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WMA11/01 Jan 2025, Question 5: Tangent and Normal Problems

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 8 marks

The question, in short

Differentiate the curve y = 4x³ + 2/x + 9 for x > 0, then use the tangent line y = k − 5x at point P to derive an equation in x and find k.

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

Part (a) · 2 marks

Find dy/dx of y = 4x³ + 2/x + 9 in simplest form.

  1. 1y = 4x³ + 2x⁻¹ + 9
  2. 2dy/dx = 12x² − 2x⁻² = 12x² − 2/x²

Answerdy/dx = 12x² − 2/x²

Part (b) · 2 marks

Show the x coordinate of P satisfies 12x⁴ + 5x² − 2 = 0.

  1. 1The tangent y = k − 5x has gradient −5
  2. 212x² − 2/x² = −5
  3. 3Multiply by x²: 12x⁴ − 2 = −5x²
  4. 412x⁴ + 5x² − 2 = 0

AnswerShown: 12x² − 2/x² = −5 ⇒ 12x⁴ − 2 = −5x² ⇒ 12x⁴ + 5x² − 2 = 0

Part (c) · 4 marks

Hence find the value of k.

  1. 1(4x² − 1)(3x² + 2) = 0
  2. 23x² + 2 = 0 has no real roots, so x² = 1/4 and x = ½ (x > 0)
  3. 3y = 4(½)³ + 2/(½) + 9 = ½ + 4 + 9 = 27/2
  4. 427/2 = k − 5(½), so k = 27/2 + 5/2 = 16

Answerk = 16

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