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WMA11/01 Jan 2021, Question 1: Equation of the Tangent Line

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

The question, in short

Differentiate a cubic-with-reciprocal curve for x > 0, then find the normal at the point where x = 1/2, giving it as ax + by + c = 0 with integer coefficients.

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

Part (a) · 3 marks

Find dy/dx of y = 2x³ − 5x² − 3/(2x) + 7, x > 0, in simplest form.

  1. 1y = 2x³ − 5x² − (3/2)x⁻¹ + 7
  2. 2dy/dx = 6x² − 10x + (3/2)x⁻²
  3. 3dy/dx = 6x² − 10x + 3/(2x²)

Answerdy/dx = 6x² − 10x + 3/(2x²)

Part (b) · 5 marks

At P with x = 1/2 on the curve, find the normal's equation as ax + by + c = 0 with integers a, b, c.

  1. 1At x = 1/2: y = 2(1/8) − 5(1/4) − 3 + 7 = 3
  2. 2dy/dx = 6(1/4) − 10(1/2) + 3/(2 × 1/4) = 1.5 − 5 + 6 = 5/2
  3. 3Normal gradient = −2/5
  4. 4y − 3 = −2/5 (x − 1/2)
  5. 55y − 15 = −2x + 1, so 2x + 5y − 16 = 0

Answer2x + 5y − 16 = 0

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