4PM1/01R Jun 2019, Question 10: Simultaneous Equations with a Curve
Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 13 marks
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The question, in short
Rewrite f(x) = 6x − x² in completed-square form D(x + E)² + F, then use it to find the maximum, intersect with y = x² − 4x + 8, and integrate to get the area between the curves.
Summarised in our own words. Use the official paper for the exact wording and any diagrams.
Part (a) · 3 marks
Express 6x − x² as D(x + E)² + F with integer D, E, F; state the three values.
- 16x − x² = −(x² − 6x)
- 2= −[(x − 3)² − 9]
- 3= −(x − 3)² + 9: D = −1, E = −3, F = 9
AnswerD = −1, E = −3, F = 9
Part (b) · 2 marks
From the completed square, give the largest value of f(x) and the x value where it happens.
- 1Maximum value = F = 9
- 2It occurs when x − 3 = 0, so x = 3
Answer(i) 9 (ii) x = 3
Part (c) · 4 marks
Solve f(x) = x² − 4x + 8 to find the coordinates of both intersection points of C and S.
- 16x − x² = x² − 4x + 8
- 22x² − 10x + 8 = 0
- 3(2x − 2)(x − 4) = 0 → x = 1, x = 4
- 4y = 5 and y = 8
Answer(1, 5) and (4, 8)
Part (d) · 4 marks
Set up and evaluate a definite integral of the difference between the two curves to find the area of R.
- 1Area = ∫ from 1 to 4 of [(6x − x²) − (x² − 4x + 8)] dx
- 2= ∫ from 1 to 4 of (−2x² + 10x − 8) dx
- 3= [−2x³/3 + 5x² − 8x] from 1 to 4
- 4= 16/3 − (−11/3) = 9
Answer9
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