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O-LevelAdditional MathematicsFunctions4037/22 · May/June 2014 · Q119 Marks

The functions f and g are defined, for real values of x greater than 2, by f(x) = 2ˣ - 1, g(x) = x(x + 1).

  1. (i)State the range of f.[1]
  2. (ii)Find an expression for f⁻¹(x), stating its domain and range.[4]
  3. (iii)Find an expression for gf(x) and explain why the equation gf(x) = 0 has no solutions.[4]

Mark scheme

Marking points from the official mark scheme (May/June 2014). The mark or mark type is shown in brackets.

  1. [B1] f(x) > 3 or [f(x)∈](3,∞) condone y > 3
  2. [M1] x + 1 = 2ʸ or y + 1 = 2ˣ
  3. [A1] f⁻¹(x) = log₂(x+1) mark final answer or log₂(y+1) = x and f⁻¹(x) = log₂(x+1) or for f⁻¹(x) = log(x+1) / log 2 (any base for this form)
  4. [B1ft] Domain x > 3 ft their range of f provided mathematically valid inequality or interval
  5. [B1] Range f⁻¹(x) > 2 condone f(x) > 2 or y > 2
  6. [B1] 2ˣ(2ˣ -1) oe isw e.g. (2ˣ−1)² + (2ˣ−1) or 2²x2ˣ +1+2ˣ −1
  7. [B1] 2ˣ(2ˣ -1)=0 leading to 2ˣ = 0, impossible oe or 2ˣ = 0 which is outside domain of gf
  8. [M1] 2ˣ = 1 ⇒ x = 0 or 2ˣ(2ˣ −1)= 2²x − 2ˣ = 0 [2²x = 2ˣ]⇒ x = 0
  9. [A1] 0 is not in the domain (and so gf(x) = 0 has no solutions)

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About this question

Topic
Functions, O-Level Additional Mathematics (syllabus 4037). Structured question, 9 marks.
Source
O-Level Additional Mathematics, May/June 2014 Paper 2 Variant 2, question 11.

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