O-LevelAdditional MathematicsLogarithmic and exponential functions4037/22 · May/June 2019 · Q78 Marks

Solve 1g(x² -3) = 0.

(a)Solve 1g(x² -3) = 0.[2]
(b)(i)Show that, for a > 0, `(ln a^(sin(2x+5)) + ln(1/a)) / (ln a)` may be written as sin(2x+5)+k, where k is an integer.[3]
(b)(ii)Hence find `∫ (ln a^(sin(2x+5)) + ln(1/a)) / (ln a) dx`.[3]

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

[M1] (a) lg(x²-3) = lg 1 soi
[A1] -2 and 2 Implies M1
[B1] (b)(i) Two separate terms in numerator: (sin(2x + 5)) lna or loga asin(2x+5) = (sin(2x + 5)) loga a OR Combines terms in numerator: ln(asin(2x+5)/a)
[B1] -Ina or loga a⁻¹ = -loga a OR (sin(2x+5) – 1) In a
[B1] sin(2x+5)-1 dep all previous marks awarded;
[B3] (b)(ii) -(1/2)cos(2x+5) + (their -1) x (+ c) (FT their numerical k. B2 for -(1/2)cos(2x + 5) seen or B1 for a cos(2x+5), a<0 or for (1/2)cos(2x + 5) or for -(1/2)cos2x + 5 seen)

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Topic: Logarithmic and exponential functions, O-Level Additional Mathematics (syllabus 4037). Structured question, 8 marks.
Source: O-Level Additional Mathematics, May/June 2019 Paper 2 Variant 2, question 7.

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