A Level Mathematics 9709 Paper 13, May/June 2026

$3x^2-18x+15$ OR $y^2-40y+256[=0]$

$3(x-1)(x-5)$ OR $(y-8)(y-32)[=0]$

$[x=]1,5$ or $[y=]8,32$

$(1,8)$ and $(5,32)$

$1\le x\le 5$

$[x\sqrt{x}=]x^{\frac{3}{2}},x^{1\frac{1}{2}}\text{ or }{x}^{1.5}$

$(x^{\frac{3}{2}}{)}^2-9(x^{\frac{3}{2}})+8$ or $x^6-65x^3+64[=0]$

$x^{\frac{3}{2}}=1,8$ or $x^3=1,64$

$[x=]1,4$

$[\text{LHS}=]\frac{1-{\cos }^2\theta }{\cos \theta }$

$\frac{{\sin }^2\theta }{\cos \theta }$

$\frac{\sin \theta \sin \theta }{\cos \theta }=\tan \theta \sin \theta [=\text{RHS}]$

Either $\tan \theta (2\sin \theta -1)=0$ or $2\sin \theta =1$ Or $[2\frac{{\sin }^2\theta }{\cos \theta }=\frac{\sin \theta }{\cos \theta }\Rightarrow ]2{\sin }^2\theta -\sin \theta =0$

And $\tan \theta =0$ and $\sin \theta =\frac{1}{2}$ (or $\sin \theta =0$ and $\sin \theta =\frac{1}{2}$)

Then $\theta =0,180,360,\theta =30,150$

$[\theta =]0,30,150,180,360$

$[\frac{\mathrm{d}y}{\mathrm{d}x}=]3-3k{x}^2$

$-3=\text{their }(3-3k(1{)}^2)\text{ or }\frac{-1}{\text{their }(3-3k(1{)}^2)}=\frac{1}{3}$

$[k=]2$

$-3=\text{their }\{3-3(\text{their }k)x^2\}\text{ or }\frac{-1}{\text{their }\{3-3(\text{refer to }k)x^2\}}=\frac{1}{3}$

$[x=]\pm 1$

$[y=]3(\text{their negative }x\text{-value})-(\text{their }k)(\text{their negative }x\text{-value}{)}^3+4$

$[Q](-1,3)$

$2^5+\left(\begin{array}{c}5\\ 1\end{array}\right)(2{)}^4(3x{)}^1+\left(\begin{array}{c}5\\ 2\end{array}\right)(2{)}^3(3x{)}^2+\left(\begin{array}{c}5\\ 3\end{array}\right)(2{)}^2(3x{)}^3\dots$ Or $2^5\{1+5\left(\frac{3x}{2}\right)+\frac{5.4}{2!}{\left(\frac{3x}{2}\right)}^2+\frac{\mathrm{5.4.3}}{3!}{\left(\frac{3x}{2}\right)}^3\dots \}$

At least three simplified terms correct in $32+240x+720x^2+1080x^3$

All four simplified terms correct in $32+240x+720x^2+1080x^3$

$(\text{their }(32+240x+720x^2+1080x^3))-(\text{their }(96x+720x^2+2160x^3\dots ))$

$32+144x[+0x^2]-1080x^3$

$x=0.01$

Replacing $x$ with $0.01$ or $0.1$ in their expansion of $(1-3x)(2+3x{)}^5$ from 5(b)

$33.43892$

${\mathrm{f}}^{\mathrm{\prime }}(x)=-\frac{48}{x^4}+3$

${\mathrm{f}}^{\mathrm{\prime }}(x)=0\Rightarrow x=\pm 2$

First correct point: $(2,13)$

Second correct point: $(-2,-3)$

$\frac{16}{(-x{)}^3}+3(-x)+5+[a]$

$-(\frac{16}{x^3}+3x+5+a)$

$a=-5$

$[\text{Arc length}=](\text{their }\frac{1}{3}\pi )\times r\text{ [m]}$

$[\text{Perimeter}=](\frac{1}{3}\pi r+\frac{1}{3}\pi r+r)[\text{m}]$

Area of sector $=\frac{1}{2}\times {18}^2\times (\text{their }\frac{1}{3}\pi )$ or triangle or segment

Second area expression (sector, triangle, or segment)

Combined area calculation: $2\times \text{area of sector}-\text{area of triangle}$ OR $\text{Triangle area}+2\times \text{segment area}$ OR $\text{Sector area}+\text{segment area}$

$108\pi -81\sqrt{3}[{\text{m}}^2]$

${\int }_1^t\frac{2}{5x^3}[\mathrm{d}x]\text{ or }{\int }_1^t\frac{2}{5}{x}^{-3}[\mathrm{d}x]=\frac{1}{9}$

$\frac{2}{5}{x}^{-2}÷(-2)[=-\frac{1}{5x^2}][+C]$

$-\frac{1}{5t^2}--\frac{1}{5}$

$[(\text{their }-\frac{1}{5t^2}+\frac{1}{5})=\frac{1}{9}\Rightarrow \frac{1}{5t^2}=\frac{4}{45}]$

$[t^2=\frac{9}{4}]t=\frac{3}{2}$

$\frac{\mathrm{d}y}{\mathrm{d}x}=a(3x+1{)}^{-\frac{1}{2}}$

$\frac{1}{2}(3x+1{)}^{-\frac{1}{2}}\times 3$

$\frac{\mathrm{d}x}{\mathrm{d}t}=6$

when $y=4$, $x=5$ only (or $\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{8}{3}$)

Use rates of change correctly with their $\frac{\mathrm{d}y}{\mathrm{d}x}$ (or $\frac{\mathrm{d}x}{\mathrm{d}y}$) to obtain an expression for $\frac{\mathrm{d}y}{\mathrm{d}t}$

Substitute their x-value from $y=4$ into their $\frac{\mathrm{d}y}{\mathrm{d}x}$ (or calculate $6÷\frac{8}{3}$)

$[\frac{\mathrm{d}y}{\mathrm{d}t}=]\frac{9}{4}$

$ar=a+3d\text{ and }a{r}^2=a+5d[\text{Or }\frac{a+5d}{a+3d}=\frac{a+3d}{a}]$

Correct equation having eliminated $d$ or $a$

Hence $3r^2-5r+2=0$

$r=\frac{2}{3}(\text{or }1)$

$\frac{a}{1-r}=54\Rightarrow a=54(1-r)[=18]$

Using $ar=a+3d\text{ or }a{r}^2=a+5d\text{ or }d=\frac{a(1-r)}{5-r}\text{ etc.}$

$[d=]-2$

Centre $(0,4)$ or $(\text{radius}{)}^2=10$

$x^2+(y-4{)}^2=10$

$x^2+y^2-8y+16=10\Rightarrow x^2+y^2-8y+6=0$

Substitute linear expression into the circle equation (or use normal/differentiation method)

Simplify to form correct quadratic in $x$ or find $x=\pm 3$

Uses $b^2-4ac=0$ with their coefficients (or find $y$-values and substitute to get $c$-values)

$-4c^2+32c+336[=0]$ (or $c^2-8c-84=0$)

$c=14\text{ or }c=-6$

$0=-3x+(\text{their positive }c\text{-value})$ and $0=-3x+(\text{their negative }c\text{-value})$

$x=\frac{14}{3}\text{ and }-2$

$[\text{Area}=]\text{e.g. }\frac{1}{2}\times (14+6)\times 2+\frac{1}{2}\times (14+6)\times \frac{14}{3}$

$\text{Area}=\frac{200}{3}\text{ or }66\frac{2}{3}$