Standard +0.3 This is a standard partial fractions question with an irreducible quadratic factor. It requires setting up the form A/(2-x) + (Bx+C)/(4+x²), equating coefficients, and solving a system of equations. While slightly more involved than linear factors only, this is a routine C4 technique with no conceptual surprises, making it slightly easier than average.
Correct form of partial fractions. Condone additional coeffs e.g. \(\frac{Ax+B}{2-x} + \frac{Cx+D}{4+x^2}\) for M1. BUT \(\frac{A}{2-x} + \frac{B}{4+x^2}\) is M0
\(3x = A(4+x^2) + (Bx+C)(2-x)\)
M1
Multiplying through oe and substituting values or equating coeffs at least as far as finding a value for one unknown (even if incorrect). Condone a sign error or single computational error for M1 but not a conceptual error
\(x=2 \Rightarrow 6=8A,\ A=\frac{3}{4}\)
A1
oe www
\(x^2\) coeffs: \(0 = A - B \Rightarrow B = \frac{3}{4}\)
A1
oe www
constants: \(0 = 4A + 2C \Rightarrow C = -1\frac{1}{2}\)
A1
oe www
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3x}{(2-x)(4+x^2)} = \frac{A}{2-x} + \frac{Bx+C}{4+x^2}$ | M1 | Correct form of partial fractions. Condone additional coeffs e.g. $\frac{Ax+B}{2-x} + \frac{Cx+D}{4+x^2}$ for M1. BUT $\frac{A}{2-x} + \frac{B}{4+x^2}$ is M0 |
| $3x = A(4+x^2) + (Bx+C)(2-x)$ | M1 | Multiplying through oe and substituting values or equating coeffs at least as far as finding a value for one unknown (even if incorrect). Condone a sign error or single computational error for M1 but not a conceptual error |
| $x=2 \Rightarrow 6=8A,\ A=\frac{3}{4}$ | A1 | oe www |
| $x^2$ coeffs: $0 = A - B \Rightarrow B = \frac{3}{4}$ | A1 | oe www |
| constants: $0 = 4A + 2C \Rightarrow C = -1\frac{1}{2}$ | A1 | oe www |
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