# Question 7:

## Part (a):

| Working | Mark | Guidance |
|---------|------|----------|
| $x = \ln(t+2),\ y = \frac{1}{t+1} \Rightarrow \frac{dx}{dt} = \frac{1}{t+2}$ | B1 | Must state $\frac{dx}{dt} = \frac{1}{t+2}$ |
| $\text{Area}(R) = \int_{\ln 2}^{\ln 4} \frac{1}{t+1}dx = \int_0^2 \left(\frac{1}{t+1}\right)\left(\frac{1}{t+2}\right)dt$ | M1 | Area $= \int \frac{1}{t+1}dx$, ignore limits |
| $\int\left(\frac{1}{t+1}\right)\times\left(\frac{1}{t+2}\right)dt$ | A1 AG | Ignore limits |
| Changing limits: $x=\ln 2 \Rightarrow t=0$; $x=\ln 4 \Rightarrow t=2$ | B1 | Changes limits $x\to t$ so that $\ln 2 \to 0$ and $\ln 4 \to 2$ |
| Hence $\text{Area}(R) = \int_0^2 \frac{1}{(t+1)(t+2)}dt$ | | **[4]** |

## Part (b):

| Working | Mark | Guidance |
|---------|------|----------|
| $\left(\frac{1}{(t+1)(t+2)}\right) = \frac{A}{(t+1)} + \frac{B}{(t+2)}$ | M1 | $\frac{A}{(t+1)}+\frac{B}{(t+2)}$ with $A$ and $B$ found |
| $1 = A(t+2) + B(t+1)$ | | |
| Let $t=-1$: $A=1$; Let $t=-2$: $B=-1$ | A1 | Finds both $A$ and $B$ correctly. Can be implied. |
| $\int_0^2 \frac{1}{(t+1)(t+2)}dt = \int_0^2 \frac{1}{(t+1)} - \frac{1}{(t+2)}dt$ | | |
| $= \left[\ln(t+1) - \ln(t+2)\right]_0^2$ | dM1 | Either $\pm a\ln(t+1)$ or $\pm b\ln(t+2)$ |
| | A1$\sqrt{}$ | Both ln terms correctly ft. |
| $= (\ln 3 - \ln 4) - (\ln 1 - \ln 2)$ | ddM1 | Substitutes **both** limits of 2 and 0 and subtracts the correct way round |
| $= \ln 3 - \ln 4 + \ln 2 = \ln 3 - \ln 2 = \ln\left(\frac{3}{2}\right)$ | A1 aef isw | $\frac{\ln 3 - \ln 4 + \ln 2}{} $ or $\ln\left(\frac{3}{4}\right)-\ln\left(\frac{1}{2}\right)$ or $\underline{\ln 3 - \ln 2}$ or $\ln\left(\frac{3}{2}\right)$ (must deal with $\ln 1$) **[6]** |

> **Note:** Writing $\frac{1}{(t+1)(t+2)} = \frac{1}{(t+1)}+\frac{1}{(t+2)}$ means first M1A0 in (b). Writing $\frac{1}{(t+1)(t+2)} = \frac{1}{(t+1)}-\frac{1}{(t+2)}$ means first M1A1 in (b).

## Part (c):

| Working | Mark | Guidance |
|---------|------|----------|
| $e^x = t+2 \Rightarrow t = e^x - 2$ | M1 | Attempt to make $t=\ldots$ the subject |
| | A1 | giving $t = e^x - 2$ |
| $y = \frac{1}{e^x-2+1} \Rightarrow y = \frac{1}{e^x-1}$ | dM1 | Eliminates $t$ by substituting in $y$ |
| | A1 | giving $y = \frac{1}{e^x-1}$ **[4]** |

**Aliter Way 2:**

| Working | Mark | Guidance |
|---------|------|----------|
| $t+1 = \frac{1}{y} \Rightarrow t = \frac{1}{y}-1$ or $t = \frac{1-y}{y}$ | M1 | Attempt to make $t=\ldots$ the subject |
| | A1 | Giving either $t=\frac{1}{y}-1$ or $t=\frac{1-y}{y}$ |
| $x = \ln\left(\frac{1}{y}-1+2\right)$ or $x=\ln\left(\frac{1-y}{y}+2\right)$ | dM1 | Eliminates $t$ by substituting in $x$ |
| $x = \ln\left(\frac{1}{y}+1\right)$, $e^x = \frac{1}{y}+1$, $e^x-1=\frac{1}{y}$ | | |
| $y = \frac{1}{e^x-1}$ | A1 | giving $y=\frac{1}{e^x-1}$ **[4]** |

**Aliter Way 3:**

| Working | Mark | Guidance |
|---------|------|----------|
| $e^x = t+2 \Rightarrow t+1 = e^x-1$ | M1 | Attempt to make $t+1=\ldots$ the subject |
| | A1 | giving $t+1=e^x-1$ |
| $y=\frac{1}{t+1} \Rightarrow y=\frac{1}{e^x-1}$ | dM1 | Eliminates $t$ by substituting in $y$ |
| | A1 | giving $y=\frac{1}{e^x-1}$ **[4]** |

**Aliter Way 4:**

| Working | Mark | Guidance |
|---------|------|----------|
| $t+1=\frac{1}{y} \Rightarrow t+2=\frac{1}{y}+1$ or $t+2=\frac{1+y}{y}$ | M1 | Attempt to make $t+2=\ldots$ the subject |
| | A1 | Either $t+2=\frac{1}{y}+1$ or $t+2=\frac{1+y}{y}$ |
| $x=\ln\left(\frac{1}{y}+1\right)$ or $x=\ln\left(\frac{1+y}{y}\right)$ | dM1 | Eliminates $t$ by substituting in $x$ |
| $e^x=\frac{1}{y}+1 \Rightarrow e^x-1=\frac{1}{y}$, $y=\frac{1}{e^x-1}$ | A1 | giving $y=\frac{1}{e^x-1}$ **[4]** |

## Part (d):

| Working | Mark | Guidance |
|---------|------|----------|
| Domain: $x \geq 0$ | B1 | $x>0$ or just $>0$ **[1]** |

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