| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2004 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and stationary points |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining routine differentiation (including negative powers), finding and classifying a stationary point using standard second derivative test, and applying the standard volume of revolution formula. All techniques are standard P1/C1 level with no novel insights required, making it slightly easier than average. |
| Spec | 1.07i 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 |
|---|---|---|
| \(dy/dx = 2x - 2/x^2\) | B1 | For \(-2/x^2\) or for \(-2x^{-2}\) |
| \(d^2y/dx^2 = 2 + 4/x^3\) | B1, B1\(\sqrt{}\) [3] | For "\(2x\)" and for "\(2\)"; for \(+4/x^3\) or \(4x^{-3}\) or for diff. his \(dy/dx\) as long as negative power of \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(dy/dx = 0\): \(2x - 2/x^2 = 0\); \(\rightarrow x^3=1 \rightarrow x=1\), \(y=3\) | M1, A1 | Putting \(dy/dx=0\) and solving for \(x\); (\(\pm 1\) gets M1A0 but can get next M1A1) |
| If \(x=1\), \(d^2y/dx^2 > 0\), Minimum | M1A1\(\sqrt{}\) [4] | Looking at sign of \(d^2y/dx^2\) or other; \(\sqrt{}\) for his \(x\) into his \(d^2y/dx^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Vol} = \pi\int y^2\,dx = \pi\int(x^4 + 4/x^2 + 4x)\,dx\) | M1 | Attempt at squaring + integration; still gets M1 if \((a+b)^2 = a^2+b^2\) |
| \(= \pi\left[x^5/5 - 4/x + 2x^2\right]\) | \(3\times\)A1 | For each term and \(\pi\); can get A1A1 for above error |
| \([\;]_2 - [\;]_1 = 71\pi/5\) or \(44.6\) | DM1A1 [6] | Use of limits; "\(-\)" needed for M1; co (no \(\pi\) loses last A1 and one of first A marks) |
# Question 10:
## Part (i)
| $dy/dx = 2x - 2/x^2$ | B1 | For $-2/x^2$ or for $-2x^{-2}$ |
| $d^2y/dx^2 = 2 + 4/x^3$ | B1, B1$\sqrt{}$ [3] | For "$2x$" and for "$2$"; for $+4/x^3$ or $4x^{-3}$ or for diff. his $dy/dx$ as long as negative power of $x$ |
## Part (ii)
| $dy/dx = 0$: $2x - 2/x^2 = 0$; $\rightarrow x^3=1 \rightarrow x=1$, $y=3$ | M1, A1 | Putting $dy/dx=0$ and solving for $x$; ($\pm 1$ gets M1A0 but can get next M1A1) |
| If $x=1$, $d^2y/dx^2 > 0$, Minimum | M1A1$\sqrt{}$ [4] | Looking at sign of $d^2y/dx^2$ or other; $\sqrt{}$ for his $x$ into his $d^2y/dx^2$ |
## Part (iii)
| $\text{Vol} = \pi\int y^2\,dx = \pi\int(x^4 + 4/x^2 + 4x)\,dx$ | M1 | Attempt at squaring + integration; still gets M1 if $(a+b)^2 = a^2+b^2$ |
| $= \pi\left[x^5/5 - 4/x + 2x^2\right]$ | $3\times$A1 | For each term and $\pi$; can get A1A1 for above error |
| $[\;]_2 - [\;]_1 = 71\pi/5$ or $44.6$ | DM1A1 [6] | Use of limits; "$-$" needed for M1; co (no $\pi$ loses last A1 and one of first A marks) |10 A curve has equation $y = x ^ { 2 } + \frac { 2 } { x }$.\\
(i) Write down expressions for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.\\
(ii) Find the coordinates of the stationary point on the curve and determine its nature.\\
(iii) Find the volume of the solid formed when the region enclosed by the curve, the $x$-axis and the lines $x = 1$ and $x = 2$ is rotated completely about the $x$-axis.
\hfill \mbox{\textit{CAIE P1 2004 Q10 [13]}}