| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2010 |
| Session | June |
| Marks | 18 |
| Topic | Taylor series |
| Type | Explicit differential equation series solution |
| Difficulty | Challenging +1.8 This is a substantial multi-part question requiring proof by induction, Taylor series construction via successive differentiation, solving a separable differential equation, and applying the result to normal distribution probability. While each component uses standard Further Maths techniques, the combination of proof, series manipulation, and cross-topic application (connecting differential equations to probability via the normal distribution) elevates this above routine exercises. The induction proof requires careful handling of derivatives, and part (iii) demands recognition that the DE solution relates to the normal CDF. However, the techniques are all within the standard Further Maths toolkit without requiring exceptional insight. |
| Spec | 2.04f Find normal probabilities: Z transformation4.01a Mathematical induction: construct proofs4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
**(i)(a)**
Differentiating $\frac{dy}{dx}+xy=0 \Rightarrow \frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=0$
[while formula gives $\frac{d^2y}{dx^2}+x\frac{d^1y}{dx^1}+1\frac{d^0y}{dx^0}=0$] and result is true for $n=1$ **B1**
MUST come from differentiating given d.e.
Assuming that $\frac{d^{k+1}y}{dx^{k+1}}+x\frac{d^ky}{dx^k}+\frac{d^{k-1}y}{dx^{k-1}}=0$ **B1** Induction hypothesis (or later round-up)
Differentiating to get $\frac{d^{k+2}y}{dx^{k+2}}+x\frac{d^{k+1}y}{dx^{k+1}}+\frac{d^ky}{dx^k}+k\frac{d^ky}{dx^k}=0$ **M1**
$\Rightarrow \frac{d^{k+2}y}{dx^{k+2}}+x\frac{d^{k+1}y}{dx^{k+1}}+(k+1)\frac{d^ky}{dx^k}=0$ **A1**
Proper explanation of induction logic **B1** [5]
**(b)**
At $x=0$, $y=1 \Rightarrow \frac{dy}{dx}=0$ **B1**
Using the given recurrence relation **M1** $\frac{d^2y}{dx^2}=-1$, $\frac{d^4y}{dx^4}=3$, $\frac{d^6y}{dx^6}=-15$
and $\frac{d^3y}{dx^3}=\frac{d^5y}{dx^5}=0$ **B1** odds all zero
Use of Maclaurin's Theorem: $y = y(0)+xy'(0)+\frac{x^2}{2!}y''(0)+\frac{x^3}{3!}y'''(0)+\ldots$ **M1**
$y = 1-\frac{1}{2}x^2+\frac{1}{8}x^4-\frac{1}{48}x^6+\ldots$ **A1 cao** [5]
**(ii)**
EITHER by the integrating factor method: $e^{\frac{1}{2}x^2}\frac{dy}{dx}+xe^{\frac{1}{2}x^2}y=0 \Rightarrow ye^{\frac{1}{2}x^2}=C$ **M1 A1**
Use of $x=0$, $y=1$ to get $y=e^{-\frac{1}{2}x^2}$ **M1 A1**
OR by separation of variables: $\int\frac{dy}{y}=-\int x\,dx \Rightarrow \ln y = -\frac{1}{2}x^2+C \Rightarrow y=Ae^{-\frac{1}{2}x^2}$ **M1 A1**
Use of $x=0$, $y=1$ to get $y=e^{-\frac{1}{2}x^2}$ **M1 A1** [4]
**(iii)**
$p(Z\leq 1) = \int_{-\infty}^{1}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\,dx \approx \frac{1}{2}+\frac{1}{\sqrt{2\pi}}\int_0^1\left(1-\frac{1}{2}x^2+\frac{1}{8}x^4-\frac{1}{48}x^6\right)dx$ **M1 M1**
$= \frac{1}{2}+\frac{1}{\sqrt{2\pi}}\left[x-\frac{1}{6}x^3+\frac{1}{40}x^5-\frac{1}{336}x^7\right]_0^1$ **A1** integration correct
$= 0.5+0.39894\times 0.85536\quad [\ ]=\frac{479}{560}$; $\frac{1}{2}+\frac{0.855357}{2.506628}$
$= 0.8412\ (4)$ **A1**
[c.f. 0.8413 from Normal tables] [4]11 (i) At all points $( x , y )$ on the curve $C , \frac { \mathrm {~d} y } { \mathrm {~d} x } + x y = 0$.
\begin{enumerate}[label=(\alph*)]
\item Prove by induction that, for all integers $n \geqslant 1$,
$$\frac { \mathrm { d } ^ { n + 1 } y } { \mathrm {~d} x ^ { n + 1 } } + x \frac { \mathrm {~d} ^ { n } y } { \mathrm {~d} x ^ { n } } + n \frac { \mathrm {~d} ^ { n - 1 } y } { \mathrm {~d} x ^ { n - 1 } } = 0$$
where $\frac { \mathrm { d } ^ { 0 } y } { \mathrm {~d} x ^ { 0 } } = y$.
\item Given that $y = 1$ when $x = 0$, determine the Maclaurin expansion of $y$ in ascending powers of $x$, up to and including the term in $x ^ { 6 }$.\\
(ii) Solve the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } + x y = 0$ given that $y = 1$ when $x = 0$.\\
(iii) Given that $Z \sim \mathrm {~N} ( 0,1 )$, use your answers to parts (i) and (ii) to find an approximation, to 4 decimal places, to the probability $\mathrm { P } ( Z \leqslant 1 )$.\\[0pt]
[Note that the probability density function of the standard normal distribution is $\mathrm { f } ( z ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - \frac { 1 } { 2 } z ^ { 2 } }$.]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2010 Q11 [18]}}