4PM1/01R Jan 2022, Question 8: 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
For curve y = 2x² − sin x, find the gradient at x = π and show the normal at A has equation x + (4π + 1)y − π(8π² + 2π + 1) = 0.
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Solution · 8 marks
- 1At x = π: y = 2π² − sin π = 2π²
- 2dy/dx = 4x − cos x
- 3At x = π: dy/dx = 4π − cos π = 4π + 1
- 4Gradient of normal = −1/(4π + 1)
- 5y − 2π² = −(x − π)/(4π + 1)
- 6(4π + 1)(y − 2π²) = −x + π
- 7x + (4π + 1)y − 8π³ − 2π² − π = 0
- 8x + (4π + 1)y − π(8π² + 2π + 1) = 0
AnswerNormal x + (4π + 1)y − π(8π² + 2π + 1) = 0 shown
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