## Question 14(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y$-intercepts: $9$ or $3+e$ | M1 | For sight of either intercept $9$ (not $10-e^0$) or $3+e$ or $3+e^1$ or $3+e^{0+1}$ |
| Distance $PQ = 6-e$ | A1 | $6-e$ (Not $9-(3+e)$) |

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## Question 14(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $3+e^{x+1} = 10-e^x$ | M1 | Equates the 2 curves |
| $e^x(e+1) = 7$ | M1 | Collects exponential terms and takes out factor of $e^x$ with correct index work |
| $x = \ln\left(\frac{7}{1+e}\right)$ or $\ln\frac{7}{1+e}$ | A1 | Correct $x$-coordinate |
| Substitutes $x = \ln\left(\frac{7}{1+e}\right)$ into $y = 10-e^x \Rightarrow y = \ldots$ | ddM1 | Dependent on both M's. Substituting their $x$ into either equation to find $y$ |
| $R = \left(\ln\left(\frac{7}{1+e}\right), \frac{3+10e}{1+e}\right)$ | A1 | Equivalent forms: $x=\ln\left(\frac{7}{1+e}\right)$, $y=10-\frac{7}{1+e}$ or $y=3+\frac{7e}{1+e}$ but **not** $y=10-e^{\ln\frac{7}{1+e}}$ |

**Way 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=10-e^x \Rightarrow e^x=10-y \Rightarrow x=\ln(10-y) \Rightarrow y=3+e^{1+\ln(10-y)}$; makes $x$ subject of equation 2 and substitutes into equation 1 | M1 | |
| $y = 3+(10-y)e$ | M1 | Uses correct index work to eliminate the "ln" |
| $y = \frac{10e+3}{1+e}$ or $10-\frac{7}{1+e}$ or $y=3+\frac{7e}{1+e}$ | A1 | Correct $y$-coordinate. **Not** $y=10-e^{\ln\frac{7}{1+e}}$ |
| $x = \ln(10-y) = \ln\left(10-\frac{10e+3}{1+e}\right)$ | ddM1 | Dependent on both M's. Substituting their $y$ into either equation to find $x$ |
| $R = \left(\ln\left(\frac{7}{1+e}\right), \frac{3+10e}{1+e}\right)$ | A1 | Allow $x=\ln\left(10-\frac{10e+3}{1+e}\right)$ |

**Way 3:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=3+e^{x+1} \Rightarrow y-3=e^{x+1} \Rightarrow e^x=\frac{y-3}{e} \Rightarrow 10-y=\frac{y-3}{e}$; makes $e^x$ subject of equation 1 and substitutes into equation 2 | M1 | |
| $10e-ye=y-3 \Rightarrow y(1+e)=10e+3$ | M1 | Uses correct algebra and factorises $y$ |
| $y=\frac{10e+3}{1+e}$ or $10-\frac{7}{1+e}$ or $y=3+\frac{7e}{1+e}$ | A1 | Correct $y$-coordinate |
| Then as Way 1 above | ddM1, A1 | |

**Way 4:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=10-e^x \Rightarrow x=\ln(10-y)$; $y=3+e^{x+1} \Rightarrow \ln(10-y)=\ln(y-3)-1$; makes $x$ and $x+1$ the subject and uses $x=x$ | M1 | |
| $\frac{y-3}{10-y}=e$ | M1 | Uses correct work to eliminate the "ln"s |
| $y=\frac{10e+3}{1+e}$ or $10-\frac{7}{1+e}$ or $y=3+\frac{7e}{1+e}$ | A1 | Correct $y$-coordinate |
| Then as above | ddM1, A1 | |

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