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WMA12/01 Jan 2025, Question 9: Stationary Points, Increasing and Optimising

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 12 marks

The question, in short

For the curve y = 9x²(5 − √x)/5 with x ≥ 0, use calculus to locate the turning point M, find where the curve meets the x-axis at P, then integrate to get the area of region R to one decimal place.

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

Part (a) · 5 marks

Differentiate the given curve and solve for the coordinates of the turning point M.

  1. 1y = 9x² − (9/5)x^(5/2)
  2. 2dy/dx = 18x − (9/2)x^(3/2)
  3. 318x − (9/2)x^(3/2) = 0, so x^(1/2) = 4 (x ≠ 0)
  4. 4x = 16
  5. 5y = 9(256)(5 − 4)/5 = 460.8, so M = (16, 460.8)

AnswerM = (16, 460.8)

Part (b) · 2 marks

Solve algebraically for the x value where the curve crosses the x-axis at P.

  1. 1y = 0 when 5 − √x = 0 (x > 0)
  2. 2√x = 5, so x = 25

Answerx = 25

Part (c) · 5 marks

Set up and evaluate a definite integral for the shaded region R, giving the area to one decimal place.

  1. 1∫(9x² − (9/5)x^(5/2)) dx = 3x³ − (18/35)x^(7/2)
  2. 2Area under curve from 16 to 25 = [3x³ − (18/35)x^(7/2)] from 16 to 25
  3. 3= (46875 − 40178.57…) − (12288 − 8426.06…) = 6696.43… − 3861.94… = 2834.49…
  4. 4Rectangle from x = 16 to 25 under y = 460.8: 9 × 460.8 = 4147.2
  5. 5Area of R = 4147.2 − 2834.49… = 1312.7 (1 d.p.)

AnswerArea of R = 1312.7

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Also searched as: WMA12 January 2025 · WMA12/01 · wma12 jan25 · IAS Maths Pure Mathematics 2 January 2025 · question 9.
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