Pure: Numerical Methods: Fluency and Exam Drill - Worksheets, Questions and Revision

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A-Level · Pure Mathematics

P9D Pure: Numerical Methods: Fluency and Exam Drill

EDEXCEL 9MA0 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

State the condition on f(a) and f(b) required for the change-of-sign test to conclude that a root of the continuous function f lies between x = a and x = b.

(Total for Question 1 is 1 mark)

Write down the general iteration formula used in the Newton-Raphson method for solving f(x) = 0.

(Total for Question 2 is 1 mark)

For the iteration x_(n+1) = g(x_n) converging to a root α, state the condition on g'(x) near α.

(Total for Question 3 is 1 mark)

Let f(x) = x³ - 6x + 2. Evaluate f(2).

(Total for Question 4 is 1 mark)

Let f(x) = x³ - 6x + 2. Evaluate f(3).

(Total for Question 5 is 1 mark)

Using x₀ = 2 and the Newton-Raphson iteration x_(n+1) = x_n - (x_n² - 7)/(2x_n), find x₁, giving your exact answer.

(Total for Question 6 is 1 mark)

Using x₀ = 1 and the iteration x_(n+1) = (2x_n + 5)^1/2, find x₁, giving your answer to 4 decimal places.

(Total for Question 7 is 1 mark)

Show that the equation x³ - 6x + 2 = 0 has a root between x = 2 and x = 3.

(Total for Question 8 is 2 marks)

Using x₀ = 2 and the Newton-Raphson iteration x_(n+1) = x_n - (x_n² - 7)/(2x_n), find x₁ and x₂, each to 4 decimal places.

(Total for Question 9 is 2 marks)

Using x₀ = 1 and the iteration x_(n+1) = (2x_n + 5)^1/2, find x₁ and x₂, each to 4 decimal places.

(Total for Question 10 is 2 marks)

Let f(x) = x³ - 6x + 2. Find f'(x), and hence find f'(2).

(Total for Question 11 is 2 marks)

Let f(x) = x³ - 6x + 2, with f'(x) = 3x² - 6. Using x₀ = 2, apply the Newton-Raphson method to find x₁, giving your answer to 4 decimal places.

(Total for Question 12 is 2 marks)

The table shows values of y = 1/(x+1) for x = 0, 1, 2, 3, 4. x: 0, 1, 2, 3, 4 y: 1, 0.5, 0.3333, 0.25, 0.2. Use the trapezium rule with all four strips (h=1) to find an estimate for the integral of 1/(x+1) from x=0 to x=4, giving your answer to 4 decimal places.

(Total for Question 13 is 2 marks)

Show that the equation x³ + 3x - 7 = 0 has a root between x = 1 and x = 2. Show that this equation can be rearranged into the iterative form x_(n+1) = (7 - 3x_n)^1/3.

(Total for Question 14 is 3 marks)

Using x₀ = 1.5, apply the iteration x_(n+1) = (7 - 3x_n)^1/3 to find x₁, x₂ and x₃, giving each value to 4 decimal places. Hence write down an estimate for the root of x³ + 3x - 7 = 0, correct to 1 decimal place.

(Total for Question 15 is 3 marks)

Let f(x) = x³ - 4x + 1. Show that f(x) = 0 has a root between x = 1 and x = 2. Using x₀ = 2 and f'(x) = 3x² - 4, apply the Newton-Raphson method to find x₁ and x₂, giving each answer to 4 decimal places.

(Total for Question 16 is 3 marks)

The curve y = ln(x+1) is defined for 0 ≤ x ≤ 4. The table gives values of y at x = 0, 1, 2, 3, 4. x: 0, 1, 2, 3, 4 y: 0, 0.6931, p, 1.3863, q. Find the values of p and q, each correct to 4 decimal places. Use the trapezium rule with all four strips (h=1) to find an estimate for the integral of ln(x+1) from x=0 to x=4, giving your answer to 4 decimal places.

(Total for Question 17 is 4 marks)

The curve y = 1/(x+2) is defined for x > -2. The table gives values of y at x = 0, 1, 2, 3, 4. x: 0, 1, 2, 3, 4 y: 0.5, 0.3333, 0.25, 0.2, 0.1667. Use the trapezium rule with all four strips (h=1) to find an estimate for the integral of 1/(x+2) from x=0 to x=4, giving your answer to 4 decimal places. By considering the sign of the second derivative of y = 1/(x+2), state, with a reason, whether this trapezium rule estimate is an overestimate or an underestimate of the true value of the integral.

(Total for Question 18 is 4 marks)

The equation x³ - 2x - 5 = 0 has a root α near x = 2. Two rearrangements are suggested: x_(n+1) = (2x_n + 5)^1/3 and x_(n+1) = (x_n³ - 5)/2. Using x₀ = 2 for each, find x₁ and x₂ for both iterations, giving each value to 4 decimal places, and hence state, with a reason, which rearrangement converges to α.

