4PM1/01R Jan 2023, Question 8: Solving Cubic Equations
Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 15 marks
اقرأ هذا الحل مشروحاً بالعربي ←
The question, in short
Given f′(x) = 17 + 2x − 3x² and that curve C passes through (a, 0), (−1, 0), (b, 0) and (0, c), work with the line through (0, c) and (b, 0) and the shaded region R.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Part (a) · 4 marks
Integrate the given gradient to obtain the cubic equation for C, using the point (−1, 0) to fix the constant and reach y = 15 + 17x + x² − x³.
- 1y = ∫ (17 + 2x − 3x²) dx = 17x + x² − x³ + k
- 2Curve passes through (−1, 0): 0 = −17 + 1 + 1 + k
- 3k = 15
- 4y = 15 + 17x + x² − x³
Answery = 15 + 17x + x² − x³ shown
Part (b) · 6 marks
Using the equation of C, determine the x-intercepts a and b and the y-intercept c.
- 1Divide by (x + 1): 15 + 17x + x² − x³ = (x + 1)(−x² + 2x + 15)
- 2−x² + 2x + 15 = (x + 3)(5 − x)
- 3x = −3, −1, 5 so a = −3, b = 5
- 4x = 0: y = 15 so c = 15
Answera = −3, b = 5, c = 15
Part (c) · 5 marks
Set up and evaluate an exact definite integral for the area between the straight line and C from x = 0 to x = b.
- 1Area R = ∫ from 0 to 5 of (15 + 17x + x² − x³) dx − area of triangle
- 2∫ = [15x + 17x²/2 + x³/3 − x⁴/4] from 0 to 5 = 2075/12
- 3Triangle under l: ½ × 5 × 15 = 37.5
- 4Area R = 2075/12 − 75/2 = 1625/12
Answer1625/12 (= 135 5/12)
Stuck on a step? Read the free lesson: Solving Cubic Equations
Practise more solving cubic equations questions
Real 4PM1 past-paper questions, marked on your working like the exam, with a worked solution for every part.
Practise free on LemmaraMore past-paper questions on solving cubic equations
Also searched as: 4PM1 1R January 2023 · 4PM1/01R · 4pm1 jan23 1r · question 8.
Lemmara is an independent study tool, not affiliated with or endorsed by Pearson Edexcel. The worked solutions are Lemmara’s own.