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WMA12/01 Jun 2019, Question 10: Stationary Points and Turning Points

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 11 marks

The question, in short

For f(x) = 36/x² + 2x − 13 with x > 0, use calculus to find where f increases, verify a definite integral equals zero, and use given integral values to find a related integral and a constant k.

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

Part (a) · 4 marks

Differentiate f(x) and solve the inequality for x where f(x) is increasing, with x > 0.

  1. 1f′(x) = −72/x³ + 2
  2. 2f(x) increasing when f′(x) > 0: 2 > 72/x³
  3. 3x³ > 36 (x > 0)
  4. 4x > ∛36 (about 3.30)

Answerx > ∛36

Part (b) · 4 marks

Integrate 36/x² + 2x − 13 from 2 to 9 and show the result is 0.

  1. 1∫(36x⁻² + 2x − 13) dx = −36/x + x² − 13x
  2. 2At x = 9: −4 + 81 − 117 = −40
  3. 3At x = 2: −18 + 4 − 26 = −40
  4. 4Integral = (−40) − (−40) = 0

AnswerShown: [−36/x + x² − 13x] from 2 to 9 = (−40) − (−40) = 0

Part (ci) · 1 mark

Given the integral from 2 to 6 equals −8, state the integral from 6 to 9.

  1. 1∫ from 2 to 9 = ∫ from 2 to 6 + ∫ from 6 to 9
  2. 20 = −8 + ∫ from 6 to 9, so ∫ from 6 to 9 = 8

Answer8

Part (cii) · 2 marks

Find constant k so that the integral from 2 to 6 of 36/x² + 2x + k equals 0.

  1. 1∫ from 2 to 6 of (36/x² + 2x + k) dx = ∫ from 2 to 6 of (36/x² + 2x − 13) dx + ∫ from 2 to 6 of (k + 13) dx
  2. 2= −8 + 4(k + 13) = 0
  3. 3k + 13 = 2
  4. 4k = −11

Answerk = −11

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Also searched as: WMA12 June 2019 · WMA12/01 · wma12 jun19 · IAS Maths Pure Mathematics 2 June 2019 · question 10.
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