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.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
- 2.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.
- 3.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.
- 4.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.
- 5.A rectangular vegetable plot has an area of 30 m² and its length is 4 m greater than its width, x metres. This gives x² + 4x − 30 = 0, which can be solved using the iterative formula xₙ₊₁ = √(30 − 4xₙ). The starting value is x₀ = 3, so x₁ is the value after the formula has been used once. Work out x₃, and use it to find an estimate for the length of the plot, giving your answer correct to 1 decimal place.
- 6.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.
- 7.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.
- 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
- 9.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.
- 10.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
- 11.f(x) = x³ − 3x² − 4. Work out the pair of consecutive integers between which the solution of f(x) = 0 lies.y = x
- 12.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.
- 13.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.
- 14.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.
- 15.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.
Answer key
- (b) 3 and 4 — Method: the graph of f(x) is continuous, so where it crosses the x-axis the value of f(x) changes sign; substitute consecutive integers until one value is negative and the next is positive. Working: f(2) = 8 − 6 − 20 = −18, f(3) = 27 − 9 − 20 = −2 and f(4) = 64 − 12 − 20 = 32. The sign changes from negative to positive between x = 3 and x = 4, so the solution lies there. Answer: 3 and 4. The distractors: 2 and 3 comes from ignoring the −3x term and solving x³ = 20, whose root is 2.71, one interval to the left; 6 and 7 comes from reading x³ as x² and solving x² − 3x − 20 = 0, whose positive root is 6.22; 4 and 5 is the interval immediately after the change of sign, named by a candidate who finds f(4) positive and quotes the interval beginning there instead of the one across which the sign actually turned.
- (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.
- (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) 4.521 — x₁ = √(3 × 4 + 7) = √19 = 4.358898944. x₂ = √(3 × 4.358898944 + 7) = √20.076696833 = 4.480702716. x₃ = √(3 × 4.480702716 + 7) = √20.442108148 = 4.521294964, which rounds to 4.521. Mislabelling the starting value x₀ as x₁, so that the working stops one iteration too early, reports the true x₂ = 4.480702716, which rounds to 4.481. Working out one iteration too many reports the true x₄ = √(3 × 4.521294964 + 7) = 4.534741987, which rounds to 4.535. Applying the formula in the wrong order, calculating √(3xₙ) + 7 at every step instead of √(3xₙ + 7), gives, from x₀ = 4: √12 + 7 = 10.464101615, then √(3 × 10.464101615) + 7 = 12.602883619, then √(3 × 12.602883619) + 7 = 13.148873950, which rounds to 13.149.
- (a) 7.9 m — Method: the iteration converges on the width of the plot, so run the formula three times from the starting value and then add 4 m, because the length is 4 m greater than the width. Working: x₁ = √(30 − 4 × 3) = √18 = 4.24264…; x₂ = √(30 − 4 × 4.24264…) = √13.02943… = 3.60963…; x₃ = √(30 − 4 × 3.60963…) = √15.56147… = 3.94480…. The estimate for the length is 3.94480… + 4 = 7.94480…, which is 7.9 m correct to 1 decimal place. Answer: 7.9 m. The iteration is still oscillating at x₃, so this is the estimate that three uses of the formula give, not a settled value. The distractors: 3.9 m is x₃ itself, the width, given by a candidate who runs the iteration correctly and then stops before the step the question actually asks for; 7.6 m uses x₂ in place of x₃, one use of the formula short, and then adds the 4 m correctly; 15.8 m multiplies the width by 4 instead of adding 4 m to it, reading greater than as a multiplier.
- (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.
- (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.
- (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) 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.
- (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.
- (d) 3 and 4 — f(3) = 3³ − 3 × 3² − 4 = 27 − 27 − 4 = −4, and f(4) = 4³ − 3 × 4² − 4 = 64 − 48 − 4 = 12. Since f(3) is negative and f(4) is positive, there is a change of sign, so the solution lies between 3 and 4. A sign slip when expanding −3x², treating it as +3x², gives f(3) = 27 + 27 − 4 = 50, a positive value; taken with the correctly negative f(2) = 8 − 12 − 4 = −8, this reads as a change of sign between 2 and 3. The same sign slip applied at x = 1 gives f(1) = 1 + 3 − 4 = 0, read as the point the sign changes, against f(0) = 0 + 0 − 4 = −4, giving 0 and 1. Shifting the correctly found interval up by one integer, an indexing slip, reports the change as lying between 4 and 5 instead of 3 and 4.
- (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) 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) 1.861 — 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.
- (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.
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