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0580/42 Feb/Mar 2022, Question 12: Stationary Points and Turning Points

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 11 marks

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The question, in short

Solve tan x = 11.43 in the range 0° to 360°, sketch y = x³ − 4x, and find a, b and k for y = x³ + ax + b with stationary points (2, k) and (−2, 10 − k).

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

Part (a) · 2 marks

Find all values of x between 0° and 360° satisfying tan x = 11.43.

  1. 1x = tan⁻¹(11.43) = 85.0°
  2. 2tan repeats every 180°: x = 85.0 + 180 = 265.0°
  3. 3x = 85.0°, 265.0°

Answer85.0, 265.0

Part (b) · 3 marks

Draw the graph of y = x³ − 4x on the given axes, showing its key features.

  1. 1y = x(x − 2)(x + 2), so the curve crosses the x-axis at −2, 0 and 2
  2. 2Draw a positive cubic through the origin: a maximum between x = −2 and 0 and a minimum between 0 and 2

Answercorrect shape and passes through origin

Part (c) · 6 marks

Using the two given stationary points, work out the values of a, b and k.

  1. 1dy/dx = 3x² + a
  2. 2At x = 2, dy/dx = 0: 12 + a = 0, so a = −12
  3. 3k = 2³ − 12 × 2 + b = b − 16
  4. 410 − k = (−2)³ − 12 × (−2) + b = b + 16
  5. 5Adding: 10 = 2b, so b = 5 and k = 5 − 16 = −11
  6. 6a = −12, b = 5, k = −11

Answera = −12, b = 5, k = −11

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Also searched as: 0580/42/F/M/22 · 0580_m22_qp_42 · 0580 m22 42 · question 12.
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