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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 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.
2.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
3.f(x) = x³ − 3x − 5. Given that f(2.2) = −0.952 and f(2.3) = 0.267, work out what this shows about the equation x³ − 3x − 5 = 0.
y = x
(a)It has a solution between x = −2.2 and x = −2.3
(b)It has exactly two solutions in this interval
(c)It has no solution between x = 2.2 and x = 2.3
(d)It has a solution between x = 2.2 and x = 2.3
4.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)
5.The equation x² − 3x − 7 = 0 can be solved using the iterative formula xₙ₊₁ = √(3xₙ + 7). The starting value is x₀ = 4, so x₁ is the value after the formula has been used once. Work out x₃ correct to 3 decimal places.
(a)4.535
(b)4.481
(c)13.149
(d)4.521
6.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
7.A cuboid has a square base of side x metres and a height that is 3 m more than x. Its volume is 150 m³. This gives the equation x³ + 3x² − 150 = 0, which can be solved using the iterative formula xₙ₊₁ = ∛(150 − 3xₙ²). Taking x₀ = 4, work out x₂ correct to 2 decimal places.
(a)84.51
(b)4.39
(c)4.67
(d)6.32
8.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
9.A water tank is a cuboid with a square base of side x metres and height (x + 1) metres. Its volume is 10 m³. This gives x³ + x² − 10 = 0, which can be solved using the iterative formula xₙ₊₁ = ∛(10 − xₙ²). Taking x₀ = 2, so that x₁ is the value found after the formula has been used once, work out x₃ correct to 3 decimal places.
(a)1.861
(b)1.885
(c)2.535
(d)1.817
10.The iterative formula xₙ₊₁ = 12 ÷ (xₙ + 2) is used repeatedly, starting from x₀ = 1. Work out the value that xₙ approaches, correct to 2 decimal places.
(a)3.00
(b)6.00
(c)2.61
(d)4.00
11.A closed cylinder has radius r cm and height (r + 5) cm. Its volume is 300 cm³, giving the equation πr²(r + 5) = 300, which can be solved using the iterative formula rₙ₊₁ = √(300 ÷ (π(rₙ + 5))). Taking r₀ = 3, work out r₃ correct to 2 decimal places.
(a)3.45
(b)3.38
(c)3.36
(d)4.82
12.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
13.The iterative formula xₙ₊₁ = 5 − 3/xₙ is used with starting value x₀ = 2.5, so that x₁ is the value after the formula has been used once. Work out x₄ correct to 3 significant figures.
(a)3.80
(b)4.29
(c)4.30
(d)2.50
14.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
15.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

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