| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Integration with differentiation context |
| Difficulty | Standard +0.3 This question requires differentiation using the chain rule (or quotient rule) to find a stationary point, then integration involving a substitution or recognition of reverse chain rule. Both parts are standard A-level techniques with straightforward algebra, though the integration requires some care. The verification in part (a) reduces computational demand, making this slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = \{-2 \times 2 \times (2x-1)^{-3} \times 2\} + \{1\}\) | B1B1 | Expect \(\frac{-8}{(2x-1)^3} + 1\). |
| Substitute \(x = \frac{3}{2}\) leading to \(\frac{dy}{dx} = \frac{-8}{8} + 1 = 0\). Hence \(x\)-coordinate of \(R\) is \(\frac{3}{2}\) | DB1 | AG. Or correct solution of \(\frac{dy}{dx} = 0\). |
| When \(x = \frac{3}{2}\), \(y = \frac{2}{4} + \frac{3}{2} = 2\) | B1 | Answer only is acceptable. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y\)-coordinate of \(P = 3\), \(y\)-coordinate of \(Q = \frac{20}{9}\) | B1 | Both required. |
| \(\left\{\frac{2(2x-1)^{-1}}{-1 \times 2}\right\} + \left\{\frac{1}{2}x^2\right\}\) | B1B1 | Area below curve. |
| \(\left[\left(-\frac{1}{3}+2\right)-\left(-1+\frac{1}{2}\right)\right] = \frac{5}{3} - \left(-\frac{1}{2}\right)\) | M1 | Apply limits \(1 \rightarrow 2\) to an integral. Expect \(\frac{13}{6}\). |
| \(\frac{1}{2}\left(3 + \frac{20}{9}\right) = \frac{47}{18}\) | M1 | Area of trapezium, only allow errors in \(y\)-coordinate of \(Q\). |
| \(\frac{47}{18} - \frac{13}{6} = \frac{4}{9}\) | A1 | Shaded region. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equation of line \(PQ\): \(\left[y = \frac{-7}{9}x + \frac{34}{9}\right]\). Integrating between 1 and 2. | M1 | Must be some evidence of use of limits. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Combining line and curve: \(\int\left(\frac{-16}{9}x + \frac{34}{9} - \frac{2}{(2x-1)^2}\right)dx\) | M1 | For area under the line if their \(\frac{34}{9}\) is seen integrated correctly and limits used. Correct first and 3rd terms. |
| \(= \frac{-8}{9}x^2 + \frac{34}{9}x + \frac{1}{(2x-1)}\) | B1B1 | |
| Use of limits on the whole integral | M1 |
## Question 11(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = \{-2 \times 2 \times (2x-1)^{-3} \times 2\} + \{1\}$ | B1B1 | Expect $\frac{-8}{(2x-1)^3} + 1$. |
| Substitute $x = \frac{3}{2}$ leading to $\frac{dy}{dx} = \frac{-8}{8} + 1 = 0$. Hence $x$-coordinate of $R$ is $\frac{3}{2}$ | DB1 | AG. Or correct solution of $\frac{dy}{dx} = 0$. |
| When $x = \frac{3}{2}$, $y = \frac{2}{4} + \frac{3}{2} = 2$ | B1 | Answer only is acceptable. |
## Question 11(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y$-coordinate of $P = 3$, $y$-coordinate of $Q = \frac{20}{9}$ | B1 | Both required. |
| $\left\{\frac{2(2x-1)^{-1}}{-1 \times 2}\right\} + \left\{\frac{1}{2}x^2\right\}$ | B1B1 | Area below curve. |
| $\left[\left(-\frac{1}{3}+2\right)-\left(-1+\frac{1}{2}\right)\right] = \frac{5}{3} - \left(-\frac{1}{2}\right)$ | M1 | Apply limits $1 \rightarrow 2$ to an integral. Expect $\frac{13}{6}$. |
| $\frac{1}{2}\left(3 + \frac{20}{9}\right) = \frac{47}{18}$ | M1 | Area of trapezium, only allow errors in $y$-coordinate of $Q$. |
| $\frac{47}{18} - \frac{13}{6} = \frac{4}{9}$ | A1 | Shaded region. |
**Alternative method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Equation of line $PQ$: $\left[y = \frac{-7}{9}x + \frac{34}{9}\right]$. Integrating between 1 and 2. | M1 | Must be some evidence of use of limits. |
**Alternative method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Combining line and curve: $\int\left(\frac{-16}{9}x + \frac{34}{9} - \frac{2}{(2x-1)^2}\right)dx$ | M1 | For area under the line if their $\frac{34}{9}$ is seen integrated correctly and limits used. Correct first and 3rd terms. |
| $= \frac{-8}{9}x^2 + \frac{34}{9}x + \frac{1}{(2x-1)}$ | B1B1 | |
| Use of limits on the whole integral | M1 | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{88c7a3f3-e129-4e9c-acf8-8c96d2668d43-14_693_782_267_669}
The diagram shows part of the curve with equation $y = x + \frac { 2 } { ( 2 x - 1 ) ^ { 2 } }$. The lines $x = 1$ and $x = 2$ intersect the curve at $P$ and $Q$ respectively and $R$ is the stationary point on the curve.
\begin{enumerate}[label=(\alph*)]
\item Verify that the $x$-coordinate of $R$ is $\frac { 3 } { 2 }$ and find the $y$-coordinate of $R$.
\item Find the exact value of the area of the shaded region.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q11 [10]}}