0606/12 Feb/Mar 2026, Question 5: Tangents, Normals and Stationary Points (Further)
Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 7 marks
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The question, in short
Differentiate y = (x² − 8)/(3 − x), locate the two stationary points P and Q, and find the exact distance between them.
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Solution · 7 marks
- 1Quotient rule: dy/dx = [2x(3 − x) − (x² − 8)(−1)]/(3 − x)²
- 2Numerator = −x² + 6x − 8, so dy/dx = 0 gives x² − 6x + 8 = 0
- 3(x − 4)(x − 2) = 0, x = 4 or x = 2
- 4x = 4: y = 8/(−1) = −8; x = 2: y = −4/1 = −4
- 5PQ = √((4 − 2)² + (−8 + 4)²) = √20 = 2√5
Answerdy/dx = (2x(3 − x) − (x² − 8)(−1))/(3 − x)²; PQ = √20 = 2√5
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