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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.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
2.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
3.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)
4.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.
5.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
6.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
7.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
8.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
9.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
10.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
11.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
12.The equation x² − x − 6 = 0 has roots x = 3 and x = −2. It can be rearranged as xₙ₊₁ = xₙ² − 6. This formula is used with starting value x₀ = 2.9, close to the root x = 3. Work out what happens to the sequence of values as n increases.
(a)The sequence settles towards x = 3
(b)The sequence stays constant at x = 2.9
(c)The sequence diverges, moving away from x = 3
(d)The sequence settles towards x = −2
13.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
14.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
15.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

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