# Question 11:

## Part (a) — Way 1

| Working | Mark | Guidance |
|---------|------|----------|
| $\{y=x^x \Rightarrow\}\ \ln y = x\ln x$ | B1 | 1.1a — Condone $\log_e y = x\log_e x$ or $\log_e y = x$ |
| $\frac{1}{y}\frac{dy}{dx} = 1 + \ln x$ | M1 | 1.1b — For either $\ln y \to \frac{1}{y}\frac{dy}{dx}$ or $x\ln x \to 1+\ln x$ or $\frac{x}{x}+\ln x$ |
| Correct differentiated equation | A1 | 2.1 — i.e. $\frac{1}{y}\frac{dy}{dx}=1+\ln x$ or $\frac{dy}{dx}=y(1+\ln x)$ or $\frac{dy}{dx}=x^x(1+\ln x)$ |
| $\left\{\frac{dy}{dx}=0\Rightarrow\right\}\ \frac{x}{x}+\ln x=0$ or $1+\ln x=0 \Rightarrow \ln x = k \Rightarrow x=\ldots$ | M1 | 1.1b — Sets $1+\ln x=0$ and rearranges; $k$ is a constant and $k\neq 0$ |
| $x = e^{-1}$ or awrt $0.368$ | A1 | 1.1b — Only solution; no other solutions for $x$. Note: $k\neq 0$ |

## Part (a) — Way 2

| Working | Mark | Guidance |
|---------|------|----------|
| $\{y=x^x\Rightarrow\}\ y=e^{x\ln x}$ | B1 | 1.1a |
| $\frac{dy}{dx}=\left(\frac{x}{x}+\ln x\right)e^{x\ln x}$ | M1 | 1.1b — For either $y=e^{x\ln x}\Rightarrow \frac{dy}{dx}=\text{f}(\ln x)e^{x\ln x}$ or $x\ln x\to 1+\ln x$ or $\frac{x}{x}+\ln x$ |
| Correct differentiated equation: $\frac{dy}{dx}=\left(\frac{x}{x}+\ln x\right)e^{x\ln x}$ or $\frac{dy}{dx}=(1+\ln x)e^{x\ln x}$ or $\frac{dy}{dx}=x^x(1+\ln x)$ | A1 | 2.1 |
| $\left\{\frac{dy}{dx}=0\Rightarrow\right\}\ \frac{x}{x}+\ln x=0$ or $1+\ln x=0 \Rightarrow \ln x=k \Rightarrow x=\ldots$ | M1 | 1.1b |
| $x=e^{-1}$ or awrt $0.368$ | A1 | 1.1b |

## Part (b) — Way 1

| Working | Mark | Guidance |
|---------|------|----------|
| Attempts both $1.5^{1.5}=1.8\ldots$ and $1.6^{1.6}=2.1\ldots$; at least one result correct to awrt 1 dp | M1 | 1.1b |
| $1.8\ldots < 2$ and $2.1\ldots > 2$; as $C$ is continuous then $1.5 < \alpha < 1.6$ | A1 | 2.1 |

## Part (b) — Way 2

| Working | Mark | Guidance |
|---------|------|----------|
| For $x^x-2$: attempts both $1.5^{1.5}-2=-0.16\ldots$ and $1.6^{1.6}-2=0.12\ldots$; at least one result correct to awrt 1 dp | M1 | 1.1b |
| $-0.16\ldots < 0$ and $0.12\ldots > 0$; as $C$ is continuous then $1.5<\alpha<1.6$ | A1 | 2.1 |

## Part (b) — Way 3

| Working | Mark | Guidance |
|---------|------|----------|
| For $\ln y = x\ln x$: attempts both $1.5\ln 1.5=0.608\ldots$ and $1.6\ln 1.6=0.752\ldots$; at least one result correct to awrt 1 dp | M1 | 1.1b |
| $0.608\ldots < 0.69\ldots$ and $0.752\ldots > 0.69\ldots$; as $C$ is continuous then $1.5<\alpha<1.6$ | A1 | 2.1 |

