Revision Library

revisionlibrary.co.uk

HIGHER

7.10 Iteration

EDEXCEL 1MA1 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

An iteration formula is x_(n+1) = 7/(x_n + 4), with x0 = 1.

(Total for Question 1 is 3 marks)

Show that the equation x³ - 4x - 2 = 0 can be rearranged to give x_(n+1) = cbrt(4x_n + 2).

(Total for Question 2 is 2 marks)

The equation x² - 5x - 3 = 0 can be rearranged into the form x = √ax + b. Which of the following is correct?

A) x = √5x + 3
B) x = (x² - 3)/5
C) x = 3/(x - 5)
D) x = 5 + 3/x

(Total for Question 3 is 1 mark)

g(x) = x³ + x - 1. Show that the equation g(x) = 0 has a root between x = 0 and x = 1.

(Total for Question 4 is 2 marks)

The equation x³ - 4x - 2 = 0 has the iteration formula x_(n+1) = cbrt(4x_n + 2). Using x0 = 2, find the values of x1, x2 and x3, giving your answers to 2 decimal places.

(Total for Question 5 is 3 marks)

f(x) = x³ + 2x - 5. Show that the equation f(x) = 0 has a root between x = 1 and x = 2.

(Total for Question 6 is 3 marks)

The equation x² - 6x - 1 = 0 has a positive root.

(a)Show that the equation x² - 6x - 1 = 0 can be rearranged to give x_(n+1) = 6 + 1/x_n.(2)

(b)Using x0 = 6, find the values of x1, x2 and x3, giving your answers to 3 decimal places.(3)

(c)Hence write down the positive root of x² - 6x - 1 = 0 correct to 2 decimal places, justifying your answer.(2)

(Total for Question 7 is 7 marks)

A designer is making an open storage box. The height, x metres, of the box must satisfy the equation x³ + 4x - 30 = 0. This equation can be rearranged to give the iteration formula x_(n+1) = cbrt(30 - 4x_n). Starting with x₀ = 3, use the iteration formula to work out the values of x₁, x₂ and x₃. Hence write down the height of the box correct to 1 decimal place.

(Total for Question 8 is 4 marks)

An iteration formula is x_(n+1) = √3x_n + 4, with x0 = -1.

(Total for Question 9 is 3 marks)

An iteration formula is x_(n+1) = √x_n + 6, with x0 = 3.

(a)Find the values of x1, x2 and x3, giving your answers to 3 decimal places.(2)

(b)Explain why every iterate in this sequence is exactly 3.(1)

(Total for Question 10 is 3 marks)

An iteration formula is x_(n+1) = 3/(x_n - 2). Which of these values of x0 would make it impossible to calculate x1?

A) x0 = 0
B) x0 = 1
C) x0 = 2
D) x0 = 5

(Total for Question 11 is 1 mark)

The equation x³ - 4x - 2 = 0 has a root at x = 2.21 (2 dp). A student instead uses the iteration formula x_(n+1) = (x_n³ - 2)/4, with x0 = 2.

(Total for Question 12 is 3 marks)

The equation x³ - 4x - 2 = 0 has a root at x = 2.21 (2 dp). Student A uses x_(n+1) = cbrt(4x_n + 2). Student B uses x_(n+1) = (x_n³ - 2)/4. Both students start with x0 = 2.

(Total for Question 13 is 3 marks)

The equation x³ - 7x + 3 = 0 has a small positive root, given by the iteration formula x_(n+1) = (x_n³ + 3)/7. Using x0 = 0.5, find the values of x1, x2 and x3, giving each answer to 3 decimal places, and hence state the root correct to 3 decimal places.

(Total for Question 14 is 4 marks)

An iteration formula is x_(n+1) = √5x_n + 2. Given that x1 = 3, find the value of x0.

(Total for Question 15 is 3 marks)

The iteration formula x_(n+1) = √2x_n + 3 converges, for a suitable x0, to the positive root of x² - 2x - 3 = 0. What is this positive root?

A) 1
B) -1
C) 3
D) 5

(Total for Question 16 is 1 mark)

The equation x³ - 3x - 1 = 0 has a root between x = 1 and x = 2.

