## Question 2:

### Part (i):
When $x = 1$, $y = 3\ln 1 + 1 - 1^2 = 0$ | E1 [1] | Verification that point lies on curve

### Part (ii):
$\frac{dy}{dx} = \frac{3}{x} + 1 - 2x$ | M1, A1cao | $\frac{d}{dx}(\ln x) = \frac{1}{x}$

At R: $\frac{dy}{dx} = 0 = \frac{3}{x} + 1 - 2x$ | M1 | Re-arranging into a quadratic $= 0$

$3 + x - 2x^2 = 0$ | M1 | Factorising or formula or completing the square

$(3-2x)(1+x) = 0$ | A1 |

$x = 1.5$ (or $x = -1$) | M1 | Substituting their $x$

$y = 3\ln 1.5 + 1.5 - 1.5^2 = 0.466$ (3 s.f.) | A1cao |

$\frac{d^2y}{dx^2} = -\frac{3}{x^2} - 2$ | B1ft | ft their $\frac{dy}{dx}$ on equivalent work

When $x = 1.5$, $\frac{d^2y}{dx^2} = -\frac{10}{3} < 0 \Rightarrow$ max | E1 [9] | www – don't need to calculate $\frac{10}{3}$; SC1 for $x = 1.5$ unsupported, SC3 if verified; condone rounding errors on 0.466

### Part (iii):
Let $u = \ln x$, $\frac{du}{dx} = \frac{1}{x}$; $\frac{dv}{dx} = 1$, $v = x$ | M1 | Integration by parts

$\int \ln x\, dx = x\ln x - \int x \cdot \frac{1}{x}\, dx$ | A1 |

$= x\ln x - \int 1\, dx$ | |

$= x\ln x - x + c$ | A1 | Condone no $c$; allow correct result to be quoted (SC3)

$A = \int_1^{2.05}(3\ln x + x - x^2)\, dx$ | B1 | Correct integral and limits (soi)

$= \left[3x\ln x - 3x + \frac{1}{2}x^2 - \frac{1}{3}x^3\right]_1^{2.05}$ | B1ft | $\left[3 \times \text{their } x\ln x - x' + \frac{1}{2}x^2 - \frac{1}{3}x^3\right]$

$= -2.5057 + 2.833\ldots$ | M1dep, A1cao [7] | Substituting correct limits dep 1st B1

$= 0.33$ (2 s.f.) | |

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