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4PM1/02 Jun 2020, Question 8: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 15 marks

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The question, in short

Two curves y = 5e^(−2x) + 4 and y = e^(2x) meet at A; find A, use tangents at A to locate B and D on the x-axis, then find the area of triangle ABD.

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Part (a) · 4 marks

Find the exact coordinates of the intersection point A of the two given curves.

  1. 15e^(−2x) + 4 = e^(2x) ⇒ 5e^(−2x) + 4 − e^(2x) = 0
  2. 2(5e^(−x) − e^x)(e^(−x) + e^x) = 0
  3. 3e^(2x) = 5 ⇒ x = (1/2)ln 5 (e^(−x) = −e^x is impossible)
  4. 4y = 5, so A = ((1/2)ln 5, 5)

AnswerA = ((1/2)ln 5, 5)

Part (b) · 5 marks

Show that the x coordinate of B, where the tangent at A to C₁ meets the x-axis, is (1/2)(5 + ln 5).

  1. 1dy/dx = −10e^(−2x)
  2. 2At A: dy/dx = −10 × 1/5 = −2
  3. 3Tangent: y − 5 = −2(x − (1/2)ln 5)
  4. 4y = 0 ⇒ x = (1/2)ln 5 + 5/2 = (1/2)(5 + ln 5)

Answerx coordinate of B = (1/2)(5 + ln 5) shown

Part (c) · 6 marks

Find the area of triangle ABD, where D is where the tangent at A to C₂ meets the x-axis.

  1. 1C₂: dy/dx = 2e^(2x) = 10 at A
  2. 2Tangent: y − 5 = 10(x − (1/2)ln 5)
  3. 3y = 0 ⇒ x_D = (1/2)(ln 5 − 1)
  4. 4BD = (1/2)(5 + ln 5) − (1/2)(ln 5 − 1) = 3
  5. 5Area = ½ × 3 × 5 = 15/2

Answer15/2 (= 7.5)

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