| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 10 |
| 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 integration of a power function after squaring) and finding a normal line (requiring differentiation using chain rule and point-slope form). Both are routine A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = 4(\pi)\int(3x-1)^{-2/3}\,dx = 4(\pi)\left[\dfrac{(3x-1)^{1/3}}{1/3}\right][\div 3]\) | M1A1A1 | Recognisable integration of \(y^2\) (M1). Independent A1, A1 for [ ][ ] |
| \(4(\pi)[2-1]\) | DM1 | Expect \(4(\pi)(3x-1)^{\frac{1}{3}}\) |
| \(4\pi\) or \(12.6\) | A1 | Apply limits \(\frac{2}{3} \to 3\). Some working must be shown. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(dy/dx = (-2/3)(3x-1)^{-4/3} \times 3\) | B1 | Expect \(-2(3x-1)^{-4/3}\) |
| When \(x = \frac{2}{3}\), \(y = 2\) so \(dy/dx = -2\) | B1B1 | 2nd B1 dep. on correct expression for \(dy/dx\) |
| Equation of normal is \(y - 2 = \frac{1}{2}\left(x - \frac{2}{3}\right)\) | M1 | Line through \((\frac{2}{3},\) their \(2)\) and with grad \(-1/m\). Dep on \(m\) from diffn |
| \(y = \frac{1}{2}x + \frac{5}{3}\) | A1 |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = 4(\pi)\int(3x-1)^{-2/3}\,dx = 4(\pi)\left[\dfrac{(3x-1)^{1/3}}{1/3}\right][\div 3]$ | M1A1A1 | Recognisable integration of $y^2$ (M1). Independent A1, A1 for [ ][ ] |
| $4(\pi)[2-1]$ | DM1 | Expect $4(\pi)(3x-1)^{\frac{1}{3}}$ |
| $4\pi$ or $12.6$ | A1 | Apply limits $\frac{2}{3} \to 3$. Some working must be shown. |
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## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $dy/dx = (-2/3)(3x-1)^{-4/3} \times 3$ | B1 | Expect $-2(3x-1)^{-4/3}$ |
| When $x = \frac{2}{3}$, $y = 2$ so $dy/dx = -2$ | B1B1 | 2nd B1 dep. on correct expression for $dy/dx$ |
| Equation of normal is $y - 2 = \frac{1}{2}\left(x - \frac{2}{3}\right)$ | M1 | Line through $(\frac{2}{3},$ their $2)$ and with grad $-1/m$. Dep on $m$ from diffn |
| $y = \frac{1}{2}x + \frac{5}{3}$ | A1 | |
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\includegraphics[max width=\textwidth, alt={}, center]{d5d94eb8-7f41-4dff-b503-8be4f20e21b7-16_648_823_262_660}
The diagram shows part of the curve $y = 2 ( 3 x - 1 ) ^ { - \frac { 1 } { 3 } }$ and the lines $x = \frac { 2 } { 3 }$ and $x = 3$. The curve and the line $x = \frac { 2 } { 3 }$ intersect at the point $A$.\\
(i) Find, showing all necessary working, the volume obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.\\
(ii) Find the equation of the normal to the curve at $A$, giving your answer in the form $y = m x + c$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q10 [10]}}