A2 October 2021 Paper 2 Q7
7 In this question you must show detailed reasoning.
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{x^3 + x^2 + 9x - 1}{x^3 + x^2 + 4x + 4} \equiv \dfrac{x^3 + x^2 + 4x + 4 + 5x - 5}{x^3 + x^2 + 4x + 4}\) \(= 1 + \dfrac{5x - 5}{x^3 + x^2 + 4x + 4}\) So \(A = 1\), \(B = 5\) and \(C = -5\) | B1 | 3.1a |
| [1] |
Notes
B1: Attempt to divide out improper fraction. Could be by symbolic division or other valid method (eg comparing coefficients or substitution of values for \(x\))
Allow embedded answers
| Scheme | Marks | AO |
|---|---|---|
| DR \(x^3 + x^2 + 4x + 4 = (x + 1)\left(x^2 + 4\right)\) | B1 | 3.1a |
| \(\dfrac{5x - 5}{x^3 + x^2 + 4x + 4} = \dfrac{D}{x + 1} + \dfrac{Ex + F}{x^2 + 4}\) \(D\left(x^2 + 4\right) + (x + 1)(Ex + F) = 5x - 5\) | M1 | 1.2 |
| \(x = -1 \Rightarrow 5D = -10 \Rightarrow D = -2\) | A1 | 1.1 |
| \(x = 0 \Rightarrow -8 + F = -5 \Rightarrow F = 3\) | A1 | 1.1 |
| \(x^2: D + E = 0 \Rightarrow E = 2\) \(1 - \dfrac{2}{x + 1} + \dfrac{2x + 3}{x^2 + 4}\) | A1 | 1.1 |
| [5] |
Notes
B1: Correct factorisation of cubic seen in working
Could be from improper fraction
M1: Correct form for partial fractions equated to their remainder rational fraction from (a). Follow through their division and factorisation.
A1: (first) Or equivalent to find \(D\) correctly.
A1: (second and third) Allow ft A1 for second and third coefficients found.
A1: (third) Or \(1 - \dfrac{2}{x + 1} + \dfrac{2x}{x^2 + 4} + \dfrac{3}{x^2 + 4}\)
(Corrected from the printed mark scheme: the \(x = 0\) line is printed as \(-2 - F = -5\); with \(D = -2\), putting \(x = 0\) gives \(4D + F = -5\), i.e. \(-8 + F = -5\), as typed above.)
| Scheme | Marks | AO |
|---|---|---|
| DR \(\displaystyle\int_0^2 \dfrac{x^3 + x^2 + 9x - 1}{x^3 + x^2 + 4x + 4}\,\mathrm{d}x = \int_0^2 1 - \frac{2}{x + 1} + \frac{2x}{x^2 + 4} + \frac{3}{x^2 + 4}\,\mathrm{d}x\) | *M1 | 3.1a |
| \(= \left[x - 2\ln(x + 1) + \ln\left(x^2 + 4\right) + \dfrac{3}{2}\tan^{-1}\left(\dfrac{x}{2}\right)\right]_0^2\) | dep*M1 | 1.1 |
| \(\left(2 - 2\ln 3 + \ln 8 + \dfrac{3\pi}{8}\right) - \ln 4\) | M1 | 1.1 |
| \(2 + \ln\left(\dfrac{2}{9}\right) + \dfrac{3}{8}\pi\) | A1 | 1.1 |
| [4] |
Notes
*M1: Split term with \(x^2 + 4\) in denominator and \(ax + b\) in numerator
dep*M1: Correctly integrate their expression (ignore limits)
M1: Correctly substitute limits to produce exact values and evaluate their \(\tan^{-1}\) term
A1: \(a = 2\), \(b = \tfrac{2}{9}\), \(c = \tfrac{3}{8}\)