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4PM1/01R Jun 2023, Question 4: Exponentials, Logarithms, e and ln

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 9 marks

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The question, in short

Fill in missing y-values for y = x/2 + 6e^(−2x) + 1 to 1 d.p., plot the curve for 0 ≤ x ≤ 6, then use a straight line to estimate roots of 2x + ln(24 − 5x) = ln 36.

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

Part (a) · 2 marks

Work out the blank y-values in the table for the given equation, rounding each to one decimal place.

  1. 1x = 1: 0.5 + 6e^(−2) + 1 = 2.31… ≈ 2.3
  2. 2x = 2: 1 + 6e^(−4) + 1 = 2.11… ≈ 2.1
  3. 3x = 3: 1.5 + 6e^(−6) + 1 = 2.51… ≈ 2.5
  4. 4x = 5: 2.5 + 6e^(−10) + 1 = 3.50… ≈ 3.5

Answerx = 1: 2.3; x = 2: 2.1; x = 3: 2.5; x = 5: 3.5

Part (b) · 2 marks

Plot the curve of the given equation on the grid for x from 0 to 6.

  1. 1Plot the 8 points from the table
  2. 2Join with a smooth curve (no ruled sections)

AnswerPoints from the table plotted and joined by a smooth curve: falls steeply from (0, 7) to a minimum near (1.5, 2.0), then rises almost linearly to (6, 4.0)

Part (c) · 5 marks

Add a suitable straight line to your graph and use intersections to estimate the equation's roots to one decimal place.

  1. 1ln((24 − 5x)/36) = −2x, so (24 − 5x)/36 = e^(−2x)
  2. 22/3 − 5x/36 = e^(−2x), so 4 − 5x/6 = 6e^(−2x)
  3. 3Add x/2 + 1 to both sides: 5 − x/3 = x/2 + 6e^(−2x) + 1
  4. 4Draw the line y = 5 − x/3
  5. 5It meets the curve at x ≈ 0.2 and x ≈ 4.8

Answerx = 0.2 to 0.25 or x = 4.75 to 4.8

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