4. (a) Express \(\frac { 2 x - 1 } { ( x - 1 ) ( 2 x - 3 ) }\) in partial fractions.
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Question 4(a):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(\frac{2x-1}{(x-1)(2x-3)} \equiv \frac{A}{(x-1)} + \frac{B}{(2x-3)}\) \(2x - 1 \equiv A(2x-3) + B(x-1)\) M1
Forming this identity. NB : A & B are not assigned in this question
Let \(x = \frac{3}{2}\): \(2 = B\left(\frac{1}{2}\right) \Rightarrow B = 4\) A1
Either one of \(A = -1\) or \(B = 4\)
Let \(x = 1\): \(1 = A(-1) \Rightarrow A = -1\) A1
Both correct for their A, B
giving \(\frac{-1}{(x-1)} + \frac{4}{(2x-3)}\)
[3 marks] Question 4(b) & (c) — Way 1:
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(\int\frac{dy}{y} = \int\frac{(2x-1)}{(2x-3)(x-1)}\,dx\) B1
Separates variables as shown. Can be implied
\(= \int\frac{-1}{(x-1)} + \frac{4}{(2x-3)}\,dx\) M1\(\checkmark\)
Replaces RHS with their partial fraction to be integrated
\(\ln y = -\ln(x-1) + 2\ln(2x-3) + c\) M1
*At least* two terms in ln's
A1\(\checkmark\) *At least* two ln terms correct
A1 All three terms correct and \(+ c\)
\(y = 10\), \(x = 2\) gives \(c = \ln 10\) B1
\(c = \ln 10\)
\(\ln y = -\ln(x-1) + \ln(2x-3)^2 + \ln 10\) M1
Using the power law for logarithms
\(\ln y = \ln\left(\frac{(2x-3)^2}{(x-1)}\right) + \ln 10\) or \(\ln y = \ln\left(\frac{10(2x-3)^2}{(x-1)}\right)\) M1
Using product and/or quotient laws to obtain a single RHS logarithmic term with/without constant \(c\)
\(y = \dfrac{10(2x-3)^2}{(x-1)}\) A1 aef
\(y = \dfrac{10(2x-3)^2}{(x-1)}\) or aef, isw
[5] & [4] marks Question 4(b) & (c) — Way 3:
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(\int\frac{dy}{y} = \int\frac{(2x-1)}{(2x-3)(x-1)}\,dx\) B1
Separates variables. Can be implied
\(= \int\frac{-1}{(x-1)} + \frac{2}{\left(x-\frac{3}{2}\right)}\,dx\) M1\(\checkmark\)
Replaces RHS with their partial fraction
\(\ln y = -\ln(x-1) + 2\ln\left(x - \frac{3}{2}\right) + c\) M1, A1\(\checkmark\), A1
At least two ln terms; at least two correct; all three correct with \(+c\)
\(y = 10\), \(x = 2\) gives \(c = \ln 10 - 2\ln\left(\frac{1}{2}\right) = \ln 40\) B1 oe
\(c = \ln 10 - 2\ln\left(\frac{1}{2}\right)\) or \(c = \ln 40\)
\(\ln y = -\ln(x-1) + \ln\left(x-\frac{3}{2}\right)^2 + \ln 40\) M1
Using power law for logarithms
\(\ln y = \ln\left(\dfrac{40\left(x-\frac{3}{2}\right)^2}{(x-1)}\right)\) M1
Using product/quotient laws
\(y = \dfrac{40\left(x-\frac{3}{2}\right)^2}{(x-1)}\) A1 aef
or aef, isw
[4 marks]
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## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2x-1}{(x-1)(2x-3)} \equiv \frac{A}{(x-1)} + \frac{B}{(2x-3)}$ | | |
| $2x - 1 \equiv A(2x-3) + B(x-1)$ | M1 | Forming this identity. **NB**: A & B are not assigned in this question |
| Let $x = \frac{3}{2}$: $2 = B\left(\frac{1}{2}\right) \Rightarrow B = 4$ | A1 | Either one of $A = -1$ or $B = 4$ |
| Let $x = 1$: $1 = A(-1) \Rightarrow A = -1$ | A1 | Both correct for their A, B |
| giving $\frac{-1}{(x-1)} + \frac{4}{(2x-3)}$ | | |
**[3 marks]**
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## Question 4(b) & (c) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{dy}{y} = \int\frac{(2x-1)}{(2x-3)(x-1)}\,dx$ | B1 | Separates variables as shown. Can be implied |
| $= \int\frac{-1}{(x-1)} + \frac{4}{(2x-3)}\,dx$ | M1$\checkmark$ | Replaces RHS with their partial fraction to be integrated |
| $\ln y = -\ln(x-1) + 2\ln(2x-3) + c$ | M1 | *At least* two terms in ln's |
| | A1$\checkmark$ | *At least* two ln terms correct |
| | A1 | All three terms correct and $+ c$ |
| $y = 10$, $x = 2$ gives $c = \ln 10$ | B1 | $c = \ln 10$ |
| $\ln y = -\ln(x-1) + \ln(2x-3)^2 + \ln 10$ | M1 | Using the power law for logarithms |
| $\ln y = \ln\left(\frac{(2x-3)^2}{(x-1)}\right) + \ln 10$ or $\ln y = \ln\left(\frac{10(2x-3)^2}{(x-1)}\right)$ | M1 | Using product and/or quotient laws to obtain a single RHS logarithmic term with/without constant $c$ |
| $y = \dfrac{10(2x-3)^2}{(x-1)}$ | A1 aef | $y = \dfrac{10(2x-3)^2}{(x-1)}$ or aef, isw |
**[5] & [4] marks**
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## Question 4(b) & (c) — Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{dy}{y} = \int\frac{(2x-1)}{(2x-3)(x-1)}\,dx$ | B1 | Separates variables. Can be implied |
| $= \int\frac{-1}{(x-1)} + \frac{2}{\left(x-\frac{3}{2}\right)}\,dx$ | M1$\checkmark$ | Replaces RHS with their partial fraction |
| $\ln y = -\ln(x-1) + 2\ln\left(x - \frac{3}{2}\right) + c$ | M1, A1$\checkmark$, A1 | At least two ln terms; at least two correct; all three correct with $+c$ |
| $y = 10$, $x = 2$ gives $c = \ln 10 - 2\ln\left(\frac{1}{2}\right) = \ln 40$ | B1 oe | $c = \ln 10 - 2\ln\left(\frac{1}{2}\right)$ or $c = \ln 40$ |
| $\ln y = -\ln(x-1) + \ln\left(x-\frac{3}{2}\right)^2 + \ln 40$ | M1 | Using power law for logarithms |
| $\ln y = \ln\left(\dfrac{40\left(x-\frac{3}{2}\right)^2}{(x-1)}\right)$ | M1 | Using product/quotient laws |
| $y = \dfrac{40\left(x-\frac{3}{2}\right)^2}{(x-1)}$ | A1 aef | or aef, isw |
**[4 marks]**
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4. (a) Express $\frac { 2 x - 1 } { ( x - 1 ) ( 2 x - 3 ) }$ in partial fractions.\\
(b) Given that $x \geqslant 2$, find the general solution of the differential equation
$$( 2 x - 3 ) ( x - 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = ( 2 x - 1 ) y$$
(c) Hence find the particular solution of this differential equation that satisfies $y = 10$ at $x = 2$, giving your answer in the form $y = \mathrm { f } ( x )$.\\
\hfill \mbox{\textit{Edexcel C4 2007 Q4 [12]}}