A Level Mathematics 9709 Paper 12, May/June 2026

Forming a three-term quadratic in $x^2$ or with a substitution (which may be undefined) or factorising the original equation. Terms do not all need to be on the same side.

$x^2=4[\text{or }-\frac{4}{3}]$

$x=\pm 2$ only

{Translation} $\left(\begin{array}{c}1\\ 10\end{array}\right)$ (or equivalent description)

{Stretch} {factor 0.5} {"parallel to y-axis" or "in y-direction" or "vertically"}

Correct order (vertical part of translation before stretch, or stretch before vertical translation $\left(\begin{array}{c}1\\ 5\end{array}\right)$)

$[g(x)=]\{0.5\}\{f(x-1)\}\{+5\}$ or $a=\frac{1}{2},b=-1\text{ and }c=5$

For attempt to find midpoint. Condone one sign error. Expect $(1,\frac{9}{2})$.

For attempt to find gradient. Condone one sign error. Expect $-\frac{1}{2}$.

For finding the negative reciprocal of their gradient.

$\frac{y-4.5}{x-1}=2$ or $y=2x+\frac{5}{2}$ OE

For forming an equation in one variable. Condone sign errors in their rearrangements.

$(-2,-\frac{3}{2})$

B1 for the second or third terms correct and simplified.

B1 for all correct and simplified.

$8+12ax+6a^2{x}^2$

Selecting the appropriate three products from their expansions.

For adding and equating their three terms to -108.

$486a^2-648a+216=0$ or $9a^2-12a+4=0$ OE

$a=\frac{2}{3}$ only

$\frac{6}{3}{x}^3+px[+q]$

For substituting $x=2$ and $-2$ into an integrated expression comprising a term in $x^3$, the term $px$ and a constant of integration and setting both equal to zero.

$p=-8[\text{and }q=0]$ OE

$x=0$

Integrating their cubic expression to reach an expression of the given form. Do not allow if both $p$ and $q$ are 0.

For substituting $(2,3)$ or $(-2,3)$ into their twice integrated expression of the given form. Allow one slip, but $r$ must be present.

$(0,11)$

${\sin }^{2}x+\frac{3}{2}\sin x={\cos }^2\frac{2}{3}\pi$ or better, e.g. $2{\sin }^{2}x+3\sin x+1=0$

$\sin x=-\frac{1}{2}\text{ and }\sin x=-1$ WWW

DB1 for any one correct angle in the range $\{0,2\pi \}$

DB1 for all three and no others in the range $\{0,2\pi \}$

$d=-4$

Finding $n$ using the correct $n^{\text{th}}$ term formula with their $a$ (but not 60, 52 and -28) and their non-zero integer $d$ (which may be positive). May be implied by $n=24$.

Use of a correct sum formula for their values with $n$ a positive integer.

$432$

Forming an equation in $k$ only (or forming an equation in $r$ or $a$ only).

$8k^2-8k-16[=0]$ or $k+4=\pm 3k$ (or finding $r$ or $a$).

$[k=]-1$

$r=\frac{1}{3}$ or $a=27$

For correctly finding $a$ using their $r$ (or vice versa) and using the sum to infinity formula for their values provided $\mathrm{\mid }r\mathrm{\mid }<1$.

$S_{\mathrm{\infty }}=\frac{81}{2}$ or $40.5$ only

$(3,6)$

$r=\sqrt{29}$ or $\frac{\sqrt{116}}{2}$ or $r^2=29$

M1 for using their centre (not $(-2,4)$ or $(8,8)$) and their labelled radius correctly.

$(x-3{)}^2+(y-6{)}^2=29$ (or $x^2-6x+y^2-12y+16=0$)

Forming a quadratic equation in one variable using their circle equation.

Simplifying to a three-term quadratic.

$2x^2-6x-20[=0]$ or $2y^2-30y+88[=0]$ OE

$x=-2,5$

$y=4,11$

For use of Pythagoras with their coordinates, but do not award if either both $x$- or both $y$-coordinates are equal.

$[PQ=]7\sqrt{2}$ or $a=7\text{ and }b=2$ only

Expect $(2x+1{)}^{\frac{3}{2}}$. (B1 for $k(2x+1{)}^{\frac{3}{2}},k\mathrm{\ne }1$.)

Use of limits 0 and 4 on their integrated expression of the form $k(2x+1{)}^{\frac{3}{2}}$. Allow $k=3$.

$=26$

Expect $3(2x+1{)}^{-\frac{1}{2}}$. (B1 for $k(2x+1{)}^{-\frac{1}{2}}$. Must be seen or used in (b).)

[Gradient at $x=4$] = 1

For equation using $(4,9)$ and the negative reciprocal of their gradient from $k(2x+1{)}^{-\frac{1}{2}}$ or finding the $y$-intercept.

Area of triangle in terms of $a$.

For equating a quadratic expression in $a$ to their 26. Not dependent on any previous marks. Expect $a^2-26a+117=0$.

$a=13-\sqrt{52}$ or $13-2\sqrt{13}$

For $b^2-4ac$ or by completing the square. Ignore $>,<,\ge \text{ or }\le$.

For completing the square or use of the quadratic formula (or factorising) on their three-term quadratic in $p$.

$-2\pm \sqrt{5}$ or $\frac{-16\pm \sqrt{320}}{8}$ or $\frac{-4\pm \sqrt{20}}{2}$ OE WWW

$[g^{\mathrm{\prime }}(x)\text{ or }\frac{\text{d}y}{\text{d}x}=]2px-2$ or $[g(x)\text{ or }y\text{ or }g=]p{(x-\frac{1}{p})}^2+\dots$ or $x=\frac{2}{2p}$

$2px-2=0⟹x=\frac{1}{p}$ or $(x-\frac{1}{p})=0⟹x=\frac{1}{p}$ or $x=\frac{2}{2p}=\frac{1}{p}$ AG WWW

Completing the square or intention to use quadratic formula. $x$ and $y$ may be interchanged at this stage. Candidates may work in terms of $p$.

Solving as far as ${(x-\frac{1}{2})}^2=\dots$ condoning subsequent sign errors in the rearrangement of their constant or using the quadratic formula. Candidates may work in terms of $p$.

For an expression for $x=\dots$, condoning sign errors only. Allow $\pm$ at this stage. Candidates may work in terms of $p$.

$\frac{1}{2}+\sqrt{\frac{x-\frac{11}{2}}{2}}$ or $\frac{1}{2}+\sqrt{\frac{2x-11}{2^2}}$ or $\frac{1+\sqrt{2x-11}}{2}$ or $\frac{2+\sqrt{8x-44}}{4}$ OE