Standard +0.3 This is a standard partial fractions question with an irreducible quadratic factor. It requires setting up the form A/(2x+1) + (Bx+C)/(x²+1), equating coefficients, and solving a simple system of equations. While slightly more involved than linear factors only, this is a routine textbook exercise that follows a well-practiced algorithm with no conceptual surprises.
Correct form of partial fractions. For omission of \(B\) or \(C\) on numerator: M0, M1, then (\(x=-\frac{1}{2}\), \(A=\frac{4}{5}\)) B1, B0, B0 is possible
\(1 = A(x^2+1) + (Bx+C)(2x+1)\)
M1
Multiply up and equating or substituting oe soi
\(x = -\frac{1}{2}:\ 1 = 1\frac{1}{4}A \Rightarrow A = \frac{4}{5}\)
B1
www
coeff of \(x^2\): \(0 = A + 2B \Rightarrow B = -\frac{2}{5}\)
B1
www
constants: \(1 = A + C \Rightarrow C = \frac{1}{5}\)
B1
www. isw for incorrect assembly of final partial fractions following correct \(A\), \(B\) & \(C\)
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{(2x+1)(x^2+1)} = \frac{A}{2x+1} + \frac{Bx+C}{x^2+1}$ | M1 | Correct form of partial fractions. For omission of $B$ or $C$ on numerator: M0, M1, then ($x=-\frac{1}{2}$, $A=\frac{4}{5}$) B1, B0, B0 is possible |
| $1 = A(x^2+1) + (Bx+C)(2x+1)$ | M1 | Multiply up and equating or substituting oe soi |
| $x = -\frac{1}{2}:\ 1 = 1\frac{1}{4}A \Rightarrow A = \frac{4}{5}$ | B1 | www |
| coeff of $x^2$: $0 = A + 2B \Rightarrow B = -\frac{2}{5}$ | B1 | www |
| constants: $1 = A + C \Rightarrow C = \frac{1}{5}$ | B1 | www. isw for incorrect assembly of final partial fractions following correct $A$, $B$ & $C$ |
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