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4PM1/01 Nov 2021, Question 10: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 15 marks

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The question, in short

For the curve y = 1/2 + sin 3x on 0 ≤ x < 2π/3, use the given diagram to locate intercepts and maximum, form a tangent at M, and integrate to find the shaded area.

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

Part (a) · 3 marks

Verify that M is at (7π/18, 0) and determine the coordinates of the other x-axis crossing N.

  1. 11/2 + sin 3x = 0, so sin 3x = −1/2
  2. 23x = 7π/6 or 11π/6 (in range 0 ≤ 3x < 2π)
  3. 3x = 7π/18 (M) and x = 11π/18 (N)

AnswerM = (7π/18, 0) shown; N = (11π/18, 0)

Part (b) · 4 marks

Determine the coordinates of the stationary point A where the curve reaches its highest value.

  1. 1dy/dx = 3cos 3x = 0
  2. 23x = π/2, x = π/6
  3. 3y = 1/2 + sin(π/2) = 3/2
  4. 4A = (π/6, 3/2)

AnswerA = (π/6, 3/2)

Part (c) · 4 marks

Obtain the tangent line at M, written as ay + b√3x − c√3π = 0 with integers a, b, c.

  1. 1At x = 7π/18: dy/dx = 3cos(7π/6) = −3√3/2
  2. 2y − 0 = −(3√3/2)(x − 7π/18)
  3. 3Multiply by 12: 12y = −18√3x + 7√3π
  4. 412y + 18√3x − 7√3π = 0

Answer12y + 18√3x − 7√3π = 0

Part (d) · 4 marks

By algebraic integration, calculate to 3 significant figures the total shaded area between the curve, y-axis and x-axis up to N.

  1. 1Area = ∫ from 0 to 7π/18 of (1/2 + sin 3x) dx + |∫ from 7π/18 to 11π/18 of (1/2 + sin 3x) dx|
  2. 2∫ (1/2 + sin 3x) dx = x/2 − (cos 3x)/3
  3. 3First integral = 1.23287…
  4. 4Second integral = −0.22828…, area 0.22828…
  5. 5Total area = 1.46

Answer1.46

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