Practise 28 exam-style questions on Iteration for Edexcel GCSE Maths (Non-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.
28 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 \(x^3 - 4x - 1 = 0\) has a solution between \(x = 2\) and \(x = 3\)
Model answer
When \(x = 2\): \(2^3 - 4(2) - 1 = 8 - 8 - 1 = -1\) When \(x = 3\): \(3^3 - 4(3) - 1 = 27 - 12 - 1 = 14\) There is a change of sign (from negative to positive), so there is a solution between \(x = 2\) and \(x = 3\).
Explanation
Work out the value of \(x^3 - 4x - 1\) at each end of the interval. A change of sign means the graph crosses the \(x\)-axis somewhere between them. You must state the conclusion, not just the two values.
Mark scheme
1 mark
M1 for substituting \(x = 2\) and \(x = 3\), e.g. \(8 - 8 - 1 = -1\) and \(27 - 12 - 1 = 14\)
1 mark
C1 for \(-1\) and 14 with a statement that there is a change of sign, so there is a solution between 2 and 3
Ella says, "\(f(1)\) and \(f(3)\) are both positive, so the equation \(f(x) = 0\) has no solutions between \(x = 1\) and \(x = 3\)."
Show that Ella is wrong.
Model answer
\(f(2) = 2(4) - 16 + 7 = -1\) \(f(1) = 1\) is positive and \(f(2) = -1\) is negative, so there is a solution between 1 and 2. \(f(2) = -1\) is negative and \(f(3) = 1\) is positive, so there is another solution between 2 and 3. So there are two solutions between 1 and 3, and Ella is wrong.
Explanation
No change of sign does not prove there is no solution: the graph can cross the \(x\)-axis twice (down and back up) between the two values. Checking a value in the middle reveals the two crossings.
Mark scheme
1 mark
M1 for finding a negative value of \(f(x)\) between 1 and 3, e.g. \(f(2) = 8 - 16 + 7 = -1\)
or \(f(1.5) = -0.5\), \(f(2.5) = -0.5\)
1 mark
C1 for explaining that there are changes of sign between 1 and 2 and between 2 and 3, so there are (two) solutions between 1 and 3
oe; or solving \(f(x) = 0\) to get \(x = 2 \pm \dfrac{\sqrt{2}}{2}\), both between 1 and 3
Question 3
3 marks
Using the iterative formula \(x_{n+1} = 1 + \dfrac{2}{x_n}\) with \(x_0 = 1\), find the values of \(x_1\), \(x_2\) and \(x_3\)
The iteration converges to one of the solutions of \(x^2 - x - 2 = 0\). Which solution? Give a reason for your answer.
Model answer
\(x_1 = 1 + \dfrac{2}{1} = 3\) \(x_2 = 1 + \dfrac{2}{3} = \dfrac{5}{3}\) \(x_3 = 1 + 2 \div \dfrac{5}{3} = 1 + \dfrac{6}{5} = \dfrac{11}{5}\) \(x^2 - x - 2 = (x - 2)(x + 1)\), so the solutions are 2 and \(-1\). The values 3, 1.67, 2.2 are on alternate sides of 2 and getting closer to it (and they are all positive), so the iteration converges to \(x = 2\).
Explanation
If the iteration settles at a value \(x\), then \(x = 1 + \dfrac{2}{x}\), which rearranges to \(x^2 - x - 2 = 0\). Of its two solutions, the iterates are clearly closing in on 2, jumping above and below it each time.
Mark scheme
1 mark
B1 for \(x_1 = 3\) and \(x_2 = \dfrac{5}{3}\)
oe
1 mark
B1 for \(x_3 = \dfrac{11}{5}\)
oe, e.g. 2.2; ft their \(x_2\)
1 mark
C1 for \(x = 2\) with a reason, e.g. the solutions are 2 and \(-1\) and the values (3, 1.67, 2.2, ...) are getting closer to 2
oe
Question 4
4 marks
Using the iterative formula \(x_{n+1} = \dfrac{12}{x_n + 1}\) with \(x_0 = 1\), find the values of \(x_1\) and \(x_2\)
The iteration converges. Find the exact value that it converges to. Give a reason for your answer.
Model answer
\(x_1 = \dfrac{12}{1 + 1} = 6\) \(x_2 = \dfrac{12}{6 + 1} = \dfrac{12}{7}\) The limit satisfies \(x = \dfrac{12}{x + 1}\) \(x^2 + x = 12\) \(x^2 + x - 12 = 0\) \((x + 4)(x - 3) = 0\) \(x = 3\) or \(x = -4\) Starting from a positive number, every value \(\dfrac{12}{x_n + 1}\) is positive, so the limit cannot be \(-4\). The iteration converges to 3.
Explanation
Replace \(x_{n+1}\) and \(x_n\) by \(x\) to find the equation the limit must satisfy, then solve it exactly. When the equation has two solutions, use the iterates to decide which one is the limit; here they are always positive.
Mark scheme
1 mark
B1 for \(x_1 = 6\) and \(x_2 = \dfrac{12}{7}\)
oe
1 mark
M1 for forming \(x = \dfrac{12}{x + 1}\) and rearranging, \(x^2 + x - 12 = 0\)
1 mark
M1 for solving, \((x + 4)(x - 3) = 0\), so \(x = 3\) or \(x = -4\)
1 mark
C1 for 3 with a reason, e.g. all the values of \(x_n\) are positive, so the limit cannot be \(-4\)
oe
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