0580/43 May/Jun 2022, Question 9: Stationary Points and Turning Points
Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 16 marks
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The question, in short
Sketch the cubic y = (x + 1)(3 − x)(3 + x) with axis intercepts marked, then expand it, locate its turning points to 2 decimal places, and find integer bounds on k for a single root.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Part (a) · 4 marks
Draw the curve y = (x + 1)(3 − x)(3 + x) and label where it meets both axes.
- 1y = 0 when x = −1, x = 3, x = −3: crosses the x-axis at −3, −1 and 3
- 2x = 0: y = 1 × 3 × 3 = 9, crosses the y-axis at 9
- 3The x³ term is −x³, so draw a negative cubic (high on the left, low on the right) through these points
AnswerCorrect sketch of negative cubic crossing the x-axis at −3, −1 and 3 and crossing the y-axis at 9
Part (bi) · 2 marks
Expand the product of the three brackets to obtain y = 9 + 9x − x² − x³.
- 1(3 − x)(3 + x) = 9 − x²
- 2(x + 1)(9 − x²) = 9x − x³ + 9 − x²
- 3y = 9 + 9x − x² − x³
AnswerCorrect completion to y = 9 + 9x − x² − x³
Part (bii) · 7 marks
Find the x-coordinates of the stationary points of y = 9 + 9x − x² − x³, showing working, to 2 decimal places.
- 1dy/dx = 9 − 2x − 3x²
- 29 − 2x − 3x² = 0, so 3x² + 2x − 9 = 0
- 3x = (−2 ± √(2² − 4 × 3 × (−9)))/(2 × 3) = (−2 ± √112)/6
- 4x = −2.097… or x = 1.430…
- 5x = −2.10 and x = 1.43
Answer−2.10 and 1.43
Part (biii) · 3 marks
For 9 + 9x − x² − x³ = k having exactly one root, determine the largest integer a and smallest integer b with k < a or k > b.
- 1At x = −2.097…: y = −5.049… (minimum)
- 2At x = 1.430…: y = 16.90… (maximum)
- 3One solution when k < −5.049… or k > 16.90…
- 4a = −6, b = 17
Answera = −6, b = 17
Stuck on a step? Read the free lesson: Stationary Points and Turning Points
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Also searched as: 0580/43/M/J/22 · 0580_s22_qp_43 · 0580 s22 43 · question 9.
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