$dI/dr = 4\pi r^2$ | B1 | or $12\pi^2/3$, condone $dr/dI, dI/dR$ |
$dV/dt = 10$ | B1 | |
$dV/dt = (dV/dr)(dr/dt)$ | M1 | a correct chain rule soi |
$\Rightarrow 10 = 4\pi \cdot 64 \cdot dr/dt$ | A1 | o.e. (soi) must be correct |
$\Rightarrow dr/dt = 0.0124 \text{ cm s}^{-1}$ | A1 [5] | $0.012$ or better or $10/256\pi$ or $5/128\pi$; mark final answer |

# Question 6(i)

$V = 20000e^{-0.2t}$ | B1 | (soi) art 16400 |
when $t = 1$, $V = 16374.615\ldots$ | B1 | condone no £, must be to nearest £100 |
so car loses (£)3600 | [2] | or B2 for correct answer |

# Question 6(ii)

When $t = 1$, $V = 13000$ | M1 | |
$\Rightarrow 13000 = 15000 e^{-k}$ | M1 | taking lns correctly or e.g. $\ln 13000 = \ln 15000 - k [ln e]$ |
$\Rightarrow -k [\ln e] = \ln(13000/15000)$ | M1 | |
$\Rightarrow k = 0.1431\ldots = 0.143 \text{ (3sf)}$ * | A1 [3] | cao NB AG must show some working if 4th d.p. not shown; e.g. $k = -\ln(13000/15000) = 0.143$ |

# Question 6(iii)

$15000e^{-0.143t} = 20000e^{-0.2t}$ | M1 * | must be correct, but could use a more accurate value for $k$ |
| | If M0, SCB1 for 5 – 5.1 years from correct calculations for each car, rot e.g. $t = 5, £7358$ (Brian), $£7338$ (Kate) or $£7334$ with more accurate $k$ |
$\Rightarrow (15000/20000) = e^{0.143 - 0.2y}$ | M1 dep | dep * |
$\Rightarrow t = \ln 0.75 / -0.057 = 5.05 \text{ years}$ | A1 | cao accept answers in the range 5 – 5.1 |
so after 5 years | [3] | o.e. e.g. $\ln 15000 - 0.143t = \ln 20000 - 0.2t$ |

# Question 7(i)

False e.g. neither 25 and 27 are prime | B1 | correct counter-example identified |
as 25 is div by 5 and 27 by 3 | B1 [2] | justified correctly; need not explicitly say 'false' |

# Question 7(ii)

True: one has factor of 2, the other 4, so | B2 | or algebraic proofs: e.g. $2n(2n+2) = 4n(n+1) = 4 \times \text{even} \times \text{odd} \text{ no so div by } 8$ |
product must have factor of 8. | [2] | B1 for stating with justification div by 4 e.g. both even, or from $4(n^2 + n)$ or $4pq$ |

# Question 8(i)

$f(-x) = \frac{-x}{\sqrt{2 + (-x)^2}}$ | M1 | substituting $-x$ for $x$ in $f(x)$ |
$= -\frac{x}{\sqrt{2 + x^2}} = -f(x)$ | A1 | 1st line must be shown, must have $f(-x) = -f(x)$ or somewhere |
Rotational symmetry of order 2 about O | B1 [3] | must have 'rotate' and 'O' and 'order 2 or 180° or ½ turn' |

# Question 8(ii)

$f'(x) = \frac{\sqrt{2+x^2} \cdot 1 - x \cdot 2 (2+x^2)^{-1/2} \cdot 2x}{(\sqrt{2+x^2})^2}$ | M1 | quotient or product rule used; QR: condone $u dv \pm v du$, but $u, v$ and denom must be correct |
| M1 | $½ u^{-1/2}$ or $-½ v^{-3/2}$ soi |
$= \frac{2 + x^2 - x^2}{(2+x^2)^{3/2}} = \frac{2}{(2+x^2)^{3/2}}$ * | A1 | correct expression; NB AG |
When $x = 0$, $f'(x) = 2/2^{3/2} = 1/\sqrt{2}$ | B1 | o.e. $\sqrt{2}/2, 2^{-1/2}, 1/2^{1/2}$, but not $2/2^{3/2}$; allow isw on these seen |
| [5] | |

# Question 8(iii)

$A = \int_1^4 \frac{x}{\sqrt{2+x^2}} [dx]$ | B1 | correct integral and limits |
let $u = 2 + x^2$, $du = 2x \, dx$ | M1 | |
$= \int \frac{1}{2\sqrt{u}} du$ | M1 | condone no $du$ or $dv$, but not $\int \frac{1}{2\sqrt{u}} dx$ |
$= [u^{1/2}]_2$ | A1 | $[u^{1/2}]$ o.e. (but not $1/u^{-1/2}$) or $[v]$ or $k=1$ |
$= \sqrt{3} - \sqrt{2}$ | A1 cao [4] | must be exact; isw approximations |

