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A-LevelMathematicsQuadratics9709/12 · May/June 2018 · Q25 Marks

The equation of a curve is y = x² – 6x + k, where k is a constant.

  1. (i)Find the set of values of k for which the whole of the curve lies above the x-axis.[2]
  2. (ii)Find the value of k for which the line y + 2x = 7 is a tangent to the curve.[3]

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

  1. [M1] (i) A complete method as far as finding a set of values for k by: Either completing the square to get (x - 3)² + k − 9 > 0, leading to k − 9 > 0 OR Differentiating and setting to 0, 2x − 6 = 0, to find x = 3, then finding y = k - 9 and using k - 9 > 0 OR Use of discriminant b² < 4ac, so 36 < 4k. Beware use of > and incorrect algebra.
  2. [A1] (i) k > 9. T&I leading to (or no working) correct answer 2/2 otherwise 0/2.
  3. [*M1] (ii) EITHER: Equates and collects terms: x² – 6x + k = 7 – 2x → x² – 4x + k − 7 (= 0).
  4. [DM1] (ii) EITHER: Correct use of discriminant = 0, involving k from a 3 term quadratic: 16 – 4(k – 7) = 0.
  5. [A1] (ii) EITHER: k = 11.
  6. [*M1] (ii) OR: Equates their dy/dx to ± 2, finds a value for x: 2x - 6 = -2 → x = 2 (y = 3).
  7. [DM1] (ii) OR: Substitutes their value(s) into the appropriate equation: (their 3) or 7 – 2(their 2) = (their 2)² – 6(their 2) + k.
  8. [A1] (ii) OR: k = 11.

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About this question

Topic
Quadratics, A-Level Mathematics (syllabus 9709). Structured question, 5 marks.
Source
A-Level Mathematics, May/June 2018 Paper 1 Variant 2, question 2.

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