0606/22 Feb/Mar 2025, Question 5: Turning Curves into Straight Lines
Cambridge IGCSE Additional Mathematics 0606 · Paper 22 (Calculator) · Calculator · 7 marks
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The question, in short
A graph of e^y against x² is a straight line with gradient −3 through (4.30, 5.85). Find y in terms of x, then the x values for which y exists.
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Part (a) · 4 marks
Use the gradient and the given point to form the linear relation, then take logs to express y in terms of x.
- 1e^y = mx² + c with m = −3
- 25.85 = −3(4.30) + c
- 3c = 18.75
- 4e^y = 18.75 − 3x², so y = ln(18.75 − 3x²)
Answery = ln(18.75 − 3x²)
Part (b) · 3 marks
Determine the range of x² for which the resulting y is defined, giving the allowed x values.
- 1y exists when 18.75 − 3x² > 0
- 2x² < 6.25
- 3Critical values ±2.5
- 4−2.5 < x < 2.5
Answer−2.5 < x < 2.5
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