Iteration - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 36 exam-style questions on Iteration for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

36 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Show that the equation \(x^3 + 5x - 3 = 0\) has a solution between \(x = 0\) and \(x = 1\)

Model answer

When \(x = 0\): \(0^3 + 5(0) - 3 = -3\)
When \(x = 1\): \(1^3 + 5(1) - 3 = 3\)
There is a change of sign (from negative to positive), so \(x^3 + 5x - 3 = 0\) has a solution between \(x = 0\) and \(x = 1\).

Explanation

Substitute both values into \(x^3 + 5x - 3\). If one result is negative and the other positive, the graph must cross zero in between (the expression is continuous), so there is a solution. You must state the sign change in your conclusion.

Mark scheme

1 mark
M1 for substituting \(x = 0\) and \(x = 1\) into \(x^3 + 5x - 3\): −3 and 3
both values needed
1 mark
A1 for a correct conclusion, e.g. there is a change of sign, so there is a solution between 0 and 1
answer only scores 0

Question 2

2 marks

The iteration formula \(x_{n+1} = 5 - \dfrac{2}{x_n}\) converges to a solution of a quadratic equation.

Find the quadratic equation in the form \(x^2 + ax + b = 0\), where \(a\) and \(b\) are integers.

Model answer

At the solution, \(x_{n+1} = x_n = x\):
\(x = 5 - \dfrac{2}{x}\)
\(x^2 = 5x - 2\)
\(x^2 - 5x + 2 = 0\)

Explanation

When the iteration has converged, the input and output are equal, so write \(x = 5 - \frac{2}{x}\). Multiply every term by \(x\) and rearrange. You could check with the formula: the larger solution is \(\frac{5 + \sqrt{17}}{2} = 4.5615...\)

Mark scheme

1 mark
M1 for \(x = 5 - \frac{2}{x}\) and multiplying through by \(x\): \(x^2 = 5x - 2\)
oe; correct answer with no working scores full marks
1 mark
A1 for \(x^2 - 5x + 2 = 0\)
cao

Question 3

3 marks

Using \(x_0 = 4\) and the iteration formula
\(x_{n+1} = 5 - \dfrac{2}{x_n}\)
find the value of \(x_3\).
Give your answer correct to 4 decimal places.

Model answer

\(x_1 = 5 - \dfrac{2}{4} = 4.5\)
\(x_2 = 5 - \dfrac{2}{4.5} = 4.55555...\)
\(x_3 = 5 - \dfrac{2}{4.55555...} = 4.56097...\)
\(x_3 = 4.5610\) (4 d.p.)

Explanation

Substitute each value into the formula in turn. Notice that the 0 must be written in 4.5610 to show 4 decimal places. The values are converging to a solution of the quadratic equation \(x^2 - 5x + 2 = 0\).

Mark scheme

1 mark
M1 for \(x_1 = 4.5\)
cao; correct answer with no working scores full marks
1 mark
M1 for \(x_2 = 4.5555...\)
accept 4.556 or \(\frac{41}{9}\)
1 mark
A1 for \(x_3 = 4.5610\)
awrt 4.5610 (4.56097...); accept \(\frac{187}{41}\)

Question 4

4 marks

Marco invests £5000.
At the end of each year, interest of 3.2% is added to his investment and then he adds a further £1200.
The value of his investment after \(n\) years, £\(V_n\), is given by
\(V_{n+1} = 1.032V_n + 1200\), with \(V_0 = 5000\)

After how many years will the value of Marco's investment first be more than £10 000?
You must show your working.

Model answer

\(V_1 = 1.032 \times 5000 + 1200 = 6360\)
\(V_2 = 1.032 \times 6360 + 1200 = 7763.52\)
\(V_3 = 1.032 \times 7763.52 + 1200 = 9211.95...\)
\(V_4 = 1.032 \times 9211.95... + 1200 = 10\,706.73...\)
\(V_3 < 10\,000 < V_4\), so after 4 years.

Explanation

Use the formula repeatedly: multiply by 1.032 to add the interest, then add £1200. Keep going until the value first exceeds £10 000, showing every value. A common error is to add the £1200 before multiplying by 1.032.

Mark scheme

1 mark
M1 for \(V_1 = 6360\)
cao
1 mark
M1 for \(V_2 = 7763.52\)
or a correct substitution of their \(V_1\)
1 mark
M1 for \(V_3 = 9211.95...\) and \(V_4 = 10\,706.73...\)
accept values to the nearest penny or pound
1 mark
A1 for 4 years
cao; must follow correct values

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