| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2010 |
| Session | June |
| Marks | 22 |
| Topic | Reduction Formulae |
| Type | Algebraic function with square root |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question combining reduction formulae, hyperbolic functions, and surface of revolution. Part (i) requires integration by parts following a hint—standard technique but multi-step. Parts (ii)(a-b) are routine verification. Part (c) requires setting up surface area formula with parametric equations—moderately challenging. Part (d) involves a non-trivial substitution with hyperbolic functions leading to algebraic manipulation. The question requires sustained reasoning across multiple techniques but follows a guided path with hints, placing it above average difficulty but not at the extreme end for Further Maths. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07b Hyperbolic graphs: sketch and properties4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07d Differentiate/integrate: hyperbolic functions4.08d Volumes of revolution: about x and y axes8.06a Reduction formulae: establish, use, and evaluate recursively8.06b Arc length and surface area: of revolution, cartesian or parametric |
**(i)**
$I_n = x^{n-1}\sqrt{1+x^2} - \int(n-1)x^{n-2}\sqrt{1+x^2}\,dx$ **M1** Use of parts; **A1**
$= x^{n-1}\sqrt{1+x^2}-(n-1)\int x^{n-2}\frac{(1+x^2)}{\sqrt{1+x^2}}\,dx$ **M1** Getting into appropriate form
$= x^{n-1}\sqrt{1+x^2}-(n-1)\{I_{n-2}+I_n\}$ **M1** for splitting 2nd integral up into $I$'s
$\Rightarrow nI_n = x^{n-1}\sqrt{1+x^2}-(n-1)I_{n-2}$ **A1** correctly shown [5]
**(ii)(a)** $(\pm 2, 0)$ **B1** [1]
**(b)**
$\frac{x^2}{4}-\frac{y^2}{16} = \cosh^2\theta-\sinh^2\theta = 1$ shown **B1**
$x = 2\cosh\theta \geq 2$ for all $\theta \Rightarrow J$ is RHS of $H$ **B1**
$y = 4\sinh\theta \geq 0$ since $\theta\geq 0 \Rightarrow J$ is 1st-quadrant part of $H$ only **B1** [3]
**(c)**
$\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2} = 2\sqrt{5\cosh^2\theta-1}$ **M1 A1**
$S = 2\pi\int y\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}\,d\theta$ **M1** for quoting surface area result...
$= 2\pi\int 4\sinh\theta\cdot 2\sqrt{5\cosh^2\theta-1}\,d\theta$ ...and using parametrisation (fully)
$= 16\pi\int_{x=2}^{x=34}\sinh\theta\sqrt{5\cosh^2\theta-1}\,d\theta$ **A1** (Legitimately: ANSWER GIVEN). Ignore limits. [4]
**(d)**
$u^2 = 5\cosh^2\theta-1 \Rightarrow 2u\,du = 10\cosh\theta\sinh\theta\,d\theta \Rightarrow \frac{u}{5\sqrt{\frac{u^2+1}{5}}}du = \sinh\theta\,d\theta$ **B1**
Limits: $x(2,34)\to\cosh\theta(1,17)\to u(2,38)$ **B1** At any stage
$S = 16\pi\frac{1}{\sqrt{5}}\frac{1}{2}\int_2^{38}\frac{u^2}{\sqrt{u^2+1}}\,du$ **M1** Complete substitution. **A1** (ignore limits) $= \frac{16\pi}{\sqrt{5}}I_2$
Now $2I_2 = x\sqrt{x^2+1}-I_0$ **M1** Use of (i)'s reduction formula or equivalent calculus
**B1** $I_0 = \sinh^{-1}u$
$\Rightarrow S = \frac{8\pi}{\sqrt{5}}\left[u\sqrt{1+u^2}-\sinh^{-1}u\right]_2^{38}$ **A1 ft**
**M1** Use of log. form for $\sinh^{-1}$
**A1** Correct answer legitimately obtained in given form from using
$\ln(38+17\sqrt{5})-\ln(2+\sqrt{5}) = \ln\left(\frac{38+17\sqrt{5}}{2+\sqrt{5}}\times\frac{\sqrt{5}-2}{\sqrt{5}-2}\right) = \ln(9+4\sqrt{5})$
$= \frac{8\pi}{\sqrt{5}}\left(644\sqrt{5}-\ln(9+4\sqrt{5})\right)$ **[9]**12 (i) Let $I _ { n } = \int \frac { x ^ { n } } { \sqrt { x ^ { 2 } + 1 } } \mathrm {~d} x$, for integers $n \geqslant 0$.\\
By writing $\frac { x ^ { n } } { \sqrt { x ^ { 2 } + 1 } }$ as $x ^ { n - 1 } \times \frac { x } { \sqrt { x ^ { 2 } + 1 } }$, or otherwise, show that, for $n \geqslant 2$,
$$n I _ { n } = x ^ { n - 1 } \sqrt { x ^ { 2 } + 1 } - ( n - 1 ) I _ { n - 2 } .$$
(ii) The diagram shows a sketch of the hyperbola $H$ with equation $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 1$.\\
\includegraphics[max width=\textwidth, alt={}, center]{32ed7cc8-3456-4cf0-952a-ee04eada1298-6_593_666_776_776}
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the points where $H$ crosses the $x$-axis.
\item The curve $J$ has parametric equations $x = 2 \cosh \theta , y = 4 \sinh \theta$, for $\theta \geqslant 0$. Show that these parametric equations satisfy the cartesian equation of $H$, and indicate on a copy of the above diagram which part of $H$ is $J$.
\item The arc of the curve $J$ between the points where $x = 2$ and $x = 34$ is rotated once completely about the $x$-axis to form a surface of revolution with area $S$. Show that
$$S = 16 \pi \int _ { \alpha } ^ { \beta } \sinh \theta \sqrt { 5 \cosh ^ { 2 } \theta - 1 } \mathrm {~d} \theta$$
for suitable constants $\alpha$ and $\beta$.
\item Use the substitution $u ^ { 2 } = 5 \cosh ^ { 2 } \theta - 1$ to show that
$$S = \frac { 8 \pi } { \sqrt { 5 } } ( 644 \sqrt { 5 } - \ln ( 9 + 4 \sqrt { 5 } ) )$$
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2010 Q12 [22]}}