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WMA11/01 Jun 2019, Question 10: Differentiating Polynomials

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

The question, in short

For the curve y = (x − 4)(2x + 1)², identify the touch point P and crossing point Q, differentiate, find a horizontal tangent at x = 5/2, and find values of a for a shifted curve through O.

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

Part (a) · 1 mark

Give the coordinates of P, where the curve touches the x-axis.

  1. 1The repeated factor (2x + 1)² gives the point where the curve touches the x-axis
  2. 22x + 1 = 0, so x = −1/2: P = (−1/2, 0)

AnswerP = (−1/2, 0)

Part (b) · 4 marks

Differentiate f(x) = (x − 4)(2x + 1)² to obtain f′(x).

  1. 1(2x + 1)² = 4x² + 4x + 1
  2. 2f(x) = (x − 4)(4x² + 4x + 1) = 4x³ − 12x² − 15x − 4
  3. 3f′(x) = 12x² − 24x − 15

Answerf′(x) = 12x² − 24x − 15

Part (c) · 3 marks

Using f′(x), show the tangent at x = 5/2 is y = k and find k.

  1. 1f′(2.5) = 12(2.5)² − 24(2.5) − 15 = 75 − 60 − 15 = 0
  2. 2The gradient is 0, so the tangent is horizontal, y = k
  3. 3f(2.5) = (2.5 − 4)(6)² = −1.5 × 36 = −54
  4. 4Tangent: y = −54, so k = −54

AnswerShown: f′(5/2) = 0, so the tangent is horizontal; y = −54 (k = −54)

Part (d) · 2 marks

For y = f(x + a) passing through the origin, state the possible values of a.

  1. 1y = f(x + a) passes through O when f(a) = 0
  2. 2The roots of f are x = 4 and x = −1/2
  3. 3So a = 4 or a = −1/2

Answera = −1/2 or a = 4

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Also searched as: WMA11 June 2019 · WMA11/01 · wma11 jun19 · IAS Maths Pure Mathematics 1 June 2019 · question 10.
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