## Question 6:

### Part (i):
$\frac{\ln x}{x}=0$
$\Rightarrow \ln x = 0$ or $\frac{\ln 1}{1}=0$ | **M1** | AO 1.1a | May be implied
$\Rightarrow x=1$ | **A1** [2] | AO 1.1 |

### Part (ii):
$y$-coordinates are $\frac{\ln 2}{2}$ and $\frac{\ln 4}{4}$

$\frac{\ln 4}{4}=\frac{2\ln 2}{4}=\frac{\ln 2}{2}$ oe | **B1\*** [2] | AO 1.1 | Allow $\frac{\ln 4}{4}=\ln 4^{\frac{1}{4}}=\ln\sqrt{2}=\frac{\ln 2}{2}$; Both $=0.346...$ B0B0
$\Rightarrow AB$ is // to $x$-axis **AG** | **B1dep\*** | AO 3.1a | Show that $\frac{\ln 4}{4}=\frac{2\ln 2}{4}$ and conclusion; use of $\frac{\ln 4}{4}-\frac{\ln 2}{2}=0$ unjustified B0B0

### Part (iii):
$\frac{dy}{dx}=\frac{x\times\frac{1}{x}-1\times\ln x}{x^2}$ or $\frac{1}{x}\times\frac{1}{x}+\ln x\times(-\frac{1}{x^2})$ oe | **M1** | AO 3.1a | Attempt diff, $\geq$ one term correct
$\frac{1}{x^2}-\frac{\ln x}{x^2}=0$ or $\frac{1-\ln x}{x^2}=0$ | **M1** | AO 1.1 | oe, their $\frac{dy}{dx}=0$
$1-\ln x=0$ oe
$x=e$ or $2.72$ or $2.7$ (2 sf) | **A1** | AO 1.1 |
Coordinates are $\left(e,\frac{1}{e}\right)$ | **A1** [4] | AO 1.1 | Allow (e, 0.368) or (e, 0.37); or (2.7, 0.37) (2 sf)

### Part (iv):
Attempt $\frac{d^2y}{dx^2}$ | **M1** | AO 2.1 | Attempt diff their $\frac{dy}{dx}$

$=\frac{x^2(-\frac{1}{x})-2x(1-\ln x)}{x^4}$ or $\frac{-3+2\ln x}{x^3}$ oe | **A1** | AO 1.2 | All correct, not necessarily simplified cao; $0.00093, -0.0038$ A1A1

Substitute $x=e$ (or 2.72) into $\frac{d^2y}{dx^2}$ | **M1** | AO 1.1 | Sub their $x$ from (iii) into their $\frac{d^2y}{dx^2}$; State grad +ve & -ve or show on diag dep A1A1 M1

$\frac{d^2y}{dx^2}=-\frac{1}{e^3}$ oe or $-0.0498$ | **A1** | AO 1.1 | cao Allow or $-0.0497$ or $-0.05$

$\frac{d^2y}{dx^2}<0$, hence maximum | **B1f** [5] | AO 3.2a | ft their result of sub their $x$ into their $\frac{d^2y}{dx^2}$; dep see result; dep M1A1A1; No proof, no marks

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