| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| 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 multi-part question combining standard techniques: finding a normal line equation and proving a midpoint property (part i), then computing a volume of revolution about the y-axis (part ii). Both parts use routine A-level methods with no novel insight required. The volume calculation involves standard rearrangement to x² in terms of y and direct application of the formula, making it slightly easier than average. |
| Spec | 1.07l 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/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{1}{2\sqrt{(9-2x^2)}} \times -4x\) | B1 | Without "\(\times -4x\)" |
| B1 | Allow even if B0 above | |
| At \(P\), \(x = 2\), \(m = -4\), Normal grad \(= \frac{1}{4}\) | M1 | For \(m_1 m_2 = -1\) calculus needed |
| Eqn \(AP\): \(y - 1 = \frac{1}{4}(x-2)\) | M1 | Normal, not tangent |
| \(\rightarrow A(-2, 0)\) or \(B(0, \frac{1}{2})\) | A1 | |
| Midpoint \(AP\) also \((0, \frac{1}{2})\) | A1 | Full justification |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int x^2 \, dy = \int \left(\frac{9}{2} - \frac{y^2}{2}\right) dy\) | M1 | Attempt to integrate \(x^2\) |
| \(= \frac{9y}{2} - \frac{y^3}{6}\) | A1 | Correct integration |
| Upper limit \(= 3\) | B1 | Evaluates upper limit |
| Uses limits 1 to 3 | DM1 | Uses both limits correctly |
| \(\rightarrow\) volume \(= 4\frac{2}{3}\pi\) | A1 | |
| [5] |
# Question 10:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{1}{2\sqrt{(9-2x^2)}} \times -4x$ | **B1** | Without "$\times -4x$" |
| | **B1** | Allow even if B0 above |
| At $P$, $x = 2$, $m = -4$, Normal grad $= \frac{1}{4}$ | **M1** | For $m_1 m_2 = -1$ calculus needed |
| Eqn $AP$: $y - 1 = \frac{1}{4}(x-2)$ | **M1** | Normal, not tangent |
| $\rightarrow A(-2, 0)$ or $B(0, \frac{1}{2})$ | **A1** | |
| Midpoint $AP$ also $(0, \frac{1}{2})$ | **A1** | Full justification |
| | **[6]** | |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int x^2 \, dy = \int \left(\frac{9}{2} - \frac{y^2}{2}\right) dy$ | **M1** | Attempt to integrate $x^2$ |
| $= \frac{9y}{2} - \frac{y^3}{6}$ | **A1** | Correct integration |
| Upper limit $= 3$ | **B1** | Evaluates upper limit |
| Uses limits 1 to 3 | **DM1** | Uses both limits correctly |
| $\rightarrow$ volume $= 4\frac{2}{3}\pi$ | **A1** | |
| | **[5]** | |10\\
\includegraphics[max width=\textwidth, alt={}, center]{9cdb00a6-1e86-4185-bb73-ed3ecab981ba-4_634_937_696_603}
The diagram shows part of the curve $y = \sqrt { } \left( 9 - 2 x ^ { 2 } \right)$. The point $P ( 2,1 )$ lies on the curve and the normal to the curve at $P$ intersects the $x$-axis at $A$ and the $y$-axis at $B$.\\
(i) Show that $B$ is the mid-point of $A P$.
The shaded region is bounded by the curve, the $y$-axis and the line $y = 1$.\\
(ii) Find, showing all necessary working, the exact volume obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $\boldsymbol { y }$-axis.
{www.cie.org.uk} after the live examination series.
}
\hfill \mbox{\textit{CAIE P1 2015 Q10 [11]}}