A Level Further Mathematics 9231 Paper 13, May/June 2026

$b=-(\alpha +\beta +\gamma )=-5$

$-19=5^2-2c$

$c=22$

$11=\frac{22}{-d}$

$d=-2$

$S_3+b{S}_2+c{S}_1+3d=0$

$S_3=-b(-19)-c(5)-3(-2)=-199$

Shear, rotation.

Correct order.

[$x$-axis fixed], with $(0,1)$ mapped to $(2,1)$.

$\frac{1}{4}\pi$ [anticlockwise] about the origin.

$\mathbf{M}=\frac{1}{2}\sqrt{2}\left(\begin{array}{cc}1& 1\\ 1& 3\end{array}\right)$

$\frac{1}{2}\sqrt{2}\left(\begin{array}{cc}1& 1\\ 1& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\frac{1}{2}\sqrt{2}\left(\begin{array}{c}x+y\\ x+3y\end{array}\right)$

$\frac{1}{2}\sqrt{2}(x+3mx)=\frac{m}{2}\sqrt{2}(x+mx)$

$x+3mx=m(x+mx)\Rightarrow 1+3m=m+m^2\Rightarrow m^2-2m-1=0$

$y=(1+\sqrt{2})x$ and $y=(1-\sqrt{2})x$

${\mathbf{P}}^{-1}=-\frac{1}{6}\left(\begin{array}{cc}3& -3\\ -7& 5\end{array}\right)=\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ \frac{7}{6}& -\frac{5}{6}\end{array}\right)$

${\mathbf{P}\mathbf{D}\mathbf{P}}^{-1}=\left(\begin{array}{cc}5& 3\\ 7& 3\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ \frac{7}{6}& -\frac{5}{6}\end{array}\right)=\left(\begin{array}{cc}5& 9\\ 7& 9\end{array}\right)\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ \frac{7}{6}& -\frac{5}{6}\end{array}\right)$

${\mathbf{P}\mathbf{D}\mathbf{P}}^{-1}=\left(\begin{array}{cc}8& -5\\ 7& -4\end{array}\right)=\mathbf{A}$

Assume that it is true for $n=k$, so ${\mathbf{A}}^k={\mathbf{P}\mathbf{D}}^k{\mathbf{P}}^{-1}$.

Then ${\mathbf{A}}^{k+1}={\mathbf{A}}^k\mathbf{A}={\mathbf{P}\mathbf{D}}^k{\mathbf{P}}^{-1}{\mathbf{P}\mathbf{D}\mathbf{P}}^{-1}={\mathbf{P}\mathbf{D}}^{k+1}{\mathbf{P}}^{-1}$ (Method)

Then ${\mathbf{A}}^{k+1}={\mathbf{A}}^k\mathbf{A}={\mathbf{P}\mathbf{D}}^k{\mathbf{P}}^{-1}{\mathbf{P}\mathbf{D}\mathbf{P}}^{-1}={\mathbf{P}\mathbf{D}}^{k+1}{\mathbf{P}}^{-1}$ (Accuracy)

So ${\mathbf{A}}^n={\mathbf{P}\mathbf{D}}^n{\mathbf{P}}^{-1}$ is true for all positive [integers] $n$ [by induction].

$16r^4+32r^3+24r^2+8r+1-(16r^4-32r^3+24r^2-8r+1)=64r^3+16r$

$64r^3+16r$

$r=13^4-1^4$ $r=25^4-3^4$ $\dots$ $r=n(2n+1{)}^4-(2n-1{)}^4$

$[64{\sum }_{r=1}^n{r}^3+16{\sum }_{r=1}^{n}r=](2n+1{)}^4-1$

$\Rightarrow 64{\sum }_{r=1}^n{r}^3=16n^4+32n^3+24n^2+8n-8n(n+1)$

$\Rightarrow 64{\sum }_{r=1}^n{r}^3=16n^2(n^2+2n+1)\Rightarrow {\sum }_{r=1}^n{r}^3=\frac{1}{4}{n}^2(n+1{)}^2$

$[S_{2n}]={\sum }_{r=1}^{2n}{r}^3-2{\sum }_{r=1}^n(2r{)}^3$

$[S_{2n}]=\frac{1}{4}(2n{)}^2(2n+1{)}^2-\frac{16}{4}{n}^2(n+1{)}^2$

$S_{2n}=n^2((2n+1{)}^2-4(n+1{)}^2)=n^2(-4n-3)$

Initial line drawn and correct parabolic shape (not vertical at furthest point), starts from the pole.

[Maximum distance of $C$ from the pole is] $2{\left(\frac{1}{3}\pi \right)}^{\frac{1}{2}}$.

