation of a curve is
$$x \ln y = 2 x + 1$$
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { y } { x ^ { 2 } }$.\\
(ii) Find the equation of the tangent to the curve at the point where $y = 1$, giving your answer in the form $a x + b y + c = 0$.
\hfill \mbox{\textit{CAIE P3 2010 Q6 [8]}}