A-LevelMathematicsNumerical methods9709/22 · Oct/Nov 2025 · Q69 Marks

Given that ∫[from -2a to a] (½ e²⁸+¼ e⁻⁸)dx = 5, where a is a positive constant, show that a = ½ ln(10+½ e⁻²⁸+¼ e⁻⁴⁸).

(a)Given that ∫[from -2a to a] (½ e²⁸+¼ e⁻⁸)dx = 5, where a is a positive constant, show that a = ½ ln(10+½ e⁻²⁸+¼ e⁻⁴⁸). [4]
(b)Hence show by calculation that the value of a lies between 1.0 and 1.2. [2]
(c)Use the iterative formula an+1 = ½ ln(10+½ e⁻ⁿ+¼ e⁻⁴ⁿ) to find the value of a correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

[4] (a) Integrate to obtain expression of form k₁e²⁸ + k₂e⁻⁸
[4] (a) Obtain correct ¼ e²⁸ −¼ e⁻⁸
[4] (a) Apply limits and arrange at least as far as ¼ e²ⁿ = ... or e²ⁿ = ...
[4] (a) Confirm a=½ln(10+½ e⁻²ⁿ+¼ e⁻⁴ⁿ)
[2] (b) Consider sign of a−½ln(10+½ e⁻ⁿ+¼ e⁻⁴ⁿ) or equivalent for values 1.0 and 1.2
[2] (b) Obtain -0.16... and 0.041... or equivalents and justify conclusion
[3] (c) Use iteration process correctly at least once
[3] (c) Obtain final answer 1.159
[3] (c) Show sufficient iterations to justify answer or show a sign change in the interval [1.1585, 1.1595]

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