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4PM1/02 Nov 2023, Question 7: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 9 marks

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The question, in short

Given 3x − y = 4 and S = 5x³ + y², show a cubic form for S, then use calculus to find the x giving minimum S, justify it is a minimum, and find that minimum value.

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Part (a) · 2 marks

Eliminate y using 3x − y = 4 to show S simplifies to 5x³ + 9x² − 24x + 16.

  1. 1y = 3x − 4
  2. 2S = 5x³ + (3x − 4)²
  3. 3= 5x³ + 9x² − 24x + 16

AnswerS = 5x³ + 9x² − 24x + 16 shown

Part (b) · 5 marks

Differentiate S with respect to x, solve for stationary x, and justify using the second derivative that S is minimised.

  1. 1dS/dx = 15x² + 18x − 24 = 0
  2. 23(5x − 4)(x + 2) = 0, x = 4/5 or −2
  3. 3d²S/dx² = 30x + 18
  4. 4At x = 4/5: 24 + 18 = 42 > 0, so minimum at x = 4/5

Answerx = 4/5 (d²S/dx² = 42 > 0, so minimum)

Part (c) · 2 marks

Substitute the x found in part b into S to calculate the minimum value of S.

  1. 1S = 5(4/5)³ + 9(4/5)² − 24(4/5) + 16
  2. 2= 128/25 = 5.12

AnswerS = 128/25 (= 5.12)

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