A Level Mathematics 9709 Paper 13, Oct/Nov 2025

$\left[\left(\begin{array}{c}6\\ 0\end{array}\right)\right]2^6\pm \left(\begin{array}{c}6\\ 1\end{array}\right)2^5\left(\frac{x}{2}\right)\pm \left(\begin{array}{c}6\\ 2\end{array}\right)2^4{\left(\frac{x}{2}\right)}^2\pm \left(\begin{array}{c}6\\ 3\end{array}\right)2^3{\left(\frac{x}{2}\right)}^3$ or $2^6(1\pm 6\left(\frac{x}{4}\right)\pm \frac{6.5}{2!}{\left(\frac{x}{4}\right)}^2\pm \frac{\mathrm{6.5.4}}{3!}{\left(\frac{x}{4}\right)}^3)$

Three correct simplified terms.

Four correct simplified terms.

$(3\times \text{their}(-20))+(-1\times \text{their }60)+(2\times \text{their }64)$

8

$5x-3=\tan (-\frac{1}{4}\pi )$ or $-\tan \left(\frac{1}{4}\pi \right)[=-1]$

$x=\frac{2}{5}$ or $0.4$

$5-5{\sin }^2\theta =4\sin \theta +4$

$5{\sin }^2\theta +4\sin \theta -1[=0][\Rightarrow (5\sin \theta -1)(\sin \theta +1)=0]$

$\sin \theta =\frac{1}{5}$ and $-1$

$\theta =2.94,4.71$

72.030004

30.3888

30[.0]

$20+(30-1)\times (\text{their }-5)$

$-125$

$r=\frac{k}{20}$ and $r=\frac{k-5}{k}$ (or $20r=k$ and $20r^2=k-5$)

$k^2-20k+100[=0]$ (or $20r^2-20r+5[=0]$)

$[r=]\frac{1}{2}$

$[S_{\mathrm{\infty }}=\frac{20}{1-\frac{1}{2}}=]40$

$\sin \left(\frac{1}{2}\theta \right)=\frac{2\sqrt{3}}{4}$ or $\cos \left(\frac{1}{2}\theta \right)=\frac{2}{4}$ or $\cos \theta =\frac{4^2+4^2-(4\sqrt{3}{)}^2}{2\times 4\times 4}[=-\frac{1}{2}]$

$[\theta =]\frac{2\pi }{3}$

$\frac{1}{2}\times 4^2\times \sin \frac{2\pi }{3}$ or $\frac{1}{2}\times 4^2\times \frac{4\pi }{3}$ or $\frac{1}{2}\times 4^2\times \frac{2\pi }{3}$

$\frac{1}{2}\times 4^2\times \frac{4\pi }{3}+\frac{1}{2}\times 4^2\times \sin \frac{2\pi }{3}$ Or $16\pi -(\frac{1}{2}\times 4^2\times \frac{2\pi }{3}-\frac{1}{2}\times 4^2\times \sin \frac{2\pi }{3})$

$\frac{32\pi }{3}+4\sqrt{3}$

{Stretch} {[scale] factor 2} {[parallel to] in/on y[-axis] or x-axis invariant}

{Translation} $\left(\begin{array}{c}3\\ -4\end{array}\right)$ or {[+]3 in/on [the] x[direction]} and {–4 in/on [the] y [direction]}

Correct transformations in the correct order without any extra transformations. Down and right can be condoned for this mark if the candidate’s intention is clear.

$a=2$

$b=-3$

$c=-4$

$[y=\frac{2}{ax-3}+\frac{1}{2}\Rightarrow ](y-\frac{1}{2})(ax-3)=2$ or $ax-3=\frac{2}{y-\frac{1}{2}}$

$ax=\frac{4}{2y-1}+3$

$[y=\text{ or }{\text{g}}^{-1}(x)=]\frac{4}{a(2x-1)}+\frac{3}{a}$ or $\frac{2}{a(x-\frac{1}{2})}+\frac{3}{a}$

Clear replacement of $a$ with 6 in ${\text{g}}^{-1}$ and, if necessary, followed by simplification to arrive at exactly the same expressions.

