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4PM1/01R Jun 2025, Question 7: Exponentials, Logarithms, e and ln

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 10 marks

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The question, in short

Complete a table for y = 3log₃(x² − x) − 2x, plot the curve for 1.3 ≤ x ≤ 3.5, then use a straight line to estimate a root of 3^((4/3)x) = 9(x² − x)³.

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 x = 1.3, 2.5, 3 and 3.5, rounding each to 2 decimal places.

  1. 1x = 1.3: 3log₃(0.39) − 2.6 = −5.17
  2. 2x = 2.5: 3log₃(3.75) − 5 = −1.39
  3. 3x = 3: 3log₃(6) − 6 = −1.11
  4. 4x = 3.5: 3log₃(8.75) − 7 = −1.08

Answerx = 1.3: y = −5.17; x = 2.5: y = −1.39; x = 3: y = −1.11; x = 3.5: y = −1.08

Part (b) · 2 marks

Draw the curve y = 3log₃(x² − x) − 2x on the grid for x from 1.3 to 3.5.

  1. 1Plot (1.3, −5.17), (1.5, −3.79), (2, −2.11), (2.5, −1.39), (3, −1.11), (3.5, −1.08)
  2. 2Join with a smooth curve

AnswerPoints from the table plotted and joined with a smooth increasing curve that levels off near y ≈ −1.1

Part (c) · 6 marks

Draw a suitable straight line and use its intersection with the curve to estimate, to 1 decimal place, the root of 3^((4/3)x) = 9(x² − x)³ between 1.3 and 3.5.

  1. 13^((4/3)x)/3² = (x² − x)³, so 3^((4/3)x − 2) = (x² − x)³
  2. 2(4/3)x − 2 = log₃(x² − x)³ = 3log₃(x² − x)
  3. 3Subtract 2x: −(2/3)x − 2 = 3log₃(x² − x) − 2x
  4. 4Draw y = −(2/3)x − 2 (through (0, −2) and (3, −4))
  5. 5Intersection with the curve: x ≈ 1.6 (or 1.7)

AnswerDraw y = −(2/3)x − 2; root x = 1.6 or 1.7

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