## Part (b) — Way 4

| Working | Mark | Guidance |
|---------|------|----------|
| For $\log y = x\log x$: attempts both $1.5\log 1.5=0.264\ldots$ and $1.6\log 1.6=0.326\ldots$; at least one result correct to awrt 2 dp | M1 | 1.1b |
| $0.264\ldots < 0.301\ldots$ and $0.326\ldots > 0.301\ldots$; as $C$ is continuous then $1.5<\alpha<1.6$ | A1 | 2.1 |

## Part (c)

| Working | Mark | Guidance |
|---------|------|----------|
| Attempts $x_{n+1}=2x_n^{1-x_n}$ at least once with $x_1=1.5$; can be implied by $2(1.5)^{1-1.5}$ or awrt $1.63$ | M1 | 1.1b |
| $\{x_4=1.67313\ldots\Rightarrow\}\ x_4=1.673$ (3 dp) **cao** | A1 | 1.1b |

## Part (d)

| Working | Mark | Guidance |
|---------|------|----------|
| Give 1st B1 for any of: oscillates; periodic; non-convergent; divergent; fluctuates; goes up and down; 1, 2, 1, 2, 1, 2; alternates (condone) | B1 | 2.5 |
| Give B1 B1 for any of: periodic sequence with period 2; oscillates between 1 and 2; fluctuates between 1 and 2; keep getting 1, 2; alternates between 1 and 2; goes up and down between 1 and 2; 1, 2, 1, 2, 1, 2, … | B1 | 2.5 |

# Question 11 (continued):

## Part (b) — Way 1
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts both $1.5^{1.5} = 1.8...$ and $1.6^{1.6} = 2.1...$ with at least one correct to 1 dp | M1 | |
| Both $1.5^{1.5} =$ awrt 1.8... and $1.6^{1.6} =$ awrt 2.1..., reason e.g. $1.8...<2$ and $2.1...>2$, or states $C$ cuts through $y=2$, $C$ continuous and conclusion | A1 | |

## Part (b) — Way 2
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts both $1.5^{1.5}-2=-0.16...$ and $1.6^{1.6}-2=0.12...$ at least one correct to 1 dp | M1 | |
| Both values correct to 1 dp, reason e.g. $-0.16...<0$ and $0.12...>0$, sign change or states $C$ cuts through $y=0$, $C$ continuous and conclusion | A1 | |

## Part (b) — Way 3
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts both $1.5\ln 1.5=0.608...$ and $1.6\ln 1.6=0.752...$ at least one correct to 1 dp | M1 | |
| Both correct to 1 dp, reason e.g. $0.608...<0.69...$ and $0.752...>0.69...$, or states they are either side of $\ln 2$, $C$ continuous and conclusion | A1 | |

## Part (b) — Way 4
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts both $1.5\log 1.5=0.264...$ and $1.6\log 1.6=0.326...$ at least one correct to 2 dp | M1 | |
| Both correct to 2 dp, reason e.g. $0.264...<0.301...$ and $0.326...>0.301...$, or states either side of $\log 2$, $C$ continuous and conclusion | A1 | |

## Part (c)
| Working | Mark | Guidance |
|---------|------|----------|
| Attempt to use formula once, implied by $2(1.5)^{1-1.5}$ or awrt 1.63 | M1 | |
| States $x_4=1.673$ cao (to 3 dp) | A1 | Give M1A1 for stating $x_4=1.673$; $x_2=1.63299...$, $x_3=1.46626...$, $x_4=1.67313...$ |

## Part (d)
| Working | Mark | Guidance |
|---------|------|----------|
| See scheme | B1 | Only marks of B1B0 or B1B1 are possible |
| See scheme | B1 | Give B0B0 for "Converges in a cob-web pattern" or "Converges up and down to $\alpha$" |

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