| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2006 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find stationary points - polynomial/exponential products |
| Difficulty | Moderate -0.8 This is a straightforward application of differentiation rules for exponentials. Finding the stationary point requires setting dy/dx = 0 and solving 6e^x - 3e^(3x) = 0, which factors easily. The second derivative test is routine. Below average difficulty as it's a standard textbook exercise with no problem-solving insight required. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks |
|---|---|
| (i) State derivative is \(6e^x - 3e^{3x}\) | B1 |
| EITHER: Equate derivative to zero and simplify to an equation of the form \(e^{2x} = a\) | M1* |
| Carry out method for calculating \(x\), where \(a > 0\) | M1(dep*) |
| Obtain answer \(x = \frac{1}{2}\ln 2\), or equivalent (0.347, or 0.346, or 0.35) | A1 |
| OR: Equate terms of the derivative and obtain a linear equation in \(x\) by taking logs correctly | M1* |
| Solve the linear equation for \(x\) | M1(dep*) |
| Obtain answer \(x = \frac{1}{2}\ln 2\), or equivalent (0.347, or 0.346, or 0.35) | A1 |
| 4 | |
| (ii) Carry out a method for determining the nature of a stationary point | M1 |
| Show that the point is a maximum with no errors seen | A1 |
| 2 |
(i) State derivative is $6e^x - 3e^{3x}$ | B1 |
EITHER: Equate derivative to zero and simplify to an equation of the form $e^{2x} = a$ | M1* |
Carry out method for calculating $x$, where $a > 0$ | M1(dep*) |
Obtain answer $x = \frac{1}{2}\ln 2$, or equivalent (0.347, or 0.346, or 0.35) | A1 |
OR: Equate terms of the derivative and obtain a linear equation in $x$ by taking logs correctly | M1* |
Solve the linear equation for $x$ | M1(dep*) |
Obtain answer $x = \frac{1}{2}\ln 2$, or equivalent (0.347, or 0.346, or 0.35) | A1 |
| | 4 |
(ii) Carry out a method for determining the nature of a stationary point | M1 |
Show that the point is a maximum with no errors seen | A1 |
| | 2 |3 The curve with equation $y = 6 \mathrm { e } ^ { x } - \mathrm { e } ^ { 3 x }$ has one stationary point.\\
(i) Find the $x$-coordinate of this point.\\
(ii) Determine whether this point is a maximum or a minimum point.
\hfill \mbox{\textit{CAIE P3 2006 Q3 [6]}}