| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and tangent/normal |
| Difficulty | Standard +0.3 This is a straightforward two-part question combining standard volume of revolution (requiring substitution u = 2x-1 and integration of u^-4) with tangent/normal geometry. Both parts use routine A-level techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.03b Straight lines: parallel and perpendicular relationships1.07m Tangents and normals: gradient and equations4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\pi]\int \frac{16}{(2x-1)^4}[dx] = [\pi]\int 16(2x-1)^{-4}[dx] = [\pi]\left(-\frac{16}{3\times2\times(2x-1)^3}\right)\) | *M1 | Integrate \(y^2\) (power incr. by 1 or div by *their* new power). M0 if more than 1 error or \(-\frac{16}{6}x(2x-1)^{-3}\) |
| \([\pi]\left(-\frac{16}{3\times2\times(2x-1)^3}\right)\) | A1 | OE e.g. \(\left(-\frac{8}{3}(2x-1)^{-3}\right)\) |
| \([\pi]\left(-\frac{16}{6\times8}+\frac{16}{6\times1}\right) = [\pi]\frac{112}{48} = [\pi]\frac{7}{3}\) | DM1 | Sub correct limits into *their* integral: \(F\left(\frac{3}{2}\right) - F(1)\). Must see at least \(\left(-\frac{1}{3}+\frac{8}{3}\right)\). Allow 1 sign error. Decimal: \(2.33\pi\) or \(7.33\) |
| Volume of cylinder \(= \pi\times1^2\times\frac{1}{2} = \frac{1}{2}\pi\) OR \([\pi]\int_1^{1.5}1[dx]=\frac{1}{2}\pi\) | B1 | \(\frac{1}{2}\pi\) or \(\pm\pi\left(\frac{3}{2}-1\right)\) seen |
| Volume of revolution \(= \frac{7}{3}\pi - \frac{1}{2}\pi = \frac{11}{6}\pi\) | A1 | A0 for 5.76 (not exact). If DM0 for insufficient substitution, or B0, SC B1 for \(\frac{11}{6}\pi\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[\frac{dy}{dx}=\right]\{-8(2x-1)^{-3}\}\{\times2\}\) | B2, 1, 0 | OE B1 for each correct element in \(\{\}\) |
| At \(B\) gradient \(= -2\) | B1 | |
| Eqn of tangent \(y-1 = \textit{their}\ {-2}\left(x-\frac{3}{2}\right)\) OR Eqn of normal \(y-1 = \textit{their}\ \frac{1}{2}\left(x-\frac{3}{2}\right)\) | M1 | SOI Following differentiation OE e.g. \(y=-2x+4\) or \(y=\frac{1}{2}x+\frac{1}{4}\). Must have \(m_N = -\frac{1}{m_T}\) for M1 |
| Tangent crosses \(x\)-axis at \(2\) or normal crosses \(x\)-axis at \(-\frac{1}{2}\) | A1 | SOI For at least one intercept correct or correct integration |
| Area \(= \frac{5}{4}\) | A1 | From intercepts: \(\frac{1}{2}\times\frac{5}{2}\times1=\frac{5}{4}\) or \(1+\frac{1}{4}=\frac{5}{4}\), from lengths: \(\frac{1}{2}\times\sqrt{5}\times\frac{\sqrt{5}}{2}=\frac{5}{4}\) or by integration |
## Question 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\pi]\int \frac{16}{(2x-1)^4}[dx] = [\pi]\int 16(2x-1)^{-4}[dx] = [\pi]\left(-\frac{16}{3\times2\times(2x-1)^3}\right)$ | *M1 | Integrate $y^2$ (power incr. by 1 or div by *their* new power). M0 if more than 1 error or $-\frac{16}{6}x(2x-1)^{-3}$ |
| $[\pi]\left(-\frac{16}{3\times2\times(2x-1)^3}\right)$ | A1 | OE e.g. $\left(-\frac{8}{3}(2x-1)^{-3}\right)$ |
| $[\pi]\left(-\frac{16}{6\times8}+\frac{16}{6\times1}\right) = [\pi]\frac{112}{48} = [\pi]\frac{7}{3}$ | DM1 | Sub correct limits into *their* integral: $F\left(\frac{3}{2}\right) - F(1)$. Must see at least $\left(-\frac{1}{3}+\frac{8}{3}\right)$. Allow 1 sign error. Decimal: $2.33\pi$ or $7.33$ |
| Volume of cylinder $= \pi\times1^2\times\frac{1}{2} = \frac{1}{2}\pi$ OR $[\pi]\int_1^{1.5}1[dx]=\frac{1}{2}\pi$ | B1 | $\frac{1}{2}\pi$ or $\pm\pi\left(\frac{3}{2}-1\right)$ seen |
| Volume of revolution $= \frac{7}{3}\pi - \frac{1}{2}\pi = \frac{11}{6}\pi$ | A1 | A0 for 5.76 (not exact). If DM0 for insufficient substitution, or B0, **SC B1** for $\frac{11}{6}\pi$ |
**Total: 5 marks**
---
## Question 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[\frac{dy}{dx}=\right]\{-8(2x-1)^{-3}\}\{\times2\}$ | B2, 1, 0 | OE B1 for each correct element in $\{\}$ |
| At $B$ gradient $= -2$ | B1 | |
| Eqn of tangent $y-1 = \textit{their}\ {-2}\left(x-\frac{3}{2}\right)$ OR Eqn of normal $y-1 = \textit{their}\ \frac{1}{2}\left(x-\frac{3}{2}\right)$ | M1 | SOI Following differentiation OE e.g. $y=-2x+4$ or $y=\frac{1}{2}x+\frac{1}{4}$. Must have $m_N = -\frac{1}{m_T}$ for M1 |
| Tangent crosses $x$-axis at $2$ **or** normal crosses $x$-axis at $-\frac{1}{2}$ | A1 | SOI For at least one intercept correct or correct integration |
| Area $= \frac{5}{4}$ | A1 | From intercepts: $\frac{1}{2}\times\frac{5}{2}\times1=\frac{5}{4}$ or $1+\frac{1}{4}=\frac{5}{4}$, from lengths: $\frac{1}{2}\times\sqrt{5}\times\frac{\sqrt{5}}{2}=\frac{5}{4}$ or by integration |
**Total: 6 marks**
---10\\
\includegraphics[max width=\textwidth, alt={}, center]{77f27b11-b931-481f-b4ef-5e549eff8086-14_631_689_274_721}
The diagram shows part of the curve with equation $y = \frac { 4 } { ( 2 x - 1 ) ^ { 2 } }$ and parts of the lines $x = 1$ and $y = 1$. The curve passes through the points $A ( 1,4 )$ and $B , \left( \frac { 3 } { 2 } , 1 \right)$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact volume generated when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.
\item A triangle is formed from the tangent to the curve at $B$, the normal to the curve at $B$ and the $x$-axis.
Find the area of this triangle.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q10 [11]}}