Standard +0.3 This is a standard partial fractions question with a linear factor and an irreducible quadratic factor, requiring the form A/(2x-1) + (Bx+C)/(x²+4). While it's an FP2 topic (making it slightly harder than core content), the method is routine and well-practiced, involving straightforward algebraic manipulation to find three constants. The 7 marks reflect the working required rather than conceptual difficulty.
$\frac{2x^3+x+12}{(2x-1)(x^2+4)} = A + \frac{B}{2x-1} + \frac{Cx+D}{x^2+4}$ | B1 | For correct form soi (A can be Px + Q, but not 0)
$2x^3+x+12 = A(2x-1)(x^2+4) + B(x^2+4) + (Cx+D)(2x-1)$ | M1 | For multiplying out from their form
$A = 1, B = 3$ | B1 M1 | For either A or B correct (dep on 1st B1); For equating at least 2 coefficients (or substitute two values for x or one of each)
$x^3: 2 = 2A$ | |
$x^2: 0 = -A+B+2C$ | |
$x^1: 1 = 8A-C+2D$ | |
$x^0: 12 = -4A+4B-D$ | |
$C = -1, D = -4$ | A1A1 | For C, D correct
$\Rightarrow 1+\frac{3}{2x-1}+\frac{-x-4}{x^2+4}$ | A1 | For correct expression www
| [7] |
**ALT:** Divide out as not proper: $\Rightarrow 1+\frac{x^2-7x+16}{(2x-1)(x^2+4)}$ | B1 | Divide out
$= 1+\frac{A}{2x-1}+\frac{Bx+C}{x^2+4}$ | B1 | Writing in this form including 1
$x^2-7x+16 = A(x^2+4) + (Bx+C)(2x-1)$ | M1 | For multiplying out from their form
$x^2: 1 = A+2B$ | M1 | For equating at least 2 coefficients (or substitute two values for x or one of each)
$x: -7 = -B+2C$ | |
$1: 16 = 4A-C$ | |
$\Rightarrow A = 3, B = -1, C = -4$ | A1 A1 | B correct; C correct
$\Rightarrow 1+\frac{3}{2x-1}+\frac{-x-4}{x^2+4}$ | A1 | For correct expression www
SC $\Rightarrow \frac{3}{2x-1}+\frac{x^2-x}{x^2+4}$
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