O-LevelAdditional MathematicsDifferentiation and integration4037/21 · May/June 2020 · Q1213 Marks

Find the x-coordinates of the stationary points of the curve y = e^(3x)(2x+3)⁶.

(a)Find the x-coordinates of the stationary points of the curve y = e^(3x)(2x+3)⁶.[6]
(b)A curve has equation y = f(x) and has exactly two stationary points. Given that f''(x) = 4x-7, f'(0.5) = 0 and f′(3) = 0, use the second derivative test to determine the nature of each of the stationary points of this curve.[2]
(c)In this question all lengths are in centimetres. [Figure 12.1] The diagram shows a solid cuboid with height h and a rectangular base measuring 4x by x. The volume of the cuboid is 40 cm³. Given that x and h can vary and that the surface area of the cuboid has a minimum value, find this value.[5]

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Mark scheme

Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.

[B1] d(e^(3x))/dx = 3e^(3x)
[M1] d(2x+3)⁶/dx = k(2x+3)⁵
[M1] their (3e^(3x)) (2x+3)⁶ + (e^(3x)) (their 12 (2x + 3)⁵)
[A1] (3e^(3x)) (2x + 3)⁶ + (e^(3x)) (12 (2x + 3)⁵)
[M1] (3e^(3x)) (2x + 3)⁵ (2x + 7) = 0
[A1] x = -1.5, -3.5
[B2] B1 for either one correct: x = 0.5 f''(0.5) [= -5] < 0 ⇒ max; x = 3 f''(3) [= 5] > 0 ⇒ min
[B1] h = 10/x²
[B1] S = 8x² + 10x (their 10/x²)
[M1] dS/dx = 16x - 100x⁻² oe
[A1] FT their dS/dx = 0 if possible; 16x - 100x⁻² = 0, x = ³√(25/4) oe
[A1] 81.4 or 81.4325... rot to four or more figs

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