EDEXCEL GCSE MATHS · HIGHER
Exam Practice: Using iteration to solve equations
Using iteration to solve equations · Equations and graphs · Lesson 6 of 6 · 17 marks · 30 minutes
|
Name |
|
Date |
|
Instructions
• Answer all the questions.
• Write your answers in the spaces provided.
• The marks for each question are shown in brackets - use this as a guide to how much to write.
• The answers are on separate pages at the back. Attempt every question before you look at them.
• Answer all questions. Show your working.
Question 1 SHOW THAT (2 marks)
Show that the equation x³ − x − 1 = 0 has a root between 1 and 2.
|
|
|
|
|
|
(Total for Question 1 = 2 marks)
Question 2 WORK OUT (3 marks)
x_(n+1) = ∛(x_n + 1) and x₀ = 1. Work out the values of x₁, x₂ and x₃.
|
|
|
|
|
|
|
|
(Total for Question 2 = 3 marks)
Question 3 SHOW THAT (2 marks)
Show that x³ − x − 1 = 0 can be rearranged to give x = ∛(x + 1).
|
|
|
|
|
|
(Total for Question 3 = 2 marks)
Question 4 FIND (4 marks)
The graph of y = x³ − x − 1 is shown. It crosses the x-axis between x = 1 and x = 2. Use a trial and improvement method to find this root correct to 1 decimal place. You must show all your working.
|
|
|
|
|
|
|
|
|
|
|
|
(Total for Question 4 = 4 marks)
Question 5 FIND (4 marks)
(a) Show that x³ + 2x − 7 = 0 has a root between 1 and 2. (b) Find this root correct to 1 decimal place.
|
|
|
|
|
|
|
|
|
|
|
|
(Total for Question 5 = 4 marks)
Question 6 EXPLAIN (2 marks)
Using x_(n+1) = ∛(x_n + 1) with x₀ = 1, the values are x₅ = 1.32463 and x₆ = 1.32470. Write down the root of x³ − x − 1 = 0 to 3 decimal places and explain how you know.
|
|
|
|
|
|
(Total for Question 6 = 2 marks)
TOTAL FOR PAPER = 17 MARKS
|
|
Answers and mark scheme
Check your answer only once you have written one.
Question 1 (2 marks)
f(1) = −1 and f(2) = 5. There is a change of sign, so a root lies between 1 and 2.
• f(1) = −1 and f(2) = 5 M1
• Change of sign conclusion C1
Question 2 (3 marks)
x₁ = 1.2599, x₂ = 1.3123, x₃ = 1.3224
• x₁ = 1.26 B1
• x₂ = 1.31 B1
• x₃ = 1.32 B1
Question 3 (2 marks)
x³ = x + 1, so x = ∛(x + 1).
• x³ = x + 1 M1
• Cube root both sides A1
Question 4 (4 marks)
f(1.3) = −0.103; f(1.4) = 0.344; the root is between 1.3 and 1.4. f(1.35) = 0.1104 (positive), so the root is below 1.35 and the answer is 1.3.
• Tests values in the interval M1
• f(1.3) < 0 and f(1.4) > 0 A1
• Tests 1.35 M1
• 1.3 A1
Question 5 (4 marks)
(a) f(1) = −4, f(2) = 5: change of sign. (b) f(1.5) = −0.625, f(1.6) = 0.296, f(1.55) = −0.176, so the root is between 1.55 and 1.6: 1.6.
• f(1) = −4 and f(2) = 5 B1
• Tests 1.5 or 1.6 M1
• Tests 1.55 M1
• 1.6 A1
Question 6 (2 marks)
1.325, because both x₅ and x₆ round to 1.325 to 3 decimal places.
• 1.325 B1
• Successive values agree to 3 d.p. C1