A2 June 2024 Paper 1 Q5
5 Express \(\dfrac{12x^3}{(2x + 1)(2x^2 + 1)}\) using partial fractions. [5]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{12x^3}{(2x + 1)(2x^2 + 1)} = A + \dfrac{B}{2x + 1} + \dfrac{Cx + D}{2x^2 + 1}\) | B1 | 1.1 |
| \(12x^3 \equiv A(2x + 1)(2x^2 + 1) + B(2x^2 + 1) + (Cx + D)(2x + 1)\) | M1* | 1.1 |
| For example: Equating coefficients of \(x^3 : A = 3\) Let \(x = 0\), gives \(A + B + D = 0\) | M1dep* | 1.1 |
| A1 | 1.1 | |
| \(3 - \dfrac{1}{2x + 1} - \dfrac{2x + 2}{2x^2 + 1}\) | A1 | 1.1 |
| [5] |
Notes
B1: Correct form (possibly implied by correct identity).
M1*: Identity without fractions. Follow through their partial fraction expression with one denominator of \(2x + 1\) and the other with \(2x^2 + 1\) - both numerators must contain at least a constant unknown. Some examples for M1 below:
\(\dfrac{B}{2x + 1} + \dfrac{Cx + D}{2x^2 + 1}\) so \(12x^3 \equiv B(2x^2 + 1) + (Cx + D)(2x + 1)\)
\(A + \dfrac{B}{2x + 1} + \dfrac{C}{2x^2 + 1}\) so \(12x^3 \equiv A(2x + 1)(2x^2 + 1) + B(2x^2 + 1) + C(2x + 1)\)
\(\dfrac{B}{2x + 1} + \dfrac{C}{2x^2 + 1}\) so \(12x^3 \equiv B(2x^2 + 1) + C(2x + 1)\).
\(\dfrac{A}{2x + 1} + \dfrac{Bx^2 + Cx + D}{2x^2 + 1}\) so \(12x^3 \equiv A(2x^2 + 1) + (Bx^2 + Cx + D)(2x + 1)\).
M1dep*: Equates coefficients or substitutes to find an equation involving only their unknowns. Do not award this mark if only two unknowns in their partial fractions (so must be at least three unknowns (or implied unknowns)).
A1: Any two (\(A = 3\), \(B = -1\), \(C = -2\), \(D = -2\)) unknowns correct from a correct partial fraction form.
A1: All four unknowns correct – condone stating the correct form of the partial fractions anywhere together with all four unknowns correctly stated without necessarily bringing both parts together as a single expression at the end.
Alternative method
| Scheme | Marks |
|---|---|
| Constant term of (\(A =\)) 3 | B1 |
| \(\dfrac{-6x^2 - 6x - 3}{(2x + 1)(2x^2 + 1)} \equiv \dfrac{B}{2x + 1} + \dfrac{Cx + D}{2x^2 + 1}\) and so \(-6x^2 - 6x - 3 \equiv B(2x^2 + 1) + (Cx + D)(2x + 1)\) | M1* |
| For example: Equating coefficients of \(x^2 : 2B + 2C = -6\) Let \(x = 0\), gives \(B + D = -3\) | M1dep* |
| A1 | |
| \(3 - \dfrac{1}{2x + 1} - \dfrac{2x + 2}{2x^2 + 1}\) | A1 |
| [5] |
B1: By polynomial division or inspection.
M1*: Re-writing \(\dfrac{\mathrm{f}(x)}{(2x + 1)(2x^2 + 1)}\), where \(\mathrm{f}(x)\) is quadratic, as \(\dfrac{B}{2x + 1} + \dfrac{Cx + D}{2x^2 + 1}\) or \(\dfrac{B}{2x + 1} + \dfrac{C}{2x^2 + 1}\) and correct identity not involving fractions following through their partial fractions and quadratic \(\mathrm{f}(x)\).
M1dep*: Equates coefficients or substitutes to find an equation involving only their unknown(s).
A1: Any (\(B = -1\), \(C = -2\), \(D = -2\)) one unknown correct from a correct partial fraction form (so must have had a correct \(\mathrm{f}(x)\)).
A1: All unknowns correct – condone stating the correct form of the partial fractions anywhere together with all four unknowns correctly stated without necessarily bringing both parts together as a single expression at the end. So must see correct partial fraction expression or \(A + \dfrac{B}{2x + 1} + \dfrac{Cx + D}{2x^2 + 1}\) stated and all correct values of \(A\), \(B\), \(C\) and \(D\) seen.