O-LevelAdditional MathematicsDifferentiation and integration4037/21 · May/June 2021 · Q97 Marks

A curve is such that d²y/dx² = sin(6x-π/2). Given that dy/dx = 1/2 at the point (π/4, 13π/12) on the curve, find the equation of the curve.

A curve is such that d²y/dx² = sin(6x-π/2). Given that dy/dx = 1/2 at the point (π/4, 13π/12) on the curve, find the equation of the curve.[7]

Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

[B2] ∫sin(6x-π/2) dx = -cos(6x-π/2)/6 + c
[B1] for -cos 6x dx = k sin 6x + c where k < 0 or k = 1/6 or -sin 6x / 6
[M1] 1/2 = -cos(6(π/4)-π/2)/6 + c (FT their k provided B1 awarded)
[A1] c = 1/3
[M2] ∫(-cos(6x-π/2)/6 + 1/3) dx = -sin(6x-π/2)/36 + x/3 + A
[M1] for msin(6x-π/2) + (their 1/3)x + A where m < 0 or m = 1/36
[M1] 13π/12 = -sin(6(π/4)-π/2)/36 + (π/4)/3 + A (FT their m and their c provided at least M1 awarded)
[A1] y = -sin(6x-π/2)/36 + x/3 + π oe cao

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Topic: Differentiation and integration, O-Level Additional Mathematics (syllabus 4037). Structured question, 7 marks.
Source: O-Level Additional Mathematics, May/June 2021 Paper 2 Variant 1, question 9.

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