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Iteration worksheet — GCSE Higher

Fifteen questions on "iteration" — DfE statement A20. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.

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Iteration worksheet — GCSE Higher

Maths · GCSE Higher · A20 · Iteration · 15 questions

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1.The equation x² − 5x − 2 = 0 can be solved using the iterative formula xₙ₊₁ = √(5xₙ + 2). The starting value is x₀ = 2, so x₁ is the value after the formula has been used once. Work out x₃ correct to 3 decimal places.
(a)4.897
(b)4.396
(c)5.146
(d)3.464
2.The equation x³ = 6x + 20 can be solved using the iterative formula xₙ₊₁ = ∛(6xₙ + 20). Taking x₀ = 3, x₁ = 3.3620 correct to 4 decimal places. Using the full unrounded value of x₁, work out x₂ correct to 3 decimal places.
(a)3.362
(b)2.722
(c)0.556
(d)3.425
3.A rectangular sheet of metal measures 20 cm by 12 cm. A square of side x cm is cut from each corner and the sides are folded up to make an open box of volume 200 cm³. This gives x³ − 16x² + 60x − 50 = 0, which can be solved using the iterative formula xₙ₊₁ = (16xₙ² − xₙ³ + 50)/60. The starting value is x₀ = 1, so x₁ is the value after the formula has been used once. Work out x₃, and use it to find the longer side of the base of the box correct to 1 decimal place.
(a)1.1 cm
(b)9.7 cm
(c)17.7 cm
(d)18.9 cm
4.The iterative formula xₙ₊₁ = √(2xₙ + 3) is used repeatedly, starting from x₀ = 1. As n increases, the values of xₙ converge to a limit, L. Work out L.
(a)−1
(b)3
(c)1
(d)−3
5.f(x) = x³ − 3x² − 4. Work out the pair of consecutive integers between which the solution of f(x) = 0 lies.
y = x
(a)4 and 5
(b)2 and 3
(c)0 and 1
(d)3 and 4
6.A student uses the iterative formula xₙ₊₁ = √(7xₙ + 3) to find an approximate solution of an equation. Work out which equation this iterative formula solves.
(a)x² + 7x − 3 = 0
(b)x² − 7x + 3 = 0
(c)x² − 7x − 9 = 0
(d)x² − 7x − 3 = 0
7.The equation x³ − 5x − 3 = 0 can be rearranged to give an iterative formula of the form xₙ₊₁ = ∛(…). Work out which one of these is a correct rearrangement.
(a)xₙ₊₁ = ∛(5xₙ + 3)
(b)xₙ₊₁ = ∛(5xₙ − 3)
(c)xₙ₊₁ = (5xₙ + 3) ÷ 3
(d)xₙ₊₁ = ∛(3xₙ + 5)
8.An allotment is in the shape of a rectangle. Its length is 5 m more than its width, x metres, and its area is 20 m². This gives x² + 5x − 20 = 0, which can be solved using the iterative formula xₙ₊₁ = 20 ÷ (xₙ + 5). Taking x₀ = 2, so that x₁ is the value found after the formula has been used once, work out x₂ correct to 2 decimal places.
(a)2.55
(b)−1.71
(c)2.00
(d)2.86
9.The equation x² + 2x − 5 = 0 can be solved using the iterative formula xₙ₊₁ = 5/(xₙ + 2). The starting value is x₀ = 1, so x₁ is the value after the formula has been used once. Work out x₂ correct to 2 decimal places.
(a)1.36
(b)1.49
(c)1.37
(d)1.67
10.f(x) = x³ − 5x − 6. Given that f(2.6) = −1.424 and f(2.7) = 0.183, work out what this shows about the equation x³ − 5x − 6 = 0.
y = x
(a)It has a solution between x = 2.6 and x = 2.7
(b)There is no root in this interval
(c)It has a solution between x = −2.6 and x = −2.7
(d)It shows that x = 2.6 is a solution of the equation
11.The equation x² = 5x − 3 is to be solved using iteration. Work out which of these iterative formulas comes from a correct rearrangement of the equation.
(a)xₙ₊₁ = 5(xₙ² + 3)
(b)xₙ₊₁ = (xₙ² + 3) ÷ 5
(c)xₙ₊₁ = (xₙ + 3) ÷ 5
(d)xₙ₊₁ = (xₙ² − 3) ÷ 5
12.Show that the equation x³ − x − 3 = 0 has a solution between x = 1 and x = 2, by working out f(1) and f(2), where f(x) = x³ − x − 3.
y = x
(a)f(1) = −3 and f(2) = −1
(b)f(1) = −3 and f(2) = 3
(c)f(1) = 3 and f(2) = −3
(d)f(1) = −2 and f(2) = 5
13.The equation x² − 7 = 0 has a positive root. Let f(x) = x² − 7. Given that x₁ = 2.6 and x₂ = 2.65, work out which of these is correct.
y = x² − 7
(a)x₁ closer: f(x₁) negative ⇒ nearer root
(b)x₁ closer: −0.24 is the smaller value
(c)x₂ is the exact root, since f(x₂) ≈ 0
(d)x₂ closer: f(x₂) = 0.0225, nearer to 0
14.The equation x³ + 4x − 9 = 0 is to be solved by iteration. Work out which one of these iterative formulas comes from a correct rearrangement of that equation.
(a)xₙ₊₁ = (9 + xₙ³)/4
(b)xₙ₊₁ = ∛(9 − 4xₙ³)
(c)xₙ₊₁ = ∛(9 − 4xₙ)
(d)xₙ₊₁ = ∛(9 + 4xₙ)
15.The iterative formula xₙ₊₁ = xₙ³ − 2 is used repeatedly, starting from x₀ = 2. Which of these describes what happens to the sequence of values as n increases?
(a)It stays constant, since x₀ = 2 is a fixed point.
(b)It oscillates between two fixed values.
(c)It converges to a fixed value close to 2.
(d)It diverges, growing rapidly without limit.

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