| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Multi-part questions combining substitution with curve/area analysis |
| Difficulty | Challenging +1.8 This is a Further Maths question requiring arc length and surface of revolution formulas with non-trivial integration. Part (a) needs arc length formula with dy/dx = tan x, leading to ∫sec x which requires knowing a non-standard result. Part (b) requires surface area formula and substitution. Both parts demand multiple techniques and careful algebraic manipulation, placing it well above average difficulty but not at the extreme end for FP1. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=18.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks |
|---|---|
| (a) \(y = \tan x\) | B1 |
| \(\sqrt{1 + y_1'^2} = \sec x\) | M1 |
| \(s = \int_0^{\pi/3} \sec x \, dx\) (AEF) | A1 |
| \(= [\ln(\sec x + \tan x)]_0^{\pi/3} = ... = \ln(2 + \sqrt{3})\) (AG) | M1A1 |
| (b) \(S = 4\pi\int_0^l (x+3)^{1/2}(1 + 1/(x+3))^{1/2} dx\) (AEF) | M1A1 |
| \(= ... = 4\pi\int_0^l (x+4)^{1/2} dx\) | A1 |
| \(= (8\pi/3)[(x+4)^{3/2}]_0^l\) | A1 |
| \(= (8\pi/3)[5\sqrt{5} - 8]\) (AG) | A1 |
(a) $y = \tan x$ | B1 |
$\sqrt{1 + y_1'^2} = \sec x$ | M1 |
$s = \int_0^{\pi/3} \sec x \, dx$ (AEF) | A1 |
$= [\ln(\sec x + \tan x)]_0^{\pi/3} = ... = \ln(2 + \sqrt{3})$ (AG) | M1A1 |
(b) $S = 4\pi\int_0^l (x+3)^{1/2}(1 + 1/(x+3))^{1/2} dx$ (AEF) | M1A1 |
$= ... = 4\pi\int_0^l (x+4)^{1/2} dx$ | A1 |
$= (8\pi/3)[(x+4)^{3/2}]_0^l$ | A1 |
$= (8\pi/3)[5\sqrt{5} - 8]$ (AG) | A1 |8
\begin{enumerate}[label=(\alph*)]
\item The curve $C _ { 1 }$ has equation $y = - \ln ( \cos x )$. Show that the length of the arc of $C _ { 1 }$ from the point where $x = 0$ to the point where $x = \frac { 1 } { 3 } \pi$ is $\ln ( 2 + \sqrt { 3 } )$.
\item The curve $C _ { 2 }$ has equation $y = 2 \sqrt { } ( x + 3 )$. The arc of $C _ { 2 }$ joining the point where $x = 0$ to the point where $x = 1$ is rotated through one complete revolution about the $x$-axis. Show that the area of the surface generated is
$$\frac { 8 } { 3 } \pi ( 5 \sqrt { } 5 - 8 )$$
\end{enumerate}
\hfill \mbox{\textit{CAIE FP1 2009 Q8 [10]}}