| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2005 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve equation from derivative (exponential/logarithmic functions) |
| Difficulty | Moderate -0.8 This is a straightforward integration question requiring standard exponential integrals and using a boundary condition to find the constant. Part (ii) adds routine stationary point analysis by setting the derivative to zero and using the second derivative test. All techniques are standard bookwork 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/dx1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(\frac{1}{2}e^{2x}\) as integral of \(e^{2x}\) | B1 | |
| State \(y = \frac{1}{2}e^{2x} + 2e^{-x} + c\) | B1 | |
| Evaluate \(c\) | M1 | |
| Obtain answer \(y = \frac{1}{2}e^{2x} + 2e^{-x} - 1\frac{1}{2}\) | A1 | 4 |
| [Condone omission of \(c\) for the second B1.] | ||
| (ii) Equate derivative to zero | M1 | |
| EITHER: Obtain \(e^{4x} = 2\) | A1 | |
| Use logarithms and obtain a linear equation in \(x\) | M1 | |
| Obtain answer \(x = 0.231\) | A1 | |
| Show that the point is a minimum with no errors seen | A1 | |
| OR: Use logarithms and obtain a linear equation in \(x\) | M1 | |
| Obtain \(2x = \ln 2 - x\) | A1 | |
| Obtain answer \(x = 0.231\) | A1 | |
| Show that the point is a minimum with no errors seen | A1 | √ 5 |
(i) State $\frac{1}{2}e^{2x}$ as integral of $e^{2x}$ | B1 |
State $y = \frac{1}{2}e^{2x} + 2e^{-x} + c$ | B1 |
Evaluate $c$ | M1 |
Obtain answer $y = \frac{1}{2}e^{2x} + 2e^{-x} - 1\frac{1}{2}$ | A1 | 4
[Condone omission of $c$ for the second B1.] |
(ii) Equate derivative to zero | M1 |
EITHER: Obtain $e^{4x} = 2$ | A1 |
Use logarithms and obtain a linear equation in $x$ | M1 |
Obtain answer $x = 0.231$ | A1 |
Show that the point is a minimum with no errors seen | A1 |
OR: Use logarithms and obtain a linear equation in $x$ | M1 |
Obtain $2x = \ln 2 - x$ | A1 |
Obtain answer $x = 0.231$ | A1 |
Show that the point is a minimum with no errors seen | A1 | √ 56 A curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { 2 x } - 2 \mathrm { e } ^ { - x }$. The point $( 0,1 )$ lies on the curve.\\
(i) Find the equation of the curve.\\
(ii) The curve has one stationary point. Find the $x$-coordinate of this point and determine whether it is a maximum or a minimum point.
\hfill \mbox{\textit{CAIE P2 2005 Q6 [9]}}