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 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.
- 2.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.
- 3.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
- 4.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.
- 5.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.
- 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.
- 7.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
- 8.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.
- 9.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?
- 10.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.
- 11.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.
- 12.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.
- 13.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
- 14.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.
- 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
- (a) xₙ₊₁ = ∛(5xₙ + 3) — Starting from x³ − 5x − 3 = 0, add 5x and 3 to both sides to get x³ = 5x + 3, then take the cube root of both sides: x = ∛(5x + 3), giving the iterative formula xₙ₊₁ = ∛(5xₙ + 3). A sign error when moving the constant term across, treating x³ − 5x − 3 = 0 as x³ = 5x − 3, gives xₙ₊₁ = ∛(5xₙ − 3). Swapping the coefficient of x with the constant term gives xₙ₊₁ = ∛(3xₙ + 5), which does not come from x³ = 5x + 3 at all. Treating cubing as meaning multiply by 3 rather than raise to the power 3, and so undoing it by dividing by 3 instead of taking a cube root, gives xₙ₊₁ = (5xₙ + 3) ÷ 3.
- (a) 1.36 — Method: put the starting value into the right-hand side to get x₁, feed that value back in to get x₂, and round only once the second value has been found. Working: x₁ = 5 ÷ (1 + 2) = 5 ÷ 3 = 1.66666…; x₂ = 5 ÷ (1.66666… + 2) = 5 ÷ 3.66666… = 1.36363…. The digit in the third decimal place is 3, so x₂ = 1.36 correct to 2 decimal places. Answer: 1.36. The distractors: 1.67 is x₁, the value after a single use of the formula, given by a candidate who counts the starting value itself as x₁; 1.49 is x₃ = 1.48648…, one use of the formula too many; 1.37 comes from writing x₁ down as 1.66, truncating the display instead of keeping it in full, and then working out 5 ÷ 3.66 = 1.36612…, which rounds up to 1.37.
- (a) It has a solution between x = 2.6 and x = 2.7 — f(2.6) is negative and f(2.7) is positive, so the graph of f crosses the x-axis between x = 2.6 and x = 2.7, meaning the equation has a solution there. Choosing 'x = 2.6 is a solution' reads an end of the interval as the root itself, but f(2.6) = −1.424, which is not zero — the change of sign locates a root between the two values, it does not land on either of them. Choosing 'between x = −2.6 and x = −2.7' confuses the negative f-VALUE at 2.6 with a negative x-value. Choosing 'no root in this interval' misapplies the rule, which needs a CHANGE of sign — and a change of sign is exactly what these two values show.
- (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) 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.
- (d) x² − 7x − 3 = 0 — Method: an iteration settles where the next value equals the one before it, so both can be written as the same letter x; replace every xₙ by x, square both sides to clear the square root, and collect all the terms on one side. Working: x = √(7x + 3) gives x² = 7x + 3 on squaring both sides; subtracting 7x and 3 from both sides gives x² − 7x − 3 = 0. Answer: x² − 7x − 3 = 0. The distractors: x² + 7x − 3 = 0 moves the 7x across the equals sign without changing its sign; x² − 7x + 3 = 0 makes that same slip on the constant instead, leaving the 3 positive as it crosses; x² − 7x − 9 = 0 squares the expression term by term, squaring the 3 to give 9 as though squaring √(7x + 3) gave 7x + 9, which is the (a + b)² = a² + b² mistake dressed as a square root.
- (d) x₂ closer: f(x₂) = 0.0225, nearer to 0 — f(2.6) = 2.6² − 7 = 6.76 − 7 = −0.24, and f(2.65) = 2.65² − 7 = 7.0225 − 7 = 0.0225. The closer a value of x is to the root, the closer f(x) is to zero — regardless of sign. Since |0.0225| = 0.0225 is much smaller than |−0.24| = 0.24, x₂ = 2.65 is closer to the root. 'x₁ closer: −0.24 is the smaller value' comes from comparing the SIGNED values of f(x) rather than their distances from zero — −0.24 is indeed less than 0.0225 as a number, but that does not mean x₁ is closer to the root. 'x₁ closer: f(x₁) negative ⇒ nearer root' invents a rule that a negative f(x) means x is closer to the root; the sign of f(x) only tells you which side of the root x is on, not how close it is. 'x₂ is the exact root, since f(x₂) ≈ 0' misreads f(x₂) = 0.0225 as zero; the true root is √7 ≈ 2.6458, so f(2.65) is close to zero but not equal to it, and x₂ is an approximation, not the exact root.
- (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.
- (d) It diverges, growing rapidly without limit. — x₁ = 2³ − 2 = 8 − 2 = 6. x₂ = 6³ − 2 = 216 − 2 = 214. x₃ = 214³ − 2 = 9800344 − 2 = 9800342. The values 6, 214, 9800342, … grow far larger at every step, so the sequence diverges rather than settling anywhere. Checking whether the sequence converges to a fixed value near 2 fails, since the terms grow enormously instead of levelling off. Checking for a repeating pair of values also fails, since 6, 214 and 9800342 are all different, with no sign of a return to 6. x₀ = 2 is a fixed point only if 2³ − 2 = 2, but 2³ − 2 = 6, not 2, so the sequence does not stay constant.
- (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.
- (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) 2.26 — Method: apply the formula repeatedly, keeping the whole display each time, and stop when two values in a row round to the same 2 decimal places; that shared rounded value is the solution to that accuracy. Working: x₁ = ∛(2 × 2 + 7) = ∛11 = 2.22398…; x₂ = ∛(2 × 2.22398… + 7) = ∛11.44796… = 2.25377…; x₃ = ∛11.50754… = 2.25767…; x₄ = ∛11.51534… = 2.25818…. Now x₃ and x₄ both round to 2.26, so the sequence has settled. Answer: 2.26. The distractors: 2.22 is x₁ rounded, quoted by a candidate who stops after one use of the formula; 2.25 is x₂ rounded, quoted by a candidate who stops as soon as two values look close instead of waiting until two consecutive values round to the same figure; 1.91 is ∛7, which comes from ignoring the 2x term and solving x³ = 7 instead.
- (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) 0.458 — x₁ = (0.4² + 3) ÷ 7 = 3.16 ÷ 7 = 0.4514 (unrounded, 0.451428...). x₂ = (x₁² + 3) ÷ 7 = (0.2038 + 3) ÷ 7 = 3.2038 ÷ 7 = 0.458 (3 d.p.). Choosing 0.632 divides only the 3 by 7 instead of dividing the whole sum x₁² + 3 by 7. Choosing 0.451 repeats the calculation for x₁ instead of moving on to x₂. Choosing 0.493 uses x₁ itself instead of x₁² inside the formula.
- (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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