$3(x - 2)^2 - 7$ isw or $a = 3, b = 2$ $c = 7$ www | 4 | B1 each for $a = 3, b = 2$ oe; and B2 for $c = 7$ oe; or M1 for $\left[-\frac{7}{3}\right.$ or for $5 - $ their $a$(their $b)^2$; or for $\frac{5 - $ their $a(their $b)^2$; soi | B0 for $(2, -7)$; may be obtained by starting again eg with calculus | may be implied by their answer; condone omission of square symbol; ignore $= 0'$

$-7$ or ft | B1 | [5]

## Question 9 (i)

$3n$ isw | 1 [1] | accept equivalent general explanation

## Question 9 (ii)

at least one of $(n - 1)^2$ and $(n + 1)^2$ correctly expanded | M1 | must be seen; M0 for just $n^2 + 1 + n^2 + n^2 + 1$

$3n^2 + 2$ | B1 | accept even if no expansions / wrong expansions seen

comment eg $3n^2$ is always a multiple of 3 so remainder after dividing by 3 is always 2 | B1 | dep on previous B1; B0 for just saying that 2 is not divisible by 3 – must comment on $3n^2$ term as well

allow B1 for $\frac{3n^2 + 2}{3} = n^2 + \frac{2}{3}$ | [3] | SC: $n, n + 1, n + 2$ used similarly can obtain first M1, and allow final B1 for similar comment on $3n^2 + 6n + 5$

## Question 10 (i)

$[\text{radius}] = \sqrt{20}$ or $2\sqrt{5}$ isw | B1 | B0 for $\pm\sqrt{20}$ oe

$[\text{centre}] = (3, 2)$ | B1 | [2] | condone lack of brackets with coordinates, here and in other questions

## Question 10 (ii)

substitution of $x = 0$ or $y = 0$ into circle equation | M1 | or use of Pythagoras with radius and a coordinate of the centre eg $20 - 2^2$ or $b^2 + 3^2 = 20$ ft their centre and/or radius; bod intent; allow M1 for $(x - 3)^2 = 20$ and/or $(y - 2)^2 = 20$

$(x - 7)(x + 1) [=0]$ | M1 | no ft from wrong quadratic; for factors giving two terms correct, or formula or completing square used with at most one error; completing square attempt must reach at least $(x - a)^2 = b$; following use of Pythagoras allow M1 for attempt to add 3 to $[\pm]4$

$(7, 0)$ and $(-1, 0)$ isw | A1 | accept $x = 7$ or $-1$ (both required); $[y =] \pm \sqrt{(-4)^2 - 4 \times 1 \times (−7)}$ oe; $\left(0, 2 \pm \sqrt{11}\right)$ or $\left(0, \frac{4 \pm \sqrt{44}}{2}\right)$ isw | A1 | accept $y = \frac{4 \pm \sqrt{44}}{2}$ oe isw; annotation is required if part marks are earned in this part: putting a tick for each mark earned is sufficient

## Question 10 (iii)

show both A and B are on circle | B1 | explicit substitution in circle equation and at least one stage of interim working required oe; or clear use of Pythagoras to show AC and BC each $= \sqrt{20}$

$(4, 5)$ | B2 | B1 each; or M1 for $\left(\frac{7+1, 6+4}{2}\right)$; from correct midpoint and centre used; B1 for $\pm\sqrt{10}$; M1 for $(4 - 3)^2 + (5 - 2)^2$ or $b^2 + 3^2$ or ft their centre and/or midpoint, or for the square root of this | may be a longer method finding length of $\frac{1}{2}$ AB and using Pythag. with radius; no ft if one coord of midpoint is same as that of centre so that distance formula/Pythag is not required eg centre correct and midpt $(3, -1)$

$\sqrt{10}$ | B2 | [5]

## Question 11 (i)

sketch of cubic the right way up, with two tps and clearly crossing the $x$ axis in 3 places | B1 | no section to be ruled; no curving back; condone slight 'flicking out' at ends but not approaching another turning point; condone some doubling (eg erased curves may continue to show); accept min tp on y-axis or in 3rd or 4th quadrant; curve must clearly extend beyond the $x$ axis at both 'ends'

crossing/reaching the $x$-axis at $-4, -2$ and 1.5 | B1 | intersections must be shown correctly labelled or worked out nearby; mark intent; accept curve crossing axis halfway between 1 and 2 if 3/2 not marked

intersection of y-axis at $-24$ | B1 | [3]