(a)Show that x³ - 3x - 1 = 0 has a root between x = 1 and x = 2.(2)

(b)Show that the equation x³ - 3x - 1 = 0 can be rearranged to give x_(n+1) = cbrt(3x_n + 1).(2)

(c)Using x0 = 2, find the values of x1, x2, x3 and x4, giving each answer to 4 decimal places.(4)

(d)Hence state the root of x³ - 3x - 1 = 0 correct to 2 decimal places, justifying your answer.(2)

(Total for Question 17 is 10 marks)

An iteration formula is x_(n+1) = 4/x_n, with x0 = 1. This formula comes from rearranging x² = 4. Show that this sequence oscillates between two values and never converges towards the positive root of x² = 4 (x = 2).

(Total for Question 18 is 3 marks)

Show that the equation 2x³ - 5x - 1 = 0 can be rearranged to give x_(n+1) = √(5x_n + 1)/(2x_n).

(Total for Question 19 is 3 marks)

Mark scheme · 7.10 Iteration

Question 1

B1 x1 = 1.400 cao
B1 x2 = 1.296 cao, ft correct use of their x1
B1 x3 = 1.322 cao, ft correct use of their x2
Answer: x1 = 1.400, x2 = 1.296, x3 = 1.322

Question 2

M1 rearranges to isolate the x³ term, e.g. x³ = 4x + 2
A1 takes the cube root of both sides to reach x = cbrt(4x + 2), cso
Answer: x_(n+1) = cbrt(4x_n + 2) (shown)

Question 3

B1 A cao
Answer: A

Question 4

B1 g(0) = -1 and g(1) = 1 both found correctly
B1 correct conclusion, e.g. since g(0) is negative and g(1) is positive (a change of sign) and g is continuous, there is a root between x = 0 and x = 1, cso
Answer: g(0) = -1, g(1) = 1; the change of sign confirms a root between x = 0 and x = 1

Question 5

B1 x1 = 2.15 cao
B1 x2 = 2.20 cao, ft correct use of their x1
B1 x3 = 2.21 cao, ft correct use of their x2
Answer: x1 = 2.15, x2 = 2.20, x3 = 2.21

Question 6

B1 f(1) = -2 cao
B1 f(2) = 7 cao
B1 correct conclusion, e.g. since f(1) is negative and f(2) is positive (a change of sign) and f is continuous, there is a root between x = 1 and x = 2, cso
Answer: f(1) = -2, f(2) = 7; the change of sign confirms a root between x = 1 and x = 2

Question 7

(a) M1 rearranges to x² = 6x + 1
(a) A1 divides both sides by x to reach x = 6 + 1/x, cso
(a) Answer: x_(n+1) = 6 + 1/x_n (shown)
(b) B1 x1 = 6.167 cao, ft from part (a)
(b) B1 x2 = 6.162 cao, ft correct use of their x1
(b) B1 x3 = 6.162 cao, ft correct use of their x2
(b) Answer: x1 = 6.167, x2 = 6.162, x3 = 6.162
(c) B1 6.16 cao, ft from part (b)
(c) B1 correct justification, e.g. x2 and x3 both round to 6.16, so the root is 6.16 to 2 dp, cso ft
(c) Answer: root = 6.16 (2 dp)

Question 8

B1 x1 = 2.621 cao
B1 x2 = 2.692 cao, ft correct use of their x1
B1 x3 = 2.679 cao, ft correct use of their x2
B1 height = 2.7 (m) stated correct to 1 decimal place, ft
Answer: x1 = 2.621, x2 = 2.692, x3 = 2.679; height = 2.7 m (1 dp)

Question 9

B1 x1 = 1.000 cao
B1 x2 = 2.646 cao, ft correct use of their x1
B1 x3 = 3.455 cao, ft correct use of their x2
Answer: x1 = 1.000, x2 = 2.646, x3 = 3.455

