Printable · GCSE Higher · ages 14-16
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.
Higher only
Iteration worksheet — GCSE Higher
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- 1.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.
- 2.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.
- 3.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.
- 4.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.
- 5.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.
- 6.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.
- 7.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.
- 8.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.
- 9.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
- 10.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.
- 11.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.
- 12.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.
- 13.The equation x² − 4x − 1 = 0 can be solved using the iterative formula xₙ₊₁ = √(4xₙ + 1). The starting value is x₀ = 1, so x₁ is the value after the formula has been used once. Work out x₂ correct to 3 decimal places.
- 14.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.
- 15.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
Answer key
- (c) The sequence diverges, moving away from x = 3 — Starting from x₀ = 2.9: x₁ = 2.9² − 6 = 2.41, x₂ = 2.41² − 6 = −0.19, x₃ = (−0.19)² − 6 = −5.96, x₄ = (−5.96)² − 6 = 29.56 — the values swing away from 3 and grow rapidly, so the sequence diverges rather than settling anywhere. Choosing 'settles towards x = 3' assumes that starting close to a root is enough for a rearrangement to converge to it, which is not always true — this rearrangement changes values too steeply near x = 3 to stay there. Choosing 'settles towards x = −2' assumes a diverging sequence must eventually land on the other root; instead it runs away to increasingly large values. Choosing 'stays constant at 2.9' ignores that applying the formula changes the value at every step.
- (a) 1.861 — x₁ = ∛(10 − 2²) = ∛6 = 1.817120593. x₂ = ∛(10 − 1.817120593²) = ∛6.698072751 = 1.885022855. x₃ = ∛(10 − 1.885022855²) = ∛6.446688837 = 1.861139399, which rounds to 1.861. Reporting x₂ instead of x₃ gives 1.885022855, which rounds to 1.885. Stopping after the first iteration and reporting x₁ instead of x₃ gives 1.817120593, which rounds to 1.817. A sign error inside the cube root, using xₙ₊₁ = ∛(10 + xₙ²) instead of ∛(10 − xₙ²), gives x₁ = ∛14 = 2.410142264, x₂ = ∛(10 + 2.410142264²) = 2.509763724, and x₃ = ∛(10 + 2.509763724²) = 2.535437381, which rounds to 2.535.
- (a) 2.55 — x₁ = 20 ÷ (2 + 5) = 20 ÷ 7 = 2.857142857. x₂ = 20 ÷ (2.857142857 + 5) = 20 ÷ 7.857142857 = 2.545454545, which rounds to 2.55. Reporting x₁ instead of x₂ gives 2.857142857, which rounds to 2.86. Dropping the +5 in the denominator, using xₙ₊₁ = 20 ÷ xₙ, gives x₁ = 20 ÷ 2 = 10 and x₂ = 20 ÷ 10 = 2, which is 2.00. A sign error in the denominator, using xₙ₊₁ = 20 ÷ (xₙ − 5), gives x₁ = 20 ÷ (2 − 5) = −6.666666667 and x₂ = 20 ÷ (−6.666666667 − 5) = −1.714285714, which rounds to −1.71.
- (a) 4.897 — Method: substitute the starting value into the right-hand side of the formula to get x₁, then feed each new value back in, keeping the whole calculator display every time and rounding only at the very end. Working: x₁ = √(5 × 2 + 2) = √12 = 3.46410…; x₂ = √(5 × 3.46410… + 2) = √19.32050… = 4.39551…; x₃ = √(5 × 4.39551… + 2) = √23.97755… = 4.89668…, which is 4.897 correct to 3 decimal places. Answer: 4.897. The distractors: 4.396 is x₂, written down by a candidate who counts the starting value x₀ as the first iterate and so stops one use of the formula early; 3.464 is x₁, the value after using the formula only once; 5.146 is x₄, one use of the formula too many — the mirror image of the first slip, made by a candidate who labels the first value worked out as x₀ rather than as x₁ and so runs the count a step long.
