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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 · Calculator · 15 questions

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1.The iterative formula xₙ₊₁ = √(2xₙ + 15) is used repeatedly, starting from x₀ = 1. Work out the value that xₙ approaches, correct to 2 decimal places.
(a)1.00
(b)5.00
(c)17.00
(d)4.99
2.The equation 7x = x² + 3 can be solved using the iterative formula xₙ₊₁ = (xₙ² + 3) ÷ 7. Taking x₀ = 0.4, x₁ = 0.4514 correct to 4 decimal places. Using the full unrounded value of x₁, work out x₂ correct to 3 decimal places.
(a)0.493
(b)0.458
(c)0.451
(d)0.632
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.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
5.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
6.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
7.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
8.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
9.f(x) = x³ − 3x − 20, and the equation f(x) = 0 has exactly one solution. Work out the pair of consecutive integers between which that solution lies.
y = x
(a)2 and 3
(b)3 and 4
(c)4 and 5
(d)6 and 7
10.The equation x³ − 2x − 7 = 0 has exactly one solution. It can be found using the iterative formula xₙ₊₁ = ∛(2xₙ + 7), with starting value x₀ = 2, so that x₁ is the value after the formula has been used once. Work out the solution correct to 2 decimal places, iterating until two consecutive values round to the same 2 decimal places.
(a)2.25
(b)1.91
(c)2.26
(d)2.22
11.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
12.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)
13.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
14.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
15.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)

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