A2 June 2025 Paper 2 Q7
7 In this question you must show detailed reasoning.
| Scheme | Marks | AO |
|---|---|---|
| DR \(1^3 - 1^2 + 3 \times 1 - 3 = 0\) so \((x - 1)\) is a factor | M1 | 2.2a |
| \(x^3 - x^2 + 3x - 3 = (x - 1)(x^2 + 3)\) | A1 | 1.1 |
| \(\dfrac{-4x^2 + 5x - 17}{x^3 - x^2 + 3x - 3} = \dfrac{A}{x - 1} + \dfrac{Bx + C}{x^2 + 3}\) | B1 | 3.1a |
| \(-4x^2 + 5x - 17\) \(= A(x^2 + 3) + (Bx + C)(x - 1)\) | M1 | 1.1 |
| \(4A = -16 \Rightarrow A = -4\) \(-4 = -4 + B \Rightarrow B = 0\) \(-17 = -12 - C \Rightarrow C = 5\) \(\left(\text{so } \dfrac{5}{x^2 + 3} - \dfrac{4}{x - 1}\right)\) | A1 A1 | 1.1 1.1 |
| [6] |
Notes
M1: Using factor theorem in attempt to factorise denominator. Must come up with \(\mathrm{f}(a) = 0\) so \((x - a)\) is a factor.
May be implied by correct factorisation seen
(corrected from the printed mark scheme, which says “factorise numerator”; it is the denominator \(x^3 - x^2 + 3x - 3\) that is factorised)
A1: soi by denominators
B1: Correct partial fraction form. B0 if e.g. \(+D\) or \((A + Ex)\) unless recovered (i.e. unnecessary constants found to be 0).
M1: Suitable method for determining constants, including comparing coefficients directly. (Provided two fractions with linear or quadratic denominators and any additional polynomial terms).
Condone minor errors e.g. \(Bx + C(x - 1)\) or denominator on one side provided intent is clear (can be determined by next step)
A1: Any constant correct from correct working (allow this mark following B0)
A1: All three constants (including \(B = 0\)) correct from correct working.
Final 3 marks independent of first two marks. ISW once constants found.
| Scheme | Marks | AO |
|---|---|---|
| DR \(\displaystyle\int \frac{\text{‘}4\text{’}}{x - 1}\,\mathrm{d}x = \text{‘}4\text{’}\ln(x - 1)\) | B1FT | 1.1 |
| \(\displaystyle\int \frac{\text{‘}5\text{’}}{x^2 + 3}\,\mathrm{d}x = k\tan^{-1}\frac{x}{\sqrt{3}}\) | M1 | 1.1 |
| \(\therefore \displaystyle\int \frac{5}{x^2 + 3} - \frac{4}{x - 1}\,\mathrm{d}x = \frac{5}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}} - 4\ln(x - 1)\) | A1 | 1.1 |
| \(\left[\dfrac{5}{\sqrt{3}}\tan^{-1}\dfrac{x}{\sqrt{3}} - 4\ln(x - 1)\right]_{\sqrt{3}}^{3}\) \(= \left(\frac{5}{\sqrt{3}}\tan^{-1}\sqrt{3} - 4\ln(2)\right) - \left(\frac{5}{\sqrt{3}}\tan^{-1}1 - 4\ln\left(\sqrt{3} - 1\right)\right)\) \(= \dfrac{5}{\sqrt{3}} \times \dfrac{\pi}{3} - 4\ln 2 - \dfrac{5}{\sqrt{3}} \times \dfrac{\pi}{4} + 4\ln\left(\sqrt{3} - 1\right)\) \(= \dfrac{5\pi}{12\sqrt{3}} + 4\ln\left(\dfrac{\sqrt{3} - 1}{2}\right)\) | A1 | 1.1 |
| [4] |
Notes
B1FT: Soi. May see \(\ln|1 - x|\). Ignore “\(+c\)”. FT their \(\int \frac{a}{bx + c}\,\mathrm{d}x = \frac{a}{b}\ln(bx + c)\)
M1: Recognising the integral as \(\tan^{-1}\) with any multiplicative constant. Condone 3 in the denominator. Could be their ‘\(C\)’.
A1: All correct. Ignore “\(+c\)”.
A1: oe e.g. \(\frac{5\pi}{6\sqrt{12}} + 2\ln\left(\frac{4 - 2\sqrt{3}}{4}\right)\) or \(\frac{5\sqrt{3}\pi}{36} + \ln\left(\frac{7 - 4\sqrt{3}}{4}\right)\) or \(\frac{5\pi}{\sqrt{432}} - \ln\left(28 + 16\sqrt{3}\right)\) or \(\frac{5\sqrt{3}\pi}{36} - 4\ln\left(1 + \sqrt{3}\right)\) but arctans must be evaluated and \(\pi\) terms must be combined into a single fraction and ln terms must be collected (allow \(\ln(\ldots)^4\) forms provided there is a single ln term). ISW once a complete, correct, acceptable form seen from correct working.
Correct answers without working is 0/4.