# Question 3:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{8x-5}{(2x-1)(4x-3)} \equiv \frac{A}{2x-1}+\frac{B}{4x-3} \Rightarrow 8x-5=A(4x-3)+B(2x-1)$ | M1 | Correct PF form with strategy to find at least one constant; look for $5-8x=A(4x-3)+B(2x-1)$ followed by values for $A$ or $B$ |
| $A=1$ or $B=2$ | A1 | One correct fraction or constant |
| $\frac{8x-5}{(2x-1)(4x-3)} \equiv \frac{1}{2x-1}+\frac{2}{4x-3}$ | A1 | Correct partial fractions; may be awarded for sight of correct PF in part (b) |

**Total: 3 marks**

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\left(\frac{1}{2x-1}+\frac{2}{4x-3}\right)dx = \frac{1}{2}\ln\|2x-1\|+\frac{1}{2}\ln\|4x-3\|(+c)$ | M1, A1ft, A1ft | M1: for $\int\frac{\alpha}{2x-1}dx=\beta\ln\|2x-1\|$ OR $\int\frac{\alpha}{4x-3}dx=\beta\ln\|4x-3\|$; allow brackets instead of moduli; condone omission of moduli; A1ft: $\frac{A}{2}\ln\|2x-1\|$; A1ft: $\frac{B}{4}\ln\|4x-3\|$; follow through their $A$ and $B$ |

**Total: 3 marks**

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[\frac{1}{2}\ln(2x-1)+\frac{1}{2}\ln(4x-3)\right]_k^{3k} = \frac{1}{2}\ln(6k-1)(12k-3)-\frac{1}{2}\ln(2k-1)(4k-3) = \frac{1}{2}\ln\frac{(6k-1)(12k-3)}{(2k-1)(4k-3)}$ | M1 | Applies limits correctly and combines logarithms |
| $\frac{1}{2}\ln\frac{(6k-1)(12k-3)}{(2k-1)(4k-3)} = \frac{1}{2}\ln 20 \Rightarrow \frac{(6k-1)(12k-3)}{(2k-1)(4k-3)}=20 \Rightarrow (6k-1)(12k-3)=20(2k-1)(4k-3) \Rightarrow 88k^2-170k+57=0$ | dM1 A1 | dM1: removes logarithms and rearranges to $88k^2-170k+57=0$; dependent on previous M |
| $88k^2-170k+57=0 \Rightarrow k=...$ | ddM1 | Solves quadratic; dependent on both previous M marks |
| $k=\frac{3}{2}$ | A1 | Correct value |

**Total: 5 marks**

# Question A1 (Integration by Partial Fractions):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{A}{2x-1}dx + \int \frac{B}{4x-3}dx = \frac{A}{2}\ln|2x-1| + \frac{B}{4}\ln|4x-3|$ | A1ft | Follow through numerical $A$ and $B$; no requirement for $+c$; also allow $\frac{1}{2}\ln(2x-1)(4x-3)(+c)$ and $\ln\sqrt{(2x-1)(4x-3)}(+c)$ |

**Part (c):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts correct limits with correct log work to combine into single ln term | M1 | Dependent on part (b) having two ln terms; condone slips/bracketing errors |
| E.g. $\frac{1}{2}\ln(6k-1)+\frac{1}{2}\ln(12k-3)-\frac{1}{2}\ln(2k-1)+\frac{1}{2}\ln(4k-3) = \frac{1}{2}\ln\frac{(6k-1)(12k-3)(4k-3)}{(2k-1)}$ | | |
| Eliminates ln's, multiplies up and collects terms to form polynomial equation in $k$ | dM1 | Dependent on previous M |
| Correct 3TQ, $= 0$ may be implied by subsequent work | A1 | |
| Solves 3TQ | ddM1 | Dependent on both previous M's; can be solved via graphical calculator |
| $x = \frac{3}{2}$, correct value and no others | A1 | |

**Special Cases (limited working):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Case I: $\frac{1}{2}\ln\frac{(6k-1)(12k-3)}{(2k-1)(4k-3)} = \frac{1}{2}\ln 20 \Rightarrow k = \frac{3}{2}$ | SC 11000 | |
| Case II: $\frac{(6k-1)(12k-3)}{(2k-1)(4k-3)} = 20 \Rightarrow k = \frac{3}{2}$ | SC 11100 | |

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