| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2017 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Normal/tangent then area with parts |
| Difficulty | Standard +0.3 This is a structured multi-part question requiring standard techniques: finding a normal line equation (routine differentiation and coordinate geometry), proving a given integration by parts result (guided proof), and applying integration by parts twice to find an area. While it requires multiple steps and careful execution, each component is a standard textbook exercise with no novel insight required. Slightly easier than average due to the scaffolding provided. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply derivative is \(2\dfrac{\ln x}{x}\) | B1 | |
| State or imply gradient of normal at \(x=\mathrm{e}\) is \(-\dfrac{1}{2}\mathrm{e}\), or equivalent | B1 | |
| Carry out complete method for finding \(x\)-coordinate of \(Q\) | M1 | |
| Obtain \(x = \mathrm{e} + \dfrac{2}{\mathrm{e}}\), or exact equivalent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Justify the given statement by integration or by differentiation | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Integrate by parts and reach \(ax(\ln x)^2 + b\displaystyle\int x\cdot\dfrac{\ln x}{x}\,\mathrm{d}x\) | M1* | |
| Complete integration and obtain \(x(\ln x)^2 - 2x\ln x + 2x\), or equivalent | A1 | |
| Use limits \(x=1\) and \(x=\mathrm{e}\) correctly, having integrated twice | DM1 | |
| Obtain exact value \(\mathrm{e}-2\) | A1 | |
| Use \(x\)-coordinate of \(Q\) found in part (i) and obtain final answer \(\mathrm{e}-2+\dfrac{1}{\mathrm{e}}\) | B1\(^\checkmark\) | f.t. on \(Q\) |
## Question 10(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply derivative is $2\dfrac{\ln x}{x}$ | B1 | |
| State or imply gradient of normal at $x=\mathrm{e}$ is $-\dfrac{1}{2}\mathrm{e}$, or equivalent | B1 | |
| Carry out complete method for finding $x$-coordinate of $Q$ | M1 | |
| Obtain $x = \mathrm{e} + \dfrac{2}{\mathrm{e}}$, or exact equivalent | A1 | |
**Total: 4**
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## Question 10(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Justify the given statement by integration or by differentiation | B1 | |
**Total: 1**
---
## Question 10(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Integrate by parts and reach $ax(\ln x)^2 + b\displaystyle\int x\cdot\dfrac{\ln x}{x}\,\mathrm{d}x$ | M1* | |
| Complete integration and obtain $x(\ln x)^2 - 2x\ln x + 2x$, or equivalent | A1 | |
| Use limits $x=1$ and $x=\mathrm{e}$ correctly, having integrated twice | DM1 | |
| Obtain exact value $\mathrm{e}-2$ | A1 | |
| Use $x$-coordinate of $Q$ found in part (i) and obtain final answer $\mathrm{e}-2+\dfrac{1}{\mathrm{e}}$ | B1$^\checkmark$ | f.t. on $Q$ |
**Total: 5**10\\
\includegraphics[max width=\textwidth, alt={}, center]{e26f21c5-3776-4c86-8440-6959c5e37486-18_337_529_260_808}
The diagram shows the curve $y = ( \ln x ) ^ { 2 }$. The $x$-coordinate of the point $P$ is equal to e, and the normal to the curve at $P$ meets the $x$-axis at $Q$.\\
(i) Find the $x$-coordinate of $Q$.\\
(ii) Show that $\int \ln x \mathrm {~d} x = x \ln x - x + c$, where $c$ is a constant.\\
(iii) Using integration by parts, or otherwise, find the exact value of the area of the shaded region between the curve, the $x$-axis and the normal $P Q$.\\
\hfill \mbox{\textit{CAIE P3 2017 Q10 [10]}}