(Total for Question 19 is 4 marks)

Mark scheme · P9D Pure: Numerical Methods: Fluency and Exam Drill

Question 1

B1 f(a) and f(b) have opposite signs (one positive, one negative)
Answer: f(a) and f(b) must have opposite signs.

Question 2

B1 x_(n+1) = x_n - f(x_n)/f'(x_n) cao
Answer: x_(n+1) = x_n - f(x_n)/f'(x_n)

Question 3

B1 |g'(α)| < 1
Answer: |g'(α)| < 1

Question 4

B1 -2 cao
Answer: -2

Question 5

B1 11 cao
Answer: 11

Question 6

B1 2.75 cao, oe 11/4
Answer: x₁ = 2.75

Question 7

B1 awrt 2.6458
Answer: x₁ = 2.6458

Question 8

M1 evaluates f(2) = -2 and f(3) = 11
A1 states f(2) < 0 and f(3) > 0, a change of sign, so a root lies between x=2 and x=3
Answer: f(2) = -2 and f(3) = 11, a sign change, so a root lies between x=2 and x=3.

Question 9

M1 finds x₁ = 2.75
A1 x₂ = awrt 2.6477
Answer: x₁ = 2.7500, x₂ = 2.6477

Question 10

M1 finds x₁ = 2.6458 (or uses the unrounded value)
A1 x₂ = awrt 3.2080
Answer: x₁ = 2.6458, x₂ = 3.2080

Question 11

M1 differentiates to find f'(x) = 3x² - 6
A1 6 cao
Answer: f'(x) = 3x² - 6; f'(2) = 6

Question 12

M1 substitutes into x₁ = x₀ - f(x₀)/f'(x₀) = 2 - (-2)/6
A1 awrt 2.3333
Answer: x₁ = 2.3333

Question 13

M1 substitutes into the trapezium rule, (1/2)[(y₀+y₄) + 2(y₁+y₂+y₃)]
A1 awrt 1.6833
Answer: 1.6833

Question 14

M1 evaluates f(1) = -3 and f(2) = 7, a change of sign, so a root lies between x=1 and x=2
M1 rearranges x³ + 3x - 7 = 0 to x³ = 7 - 3x
A1 takes the cube root of both sides to give x = (7-3x)^1/3 (shown)
Answer: Root between x=1 and x=2 (shown); rearrangement x_(n+1) = (7-3x_n)^1/3 (shown)

Question 15

M1 finds x₁ = awrt 1.3572
A1 finds x₂ = awrt 1.4307 and x₃ = awrt 1.3938
A1 1.4 cao (ft from x₃, or from the known trend of the iteration)
Answer: x₁ = 1.3572, x₂ = 1.4307, x₃ = 1.3938; root is approximately 1.4

Question 16

M1 evaluates f(1) = -2 and f(2) = 1, a change of sign, so a root lies between x=1 and x=2
M1 substitutes into x₁ = x₀ - f(x₀)/f'(x₀) = 2 - 1/8
A1 x₁ = 1.8750, x₂ = awrt 1.8610
Answer: Root between x=1 and x=2 (shown); x₁ = 1.8750, x₂ = 1.8610

Question 17

B1 p = ln(3) = awrt 1.0986
B1 q = ln(5) = awrt 1.6094
M1 substitutes into the trapezium rule, (1/2)[(y₀+y₄) + 2(y₁+y₂+y₃)]
A1 awrt 3.9828
Answer: p = 1.0986, q = 1.6094; trapezium estimate = 3.9828

Question 18

M1 substitutes into the trapezium rule, (1/2)[(y₀+y₄) + 2(y₁+y₂+y₃)]
A1 awrt 1.1167
M1 finds the second derivative, d²y/dx² = 2/(x+2)³, and notes it is positive for x > -2
A1 states the curve is convex (concave up), so the chords lie above the curve, and the trapezium rule gives an overestimate
Answer: Trapezium estimate = 1.1167; this is an overestimate since d²y/dx² > 0 for x > -2

Question 19

M1 finds x₁ = awrt 2.0801 and x₂ = awrt 2.0924 for x_(n+1) = (2x_n+5)^1/3
A1 finds x₁ = 1.5000 and x₂ = -0.8125 for x_(n+1) = (x_n³-5)/2
M1 compares the two sequences: the first is settling towards a value near 2.09, the second is moving further away from x₀
A1 concludes x_(n+1) = (2x_n+5)^1/3 converges to α (the values are approaching a fixed value), while x_(n+1) = (x_n³-5)/2 diverges (the values move away from the root)
Answer: First iteration converges to α (approximately 2.09); second iteration diverges

Question 19

4 marks

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Pure: Numerical Methods: Fluency and Exam Drill - Worksheets, Questions and Revision

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