June 2025 Paper 3 Q8
8
(a) The expression\[\frac{x}{2x^2 + 3x + 1}\]can be written in the form\[\frac{A}{x + 1} + \frac{B}{2x + 1}\]
Find the value of \(A\) and the value of \(B\)
[3 marks](b) Use your answer to part (a) to show that\[\int_0^4 \frac{x}{2x^2 + 3x + 1}\,\mathrm{d}x = \ln q\]
where \(q\) is a rational number to be found.
[4 marks]| Scheme | Marks | AO |
|---|---|---|
| Uses a suitable method and finds a value for \(A\) or \(B\). eg rearranges and substitutes values or compares coefficients or uses cover-up method or uses inspection PI by \(A = 1\) or \(B = -1\) | M1 | 1.1a |
| Obtains \(A = 1\) Allow if seen in \(\dfrac{1}{x + 1}\) | A1 | 1.1b |
| Obtains \(B = -1\) Allow if seen in \(\dfrac{-1}{2x + 1}\) | A1 | 1.1b |
| (3) |
Typical solution
\[x = A(2x + 1) + B(x + 1)\]\[x = -1 \Rightarrow -1 = -A\]\[A = 1\]\[x = -\frac{1}{2} \Rightarrow -\frac{1}{2} = \frac{1}{2}B\]\[B = -1\]| Scheme | Marks | AO |
|---|---|---|
| Uses partial fractions with their \(A\) and \(B\) to obtain \(A\ln(x + 1)\) or \(\dfrac{B}{2}\ln(2x + 1)\) OE Condone \(B\) rather than \(\dfrac{B}{2}\) Condone missing bracket or sign error of \(A\) or \(B\) | M1 | 1.1a |
| Obtains \(\ln(x + 1) - \dfrac{1}{2}\ln(2x + 1)\) OE | A1 | 1.1b |
| Substitutes both limits correctly into their integrated expression of two \(\ln\) terms (the subtraction does not need to be seen) Condone 0 for \(\left(\ln 1 - \dfrac{1}{2}\ln 1\right)\) | M1 | 1.1a |
| Completes reasoned argument to obtain \(\ln\dfrac{5}{3}\) Allow unsimplified fraction or correct recurring decimal for \(\dfrac{5}{3}\) CSO | R1 | 2.1 |
| (4) | ||
| (7 marks) |