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O-LevelAdditional MathematicsFunctions4037/12 · Oct/Nov 2013 · Q129 Marks

The function f is such that f(x) = 3x² – 1 for –10 ≤ x ≤ 8. Find the range of f.

  1. (a)(i)The function f is such that f(x) = 3x² – 1 for –10 ≤ x ≤ 8. Find the range of f.[3]
  2. (a)(ii)Write down a suitable domain for f for which f⁻¹ exists.[1]
  3. (b)Functions g and h are defined by g(x) = 4eˣ - 2 for x ∈ R, h(x) = ln 5x for x > 0.
  4. (b)(i)Find g⁻¹(x).[2]
  5. (b)(ii)Solve gh(x) = 18.[3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2013). The mark or mark type is shown in brackets.

  1. [M1] (a) (i) for substitution of either x = -10 or x = 8, may be seen on diagram
  2. [B1] Min point at (0, -1) or when y = -1
  3. [A1] Must have ≤ for A1, do not allow x (Range: -1 ≤ y ≤ 299)
  4. [B1] (a) (ii) Allow any domain which will make f a one-one function (e.g. x ≥ 0 or equivalent)
  5. [M1] (b) (i) for complete method to find the form inverse function, must involve ln or lg if appropriate. May still be in terms of y.
  6. [A1] must be in terms of x (g⁻¹(x) = ln((x+2)/4))
  7. [M1] (b) (ii) for correct order (gh(x) = g(ln 5x))
  8. [A1] for correct expression 4e^(ln 5x) - 2
  9. [A1] for correct solution from correct working (x = 1)
  10. Or:
  11. [M1] for correct order (h(x) = g⁻¹(18))
  12. [A1] for correct equation (ln 5x = ln 5)
  13. [A1] for correct solution from correct working (x = 1)

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About this question

Topic
Functions, O-Level Additional Mathematics (syllabus 4037). Structured question, 9 marks.
Source
O-Level Additional Mathematics, Oct/Nov 2013 Paper 1 Variant 2, question 12.

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