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4PM1/01 Jun 2022, Question 6: Exponentials, Logarithms, e and ln

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 11 marks

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The question, in short

Complete a table for y = 1 + 3e^(−x), plot its graph for 0 ≤ x ≤ 3, then use straight lines on the grid to estimate roots of two equations.

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

Part (a) · 2 marks

Fill in the missing y values for y = 1 + 3e^(−x) at the given x values, rounding to 2 decimal places where needed.

  1. 1x = 0: 1 + 3 = 4
  2. 2x = 1: 1 + 3e^(−1) = 2.10
  3. 3x = 2: 1 + 3e^(−2) = 1.41

Answerx = 0: 4; x = 1: 2.10; x = 2: 1.41

Part (b) · 2 marks

Draw the curve y = 1 + 3e^(−x) on the grid for x from 0 to 3.

  1. 1Plot the seven points from the table
  2. 2Join them with a smooth curve

AnswerSmooth decreasing curve through (0, 4), (0.25, 3.34), (0.5, 2.82), (1, 2.10), (1.5, 1.67), (2, 1.41), (3, 1.15)

Part (c) · 3 marks

Draw a suitable straight line and estimate, to 1 decimal place, the root of x = e^(−x) for 0 ≤ x ≤ 3.

  1. 1x = e^(−x) ⇒ 3x = 3e^(−x) ⇒ 3x + 1 = 1 + 3e^(−x)
  2. 2Draw y = 3x + 1
  3. 3The line meets the curve at x ≈ 0.6

Answerx = 0.5 or 0.6

Part (d) · 4 marks

Draw a suitable straight line and estimate, to 1 decimal place, the root of ln(x − 1)³ = −3x for 0 ≤ x ≤ 3.

  1. 1ln (x − 1)³ = −3x ⇒ 3 ln (x − 1) = −3x ⇒ ln (x − 1) = −x
  2. 2x − 1 = e^(−x) ⇒ 3x − 3 = 3e^(−x) ⇒ 3x − 2 = 1 + 3e^(−x)
  3. 3Draw y = 3x − 2
  4. 4The line meets the curve at x ≈ 1.3

Answerx = 1.2 or 1.3

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