Let f(x) = (21 – 8x – 2x²) / ((1 + 2x)(3−x)²).

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [B1] State or imply the form A/(1+2x) + B/(3-x) + C/((3-x)²); Alternative form: A/(1+2x) + (Dx+E)/((3-x)²)
2. [M1] Use a correct method to find a constant; Incorrect format for partial fractions: Allow M1 and a possible A1 if obtain one of these correct values. Max 2/5 Allow M1 even if multiply up by (1 + 2x)(3 - x)³.
3. [A1] Obtain one of A = 2, B = 2 and C = -3; Alternative form: obtain one of A = 2, D = −2 and E = 3 .
4. [A1] Obtain a second value
5. [A1] Obtain the third value; Do not need to substitute values back into original form. If A/(1+2x) + B/(3-x) + (Cx+D)/((3-x)²) B0 but M1 A1 for A, A1 for B and A1 for C and D. If C = 0 then recovers B1 from above.
6. [M1] Use a correct method to obtain the first two terms of one of the unsimplified expansions; (1+2x)⁻¹, (1-(x/3))⁻¹, (1-(x/3))⁻², (3-x)⁻¹, (3-x)⁻²
7. [A1 FT] Obtain the correct unsimplified expansions up to the term in x² for each partial fraction; Follow through on their A, B, C A(1 + (−1)(2x) + ((-1)(-2)/2)(2x)² + ...) B(1 + (−1)(-x/3) + ((-1)(-2)/2)(-x/3)² + ...) C(1 + (−2)(-x/3) + ((-2)(-3)/2)(-x/3)² + ...). Must be their coefficients from (a) but may be unsimplified expansions for FT marks. If correct, expect to see 2(1 - 2x + (2x)²) or 2 – 4x + 8x² (1/3)(1 + (x/3) + (x/3)²) or (1/3) + (x/9) + (x²/27) (-2x+3)/9 * (1 + (2x/3) + (x²/3)) or (-2x+3)/9 * (1 + (x/3) + (x²/9))
8. [A1 FT] If correct, should be working with 2/(1+2x) + 2/(3-x) + -3/((3-x)² ) or 2/(1+2x) + (-2x+3)/((3-x)² )
9. [A1] Obtain final answer 7/3 - 4x + (215/27)x²; Accept 2 (1/3) - 4x + 7 (26/27)x² . No ISW.