$\frac{1}{2}{\int }_0^{\frac{1}{3}\pi }\theta {\sec }^2\theta \mathrm{d}\theta$

Integrates $\theta {\sec }^2\theta$ by parts.

$=\frac{1}{2}[\theta \tan \theta {]}_0^{\frac{1}{3}\pi }-\frac{1}{2}{\int }_0^{\frac{1}{3}\pi }\tan \theta \mathrm{d}\theta$

$=\frac{1}{2}[\theta \tan \theta -\ln \sec \theta {]}_0^{\frac{1}{3}\pi }$

$=\frac{1}{2}(\frac{1}{3}\pi \sqrt{3}-\ln 2)$

$r^2{\cos }^2\theta =\theta \Rightarrow x^2=\theta$

$x^2={\tan }^{-1}\frac{y}{x}$ or $\tan (x^2)=\frac{y}{x}$

$y=x\tan (x^2)$

$\frac{1}{2}(2{\left(\frac{\pi }{3}\right)}^{\frac{1}{2}}\cos \frac{1}{3}\pi )(2{\left(\frac{\pi }{3}\right)}^{\frac{1}{2}}\sin \frac{1}{3}\pi )=\frac{1}{6}\pi \sqrt{3}$

$\frac{1}{6}\pi \sqrt{3}-\frac{1}{2}(\frac{1}{3}\pi \sqrt{3}-\ln 2)$

$\frac{1}{2}\ln 2$

$x^2+x+1=0$ has no real roots as discriminant is $[1^2-4=]-3$ and this is less than zero, hence shown.

$y=0$

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

$x^2+2x=0$

$(0,1)$ and $(-2,-\frac{1}{3})$

Axes and approach to asymptote.

Correct shape and position, and maximum apparently on the axis

$(0,1)$ and $(-1,0)$

Correct shape with reflection in $x$-axis for $y<0$.

Correct shape at $(-1,0)$ and must have a correct graph in part (c)

$\frac{x+1}{x^2+x+1}=\frac{1}{3}$ or $\frac{x+1}{x^2+x+1}<\frac{1}{3}$

$\frac{x+1}{x^2+x+1}=-\frac{1}{3}$ or $\frac{x+1}{x^2+x+1}>-\frac{1}{3}$

$x<1-\sqrt{3}$, $x>1+\sqrt{3}$

$x\mathrm{\ne }-2$

$\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& -1& 0\\ -3& 1& -1\end{array}\mid$

$=\left(\begin{array}{c}1\\ 1\\ -2\end{array}\right)$

$(0)+(1)-2(2)=-3$

$x+y-2z=-3$

$\frac{1}{\sqrt{6}}\mid \left(\begin{array}{c}1\\ 1\\ -2\end{array}\right)\cdot \left(\begin{array}{c}6+6\theta \\ 1+\theta \\ 2+5\theta \end{array}\right)-(-3)\mid$

$=\frac{\mathrm{\mid }3\theta -6\mathrm{\mid }}{\sqrt{6}}$

$\mathrm{\mid }3\theta -6\mathrm{\mid }<6$

$0<\theta <4$

$\overrightarrow{OQ}=\left(\begin{array}{c}0\\ \lambda \\ 2\end{array}\right),\overrightarrow{OR}=\left(\begin{array}{c}1-3\mu \\ \mu \\ 2-\mu \end{array}\right)\Rightarrow \overrightarrow{QR}=\left(\begin{array}{c}1-3\mu \\ \mu -\lambda \\ 2-\mu -2\lambda \end{array}\right)$

Finds $\overrightarrow{QR}$ or direction of common perpendicular.

$\left(\begin{array}{c}1-3\mu \\ \mu -\lambda \\ 2-\mu -2\lambda \end{array}\right)\cdot \left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)=0$

$\mu +5\lambda =4$

$\left(\begin{array}{c}1-3\mu \\ \mu -\lambda \\ 2-\mu -2\lambda \end{array}\right)\cdot \left(\begin{array}{c}-3\\ 1\\ -1\end{array}\right)=0\Rightarrow 11\mu +\lambda =5$

Solves for $\mu$ and substitutes into $\overrightarrow{OR}$ and $\overrightarrow{QR}$

$\mu =\frac{7}{18}\Rightarrow \overrightarrow{OR}=\frac{1}{18}\left(\begin{array}{c}-3\\ 7\\ 29\end{array}\right)$ and $\overrightarrow{QR}=\left(\begin{array}{c}-\frac{1}{6}\\ -\frac{1}{3}\\ \frac{1}{6}\end{array}\right)$

$\mathbf{r}=\frac{1}{18}\left(\begin{array}{c}-3\\ 7\\ 29\end{array}\right)+k\left(\begin{array}{c}1\\ 2\\ -1\end{array}\right)$