| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2009 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find stationary points - logarithmic functions |
| Difficulty | Moderate -0.8 This is a straightforward application of the product rule to find dy/dx = ln x + 1, setting equal to zero gives x = 1/e immediately, then using the second derivative test. It's simpler than average A-level questions as it requires only routine differentiation and a single-step algebraic solution with no complex manipulation. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use product rule | M1* | |
| Obtain derivative in any correct form | A1 | |
| Equate derivative to zero and solve for \(x\) | M1(dep*) | |
| Obtain \(x = \frac{1}{e}\), or exact equivalent | A1 | |
| Obtain \(y = -\frac{1}{e}\), or exact equivalent | A1 | [5 marks] |
| (ii) Carry out complete method for determining the nature of a stationary point | M1 | |
| Show that at \(x = \frac{1}{e}\) there is a minimum point, with no errors seen | A1 | [2 marks] |
**(i)** Use product rule | M1* |
Obtain derivative in any correct form | A1 |
Equate derivative to zero and solve for $x$ | M1(dep*) |
Obtain $x = \frac{1}{e}$, or exact equivalent | A1 |
Obtain $y = -\frac{1}{e}$, or exact equivalent | A1 | [5 marks]
**(ii)** Carry out complete method for determining the nature of a stationary point | M1 |
Show that at $x = \frac{1}{e}$ there is a minimum point, with no errors seen | A1 | [2 marks]6 The curve with equation $y = x \ln x$ has one stationary point.\\
(i) Find the exact coordinates of this point, giving your answers in terms of e .\\
(ii) Determine whether this point is a maximum or a minimum point.
\hfill \mbox{\textit{CAIE P2 2009 Q6 [7]}}