| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2004 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Stationary points then area/volume |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on exponential functions requiring basic differentiation, tangent line equations, and integration of exponentials. All techniques are standard A-level procedures with no novel problem-solving required, though the multi-step nature and integration of e^(-2x) elevates it slightly above pure recall. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07m Tangents and normals: gradient and equations1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | State coordinates \((0, 5)\) | B1 |
| (ii) | State first derivative of the form \(ke^x + me^{-2x}\), where \(km \neq 0\) | M1 |
| Obtain correct first derivative \(2e^x - 6e^{-2x}\) | A1 | |
| Substitute \(x = 0\), obtaining gradient of \(-4\) | A1\(\sqrt{}\) | |
| Form equation of line through \(A\) with this gradient (NOT the normal) | M1 | |
| Obtain equation in any correct form e.g. \(y - 5 = -4x\) | A1 | |
| Obtain coordinates \((1.25, 0)\) or equivalent | A1 | [6] |
| (iii) | Integrate and obtain \(2e^x - \frac{3}{2}e^{-2x}\), or equivalent | B1 + B1 |
| Use limits \(x = 0\) and \(x = 1\) correctly | M1 | |
| Obtain answer \(4.7\) | A1 | [4] |
# Question 7:
**(i)** | State coordinates $(0, 5)$ | B1 | **[1]**
**(ii)** | State first derivative of the form $ke^x + me^{-2x}$, where $km \neq 0$ | M1 | |
Obtain correct first derivative $2e^x - 6e^{-2x}$ | A1 | |
Substitute $x = 0$, obtaining gradient of $-4$ | A1$\sqrt{}$ | |
Form equation of line through $A$ with this gradient (NOT the normal) | M1 | |
Obtain equation in any correct form e.g. $y - 5 = -4x$ | A1 | |
Obtain coordinates $(1.25, 0)$ or equivalent | A1 | **[6]**
**(iii)** | Integrate and obtain $2e^x - \frac{3}{2}e^{-2x}$, or equivalent | B1 + B1 | |
Use limits $x = 0$ and $x = 1$ correctly | M1 | |
Obtain answer $4.7$ | A1 | **[4]**
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{25dffd43-9456-449b-be77-8402109ee603-3_608_672_283_733}
The diagram shows the curve $y = 2 \mathrm { e } ^ { x } + 3 \mathrm { e } ^ { - 2 x }$. The curve cuts the $y$-axis at $A$.\\
(i) Write down the coordinates of $A$.\\
(ii) Find the equation of the tangent to the curve at $A$, and state the coordinates of the point where this tangent meets the $x$-axis.\\
(iii) Calculate the area of the region bounded by the curve and by the lines $x = 0 , y = 0$ and $x = 1$, giving your answer correct to 2 significant figures.
\hfill \mbox{\textit{CAIE P2 2004 Q7 [11]}}