A-LevelMathematicsNumerical methods9709/22 · Feb/Mar 2018 · Q59 Marks

It is given that ∫₋ₐ²ᵃ 4e⁻²ˣ dx = 25, where a is a positive constant.

(i)Show that a = ½ ln(12.5 + e⁻⁴ᵃ).[4]
(ii)Use the equation in part (i) to show by calculation that 1.0 < a < 1.5.[2]
(iii)Use an iterative formula based on the equation in part (i) to find the value of a correct to 4 significant figures. Show the result of each iteration to 6 significant figures.[3]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2018). The mark or mark type is shown in brackets.

[B1] Integrate to obtain -2e-2x
[M1] Apply limits correctly to integral of form ke-2x
[A1] Obtain -2e-4a + 2e2a = 25
[A1] Rearrange to confirm a = ½ln(12.5 + e-4a) AG; necessary detail needed
[M1] Consider sign of a - ½ln(12.5 + e-4a) or equivalent for 1.0 and 1.5
[A1] Obtain -0.26 and 0.24 or equivalent and justify conclusion AG; necessary detail needed
[M1] Use iterative process correctly at least once
[A1] Obtain final answer 1.263
[A1] Show sufficient iterations to 6 sf to justify answer or show a sign change in the interval (1.2625,1.2635)

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