Question 10

(a) B1 x1 = 3.000 cao
(a) B1 x2 = 3.000 and x3 = 3.000, ft
(a) Answer: x1 = 3.000, x2 = 3.000, x3 = 3.000
(b) B1 correct explanation, e.g. x0 = 3 already satisfies x² - x - 6 = 0 (i.e. x = 3 is already an exact root of the equation), so substituting it into the formula returns 3 every time, cso
(b) Answer: x0 = 3 is already the exact root, so the iteration reproduces 3 at every stage

Question 11

B1 C cao
Answer: C

Question 12

B1 x1 = 1.5 cao
B1 x2 = 0.344 cao (or awrt 0.34), ft correct use of their x1
B1 correct conclusion, e.g. the values are moving further away from x = 2.21 rather than closer to it, so this rearrangement diverges and does not find the root, cso
Answer: x1 = 1.5, x2 = 0.344; the sequence moves away from x = 2.21, so this rearrangement diverges

Question 13

B1 Student A: x1 = 2.15 and x2 = 2.20, both correctly found
B1 Student B: x1 = 1.500 and x2 = 0.344, both correctly found
B1 correct conclusion, e.g. Student A's values get closer to 2.21 at each step, so the sequence converges to the root, whereas Student B's values move further away from 2.21, so the sequence diverges, cso
Answer: Student A converges towards 2.21; Student B diverges away from 2.21

Question 14

B1 x1 = 0.446 cao
B1 x2 = 0.441 cao, ft correct use of their x1
B1 x3 = 0.441 cao, ft correct use of their x2
B1 root = 0.441 (3 dp) stated, justified since x2 and x3 agree to 3 dp, ft
Answer: x1 = 0.446, x2 = 0.441, x3 = 0.441; root = 0.441 (3 dp)

Question 15

M1 squares both sides correctly, e.g. 3² = 5x0 + 2, i.e. 9 = 5x0 + 2
M1 rearranges correctly, e.g. 5x0 = 7
A1 x0 = 1.4 cao
Answer: x0 = 1.4

Question 16

B1 C cao
Answer: C

Question 17

(a) B1 h(1) = -3 and h(2) = 1 both found correctly, where h(x) = x³ - 3x - 1
(a) B1 correct conclusion, e.g. since h(1) is negative and h(2) is positive (a change of sign), there is a root between x = 1 and x = 2, cso
(a) Answer: h(1) = -3, h(2) = 1; sign change confirms a root between x = 1 and x = 2
(b) M1 rearranges to isolate the x³ term, e.g. x³ = 3x + 1
(b) A1 takes the cube root of both sides to reach x = cbrt(3x + 1), cso
(b) Answer: x_(n+1) = cbrt(3x_n + 1) (shown)
(c) B1 x1 = 1.9129 cao, ft from part (b)
(c) B1 x2 = 1.8888 cao, ft correct use of their x1
(c) B1 x3 = 1.8821 cao, ft correct use of their x2
(c) B1 x4 = 1.8801 cao, ft correct use of their x3
(c) Answer: x1 = 1.9129, x2 = 1.8888, x3 = 1.8821, x4 = 1.8801
(d) B1 1.88 cao, ft from part (c)
(d) B1 correct justification, e.g. x3 and x4 both round to 1.88, so the root is 1.88 to 2 dp, cso ft
(d) Answer: root = 1.88 (2 dp)

Question 18

B1 x1 = 4 cao
B1 x2 = 1 cao, and states the pattern then repeats indefinitely (x3 = 4, x4 = 1, ...)
B1 correct conclusion, e.g. the sequence oscillates between 1 and 4 forever and never settles near x = 2, so this rearrangement does not converge to a root of x² = 4, cso
Answer: x1 = 4, x2 = 1, x3 = 4, x4 = 1, ...; the sequence oscillates between 1 and 4 and never converges to x = 2

Question 19

M1 rearranges to isolate the terms in x, e.g. 2x³ = 5x + 1
M1 divides both sides by 2x (x is not 0) to reach x² = (5x + 1)/(2x)
A1 takes the square root of both sides to reach x = √(5x + 1)/(2x), cso
Answer: x_(n+1) = √(5x_n + 1)/(2x_n) (shown)

Iteration - Worksheets, Questions and Revision