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WMA11/01 Jan 2025, Question 6: Integrating with Two Conditions

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 10 marks

The question, in short

Use the given derivative, the point P(4, −5) on curve C, and the tangent gradient 7 at P to find constants and the equation of the curve, then locate the transformed point Q.

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

Part (a) · 2 marks

Substitute x = 4 and the gradient 7 into f′(x) to derive the stated equation linking a and b.

  1. 1f′(4) = (2(4)² + 4a + b)/(4√4) = 7
  2. 2(32 + 4a + b)/8 = 7
  3. 332 + 4a + b = 56
  4. 44a + b = 24

AnswerShown: (2(16) + 4a + b)/(4 × 2) = 7 ⇒ 32 + 4a + b = 56 ⇒ 4a + b = 24

Part (b) · 7 marks

Combine with a + b = −9 to find a and b, then integrate f′(x) and use P to determine f(x).

  1. 14a + b = 24 and a + b = −9 give 3a = 33: a = 11, b = −20
  2. 2f′(x) = (2x² + 11x − 20)/(4√x) = ½x^(3/2) + (11/4)x^(1/2) − 5x^(−1/2)
  3. 3f(x) = (1/5)x^(5/2) + (11/6)x^(3/2) − 10x^(1/2) + c
  4. 4At (4, −5): (1/5)(32) + (11/6)(8) − 10(2) + c = −5
  5. 532/5 + 44/3 − 20 + c = −5, so c = 15 − 32/5 − 44/3 = −91/15
  6. 6f(x) = (1/5)x^(5/2) + (11/6)x^(3/2) − 10√x − 91/15

Answerf(x) = (1/5)x^(5/2) + (11/6)x^(3/2) − 10√x − 91/15

Part (c) · 1 mark

State the coordinates of Q after P is mapped by the transformation y = f(x − 3).

  1. 1y = f(x − 3) is a translation 3 units in the positive x-direction
  2. 2P(4, −5) moves to Q(7, −5)

AnswerQ(7, −5)

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Also searched as: WMA11 January 2025 · WMA11/01 · wma11 jan25 · IAS Maths Pure Mathematics 1 January 2025 · question 6.
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