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A-LevelMathematicsFunctions9709/12 · Oct/Nov 2014 · Q1110 Marks

The function f : x → 6 − 4 cos(½x) is defined for 0 ≤ x ≤ 2π.

  1. (i)Find the exact value of x for which f(x) = 4.[3]
  2. (ii)State the range of f.[2]
  3. (iii)Sketch the graph of y = f(x).[2]
  4. (iv)Find an expression for f⁻¹(x).[3]

Mark scheme

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

  1. [1] Set f(x) = 4: 6 − 4cos(½x) = 4 → 4cos(½x) = 2. Makes cos(½x) the subject.
  2. [1] cos(½x) = ½. Looks up “½x” before ×2
  3. [1] ½x = π/3 → x = 2π/3. co (120° gets A0 – decimals A0)
  4. [2] The range of f is 2 ≤ f(x) ≤ 10. condone <
  5. [1] The sketch graph should show a point of inflexion at x = π.
  6. [1] The sketch graph should be fully correct.
  7. [1] Let y = 6 − 4cos(½x). Rearrange to make cos(½x) the subject: cos(½x) = (6-y)/4. Makes cos⁻¹(½x) the subject
  8. [1] Rearrange further: ½x = cos⁻¹((6-y)/4). Order of operations correct (M marks allowed if + for –)
  9. [1] x = 2cos⁻¹((6-y)/4). Replace y with x for inverse function: f⁻¹(x) = 2cos⁻¹((6-x)/4). oe - needs to be a function of x not y

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About this question

Topic
Functions, A-Level Mathematics (syllabus 9709). Structured question, 10 marks.
Source
A-Level Mathematics, Oct/Nov 2014 Paper 1 Variant 2, question 11.

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