Centre is $(2,4)$ (or Either $\frac{y-1}{x-6}$ or $\frac{y-7}{x+2}$)

$[(\text{Radius}{)}^2=](6-2{)}^2+(1-4{)}^2$ or ${\left(\frac{1}{2}\sqrt{(6-(-2){)}^2+(1-7{)}^2}\right)}^2[=25]$ (or $\frac{y-1}{x-6}\times \frac{y-7}{x+2}=-1$)

$(x-2{)}^2+(y-4{)}^2=(5{)}^2$ (or $x^2+y^2-4x-8y-5=0$)

$[\text{Gradient of normal through centre}=]2$

$y-4=(\text{their gradient})(x-2)$ Or $4=(\text{their }2)(2)+c\Rightarrow c=[0]$

$(x-2{)}^2+(2x-4{)}^2=25$

$[5(x-2{)}^2=25\Rightarrow ](x-2{)}^2=5$

$[x=]2\pm \sqrt{5}$

$\frac{1}{2}x+\frac{4}{x}=4.5\Rightarrow x^2-9x+8=0$

$[x=]1,8$

$\pi \times {4.5}^2\times (\text{their }8-\text{their }1)=\left[\frac{567}{4}\pi \text{ or }445.3\right]$

$[\pi ]\int {(\frac{1}{2}x+\frac{4}{x})}^2[\text{d}x]$

First two correct terms in $[\pi ](\frac{1}{12}{x}^3+4x-\frac{16}{x})$

All three correct terms in $[\pi ](\frac{1}{12}{x}^3+4x-\frac{16}{x})$

$[\pi ](\frac{218}{3}+\frac{143}{12})=[\pi ]\frac{1015}{12}$

$\text{Volume of solid}=\frac{343}{6}\pi$

$\{2\}$

$\{(x+1{)}^2\}$

$\{+8\}$

$[\text{Range or f}[(x)]\text{ or }y]\ge 8$

$p(5m{)}^2+4(5m)-5[=0]$ and $p(-9m{)}^2+4(-9m)-5[=0]$

Eliminate $p$ by substitution or equating two expressions for $p$.

$2520m=280$ (or simplify as far as a quadratic where $m=0$ is ignored/dismissed and substitute for $m$ or $p$)

$m=\frac{1}{9}$

$p=9$

$\left\{\frac{-8\times 3}{(3x-8{)}^2}\right\}$

$\{+\frac{6}{(x-1{)}^2}\}$

$\left\{\frac{-8\times 3}{(3.3-8{)}^2}\right\}\{+\frac{6}{(3-1{)}^2}\}[=\frac{-45}{2}]$

$y-5=\left(\text{their }\frac{-45}{2}\right)(x-3)$ or $5=\left(\text{their }\frac{-45}{2}\right)(3)+c\Rightarrow c=\left[\frac{145}{2}\right]$

$[-8x-5=\frac{-45}{2}(x-3)\Rightarrow ]29x=145$

$x=5,y=-40$

$\frac{-24}{(3x-8{)}^2}+\frac{6}{(x-1{)}^2}=0\Rightarrow 24(x-1{)}^2=6(3x-8{)}^2[\Rightarrow 4(x-1{)}^2=(3x-8{)}^2]$

$5x^2-40x+60[=0]$ or $\pm 2(x-1)=3x-8$

$2,6$

$\left\{\frac{-24\times -2\times 3}{(3x-8{)}^3}\right\}$

$\left\{\frac{6\times -2}{(x-1{)}^3}\right\}$

$\text{At }x=2,\frac{{\text{d}}^{2}y}{\text{d}{x}^2}=\frac{\text{their }144}{(6-8{)}^3}-\frac{\text{their }12}{(2-1{)}^3}$ and at $x=6,\frac{{\text{d}}^{2}y}{\text{d}{x}^2}=\frac{\text{their }144}{(18-8{)}^3}-\frac{\text{their }12}{(6-1{)}^3}$

$\text{At }x=2,\frac{{\text{d}}^{2}y}{\text{d}{x}^2}=-30<0,\text{ therefore max[imum]}$; $\text{At }x=6,\frac{{\text{d}}^{2}y}{\text{d}{x}^2}=\frac{48}{{10}^3}\left[\frac{6}{125}\right]>0,\text{ therefore min[imum]}$