| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2006 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and related rates |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard techniques: differentiation using chain rule, related rates with direct substitution, and volume of revolution with a simple rational function. All parts follow routine procedures with no conceptual challenges or novel insights required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = -6(5 - 2x)^{-2} \times (-2)\) \(= \frac{12}{(5-2x)^2} \rightarrow \frac{1}{3}\) | B1 M1 A1 [3] | B1 for \(\frac{dy}{dx} = -6(5-2x)^{-2}\). M1 for (\(x-2\)). Co. |
| (ii) Use of chain rule. \(\frac{dx}{dt} = 0.02 + (i) = 0.015\) | M1 A1∨ [2] | M1 for dividing 0.02 by answer to (i). |
| (iii) \(V = \pi \int \frac{16}{(5-2x)^2} dx\) \(= 36\pi \left[\frac{(5-2x)^{-1}}{-1}\right] + (-2)\) \([]^1 - []^0 = \frac{12\pi}{5}\) | M1 A1 M1 DM1 A1 [5] | Attempt at integration of \(y^2\) (ignore \(\pi\)). For \(36\pi \left[\frac{(5-2x)^{-1}}{-1}\right]\). M1 for \(-(-2)\). DM1 for correct use of limits- not earned if \([]^1\) ignored or put to 0. |
$y = \frac{6}{5 - 2x}$
(i) $\frac{dy}{dx} = -6(5 - 2x)^{-2} \times (-2)$ $= \frac{12}{(5-2x)^2} \rightarrow \frac{1}{3}$ | B1 M1 A1 [3] | B1 for $\frac{dy}{dx} = -6(5-2x)^{-2}$. M1 for ($x-2$). Co.
(ii) Use of chain rule. $\frac{dx}{dt} = 0.02 + (i) = 0.015$ | M1 A1∨ [2] | M1 for dividing 0.02 by answer to (i).
(iii) $V = \pi \int \frac{16}{(5-2x)^2} dx$ $= 36\pi \left[\frac{(5-2x)^{-1}}{-1}\right] + (-2)$ $[]^1 - []^0 = \frac{12\pi}{5}$ | M1 A1 M1 DM1 A1 [5] | Attempt at integration of $y^2$ (ignore $\pi$). For $36\pi \left[\frac{(5-2x)^{-1}}{-1}\right]$. M1 for $-(-2)$. DM1 for correct use of limits- not earned if $[]^1$ ignored or put to 0.
---8 The equation of a curve is $y = \frac { 6 } { 5 - 2 x }$.\\
(i) Calculate the gradient of the curve at the point where $x = 1$.\\
(ii) A point with coordinates $( x , y )$ moves along the curve in such a way that the rate of increase of $y$ has a constant value of 0.02 units per second. Find the rate of increase of $x$ when $x = 1$.\\
(iii) The region between the curve, the $x$-axis and the lines $x = 0$ and $x = 1$ is rotated through $360 ^ { \circ }$ about the $x$-axis. Show that the volume obtained is $\frac { 12 } { 5 } \pi$.
\hfill \mbox{\textit{CAIE P1 2006 Q8 [10]}}