Mathematics · Pearson Edexcel 9MA0 · Topic 9

Numerical methods

Finding approximate answers when exact ones are not possible: sign changes, iteration, Newton–Raphson and the trapezium rule.

5 sub-topics
Taught in Year 13
61 practice questions
5 animated lessons

Practice questions (61) Lessons Worksheets All Edexcel Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

9.1Locate roots by change of signY13
9.2Iteration, cobweb and staircase diagramsY13
9.3Newton–Raphson methodY13
9.4Trapezium ruleY13
9.5Numerical methods in contextY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 61 questions in this topic, free, no account needed.

Numerical methods

Not needed for Year 12Easy

Locating a root · Year 13 · 2 marks

\mathrm{f}(x) = x^3 - 2x - 5

Show that $\mathrm{f}(x) = 0$ has a root between $x = 2$ and $x = 3$ .
(2)

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Work out both values. $\mathrm{f}(2) = -1$ , $\mathrm{f}(3) = 16$
Mention both the sign change and continuity.There is a change of sign and $\mathrm{f}$ is continuous, so there is a root between 2 and 3

Total2

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Numerical methods

Not needed for Year 12Medium

Rearranging for an iteration · Year 13 · 4 marks

(a)
Show that $x^3 - 2x - 5 = 0$ can be written as $x = \sqrt{2 + \dfrac{5}{x}}$ .
(2)
(b)
Use $x_{n+1}$ ${}= \sqrt{2 + \dfrac{5}{x_n}}$ with $x_0 = 2$ to find $x_1$ , $x_2$ and $x_3$ .
(2)

Check your answer before you look at the solution

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Part (a)2 marks

Divide by $x$ . $x^3$ ${}= 2x + 5$ ${}\Rightarrow x^2$ ${}= 2 + \dfrac{5}{x}$
Take the positive square root (the root is positive)

Part (b)2 marks

$x_1 = 2.1213$ , $x_2 = 2.0873$
The values swing either side of the root as they close in. $x_3 = 2.0965$

Total4

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Numerical methods

Not needed for Year 12Hard

When Newton–Raphson fails · Year 13 · 5 marks

\mathrm{f}(x) = x^3 - 3x + 1

(a)
Explain why the Newton–Raphson method fails with $x_0 = 1$ .
(2)
(b)
Use $x_0 = 1.5$ to find $x_1$ .
(3)

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Part (a)2 marks

$\mathrm{f}'(1) = 3 - 3 = 0$
The tangent is horizontal, so it never meets the $x$ -axis (we would divide by zero)

Part (b)3 marks

$\mathrm{f}(1.5) = -0.125$ , $\mathrm{f}'(1.5) = 3.75$
$x_1 = 1.5 - \dfrac{-0.125}{3.75}$
$= 1.5333$

Total5

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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