0606/12 May/Jun 2025, Question 6: Turning Curves into Straight Lines
Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 6 marks
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The question, in short
Given a straight line through (2, 5) and (−8, 25) on a graph of ln y against x³, find y as a function of x, then find x when y = e²⁵.
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Part (a) · 4 marks
Use the two points to find the gradient and intercept of the line, then convert the ln y versus x³ relationship into an equation for y in terms of x.
- 1ln y = mx³ + c
- 25 = 2m + c and 25 = −8m + c
- 3m = −2, c = 9
- 4ln y = 9 − 2x³, so y = e^(9 − 2x³)
Answery = e^(9 − 2x³)
Part (b) · 2 marks
Substitute y = e²⁵ into your expression for y and solve for the value of x.
- 1e²⁵ = e^(9 − 2x³), so 25 = 9 − 2x³
- 2x³ = −8
- 3x = −2
Answerx = −2
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Also searched as: 0606/12/M/J/25 · 0606_s25_qp_12 · 0606 s25 12 · question 6.
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