0606/21 Oct/Nov 2025, Question 7: Tangents, Normals and Stationary Points (Further)
Cambridge IGCSE Additional Mathematics 0606 · Paper 21 (Calculator) · Calculator · 6 marks
اقرأ هذا الحل مشروحاً بالعربي ←
The question, in short
For the curve y = x − ln(x² + 2x), demonstrate that only one stationary point exists, then determine the x-value at that point.
Summarised in our own words. Use the official paper for the exact wording and any diagrams.
Solution · 6 marks
- 1dy/dx = 1 − (2x + 2)/(x² + 2x)
- 2At a stationary point: x² + 2x = 2x + 2
- 3x² − 2 = 0, so x = ±√2
- 4x = −√2 gives x² + 2x = 2 − 2√2 < 0, so ln(2 − 2√2) is undefined
- 5So there is exactly one stationary point, at x = √2
Answerx = √2 (only stationary point; x = −√2 rejected)
Stuck on a step? Read the free lesson: Tangents, Normals and Stationary Points (Further)
Practise more tangents, normals and stationary points (further) questions
Real 0606 past-paper questions, marked on your working like the exam, with a worked solution for every part.
Practise free on LemmaraMore past-paper questions on tangents, normals and stationary points (further)
Also searched as: 0606/21/O/N/25 · 0606_w25_qp_21 · 0606 w25 21 · question 7.
Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education. The worked solutions are Lemmara’s own.