| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 15 |
| Topic | Reduction Formulae |
| Type | Volume and surface area of revolution |
| Difficulty | Hard +2.3 This is a challenging Pre-U Further Maths question requiring integration by parts to derive a reduction formula for sec^n(t), then applying it to a surface of revolution problem. Part (i) demands careful manipulation and evaluation at π/3, while part (ii) requires recognizing the surface area formula in parametric form and connecting it to I_5, followed by recursive calculation using the reduction formula. The multi-step nature, proof requirement, and non-standard integral make this significantly harder than typical A-level questions. |
| Spec | 4.08d Volumes of revolution: about x and y axes4.08f Integrate using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(I_n = \int_0^{\frac{1}{6}\pi} \sec^n t \, dt = \int_0^{\frac{1}{6}\pi} \sec^{n-2}t \cdot \sec^2 t \, dt\) | M1 | Correct splitting and use of parts |
| \(= [\sec^{n-2}t \tan t]_0^{\frac{1}{6}\pi} - \int_0^{\frac{1}{6}\pi} \tan t(n-2)\sec^{n-3}t \sec t \tan t \, dt\) | A1, A1 | |
| \(I_n = \left(\frac{2}{\sqrt{3}}\right)^{n-2} \cdot \frac{1}{\sqrt{3}} - (n-2)\int_0^{\frac{1}{6}\pi} \sec^{n-2}t(\sec^2 t - 1) dt\) | M1 | Subst'. for \(\tan\) … |
| \(= \frac{2^{n-2}}{(\sqrt{3})^{n-1}} - (n-2)\{I_n - I_{n-2}\}\) | M1 | … and reverting to \(I\)'s |
| \((n-1)I_n = \frac{2^{n-2}}{(\sqrt{3})^{n-1}} + (n-2)I_{n-2}\) ANSWER GIVEN | A1 | |
| [5] | ||
| (ii) \(\dot{x}^2 + \dot{y}^2 = (\sec^2 t)^2 + (\sec^2 t \tan t)^2\) | M1 | |
| \(= \sec^4 t(1 + \tan^2 t) = \sec^6 t\) | A1 | |
| Use of \(S = \int 2\pi y\sqrt{\dot{x}^2 + \dot{y}^2} \, dt\) with \(y = \frac{1}{2}\sec^2 t\) and their \(\dot{x}\) and \(\dot{y}\) | M1 | |
| \(= \int 2\pi \cdot \frac{1}{2}\sec^2 t \sec^3 t \, dt = \pi I_5\) | A1 | |
| [4] | ||
| \(I_1 = \ln[\sec t + \tan t]\) | M1 | |
| \(= \ln\left(\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}}\right) - \ln 1 = \frac{1}{2}\ln 3\) | A1 | |
| Then, using the R.F., \(2I_3 = \frac{2}{3} + \frac{1}{2}\ln 3 \Rightarrow I_3 = \frac{1}{3} + \frac{1}{4}\ln 3\) | M1 | |
| Using the R.F. again, \(4I_5 = \frac{8}{9} + 3\left(\frac{1}{3} + \frac{1}{4}\ln 3\right)\) | M1, A1 | |
| \(\ldots\) leading to \(S = \frac{1}{144}(68 + 27\ln 3)\pi\) or exact equivalent | A1 | |
| [6] |
**(i)** $I_n = \int_0^{\frac{1}{6}\pi} \sec^n t \, dt = \int_0^{\frac{1}{6}\pi} \sec^{n-2}t \cdot \sec^2 t \, dt$ | M1 | Correct splitting and use of parts
$= [\sec^{n-2}t \tan t]_0^{\frac{1}{6}\pi} - \int_0^{\frac{1}{6}\pi} \tan t(n-2)\sec^{n-3}t \sec t \tan t \, dt$ | A1, A1 |
$I_n = \left(\frac{2}{\sqrt{3}}\right)^{n-2} \cdot \frac{1}{\sqrt{3}} - (n-2)\int_0^{\frac{1}{6}\pi} \sec^{n-2}t(\sec^2 t - 1) dt$ | M1 | Subst'. for $\tan$ …
$= \frac{2^{n-2}}{(\sqrt{3})^{n-1}} - (n-2)\{I_n - I_{n-2}\}$ | M1 | … and reverting to $I$'s
$(n-1)I_n = \frac{2^{n-2}}{(\sqrt{3})^{n-1}} + (n-2)I_{n-2}$ **ANSWER GIVEN** | A1 |
| [5] |
**(ii)** $\dot{x}^2 + \dot{y}^2 = (\sec^2 t)^2 + (\sec^2 t \tan t)^2$ | M1 |
$= \sec^4 t(1 + \tan^2 t) = \sec^6 t$ | A1 |
Use of $S = \int 2\pi y\sqrt{\dot{x}^2 + \dot{y}^2} \, dt$ with $y = \frac{1}{2}\sec^2 t$ and their $\dot{x}$ and $\dot{y}$ | M1 |
$= \int 2\pi \cdot \frac{1}{2}\sec^2 t \sec^3 t \, dt = \pi I_5$ | A1 |
| [4] |
$I_1 = \ln[\sec t + \tan t]$ | M1 |
$= \ln\left(\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}}\right) - \ln 1 = \frac{1}{2}\ln 3$ | A1 |
Then, using the R.F., $2I_3 = \frac{2}{3} + \frac{1}{2}\ln 3 \Rightarrow I_3 = \frac{1}{3} + \frac{1}{4}\ln 3$ | M1 |
Using the R.F. again, $4I_5 = \frac{8}{9} + 3\left(\frac{1}{3} + \frac{1}{4}\ln 3\right)$ | M1, A1 |
$\ldots$ leading to $S = \frac{1}{144}(68 + 27\ln 3)\pi$ or exact equivalent | A1 |
| [6] |
---\begin{enumerate}[label=(\roman*)]
\item Let $I_n = \int_0^{\frac{\pi}{2}} \sec^n t \, dt$ for positive integers $n$. Prove that, for $n \geqslant 2$,
$$(n - 1)I_n = \frac{2^{n-2}}{(\sqrt{3})^{n-1}} + (n - 2)I_{n-2}.$$ [5]
\item The curve with parametric equations $x = \tan t$, $y = \frac{1}{4}\sec^2 t$, for $0 \leqslant t \leqslant \frac{1}{4}\pi$, is rotated through $2\pi$ radians about the $x$-axis to form a surface of revolution of area $S$. Show that $S = \pi I_5$ and evaluate $S$ exactly. [10]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2011 Q11 [15]}}