### 8(a)
$e^{-2x} = 3$ | M1 | |

$-2x = \ln 3$ | | |

$x = -\frac{1}{2} \ln 3$ | A1 | OE ISW | **Total: 2**

### 8(b)
$\int xe^{-2x} dx$ | M1 | differentiating and integrating |

$u = x$ $\frac{dv}{dx} = e^{-2x}$ | | |

$\frac{du}{dx} = 1$ $v = -\frac{1}{2}e^{-2x}$ | | |

$\int = -\frac{1}{2}xe^{-2x} + \int \frac{1}{2}e^{-2x}(dx)$ | m1 | correct subs of their values into parts formula |

$= -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + c$ | A1 A1 | No further incorrect working | **Total: 4**

### 8(c)(i)
$y = e^{-2x} + 6x$ | M1 | $ke^{-2x} + 6 = 0$ |

$\frac{dy}{dx} = -2e^{-2x} + 6 = 0$ | | |

$\frac{dy}{dx} = 0 \Rightarrow -2(e^{-2x} - 3) = 0$ | | |

$x = -\frac{1}{2}\ln 3$ | A1 | OE |

$y = 3 + 6\left(-\frac{1}{2}\ln 3\right)$ | M1 | Correct substitute of their valid $x$ |

$= 3 - 3\ln 3$ | A1 | OE ISW | **Total: 4**

### 8(c)(ii)
$\frac{d^2y}{dx^2} = 4e^{-2x} \begin{cases} = 12 \\ > 0 \end{cases}$ | M1 | Other methods need justification. Allow error in $\frac{d^2y}{dx^2}$ or $x$-value, but not both |

$\therefore$ minimum | A1 | | **Total: 2**

### 8(c)(iii)
$(V) = \pi \int_0^{(1)} y^2 dx = (\pi) \int_{(0)}^{(1)} (e^{-2x} - 6x)^2 (dx)$ | M1 | Either |

$= (\pi) \int_{(0)}^{(1)} (e^{-4x} + 12xe^{-2x} + 36x^2) dx$ | B1 | Correct expansion |

$= (\pi) \left[-\frac{1}{4}e^{-4x} - 6xe^{-2x} - 3e^{-2x} + 12x^3\right]_{(0)}^{(1)}$ | A1 | 3 correct terms; '−6', '−3' correct or $12 \times$ their (b) |

$= \pi \left[\left(-\frac{1}{4}e^{-4} - 9e^{-2} + 12\right) - \left(-\frac{1}{4} - 3\right)\right]$ | A1 | All correct |

$= \pi \left[15\frac{1}{4} - 9e^{-2} - \frac{1}{4}e^{-4}\right]$ | | |

$= 44.1$ | B1 | AWRT | **Total: 5**

---

# TOTALS BY QUESTION

| Q | Total |
|---|-------|
| 1 | 7 |
| 2 | 8 |
| 3 | 7 |
| 4 | 9 |
| 5 | 9 |
| 6 | 6 |
| 7 | 12 |
| 8 | 17 |
| **GRAND TOTAL** | **75** |