- (b) 3 — At the limit, L = √(2L + 3). Squaring both sides: L² = 2L + 3, so L² − 2L − 3 = 0, which factorises as (L − 3)(L + 1) = 0, giving L = 3 or L = −1. Since the sequence of iterates stays positive throughout, the limit is L = 3. Taking the other, negative root without rejecting it gives −1. Treating the equation L = 2L + 3 as already linear, forgetting to square both sides first, gives −L = 3, so L = −3. A sign error when factorising, writing (L + 3)(L − 1) = 0 instead of (L − 3)(L + 1) = 0, gives L = 1.
- (b) 5.00 — Continuing the iteration: x₁ = √(2 × 1 + 15) = √17 = 4.1231, x₂ = √(2 × 4.1231 + 15) = √23.2462 = 4.8214, x₃ = √(2 × 4.8214 + 15) = √24.6428 = 4.9642, x₄ = √(2 × 4.9642 + 15) = √24.9284 = 4.9928, and the values keep climbing towards 5.00 as n increases (the limit L satisfies L² = 2L + 15, so L² − 2L − 15 = 0, giving L = 5). Choosing 4.99 stops after x₄, one iteration before the value has settled fully to 5.00. Choosing 17.00 uses the value under the very first square root (2 × 1 + 15 = 17) as if that number itself were the limit. Choosing 1.00 assumes the sequence never moves from the starting value x₀.
- (b) 3.38 — r₁ = √(300 ÷ (π × 8)) = √11.9366 = 3.4550. r₂ = √(300 ÷ (π × 8.4550)) = √11.2947 = 3.3608. r₃ = √(300 ÷ (π × 8.3608)) = √11.4232 = 3.3798, which rounds to 3.38. Choosing 3.36 stops at r₂, one iteration too early. Choosing 4.82 leaves out the '+ 5' inside the bracket, dividing by π × rₙ instead of π × (rₙ + 5). Choosing 3.45 comes from using π ≈ 3 instead of the calculator's π key throughout.
- (c) 17.7 cm — Method: the iteration converges on x, the depth of the box, which is also the side of each square cut away; a square is removed from both ends of the 20 cm side, so the longer side of the base is 20 − 2x. Run the formula three times, then carry out that subtraction. Working: x₁ = (16 × 1² − 1³ + 50) ÷ 60 = 65 ÷ 60 = 1.08333…; x₂ = 67.50636… ÷ 60 = 1.12510…; x₃ = 68.82959… ÷ 60 = 1.14715…. The longer side of the base is 20 − 2 × 1.14715… = 17.70568…, which is 17.7 cm correct to 1 decimal place. Answer: 17.7 cm. The distractors: 1.1 cm is x₃ itself rounded, the depth of the box, given by a candidate who runs the iteration correctly and then stops before the step the question actually asks for; 18.9 cm is 20 − 1.14715…, taking away one square instead of two and forgetting that a corner is cut from each end of that side; 9.7 cm is 12 − 2 × 1.14715…, the shorter side of the base, which measures the wrong edge of the sheet.
- (b) f(1) = −3 and f(2) = 3 — f(1) = 1³ − 1 − 3 = 1 − 1 − 3 = −3. f(2) = 2³ − 2 − 3 = 8 − 2 − 3 = 3. Since f(1) is negative and f(2) is positive, there is a change of sign, so a solution lies between x = 1 and x = 2. Dropping the −x term entirely gives f(1) = 1 − 3 = −2 and f(2) = 8 − 3 = 5. Using x² instead of x³ throughout gives f(1) = 1 − 1 − 3 = −3, which happens to coincide with the correct value at x = 1, but f(2) = 4 − 2 − 3 = −1, which does not show a change of sign at all. Working out −f(x) instead of f(x), a sign-flip error, gives f(1) = 3 and f(2) = −3, the correct sizes but with both signs reversed.
