Lemmara logoLemmaraWorked solution

4PM1/02 Nov 2021, Question 4: Equation of the Tangent Line

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 11 marks

اقرأ هذا الحل مشروحاً بالعربي ←

The question, in short

For curve y = 2x² + px + q with a stationary point at (3, −5), find p and q, identify the stationary point's nature, and find the normal at x = 1.

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

Part (a) · 4 marks

Show that p = −12 and determine the integer q using the stationary point at (3, −5).

  1. 1dy/dx = 4x + p = 0 at x = 3, so 12 + p = 0, p = −12
  2. 2−5 = 2(3²) − 12(3) + q
  3. 3q = 13
  4. 4(Or: y = 2(x − 3)² − 5 = 2x² − 12x + 13)

Answerp = −12 shown; q = 13

Part (b) · 1 mark

State whether the stationary point is a maximum or minimum, giving a reason.

  1. 1d²y/dx² = 4 > 0, so the stationary point is a minimum

AnswerMinimum, since the coefficient of x² is positive (d²y/dx² = 4 > 0)

Part (c) · 6 marks

Find the normal to C at x = 1, giving the equation in the form ax + by + c = 0.

  1. 1dy/dx = 4x − 12 = −8 at x = 1
  2. 2Gradient of normal = 1/8
  3. 3At x = 1: y = 2 − 12 + 13 = 3
  4. 4y − 3 = (1/8)(x − 1)
  5. 5x − 8y + 23 = 0

Answerx − 8y + 23 = 0

Stuck on a step? Read the free lesson: Equation of the Tangent Line

Practise more equation of the tangent line questions

Real 4PM1 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 equation of the tangent line

Also searched as: 4PM1 2 November 2021 · 4PM1/02 · 4pm1 nov21 2 · question 4.
Lemmara is an independent study tool, not affiliated with or endorsed by Pearson Edexcel. The worked solutions are Lemmara’s own.