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O-LevelAdditional MathematicsStraight line graphs4037/11 · May/June 2024 · Q1110 Marks

11 The tangent to the curve y = (3x-1)¹/² at the point where x = 3 meets the coordinate axes at the points A and B. The point with coordinates (a, a) lies on the perpendicular bisector of the line AB. Find the exact value of a.

  1. The tangent to the curve y = (3x-1)¹/² at the point where x = 3 meets the coordinate axes at the points A and B. The point with coordinates (a, a) lies on the perpendicular bisector of the line AB. Find the exact value of a.[10]

Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

  1. [10] B1 (When x = 3, y = 2)
  2. [10] M1 Allow for k(3x-1)^(-1/2) (dy/dx = 3/2 (3x-1)^(-1/2))
  3. [10] A1 (When x = 3, dy/dx = 1/4)
  4. [10] M1 Allow using their y and their dy/dx (Tangent equation: y-2 = 1/4(x-3))
  5. [10] A1 For both (Intercepts on the axes: x = -5, y = 5/4)
  6. [10] M1 Dep on M mark for equation of tangent FT on their intercepts (Midpoint of AB: (-5/2, 5/8))
  7. [10] M1 FT on their dy/dx or from their x and y intercept (Grad of perp bisector = -4)
  8. [10] M1 Allow using their – 4 and their midpoint (Perp bisector equation: y-5/8 = -4(x+5/2))
  9. [10] M1 Dep on previous M mark for use of (a, a) in their perp bisector equation
  10. [10] A1 a = -15/8 or -1.875

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About this question

Topic
Straight line graphs, O-Level Additional Mathematics (syllabus 4037). Structured question, 10 marks.
Source
O-Level Additional Mathematics, May/June 2024 Paper 1 Variant 1, question 11.

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