| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and area |
| Difficulty | Standard +0.3 This is a straightforward volumes of revolution question with standard techniques. Part (i) is simple algebraic rearrangement, part (ii) requires basic integration of polynomial terms, and part (iii) applies the standard formula for rotation about the y-axis. All steps are routine for P1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = \dfrac{4}{y^2} - 1\) | B1 [1] | AG At least 1 step of working needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\displaystyle\int\left(\frac{4}{y^2}-1\right)dy = \left[-\frac{4}{y}-y\right]\) | B1B1 | For \(-\dfrac{4}{y}, -y\) |
| Upper limit \(= 2\) | B1 | |
| \(\left[\left(-\dfrac{4}{2}-2\right)-(-4-1)\right]\) | M1 | Apply limits 1 and *their* 2 'correctly'; SC B2 for \(\int 2(x+1)^{-\frac{1}{2}}\,dx - 3 \to 1\) |
| \(1\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((\pi)\displaystyle\int x^2\,dy = (\pi)\int\left(\frac{16}{y^4}-\frac{8}{y^2}+1\right)dy\) | B1B1 | |
| \((\pi)\left[\dfrac{-16}{3y^3}+\dfrac{8}{y}+y\right]\) | B1 | |
| \((\pi)\left[\left(\dfrac{-16}{24}+4+2\right)-\left(\dfrac{-16}{3}+8+1\right)\right]\) | M1 | Apply limits 1 and *their* 2 'correctly' |
| \(\dfrac{5\pi}{3}\) | A1 [5] |
# Question 11:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = \dfrac{4}{y^2} - 1$ | B1 [1] | AG At least 1 step of working needed |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\displaystyle\int\left(\frac{4}{y^2}-1\right)dy = \left[-\frac{4}{y}-y\right]$ | B1B1 | For $-\dfrac{4}{y}, -y$ |
| Upper limit $= 2$ | B1 | |
| $\left[\left(-\dfrac{4}{2}-2\right)-(-4-1)\right]$ | M1 | Apply limits 1 and *their* 2 'correctly'; SC B2 for $\int 2(x+1)^{-\frac{1}{2}}\,dx - 3 \to 1$ |
| $1$ | A1 [5] | |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(\pi)\displaystyle\int x^2\,dy = (\pi)\int\left(\frac{16}{y^4}-\frac{8}{y^2}+1\right)dy$ | B1B1 | |
| $(\pi)\left[\dfrac{-16}{3y^3}+\dfrac{8}{y}+y\right]$ | B1 | |
| $(\pi)\left[\left(\dfrac{-16}{24}+4+2\right)-\left(\dfrac{-16}{3}+8+1\right)\right]$ | M1 | Apply limits 1 and *their* 2 'correctly' |
| $\dfrac{5\pi}{3}$ | A1 [5] | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{4d8fcc3d-a2da-4d98-8500-075d10847be3-4_636_951_255_596}
The diagram shows the line $y = 1$ and part of the curve $y = \frac { 2 } { \sqrt { } ( x + 1 ) }$.\\
(i) Show that the equation $y = \frac { 2 } { \sqrt { } ( x + 1 ) }$ can be written in the form $x = \frac { 4 } { y ^ { 2 } } - 1$.\\
(ii) Find $\int \left( \frac { 4 } { y ^ { 2 } } - 1 \right) \mathrm { d } y$. Hence find the area of the shaded region.\\
(iii) The shaded region is rotated through $360 ^ { \circ }$ about the $\boldsymbol { y }$-axis. Find the exact value of the volume of revolution obtained.
\hfill \mbox{\textit{CAIE P1 2012 Q11 [11]}}