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A-LevelMathematicsFunctions9709/12 · May/June 2017 · Q1011 Marks

The function f is defined by f(x) = 3 tan(1/2 x) - 2, for -1/2 π ≤ x ≤ 1/2 π.

  1. (i)Solve the equation f(x) + 4 = 0, giving your answer correct to 1 decimal place.[3]
  2. (ii)Find an expression for f⁻¹(x) and find the domain of f⁻¹.[5]
  3. (iii)Sketch, on the same diagram, the graphs of y = f(x) and y = f⁻¹(x).[3]

Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

  1. 3tan(1/2 x) = -2 → tan(1/2 x) = -2/3
  2. [M1] Attempt to obtain tan(1/2 x) = k from 3tan(1/2 x) + 2 = 0
  3. 1/2 x = -0.6 (-0.588) → x = -1.2
  4. [M1 A1] tan⁻¹ k. Seeing 1/2 x = -33.69° or x= -67.4° implies M1M1. Extra answers between -1.57 &1.57 lose the A1. Multiples of π are acceptable (eg – 0.374π)
  5. y + 2 / 3 = tan(1/2 x)
  6. [M1] Attempt at isolating tan(1/2x)
  7. → f⁻¹(x) = 2tan⁻¹((x+2)/3)
  8. [M1 A1] Inverse tan followed by × 2. Must be function of x for A1.
  9. Domain = -5, 1
  10. [B1 B1] Values stated B1 for -5, B1 for 1.
  11. A tan graph through the first, third and fourth quadrants. (B1)
  12. An invtan graph through the first, second and third quadrants.(B1)
  13. Two curves clearly symmetrical about y = x either by sight or by exact end points. Line not required. Approximately in correct domain and range. (Not intersecting.) (B1)
  14. Labels on axes not required.

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About this question

Topic
Functions, A-Level Mathematics (syllabus 9709). Structured question, 11 marks.
Source
A-Level Mathematics, May/June 2017 Paper 1 Variant 2, question 10.

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