O-LevelAdditional MathematicsDifferentiation and integration4037/12 · May/June 2015 · Q89 Marks

(i) Find ∫(10e²ˣ + e⁻²ˣ)dx.

(i)Find ∫(10e²ˣ + e⁻²ˣ)dx.[2]
(ii)Hence find ∫ᵏ₋ₖ(10e²ˣ + e⁻²ˣ)dx in terms of the constant k.[2]
(iii)Given that ∫ᵏ₋ₖ(10e²ˣ + e⁻²ˣ)dx = -60, show that 11e²ᵏ – 11e⁻²ᵏ + 120 = 0.[2]
(iv)Using a substitution of y = e²ᵏ or otherwise, find the value of k in the form alnb, where a and b are constants.[3]

Mark scheme

Marking points from the official mark scheme (May/June 2015). The mark or mark type is shown in brackets.

[2] (i) B1,B1 B1 for each term, allow unsimplified
[2] (ii) M1 use of limits provided integration has taken place. Signs must be correct if brackets are not included.
[2] (ii) A1 allow any correct form
[2] (iii) B1 correct expression from (ii) either simplified or unsimplified equated to -60, must be first line seen.
[2] (iii) DB1 must be convinced as AG
[3] (iv) M1 attempt to obtain a quadratic equation in y or e²ᵏ and solve to get y or e²ᵏ (only need positive solution)
[3] (iv) DM1 attempt to deal with e to get k =.
[3] (iv) A1 any of given answers only

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