| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Integration with substitution given |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard algebraic manipulation of logarithms (part i) and then a routine substitution integral (part ii). Both parts are guided ('show that'), requiring no problem-solving insight, just careful execution of A-level techniques. Slightly easier than average due to the scaffolding provided. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Remove logarithms correctly and obtain \(e^x = \frac{1-y}{y}\) | B1 | |
| Obtain the given answer \(y = \frac{e^{-x}}{1+e^{-x}}\) following full working | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State integral \(k\ln(1+e^{-x})\) where \(k = \pm1\) | *M1 | |
| State correct integral \(-\ln(1+e^{-x})\) | A1 | |
| Use limits correctly | DM1 | |
| Obtain the given answer \(\ln\!\left(\dfrac{2e}{e+1}\right)\) following full working | A1 |
## Question 3(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Remove logarithms correctly and obtain $e^x = \frac{1-y}{y}$ | B1 | |
| Obtain the **given answer** $y = \frac{e^{-x}}{1+e^{-x}}$ following full working | B1 | |
**Total: 2**
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## Question 3(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| State integral $k\ln(1+e^{-x})$ where $k = \pm1$ | *M1 | |
| State correct integral $-\ln(1+e^{-x})$ | A1 | |
| Use limits correctly | DM1 | |
| Obtain the **given answer** $\ln\!\left(\dfrac{2e}{e+1}\right)$ following full working | A1 | |
**Total: 4**
---3 It is given that $x = \ln ( 1 - y ) - \ln y$, where $0 < y < 1$.\\
(i) Show that $y = \frac { \mathrm { e } ^ { - x } } { 1 + \mathrm { e } ^ { - x } }$.\\
(ii) Hence show that $\int _ { 0 } ^ { 1 } y \mathrm {~d} x = \ln \left( \frac { 2 \mathrm { e } } { \mathrm { e } + 1 } \right)$.\\
\hfill \mbox{\textit{CAIE P3 2017 Q3 [6]}}