Edexcel F1 2022 January — Question 8 10 marks

Exam BoardEdexcel
ModuleF1 (Further Pure Mathematics 1)
Year2022
SessionJanuary
Marks10
PaperDownload PDF ↗
TopicSign Change & Interval Methods
TypePure Interval Bisection Only
DifficultyStandard +0.3 This is a straightforward application of standard numerical methods (sign change, interval bisection, linear interpolation) with a given function and table. Parts (a)-(c) involve routine substitution and bisection algorithm, while part (d) applies a standard linear interpolation formula. The function is complex but all calculations are guided, requiring no novel insight—slightly easier than average A-level due to its procedural nature.
Spec1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

$$f(x) = 2x^{-\frac{2}{3}} + \frac{1}{2}x - \frac{1}{3x - 5} - \frac{5}{2} \quad x \neq \frac{5}{3}$$ The table below shows values of \(f(x)\) for some values of \(x\), with values of \(f(x)\) given to 4 decimal places where appropriate.
\(x\)12345
\(f(x)\)0.5\(-0.2885\)0.5834
  1. Complete the table giving the values to 4 decimal places. [2]
The equation \(f(x) = 0\) has exactly one positive root, \(\alpha\). Using the values in the completed table and explaining your reasoning,
  1. determine an interval of width one that contains \(\alpha\). [2]
  2. Hence use interval bisection twice to obtain an interval of width 0.25 that contains \(\alpha\). [3]
Given also that the equation \(f(x) = 0\) has a negative root, \(\beta\), in the interval \([-1, -0.5]\)
  1. use linear interpolation once on this interval to find an approximation for \(\beta\). Give your answer to 3 significant figures. [3]

$$f(x) = 2x^{-\frac{2}{3}} + \frac{1}{2}x - \frac{1}{3x - 5} - \frac{5}{2} \quad x \neq \frac{5}{3}$$

The table below shows values of $f(x)$ for some values of $x$, with values of $f(x)$ given to 4 decimal places where appropriate.

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$f(x)$ & 0.5 & & $-0.2885$ & & 0.5834 \\
\hline
\end{tabular}
\end{center}

\begin{enumerate}[label=(\alph*)]
\item Complete the table giving the values to 4 decimal places.
[2]
\end{enumerate}

The equation $f(x) = 0$ has exactly one positive root, $\alpha$.

Using the values in the completed table and explaining your reasoning,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item determine an interval of width one that contains $\alpha$.
[2]

\item Hence use interval bisection twice to obtain an interval of width 0.25 that contains $\alpha$.
[3]
\end{enumerate}

Given also that the equation $f(x) = 0$ has a negative root, $\beta$, in the interval $[-1, -0.5]$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item use linear interpolation once on this interval to find an approximation for $\beta$.

Give your answer to 3 significant figures.
[3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel F1 2022 Q8 [10]}}