## Question 11 (ii)

$-2, 0$ and $7/2$ oe isw or ft their intersections | 2 | B1 for 2 correct or ft for $(-2, 0) (0, 0)$ and $(3.5, 0)$ or M1 for $(x + 2) \times (2x - 7)$ oe or SC1 for $-6, -4$ and $-1/2$ oe

[2]

## Question 11 (iii) (A)

correct expansion of product of 2 brackets of $f(x)$ | M1 | need not be simplified; condone lack of brackets for M1; or allow M1 for expansion of all 3 brackets, showing all terms, with at most one error: $2x^3 + 4x^2 + 8x^2 - 3x^2 + 16x - 12x - 6x - 24$ | eg $2x^2 + 5x - 12$ or $2x^2 + x - 6$ or $x^2 + 6x + 8$; may be seen in (i) – allow the M1; the part (i) work appears at the foot of the image for (iii)A, so mark this rather than in (i)

correct expansion of quadratic and linear and completion to given answer | A1 | for correct completion if all 3 brackets already expanded, with some reference to show why $-24$ changes to $-9$; condone lack of brackets if they have gone on to expand correctly; condone $+15'$ appearing at some stage | [2] | NB answer given; mark the whole process

## Question 11 (iii) (B)

$g(1) = 2 + 9 - 2 - 9 [=0]$ | B1 | allow this mark for $(x - 1)$ shown to be a factor and a statement that this means that $x = 1$ is a root [of $g(x) = 0$] oe

attempt at division by $(x - 1)$ as far as $2x^2 - 2x^2$ in working | M1 | or inspection with at least two terms of quadratic factor correct; M0 for division by $x + 1$ after $g(1) = 0$ unless further working such as $g(-1) = 0$ shown, but this can go on to gain last M1A1

correctly obtaining $2x^2 + 11x + 9$ | A1 | allow B2 for another linear factor found by the factor theorem; NB mixture of methods may be seen in this part – mark equivalently eg three uses of factor theorem, or two uses plus inspection to get last factor;

factorising a correct quadratic factor | M1 | for factors giving two terms correct; eg allow M1 for factorising $2x^2 + 7x - 9$ after division by $x + 1$

$(2x + 9)(x + 1)(x - 1)$ isw | A1 | allow $2(x + 9/2)(x + 1)(x - 1)$ oe; dependent on 2nd M1 only; condone omission of first factor found; ignore $'= 0'$ seen | [5] | SC alternative method for last 4 marks: allow first M1A1 for $(2x + 9)(x^2 - 1)$ and then second M1A1 for full factorisation

## Question 12 (i)

$y = 2x + 3$ drawn accurately | M1 | at least as far as intersecting curve twice; ruled straight line and within 2mm of $(2, 7)$ and $(-1, 1)$

$(-1.6$ to $-1.7, -0.2$ to $-0.3)$ | B1 | intersections may be in form $x = ..., y = ...$

$(2.1$ to $2.2, 7.2$ to $7.4)$ | B1 | [3] | if marking by parts and you see work relevant to (ii), put a yellow line here and in (ii) to alert you to look

## Question 12 (ii)

$\frac{1}{x - 2} = 2x + 3$ | M1 | or attempt at elimination of $x$ by rearrangement and substitution; may be seen in (i) – allow marks; the part (i) work appears at the foot of the image for (ii) so show marks there rather than in (i)

$1 = (2x + 3)(x - 2)$ | M1 | condone lack of brackets; implies first M1 if that step not seen

$1 = 2x^2 - x - 6$ oe | A1 | for correct expansion; need not be simplified; NB A0 for $2x^2 - x - 7 = 0$ without expansion seen [given answer]; after completing square, accept $\frac{1 \pm \sqrt{57}}{4}$ or better

$1 \pm \sqrt{12 - 4 \times 2 \times (-7)}$ oe | M1 | use of formula or completing square on given equation, with at most one error; completing square attempt must reach at least $[2](x - a)^2 = b$ or $(2x - c)^2 = d$ stage oe with at most one error

$\frac{1 \pm \sqrt{57}}{4}$ isw | A1 | [5]

## Question 12 (iii)

$\frac{1}{x - 2} = -x + k$ and attempt at rearrangement | M1 |

$x^2 - (k + 2) x + 2k + 1 [=0]$ | M1 | for simplifying and rearranging to zero; condone one error; collection of $x$ terms with bracket not required

$b^2 - 4ac = 0$ oe seen or used | M1 | | = 0 may not be seen, but may be implied by their final values of $k$

$[k =] 0$ or $4$ as final answer, both required | A1 | SC1 for 0 and 4 found if 3rd M1 not earned (may or may not have earned first two Ms); eg obtained graphically or using calculus and/or final answer given as a range

[4]