1.
$$\frac { 9 x ^ { 2 } } { ( x - 1 ) ^ { 2 } ( 2 x + 1 ) } = \frac { A } { ( x - 1 ) } + \frac { B } { ( x - 1 ) ^ { 2 } } + \frac { C } { ( 2 x + 1 ) }$$
Find the values of the constants \(A , B\) and \(C\).
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\(9x^2 = A(x-1)(2x+1) + B(2x+1) + C(x-1)^2\) B1
\(x \to 1\): \(9 = 3B \Rightarrow B = 3\) M1
\(x \to -\frac{1}{2}\): \(\frac{9}{4} = \left(-\frac{3}{2}\right)^2 C \Rightarrow C = 1\) A1
Any two of \(A, B, C\)
\(x^2\) terms: \(9 = 2A + C \Rightarrow A = 4\) A1
All three correct
Alternatives for finding \(A\):
\(x\) terms: \(0 = -A + 2B - 2C \Rightarrow A = 4\)
Answer Marks
Constant terms: \(0 = -A + B + C \Rightarrow A = 4\) A1
(4) [4]
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$9x^2 = A(x-1)(2x+1) + B(2x+1) + C(x-1)^2$ | B1 |
$x \to 1$: $9 = 3B \Rightarrow B = 3$ | M1 |
$x \to -\frac{1}{2}$: $\frac{9}{4} = \left(-\frac{3}{2}\right)^2 C \Rightarrow C = 1$ | A1 | Any two of $A, B, C$
$x^2$ terms: $9 = 2A + C \Rightarrow A = 4$ | A1 | All three correct
Alternatives for finding $A$:
$x$ terms: $0 = -A + 2B - 2C \Rightarrow A = 4$
Constant terms: $0 = -A + B + C \Rightarrow A = 4$ | A1 |
| (4) [4] |
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1.
$$\frac { 9 x ^ { 2 } } { ( x - 1 ) ^ { 2 } ( 2 x + 1 ) } = \frac { A } { ( x - 1 ) } + \frac { B } { ( x - 1 ) ^ { 2 } } + \frac { C } { ( 2 x + 1 ) }$$
Find the values of the constants $A , B$ and $C$.\\
\hfill \mbox{\textit{Edexcel C4 2011 Q1 [4]}}