Growth, Decay & Iteration

The equation x³ - 3x - 5 = 0 has a root near x = 2. Using the iteration x_(n+1) = (3x_n + 5)^(1/3) with x₀ = 2, find x₂, correct to 3 decimal places.

• Ax₂ ≈ 2.224
• Bx₂ ≈ 2.268✓
• Cx₂ ≈ 2.301
• Dx₂ ≈ 2.259

Answer: Starting from x₀ = 2, the iteration gives x₁ = (3(2) + 5)^(1/3) ≈ 2.224, and substituting this back gives x₂ = (3(2.224) + 5)^(1/3) ≈ 2.268. Stopping after only one iteration, or a rounding slip inside the cube root, produces the other values.

The equation x³ + x - 9 = 0 has a root near x = 2. Using the iteration x_(n+1) = (9 - x_n)^(1/3) with x₀ = 2, find x₂, correct to 3 decimal places.

• Ax₂ ≈ 1.920✓
• Bx₂ ≈ 1.927
• Cx₂ ≈ 1.900
• Dx₂ ≈ 1.913

Answer: Starting from x₀ = 2, the iteration gives x₁ = (9 − 2)^(1/3) ≈ 1.913, and substituting this back gives x₂ = (9 − 1.913)^(1/3) ≈ 1.920. Stopping after only the first iteration, or a rounding error inside the cube root, produces the other values.

A student uses the iteration x_(n+1) = √(2x_n + 3), with x₀ = 4, to approximate a root of x² - 2x - 3 = 0. Find x₂, correct to 3 decimal places.

• Ax₂ ≈ 3.146
• Bx₂ ≈ 3.317
• Cx₂ ≈ 3.082
• Dx₂ ≈ 3.104✓

Answer: Starting from x₀ = 4, the iteration gives x₁ = √(2(4) + 3) = √11 ≈ 3.317, and substituting this back gives x₂ = √(2(3.317) + 3) ≈ 3.104. Stopping after the first iteration, or a rounding slip inside the square root, produces the other values.

A savings account balance follows the iteration V_(n+1) = 1.035V_n, with V₀ = £2,000, modelling 3.5% annual compound interest. Find V₄, to the nearest penny.

• A£2,280.00
• B£2,295.00
• C£2,295.05✓
• D£2,300.00

Answer: The recurrence V_(n+1) = 1.035V_n applies the 3.5% growth once per step. Iterating from V₀ = 2000 gives V₁ = 2070, V₂ = 2142.45, V₃ = 2217.44, and V₄ = 2295.05 to the nearest penny. Using a rounded multiplier of 1.1475 at each stage instead of iterating exactly gives £2,295.00, and the remaining values come from other rounding slips.

A car's value follows the iteration V_(n+1) = 0.88V_n, with V₀ = £18,000, modelling 12% annual depreciation. Find V₃, to the nearest pound.

• A£12,672
• B£11,865
• C£12,266✓
• D£15,840

Answer: The recurrence V_(n+1) = 0.88V_n retains 88% of the previous value at each step. Iterating from V₀ = 18,000 gives V₁ = 15,840, V₂ = 13,939.20, and V₃ = 12,266 to the nearest pound. Stopping after only one iteration gives £15,840, and the remaining values come from other rounding slips.

£5,000 is invested at a compound interest rate of r% per year. After 2 years the investment is worth £5,304.50. Find r.

• Ar = 6%
• Br = 3%✓
• Cr = 3.1%
• Dr = 1.5%

Answer: Dividing 5304.50 by 5000 gives (1 + r/100)² = 1.0609, and taking the square root gives 1 + r/100 = 1.03, so r = 3%. Doubling the rate instead of taking a square root, or a rounding slip, produces the other values.

Priya invests £3,000 at a compound interest rate that exactly doubles her money in 12 years. Find this annual rate, to 1 decimal place.

• A8.3%
• B5.9%✓
• C16.7%
• D6.0%

Answer: Doubling means (1 + r)¹² = 2, so 1 + r = 2^(1/12) ≈ 1.059, giving r ≈ 5.9%. Rounding too early, or approximating the doubling with simple division instead of a twelfth root, produces the other values.

