# Question 6:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{k}{2k-15}=\frac{k+4}{k} \Rightarrow k^2=2k^2+8k-15k-60 \Rightarrow k^2-7k-60=0$ | M1A1* | Forms correct equation linking three terms; rearranges to given quadratic; must show intermediate working; no errors seen |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k=12$ | B1 | Ignore $-5$; may be implied by value of second term |
| $r=\frac{"12"}{2\times"12"-15}=\frac{4}{3}$ | M1 | Substitutes positive $k$ into expressions for two terms to find $r$ |
| $("12"+4)\times\frac{4}{3}=\ldots$ or $(2\times"12"-15)\times\left(\frac{4}{3}\right)^3=\ldots$ | M1 | Attempts 4th term using positive $k$ and positive $r$ |
| $=21300$ | A1 | Or 21330 or 21333 only, or equivalences e.g. 21.3 thousand |

## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{9\left(1-\left(\frac{4}{3}\right)^N\right)}{1-\frac{4}{3}}>3000 \Rightarrow \left(\frac{4}{3}\right)^N>\frac{1009}{9}$ | M1 | Uses $a="9"$ or $a="9000"$ and $r=\frac{4}{3}$ in sum formula; proceeds to $A\left(\frac{4}{3}\right)^N \ldots B$ |
| $\log\left(\frac{4}{3}\right)^N>\log\left(\frac{1009}{9}\right) \Rightarrow N>\dfrac{\log\left(\frac{1009}{9}\right)}{\log\left(\frac{4}{3}\right)}$ | dM1 | Attempts to find $N$ using logarithms correctly; dependent on previous M mark |
| $N>16.405\ldots \Rightarrow N=17$ | A1 | cso; no arithmetical errors and correct log work required; trial and improvement scores max M1dM0A0 |

# Question 7a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $H = t^{\frac{1}{2}} + \frac{3t^{\frac{3}{2}}}{10} - \frac{t^{\frac{5}{2}}}{20} + 17 \Rightarrow \frac{dH}{dt} = \frac{1}{2}t^{-\frac{1}{2}} + \frac{9}{20}t^{\frac{1}{2}} - \frac{1}{8}t^{\frac{3}{2}}$ | M1A1A1 | M1: Multiplies out and attempts to differentiate; award for one correct index from correct working. A1: One of $\frac{1}{2}t^{-\frac{1}{2}}$, $\frac{9}{20}t^{\frac{1}{2}}$, $-\frac{1}{8}t^{\frac{3}{2}}$ simplified or unsimplified. A1: Full correct derivative. |
| e.g. $\frac{1}{2}t^{-\frac{1}{2}} + \frac{9}{20}t^{\frac{1}{2}} - \frac{1}{8}t^{\frac{3}{2}} = 0 \Rightarrow \frac{1}{2} + \frac{9}{20}t - \frac{1}{8}t^2 = 0 \Rightarrow 5\alpha^2 - 18\alpha - 20 = 0$ * | dM1A1* | dM1: Sets derivative to zero, proceeds via correct method to a 3TQ in $t$ or $\alpha$; powers and coefficients must be integers. A1*: Achieves $5\alpha^2 - 18\alpha - 20 = 0$ with no errors, all previous marks scored. |

# Question 7b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha = \frac{9 + \sqrt{181}}{5} = \text{awrt } 4.491$ | B1 | $\alpha = \text{awrt } 4.491$ only. If both roots stated, awrt 4.491 must be the only one used in $\frac{d^2H}{dt^2}$. |

# Question 7c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^2H}{dt^2} = -\frac{1}{4}t^{-\frac{3}{2}} + \frac{9}{40}t^{-\frac{1}{2}} - \frac{3}{16}t^{\frac{1}{2}} \Rightarrow -\frac{(``4.49")^{-\frac{3}{2}}}{4} + \frac{9(``4.49")^{-\frac{1}{2}}}{40} - \frac{3(``4.49")^{\frac{1}{2}}}{16}$ | M1 | Attempts to differentiate $\frac{dH}{dt}$ to achieve second derivative of form $...t^{-\frac{3}{2}} + ...t^{-\frac{1}{2}} + ...t^{\frac{1}{2}}$; substitutes value of $\alpha$ between 0 and 7, or considers sign of second derivative. |
| $\frac{d^2H}{dt^2} = (-0.3...) < 0 \Rightarrow \text{max}$ * | A1* | $\frac{d^2H}{dt^2} = -0.3$ or better $(-0.3174...)$; must be $-0.3$ or better when evaluated. Second derivative must be correct but condone in terms of $\alpha$. |

---