| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 9 |
| 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 differentiation (finding tangent equation using quotient/chain rule), simple coordinate comparison, and volume of revolution using a standard integral. All techniques are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.08h Integration by substitution4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{-9}{(2x+3)^2} \times 2\) | B1 B1 | Correct without the \(\times 2\); for \(\times 2\), independent of first part |
| \(\rightarrow m = -\frac{2}{9}\) | M1 | Correct form of tangent: numerical \(dy/dx\) |
| \(\rightarrow y - 1 = -\frac{2}{9}(x-3)\) | A1\(\checkmark\) | For his \(m\) following use of \(dy/dx\); (normal \(\rightarrow\) max 2/4, no calculus 0/4) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Meets the \(y\)-axis when \(x = 0\), \(y = 1\frac{2}{3}\) | B1 | Sets \(x\) to \(0\) in his tangent |
| This is nearer to \(B\) than to \(O\) | The \(1\frac{2}{3}\) and part (i) must be correct | |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Integral of \(\frac{81}{(2x+3)^2} = \frac{-81}{2x+3} \div 2\) | B1 B1 | Correct without the \(\div 2\); for \(\div 2\) |
| Uses limits 0 to 3 \(\rightarrow \frac{-9}{2} - \frac{-81}{6} = 9\pi\) | M1 | Use of limits with integral of \(y^2\) only |
| A1 | no \(\pi\) – max \(\frac{3}{4}\); use of area \(- 0/4\) | |
| [4] |
## Question 9:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{-9}{(2x+3)^2} \times 2$ | B1 B1 | Correct without the $\times 2$; for $\times 2$, independent of first part |
| $\rightarrow m = -\frac{2}{9}$ | M1 | Correct form of tangent: numerical $dy/dx$ |
| $\rightarrow y - 1 = -\frac{2}{9}(x-3)$ | A1$\checkmark$ | For his $m$ following use of $dy/dx$; (normal $\rightarrow$ max 2/4, no calculus 0/4) |
| **[4]** | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Meets the $y$-axis when $x = 0$, $y = 1\frac{2}{3}$ | B1 | Sets $x$ to $0$ in his tangent |
| This is nearer to $B$ than to $O$ | | The $1\frac{2}{3}$ and part (i) must be correct |
| **[1]** | | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Integral of $\frac{81}{(2x+3)^2} = \frac{-81}{2x+3} \div 2$ | B1 B1 | Correct without the $\div 2$; for $\div 2$ |
| Uses limits 0 to 3 $\rightarrow \frac{-9}{2} - \frac{-81}{6} = 9\pi$ | M1 | Use of limits with integral of $y^2$ only |
| | A1 | no $\pi$ – max $\frac{3}{4}$; use of area $- 0/4$ |
| **[4]** | | |9\\
\includegraphics[max width=\textwidth, alt={}, center]{11bfe5bd-604c-43e5-81e7-4c1f5676bcbb-4_502_663_255_740}
The diagram shows part of the curve $y = \frac { 9 } { 2 x + 3 }$, crossing the $y$-axis at the point $B ( 0,3 )$. The point $A$ on the curve has coordinates $( 3,1 )$ and the tangent to the curve at $A$ crosses the $y$-axis at $C$.\\
(i) Find the equation of the tangent to the curve at $A$.\\
(ii) Determine, showing all necessary working, whether $C$ is nearer to $B$ or to $O$.\\
(iii) Find, showing all necessary working, the exact volume obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.
\hfill \mbox{\textit{CAIE P1 2012 Q9 [9]}}