# Question 8(iv)(A)

$y^2 = \frac{x^2}{2+x^2}$ | M1 | squaring (correctly) |
$\Rightarrow 1/y^2 = (2 + x^2)\lambda^2 = 2\lambda^2 + 1$ * | A1 | or equivalent algebra NB AG; if argued backwards from given result without error, SCB1 |
| [2] | must show $\sqrt{(2+x^2)^2} + 2 + x^2$ (o.e.) |

# Question 8(iv)(B)

$-2y^3 dy/dx = -4x^{-3}$ | B1 B1 | LHS, RHS; condone $dy/dx -2y^3$ unless pursued |
$\Rightarrow dy/dx = -4x^{-3}/(-2y^3) = 2y^3/x^3$ * | B1 | NB AG |
Not possible to substitute $x = 0$ and $y = 0$ into | B1 | soi (e.g. mention of 0/0); condone 'can't substitute $x = 0$' o.e. (i.e. need not mention $y = 0$). Condone also 'division by 0 is infinite' |
this expression | [4] | |

# Question 9(i)

$xe^{-2x} = mx$ | M1 | may be implied from 2nd line |
$\Rightarrow e^{-2x} = m$ | M1 | dividing by $x$, or subtracting $\ln x$ |
$\Rightarrow -2x = \ln m$ | M1 | o.e. $[\ln x] - 2x = \ln m + [\ln x]$ or factorising: $x(e^{-2x} - m) = 0$ |
$\Rightarrow x = -1/2 \ln m$ * | A1 [3] | NB AG |
or If $x = -1/2 \ln m$, $y = -1/2 \ln m \times e^{\ln m} = -1/2 \ln m \times m$ | M1 | substituting correctly |
$= -1/2 \ln m \times m$ | A1 | |
so P lies on $y = mx$ | A1 | |

# Question 9(ii)

let $u = x$, $u' = 1$, $v = e^{-2x}$, $v' = -2e^{-2x}$ | M1 * | product rule consistent with their derivs |
$dy/dx = e^{-2x} - 2xe^{-2x}$ | A1 | o.e. correct expression |
$= e^{-2(- 1/2 \ln m)} - 2(- \frac{1}{2} \ln m) e^{-2(- 1/2 \ln m)}$ | M1 dep | subst $x = -1/2 \ln m$ into their deriv dep M1 * |
$= e^{\ln m} + e^{\ln m} \ln m \quad [= m + m \ln m]$ | A1 cao | condone $e^{\ln m}$ not simplified; but not $2(- 1/2 \ln m)$, but mark final ans |
| [4] | |

# Question 9(iii)

$m + m \ln m = -m$ | M1 | their gradient from (ii) = $-m$ |
$\Rightarrow \ln m = -2$ | A1 | NB AG |
$\Rightarrow m = e^{-2}$ * | A1 | |
or $y + 1/2 m \ln m = m(1 + \ln m)(x + 1/2 \ln m)$ | B2 | for fully correct methods finding x-intercept of equation of tangent and equating to $-\ln m$ |
| | |
$y = 0 \Rightarrow 1/2 m \ln m = m(1 + \ln m)(- 1/2 \ln m)$ | | |
$\Rightarrow 1 + \ln m = -1, \ln m = -2, m = e^{-2}$ | A1 | |
At P, $x = 1$ | B1 | |
$\Rightarrow y = e^{-2}$ | B1 | isw approximations; not $e^{-2} \times 1$ |
| [4] | |

# Question 9(iv)

Area under curve $= \int_0^1 xe^{-2x} dx$ | M1 | parts, condone $v = k e^{-2x}$, provided it is used consistently in their parts formula |
$u = x, u' = 1, v = e^{-2x}, v' = -1/2 e^{-2x}$ | | |
$= \left[- \frac{1}{2} xe^{-2x}\right] + \int_0^1 \frac{1}{2} e^{-2x} dx$ | A1 ft | ft their $v$ |
$= \left[- \frac{1}{2} xe^{-2x} - \frac{1}{4} e^{-2x}\right]_0$ | A1 | |
$= (- 1/2 e^{-2} - 1/4 e^{-2}) - (0 - 1/4 e^0)$ | A1 | correct expression; need not be simplified |
$[= 1/4 - 3/4 e^{-2}]$ | | |
Area of triangle $= 1/2 \text{ base} \times \text{height}$ | M1 | ft their 1, $e^{-2}$ or $[e^{-2} x^2/2]$ |
$= 1/2 \times 1 \times e^{-2}$ | A1 | |
So area enclosed $= 1/4 - 3/4 e^{-2}$ | A1 cao [7] | o.e. must be exact, two terms only; isw |