- (b) xₙ₊₁ = (xₙ² + 3) ÷ 5 — Starting from x² = 5x − 3, add 3 to both sides: x² + 3 = 5x. Divide both sides by 5: x = (x² + 3) ÷ 5. Writing this as an iteration gives xₙ₊₁ = (xₙ² + 3) ÷ 5. xₙ₊₁ = (xₙ² − 3) ÷ 5 comes from a sign error when moving the −3 across the equals sign — it should become +3, not stay as −3. xₙ₊₁ = 5(xₙ² + 3) comes from multiplying by 5 instead of dividing by 5 when isolating x. xₙ₊₁ = (xₙ + 3) ÷ 5 comes from dropping the index on x², using xₙ instead of xₙ².
- (c) xₙ₊₁ = ∛(9 − 4xₙ) — Method: a formula xₙ₊₁ = f(xₙ) is a correct rearrangement when the equation x = f(x) turns back into the equation you started with, so rearrange x³ + 4x − 9 = 0 by making the cube the subject. Working: x³ + 4x − 9 = 0 gives x³ = 9 − 4x, because the 4x and the 9 each change sign as they cross the equals sign; taking the cube root of both sides gives x = ∛(9 − 4x), which is the formula xₙ₊₁ = ∛(9 − 4xₙ). Answer: xₙ₊₁ = ∛(9 − 4xₙ). The distractors: ∛(9 + 4xₙ) comes from writing x³ = 9 + 4x, moving the 4x across the equals sign without changing its sign; (9 + xₙ³)/4 comes from making the linear term the subject but keeping the sign of the cube, writing 4x = 9 + x³ when the equation gives 4x = 9 − x³; ∛(9 − 4xₙ³) cubes the x in the linear term as well, changing a term the original equation never cubed.
- (b) 4.39 — x₁ = ∛(150 − 3 × 4²) = ∛(150 − 48) = ∛102 = 4.672 (unrounded). x₂ = ∛(150 − 3 × 4.672²) = ∛(150 − 65.49) = ∛84.51 = 4.39 (2 d.p.). Choosing 4.67 stops after only one iteration, giving x₁ instead of x₂. Choosing 84.51 finds the value inside the cube root for x₂ but never takes the cube root. Choosing 6.32 comes from adding 3xₙ² instead of subtracting it inside the root, which does not match the given formula.
- (c) 3.153 — x₁ = √(4 × 1 + 1) = √5 = 2.236067977. x₂ = √(4 × 2.236067977 + 1) = √9.944271908 = 3.153453965, which rounds to 3.153. Reporting x₁ instead of x₂ gives 2.236067977, which rounds to 2.236. A sign error inside the root, using xₙ₊₁ = √(4xₙ − 1) instead of √(4xₙ + 1), gives x₁ = √3 = 1.732050808 and x₂ = √(4 × 1.732050808 − 1) = √5.928203232 = 2.434790182, which rounds to 2.435. Applying the formula in the wrong order, working out √(4xₙ) + 1 at every step instead of √(4xₙ + 1), gives x₁ = √4 + 1 = 3 and x₂ = √(4 × 3) + 1 = 4.464101615, which rounds to 4.464.
- (d) 3.425 — x₁ = ∛(6 × 3 + 20) = ∛38 = 3.3620 (unrounded, 3.36198...). x₂ = ∛(6 × 3.3620 + 20) = ∛40.172 = 3.425 (3 d.p.). Choosing 3.362 stops at x₁ instead of continuing to x₂. Choosing 2.722 leaves out the '+ 20' inside the root, working out ∛(6 × 3.3620) = ∛20.172 = 2.722. Choosing 0.556 subtracts 20 instead of adding it, working out ∛(6 × 3.3620 − 20) = ∛0.172 = 0.556.
- (d) It has a solution between x = 2.2 and x = 2.3 — Since f(2.2) is negative and f(2.3) is positive, the graph of f crosses the x-axis somewhere between x = 2.2 and x = 2.3, so the equation has a solution in that interval. Choosing 'between x = −2.2 and x = −2.3' confuses the negative f-VALUE at 2.2 with a negative x-value. Choosing 'no solution' misreads a change of sign as meaning the opposite of what it shows. Choosing 'exactly two solutions' assumes a single change of sign must give two roots, which is not what the rule guarantees.
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