$n(n+1)(n+2)$ | M1 | condone division by $n$ and then $(n+1)(n+2)$ seen, or separate factors shown after factor theorem used

argument from general consecutive numbers leading to: at least one must be even | A1 | or divisible by 2

[exactly] one must be multiple of 3 | A1 | if M0: allow SC1 for showing given expression always even

## Question 11(i)

$x = -1$ or $-21/4$ oe isw | A1 | or A1 for $(-1, 11)$ and A1 for $(-21/4, 61/4)$ oe

$y = 11$ or $61/4$ oe isw | A1 | from formula: accept $x = -1$ or $-42/8$ oe isw

**Section Notes:**
- condone spurious $y = 4x^2 + 25x + 21$ as one error (and then count as eligible for 3rd M1)

### Algebraic Working

$4x^2 + 25x + 21 = [0]$ | M1 | for collection of terms and rearrangement to zero; condone one error

$(4x + 21)(x + 1)$ | M1 | for factors giving at least two terms of their quadratic correct or for subst into formula with no more than two errors [dependent on attempt to rearrange to zero]

| For full use of completing square with no more than two errors allow 2nd and 3rd M1s simultaneously

## Question 11(ii)

$4(x+3)^2 - 5$ isw | 4 | B1 for $a = 4$, B1 for $b = 3$, B2 for $c = -5$ or M1 for $31 - 4 \times$ their $b^2$ soi or for $-5/4$ or for $31/4 - $ their $b^2$ soi | eg an answer of $(x+3)^2 - \frac{5}{4}$ earns B0 B1 M1; $1(2x+6)^2 - 5$ earns B0 B0 B2; $4($ earns first B1; condone omission of square symbol

## Question 11(iii)(A)

$x = -3$ or ft (−their b) from (ii) | 1 | **0** for just $-3$ or ft; **0** for $x = -3, y = -5$ or ft

## Question 11(iii)(B)

$-5$ or ft their $c$ from (ii) | 1 | allow $y = -5$ or ft | **0** for just $(-3, -5)$; bod 1 for $x = -3$ stated then $y = -5$ or ft

## Question 12(i)

$y = 2x + 5$ drawn | M1 | condone unruled and some doubling; tolerance: must pass within/touch at least two circles on overlay; the line must be drawn long enough to intersect curve at least twice

$-2, -1.4$ to $-1.2, 0.7$ to $0.85$ | A2 | A1 for two of these correct | condone coordinates or factors

## Question 12(ii)

$4 = 2x^3 + 5x^2$ or $2x + 5 - \frac{4}{x^2} = 0$ | B1 | condone omission of final '= 0'

and completion to given answer | B1 | or correct division / inspection showing that $x + 2$ is factor

$f(-2) = -16 + 20 - 4 = 0$ | B1 | or correct division / inspection showing that $x + 2$ is factor

use of $x + 2$ as factor in long division of given cubic as far as $2x^2$ + $4x^2$ in working | M1 | or inspection or equating coefficients, with at least two terms correct; may be set out in grid format

$2x^2 + x - 2$ obtained | A1 | condone omission of + sign (eg in grid format)

$\left[x = \right]\frac{-1 \pm \sqrt{1-4 \times 2 \times -2}}{2 \times 2}$ oe | M1 | dep on previous M1 earned; for attempt at formula or full attempt at completing square; allow even if their 'factor' has a remainder shown in working; M0 for just an attempt to factorise

$\frac{-1 \pm \sqrt{17}}{4}$ oe isw | A1 |

## Question 12(iii)

$\frac{4}{x^2} = x + 2$ or $y = x + 2$ soi | M1 | eg is earned by correct line drawn

$y = x + 2$ drawn | A1 | condone unruled; need drawn for $-1.5 \leq x \leq 1.2$; to pass through/touch relevant circle(s) on overlay

1 real root | A1 |

## Question 13(i)

[radius] = 4 | B1 | B0 for $\pm 4$

[centre] $(4, 2)$ | B1 | condone omission of brackets

## Question 13(ii)

$(x-4)^2 + (-2)^2 = 16$ oe | M1 | for subst $y = 0$ in circle eqn; NB candidates may expand and rearrange eqn first, making errors – they can still earn this M1 when they subst $y = 0$ in their circle eqn; condone omission of $(-2)^2$ for this first M1 only; not for second and third M1s; do not allow substitution of $x = 0$ for any Ms in this part

$(x-4)^2 = 12$ or $x^2 - 8x + 4 = [0]$ | M1 | putting in form ready to solve by comp sq, or for rearrangement to zero; condone one error; eg allow M1 for $x^2 + 4 = 0$ [but this two-term quadratic is not eligible for 3rd M1]

$x - 4 = \pm\sqrt{12}$ or $\left[x = \right]\frac{8 \pm \sqrt{8^2-4 \times 1 \times 4}}{2 \times 1}$ | M1 | for attempt at comp square or formula; dep on previous M2 earned and on three-term quadratic; not more than two errors in formula / substitution; allow M1 for $x - 4 = \sqrt{12}$; M0 for just an attempt to factorise

$\left[x = \right]4 \pm \sqrt{12}$ or $4 \pm 2\sqrt{3}$ or $\pm 2\sqrt{3}$ or $\frac{8 \pm \sqrt{48}}{2}$ oe isw | A1 | A1 for one solution

or

sketch showing centre $(4, 2)$ and triangle with hyp 4 and ht 2 | M1 |

$4^2 - 2^2 = 12$ | M1 | or the square root of this; implies previous M1 if no sketch seen

$\left[x = \right]4 \pm \sqrt{12}$ oe | A2 | A1 for one solution

## Question 13(iii)

subst $\left(4+2\sqrt{2}, 2+2\sqrt{2}\right)$ into circle eqn and showing at least one step in correct completion | B1 | or showing sketch of centre C and A and using Pythag: $\left(2\sqrt{2}\right)^2 + \left(2\sqrt{2}\right)^2 = 8 + 8 = 16$

Sketch of both tangents | M1 | need not be ruled; must have negative gradients with tangents intended to be parallel and one touching above and to right of centre; mark intent to touch – allow just missing or just crossing circle twice; condone A not labelled

grad tgt = $-1$ or $-1/$their grad CA | M1 | allow ft after correct method seen for grad $CA = \frac{2+2\sqrt{2}-2}{4+2\sqrt{2}-4} = \frac{2\sqrt{2}}{2\sqrt{2}}$ oe (may be on/ near sketch)

$y - (2+2\sqrt{2}) = $ their $m(x-(4+2\sqrt{2}))$ | M1 | or $y = $ their $mx + c$ and subst of $\left(4+2\sqrt{2}, 2+2\sqrt{2}\right)$; for intent; condone lack of brackets for M1; independent of previous Ms; condone grad of CA used

$y = -x + 6 + 4\sqrt{2}$ oe isw | A1 | accept simplified equivs eg $x + y = 6 + 4\sqrt{2}$

parallel tgt goes through $\left(4-2\sqrt{2}, 2-2\sqrt{2}\right)$ | M1 | or it wrong centre; may be shown on diagram; may be implied by correct equation for the tangent (allow ft their gradient)

eqn is $y = -x + 6 - 4\sqrt{2}$ oe isw | A1 | accept simplified equivs eg $x + y = 6 - 4\sqrt{2}$ | A0 if obtained as eqn of other tangent instead of the tangent at A (eg after omission of brackets); A0 if this is given as eqn of the tangent at A instead of other tangent (eg after omission of brackets)