Lemmara logoLemmaraWorked solution

0606/13 May/Jun 2026, Question 12: Tangents, Normals and Stationary Points (Further)

Cambridge IGCSE Additional Mathematics 0606 · Paper 13 (Non-calculator) · Non-calculator · 7 marks

اقرأ هذا الحل مشروحاً بالعربي ←

The question, in short

A curve is defined by y = e^(2a/x) with a positive constant a. Find the gradient function, then use the normal at x = a through the origin to determine a.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 2 marks

Differentiate y = e^(2a/x) with respect to x, giving the result containing a.

  1. 1Let u = 2a/x, so du/dx = −2a/x²
  2. 2dy/dx = e^u × du/dx
  3. 3dy/dx = −(2a/x²) e^(2a/x)

Answerdy/dx = −(2a/x²) e^(2a/x)

Part (b) · 5 marks

Given the normal at x = a passes through the origin, work out the value of a.

  1. 1At x = a: y = e², gradient = −(2/a)e²
  2. 2Gradient of normal = (a/2)e⁻²
  3. 3Normal: y − e² = (a/2)e⁻²(x − a)
  4. 4Through (0, 0): −e² = −(a²/2)e⁻²
  5. 5a² = 2e⁴, a = √2 e² (a > 0)

Answera = √2 e²

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 Lemmara

More past-paper questions on tangents, normals and stationary points (further)

Also searched as: 0606/13/M/J/26 · 0606_s26_qp_13 · 0606 s26 13 · question 12.
Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education. The worked solutions are Lemmara’s own.