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4PM1/01R Jun 2019, Question 8: Exponentials, Logarithms, e and ln

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

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The question, in short

Complete a table for y = 2 + ln(2x + 1), plot the curve for 0 ≤ x ≤ 3, then use straight-line intersections 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 = 2 + ln(2x + 1) at x = 0.25, 0.5 and 3, rounding each to 2 decimal places.

  1. 1x = 0.25: 2 + ln 1.5 = 2.41
  2. 2x = 0.5: 2 + ln 2 = 2.69
  3. 3x = 3: 2 + ln 7 = 3.95

Answer2.41, 2.69, 3.95

Part (b) · 2 marks

Plot the graph of y = 2 + ln(2x + 1) on the grid for 0 ≤ x ≤ 3.

  1. 1Plot the 7 points from the table
  2. 2Join them with a smooth increasing curve

AnswerSmooth curve through (0, 2), (0.25, 2.41), (0.5, 2.69), (1, 3.10), (1.5, 3.39), (2, 3.61), (3, 3.95)

Part (c) · 3 marks

Draw a suitable straight line to estimate, to 1 decimal place, the root of ln(2x + 1) = 3x − 4 for 0 ≤ x ≤ 3.

  1. 1ln(2x + 1) = 3x − 4 → 2 + ln(2x + 1) = 3x − 2
  2. 2Draw y = 3x − 2
  3. 3It meets the curve at x ≈ 1.8 (accept 1.9)

Answerx = 1.8 or 1.9

Part (d) · 4 marks

Draw a suitable straight line to estimate, to 1 decimal place, the root of e^(6 − x) − (2x + 1)² = 0 for 0 ≤ x ≤ 3.

  1. 1e^(6 − x) = (2x + 1)² → 6 − x = 2 ln(2x + 1)
  2. 2ln(2x + 1) = 3 − x/2 → 2 + ln(2x + 1) = 5 − x/2
  3. 3Draw y = 5 − x/2
  4. 4It meets the curve at x ≈ 2.4 (accept 2.5)

Answerx = 2.4 or 2.5

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