| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area under curve with fractional/negative powers or roots |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard techniques: (a) differentiation using chain rule and finding tangent equation, (b) volume of revolution using the standard formula. Both are routine P1/C2 level exercises with no conceptual challenges, though the chain rule with the cube root and the algebra require care. Slightly above average due to the two-part nature and algebraic manipulation needed. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Differentiate to obtain form \(kx^2(2x^3+10)^{-\frac{1}{3}}\) | M1 | OE |
| \(3x^2(2x^3+10)^{-\frac{1}{2}}\) | A1 | Or unsimplified equivalent |
| Substitute \(x = 3\) in first derivative and evaluate to find gradient | *M1 | Expect \(\frac{27}{8}\). Allow if first derivative of forms \(k(2x^3+10)^{\frac{-1}{2}}\), \(kx(2x^3+10)^{\frac{-1}{2}}\) or \(kx^2(2x^3+10)^{\frac{-1}{2}}\) |
| Attempt equation of tangent at \((3, 8)\) with numerical gradient | DM1 | Use of gradient of the normal is DM0 |
| \([\pm](27x - 8y - 17) = 0\) or integer multiples | A1 | |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply volume is \(\pi\int(2x^3+10)\,dx\) | B1 | Implied if \(\pi\) appears only at the end. Do not allow an unsimplified: \(\pi\int\left((2x^3+10)^{1/2}\right)^2\) |
| Integrate to obtain \(k_1x^4 + k_2x\) and evaluate using limits 1 and 3 | M1 | Where \(k_1k_2 \neq 0\) |
| \(60\pi\) | A1 | OE. Allow from a correct integral and sight of limits. Allow numerical answers in the range 188–189 |
| Total | 3 |
## Question 9(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate to obtain form $kx^2(2x^3+10)^{-\frac{1}{3}}$ | M1 | OE |
| $3x^2(2x^3+10)^{-\frac{1}{2}}$ | A1 | Or unsimplified equivalent |
| Substitute $x = 3$ in first derivative and evaluate to find gradient | *M1 | Expect $\frac{27}{8}$. Allow if first derivative of forms $k(2x^3+10)^{\frac{-1}{2}}$, $kx(2x^3+10)^{\frac{-1}{2}}$ or $kx^2(2x^3+10)^{\frac{-1}{2}}$ |
| Attempt equation of tangent at $(3, 8)$ with numerical gradient | DM1 | Use of gradient of the normal is DM0 |
| $[\pm](27x - 8y - 17) = 0$ or integer multiples | A1 | |
| **Total** | **5** | |
---
## Question 9(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply volume is $\pi\int(2x^3+10)\,dx$ | B1 | Implied if $\pi$ appears only at the end. Do not allow an unsimplified: $\pi\int\left((2x^3+10)^{1/2}\right)^2$ |
| Integrate to obtain $k_1x^4 + k_2x$ and evaluate using limits 1 and 3 | M1 | Where $k_1k_2 \neq 0$ |
| $60\pi$ | A1 | OE. Allow from a correct integral and sight of limits. Allow numerical answers in the range 188–189 |
| **Total** | **3** | |
---9\\
\includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-12_764_967_292_555}
The diagram shows the curve with equation $y = \sqrt { 2 x ^ { 3 } + 10 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the tangent to the curve at the point where $x = 3$. Give your answer in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers.\\
\includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-12_2716_35_141_2013}
\item The region shaded in the diagram is enclosed by the curve and the straight lines $x = 1 , x = 3$ and $y = 0$.
Find the volume of the solid obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2024 Q9 [8]}}