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O-LevelAdditional MathematicsDifferentiation and integration4037/12 · Oct/Nov 2024 · Q29 Marks

It is given that y = f(x), where f(x) = (2x-5)(x-1)2.

  1. (a)Find the coordinates of the stationary points on the curve y = f(x).[4]
  2. (b)On the axes, sketch the graph of y = f(x), stating the intercepts with the axes.[3]
  3. (c)Hence find the values of k for which f(x) = k has exactly one solution.[2]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.

  1. [M1] (f'(x)) = 2(x-1)(2x-5)+2(x−1)² oe or 6x²-18x+12
  2. [M1] Dep for equating their quadratic f'(x) to zero and attempt to solve to obtain x = ...
  3. [A1] x = 1, y = 0
  4. [A1] x = 2, y = -1
  5. [B1] for a correct cubic shape
  6. [B1] for a correct cubic shape in the correct position, touching the x-axis once in the 4th quadrant and intersecting once with the positive x-axis
  7. [B1] for all intercepts and no extras
  8. [B1] k<-1
  9. [B1] k>0

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About this question

Topic
Differentiation and integration, O-Level Additional Mathematics (syllabus 4037). Structured question, 9 marks.
Source
O-Level Additional Mathematics, Oct/Nov 2024 Paper 1 Variant 2, question 2.

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