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4PM1/01R Jun 2023, Question 7: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 8 marks

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The question, in short

Curve y = mx² + 64√x + 39 has a stationary point at (4, n), with m and n integers. Using calculus, find m and n, then determine the stationary point's nature.

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Part (a) · 6 marks

Differentiate y = mx² + 64√x + 39, set the derivative to zero at x = 4, and solve for the integer m, then find n.

  1. 1dy/dx = 2mx + 32x^(−1/2)
  2. 2At x = 4: 8m + 32/2 = 0, so 8m + 16 = 0
  3. 3m = −2
  4. 4n = −2(16) + 64√4 + 39 = −32 + 128 + 39 = 135

Answerm = −2, n = 135

Part (b) · 2 marks

Use the second derivative or a sign test to establish whether the stationary point at x = 4 is a maximum or minimum.

  1. 1d²y/dx² = 2m − 16x^(−3/2) = −4 − 16x^(−3/2)
  2. 2At x = 4: −4 − 16/8 = −6
  3. 3−6 < 0, so the point is a maximum

Answerd²y/dx² = −6 < 0, so the stationary point is a maximum

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