Moderate -0.5 This is a straightforward partial fractions question with a linear factor and an irreducible quadratic factor, requiring standard technique taught in C4. The setup is routine (A/(2x+1) + (Bx+C)/(x²+1)), and solving for coefficients involves basic algebra with no conceptual challenges beyond applying the method correctly.
Multiply up and equating or substituting: \(1 = A(x^2+1) + (Bx+C)(2x+1)\)
\(x = -\frac{1}{2}: 1 = \frac{5}{4}A \Rightarrow A = \frac{4}{5}\)
Coeff of \(x^2\): \(0 = A + 2B \Rightarrow B = -\frac{2}{5}\)
Constants: \(1 = A + C \Rightarrow C = \frac{1}{5}\)
Answer
Marks
M1
correct form of partial fractions
M1
mult up and equating or substituting oe soi www
B1
www
B1
www
B1
www
For omission of \(B\) or \(C\) on numerator, M0, M1, then (x= -1/2, A= 4/5) B1, B0, B0 is possible.
For \(\frac{A+Dx}{2x+1} + \frac{Bx+C}{x^2+1}\), M1,M1 then B1 for both A=4/5 and D=0, B1 is possible.
isw for incorrect assembly of final partial fractions following correct A,B & C.
condone omission of brackets for second M1 only if the brackets are implied by subsequent working.
Total: [5]
**Answer:** $\frac{1}{(2x+1)(x^2+1)} = \frac{A}{2x+1} + \frac{Bx+C}{x^2+1}$
Multiply up and equating or substituting: $1 = A(x^2+1) + (Bx+C)(2x+1)$
$x = -\frac{1}{2}: 1 = \frac{5}{4}A \Rightarrow A = \frac{4}{5}$
Coeff of $x^2$: $0 = A + 2B \Rightarrow B = -\frac{2}{5}$
Constants: $1 = A + C \Rightarrow C = \frac{1}{5}$
| M1 | correct form of partial fractions |
| M1 | mult up and equating or substituting oe soi www |
| B1 | www |
| B1 | www |
| B1 | www |
| | For omission of $B$ or $C$ on numerator, M0, M1, then (x= -1/2, A= 4/5) B1, B0, B0 is possible. |
| | For $\frac{A+Dx}{2x+1} + \frac{Bx+C}{x^2+1}$, M1,M1 then B1 for both A=4/5 and D=0, B1 is possible. |
| | isw for incorrect assembly of final partial fractions following correct A,B & C. |
| | condone omission of brackets for second M1 only if the brackets are implied by subsequent working. |
**Total: [5]**
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