June 2025 Paper 1 Q2
2 Express \(\dfrac{7x-25}{(x-1)(x-4)^2}\) in partial fractions. [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{7x-25}{(x-1)(x-4)^2} = \dfrac{A}{x-1} + \dfrac{B}{x-4} + \dfrac{C}{(x-4)^2}\) | M1 | 1.1a |
| \(7x-25 = A(x-4)^2 + B(x-1)(x-4) + C(x-1)\) | M1 | 1.1 |
| When \(x=4,\ 28-25=3C\) giving \(C=1\) When \(x=1,\ 7-25=9A\) giving \(A=-2\) Equate coefficients of \(x^2\): \(0=A+B\) giving \(B=2\) | M1 | 1.1 |
| \(\dfrac{-2}{x-1} + \dfrac{2}{x-4} + \dfrac{1}{(x-4)^2}\) | A1 | 1.1 |
| [4] |
Notes
M1: Sets up partial fractions of the correct form
M1: Clears the denominator (for their form with at least two fractions). Not dependent on previous M1
M1: Uses at least one value of \(x\) leading to a value for one of the constants (for their form). Allow for equating coefficients for at least one power of \(x\) leading to a value for one of the constants
A1: Fully correct expression must be seen
SC: If \(\dfrac{7x-25}{(x-1)(x-4)^2} = \dfrac{A}{x-1} + \dfrac{Bx+C}{(x-4)^2}\) used, award M0 and up to M1 M1 SC B1 for \(\dfrac{7x-25}{(x-1)(x-4)^2} = \dfrac{-2}{x-1} + \dfrac{2x-7}{(x-4)^2}\)