A bacteria population follows the iteration P_(n+1) = 1.15P_n, with P₀ = 200 (n measured in hours). Find P₅, to the nearest whole number.

• A230
• B402✓
• C460
• D375

Answer: The recurrence P_(n+1) = 1.15P_n applies 15% growth at each step. Iterating from P₀ = 200 five times gives P₅ ≈ 402 (nearest whole number). Using fewer growth steps, or a rounding slip in the iteration, produces the other values.

A radioactive substance's mass follows the iteration M_(n+1) = 0.92M_n grams, with M₀ = 80 (n measured in years). Find M₆, to 1 decimal place.

• A48.5 g✓
• B44.0 g
• C46.2 g
• D50.9 g

Answer: The recurrence M_(n+1) = 0.92M_n retains 92% of the previous mass at each step. Iterating from M₀ = 80 six times gives M₆ ≈ 48.5 g (1 d.p.). Using the wrong number of decay steps, or a rounding slip in the iteration, produces the other values.

A rare stamp's value follows the iteration V_(n+1) = 1.08V_n, with V₀ = £45 in 2020. Find V₅, the value in 2025, to the nearest penny.

• A£66.12✓
• B£68.90
• C£61.20
• D£64.80

Answer: The recurrence V_(n+1) = 1.08V_n applies 8% growth at each step. Iterating from V₀ = 45 five times (2020 to 2025) gives V₅ ≈ £66.12. Using 4 iterations instead of 5, or a rounding slip, produces the other values.

A forest's tree population follows the iteration P_(n+1) = 0.96P_n, with P₀ = 12,500 in 2022. Find P₅, the population in 2027, to the nearest 10 trees.

• A10,000 trees
• B9,800 trees
• C10,190 trees✓
• D10,200 trees

Answer: The recurrence P_(n+1) = 0.96P_n retains 96% of the previous population at each step. Iterating from P₀ = 12,500 five times (2022 to 2027) gives P₅ ≈ 10,192, which rounds to 10,190 trees (nearest 10). Using the wrong number of iterations, or a rounding slip, produces the other values.

A sequence follows the iteration x_(n+1) = x_n/2 + 3, with x₁ = 10. Find x₃.

• Ax₃ = 7✓
• Bx₃ = 8
• Cx₃ = 9
• Dx₃ = 6.5

Answer: Applying the rule to x₁ = 10 gives x₂ = 10/2 + 3 = 8, and applying it again gives x₃ = 8/2 + 3 = 7. Stopping after only one iteration, or a slip halving the value, produces the other results.

The iterative process x_(n+1) = √(x_n + 6) is used starting with x₀ = 1. Find x₂, correct to 2 decimal places.

• Ax₂ ≈ 2.83
• Bx₂ ≈ 3.00
• Cx₂ ≈ 2.65
• Dx₂ ≈ 2.94✓

Answer: Starting from x₀ = 1, the rule gives x₁ = √(1 + 6) = √7 ≈ 2.65, and applying it again gives x₂ = √(2.65 + 6) ≈ 2.94. Stopping after the first iteration, or a rounding slip inside the square root, produces the other values.

A savings model uses the iteration A_(n+1) = 1.01A_n + 50, with A₀ = £200. Find A₂, to the nearest penny.

• A£302.00
• B£307.07
• C£254.52
• D£304.52✓

Answer: Applying the rule to A₀ = 200 gives A₁ = 1.01(200) + 50 = £252, and applying it again gives A₂ = 1.01(252) + 50 = £304.52. Stopping after the first iteration, or a rounding slip in the multiplication, produces the other values.

The iteration x_(n+1) = (x_n² + 5)/(2x_n) is used to approximate √5, starting with x₀ = 2. Find x₂, correct to 4 decimal places.

• Ax₂ ≈ 2.2400
• Bx₂ ≈ 2.2500
• Cx₂ ≈ 2.2361✓
• Dx₂ ≈ 2.2306

Answer: Starting from x₀ = 2, the rule gives x₁ = (4 + 5)/4 = 2.25, and applying it again gives x₂ = (2.25² + 5)/4.5 ≈ 2.2361, very close to √5. Stopping after the first iteration, or a rounding slip in the division, produces the other values.