| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Topic | Taylor series |
| Type | Taylor series about x=1: differential equation with given conditions at x=1 |
| Difficulty | Challenging +1.2 This is a structured Taylor series question with clear scaffolding. Part (i) requires substituting given values into the differential equation (routine), then differentiating the equation to find the third derivative (standard technique). Part (ii) applies the Taylor series formula mechanically with provided derivatives. While it involves a second-order ODE and requires careful differentiation using the product rule, the question guides students through each step explicitly, making it more accessible than it initially appears. This is moderately above average difficulty due to the Further Maths context and multi-step calculus, but the scaffolding prevents it from being truly challenging. |
| Spec | 4.08a Maclaurin series: find series for function4.10a General/particular solutions: of differential equations |
The function $y$ satisfies $\frac{d^2y}{dx^2} + x^2y = x$, and is such that $y = 1$ and $\frac{dy}{dx} = 1$ when $x = 1$.
\begin{enumerate}[label=(\roman*)]
\item Using the given differential equation
\begin{enumerate}[label=(\alph*)]
\item state the value of $\frac{d^2y}{dx^2}$ when $x = 1$, [1]
\item find, by differentiation, the value of $\frac{d^3y}{dx^3}$ when $x = 1$. [2]
\end{enumerate}
\item Hence determine the Taylor series for $y$ about $x = 1$ up to and including the term in $(x-1)^3$ and deduce, correct to 4 decimal places, an approximation for $y$ when $x = 1.1$. [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2018 Q7 [6]}}