| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Rotation about x-axis: rational or reciprocal function |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard techniques. Part (i) requires basic differentiation of a rational function and recognizing no stationary points exist. Part (ii) is a routine volume of revolution calculation using the standard formula. Part (iii) involves forming a quadratic from the intersection condition and applying the discriminant criterion. All parts are textbook exercises requiring no novel insight, making this slightly easier than average. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dy}{dx} = -9(2-x)^{-2} \times -1\) | B1 | Without the "\(\times -1\)" — Independent |
| B1 | With the "\(\times -1\)" — Independent | |
| \(\dfrac{9}{(2-x)^2} \neq 0\). No turning points. | B1\(\sqrt{}\) | \(\sqrt{}\) provided of form \(k \div (2-x)^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\displaystyle\int y^2\,dx = -81(2-x)^{-1} \div (-1)\) | B1 | Answer without the "\(\div -1\) including \(\pi\)" |
| B1 | For "\(\div -1\)" | |
| Use of limits 0 to 1 | M1 | Uses both limits in an integral of \(y^2\) — if "0" ignored, M0 |
| \(\rightarrow \dfrac{81\pi}{2}\) (or 127) | A1 | co (If \(\pi\) omitted — max 3/4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{9}{2-x} = x + k\) | M1 | Elimination of \(y\) |
| \(\rightarrow x^2 - 2x + kx - 2k + 9 = 0\) | ||
| Uses \(b^2 - 4ac\) \(\rightarrow k^2 + 4k - 32\) | M1 | Uses discriminant |
| \(\rightarrow\) end-points of 4 and \(-8\) | A1 | End-values correct |
| Range for 2 points of intersection: \(\rightarrow k < -8,\ k > 4\) | A1 | Accept \(\leq,\ \geq\) |
# Question 11:
**Given:** $y = \dfrac{9}{2-x}$
---
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} = -9(2-x)^{-2} \times -1$ | B1 | Without the "$\times -1$" — Independent |
| | B1 | With the "$\times -1$" — Independent |
| $\dfrac{9}{(2-x)^2} \neq 0$. No turning points. | B1$\sqrt{}$ | $\sqrt{}$ provided of form $k \div (2-x)^2$ |
**[3 marks]**
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## Part (ii):
$V = \pi \displaystyle\int_0^1 \dfrac{81}{(2-x)^2}\,dx$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\displaystyle\int y^2\,dx = -81(2-x)^{-1} \div (-1)$ | B1 | Answer without the "$\div -1$ including $\pi$" |
| | B1 | For "$\div -1$" |
| Use of limits 0 to 1 | M1 | Uses both limits in an integral of $y^2$ — if "0" ignored, M0 |
| $\rightarrow \dfrac{81\pi}{2}$ (or 127) | A1 | co (If $\pi$ omitted — max 3/4) |
**[4 marks]**
---
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{9}{2-x} = x + k$ | M1 | Elimination of $y$ |
| $\rightarrow x^2 - 2x + kx - 2k + 9 = 0$ | | |
| Uses $b^2 - 4ac$ $\rightarrow k^2 + 4k - 32$ | M1 | Uses discriminant |
| $\rightarrow$ end-points of 4 and $-8$ | A1 | End-values correct |
| Range for 2 points of intersection: $\rightarrow k < -8,\ k > 4$ | A1 | Accept $\leq,\ \geq$ |
**[4 marks]**11 The equation of a curve is $y = \frac { 9 } { 2 - x }$.\\
(i) Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and determine, with a reason, whether the curve has any stationary points.\\
(ii) Find the volume obtained when the region bounded by the curve, the coordinate axes and the line $x = 1$ is rotated through $360 ^ { \circ }$ about the $x$-axis.\\
(iii) Find the set of values of $k$ for which the line $y = x + k$ intersects the curve at two distinct points.
\hfill \mbox{\textit{CAIE P1 2010 Q11 [11]}}