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4PM1/02 Jun 2020, Question 9: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 18 marks

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The question, in short

Rearrange the curve y = (2 + 4x − x²)/(2x + 1) into a given quadratic form, use it to find where x is real, locate stationary points, and sketch the curve.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 3 marks

Rewrite the curve equation as ax² + (by − 4)x + (y − c) = 0 and find the integer values of a, b and c.

  1. 1y(2x + 1) = 2 + 4x − x²
  2. 22xy + y = 2 + 4x − x²
  3. 3x² + (2y − 4)x + (y − 2) = 0

Answerx² + (2y − 4)x + (y − 2) = 0, so a = 1, b = 2, c = 2

Part (b) · 4 marks

Using the quadratic from part a, show that x is real only when y ≤ 2 or y ≥ 3.

  1. 1x real ⇒ b² ≥ 4ac: (2y − 4)² ≥ 4(y − 2)
  2. 24y² − 16y + 16 − 4y + 8 ≥ 0 ⇒ 4y² − 20y + 24 ≥ 0
  3. 34(y − 2)(y − 3) ≥ 0
  4. 4y ≤ 2 or y ≥ 3

Answerx real ⇒ y ≤ 2 or y ≥ 3 shown

Part (c) · 6 marks

Find the coordinates of the stationary points of the curve C.

  1. 1dy/dx = [(4 − 2x)(2x + 1) − 2(2 + 4x − x²)]/(2x + 1)²
  2. 2dy/dx = 0 ⇒ (4 − 2x)(2x + 1) − 2(2 + 4x − x²) = 0
  3. 3−2x² − 2x = 0 ⇒ 2x(x + 1) = 0 ⇒ x = 0 or −1
  4. 4Stationary points (0, 2) and (−1, 3)

Answer(0, 2) and (−1, 3)

Part (d) · 5 marks

Sketch C, marking the exact x-axis crossing points, the vertical asymptote, and the stationary point coordinates.

  1. 12 + 4x − x² = 0 ⇒ x = 2 ± √6 (≈ −0.45 and 4.45)
  2. 2Vertical asymptote x = −1/2
  3. 3Left of x = −1/2 the curve lies above y = 3 with a minimum at (−1, 3)
  4. 4Right of x = −1/2 the curve lies below y = 2 with a maximum at (0, 2)

AnswerTwo branches with vertical asymptote x = −1/2: left branch above y = 3 with minimum (−1, 3); right branch below y = 2 with maximum (0, 2), crossing the x-axis at (2 − √6, 0) and (2 + √6, 0)

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