A Level Further Mathematics 9231 Paper 11, May/June 2025

$(2-3r)(5-3r)=9r^2-21r+10$

$9(\frac{1}{6}n(n+1)(2n+1))-21(\frac{1}{2}n(n+1))+10n$

$=3n^3-6n^2+n$

$\frac{1}{(2-3r)(5-3r)}=\frac{1}{3}(\frac{1}{2-3r}-\frac{1}{5-3r})$

${\sum }_{r=1}^n\frac{1}{(2-3r)(5-3r)}=\frac{1}{3}(-\frac{1}{1}-\frac{1}{2})+\frac{1}{3}(-\frac{1}{4}-(-\frac{1}{1}))+\frac{1}{3}(-\frac{1}{7}-(-\frac{1}{4}))+\cdots +\frac{1}{3}(\frac{1}{2-3n}-\frac{1}{5-3n})$ (method)

${\sum }_{r=1}^n\frac{1}{(2-3r)(5-3r)}=\frac{1}{3}(-\frac{1}{1}-\frac{1}{2})+\frac{1}{3}(-\frac{1}{4}-(-\frac{1}{1}))+\frac{1}{3}(-\frac{1}{7}-(-\frac{1}{4}))+\cdots +\frac{1}{3}(\frac{1}{2-3n}-\frac{1}{5-3n})$ (accuracy)

$=-\frac{1}{6}+\frac{1}{3(2-3n)}$

${\sum }_{r=1}^{\mathrm{\infty }}\frac{1}{(2-3r)(5-3r)}=-\frac{1}{6}$

$y=x^3-1$

$(y+1{)}^{\frac{2}{3}}+2(y+1{)}^{\frac{1}{3}}+1=0\Rightarrow 2(y+1{)}^{\frac{1}{3}}=-y-2$ $8(y+1)=-(y+2{)}^3\Rightarrow 8y+8=-y^3-6y^2-12y-8$

$y^3+6y^2+20y+16=0$

$({\alpha }^3-1{)}^2+({\beta }^3-1{)}^2+({\gamma }^3-1{)}^2=6^2-2(20)$

$-4$

$({\alpha }^3-1{)}^3+({\beta }^3-1{)}^3+({\gamma }^3-1{)}^3=-6(-4)-20(-6)-16(3)$

$96$

$u_1=5=6^1-1$

Assume that it is true for $n=k$, so $u_k=6^k-1$.

Then $u_{k+1}=6(6^k-1)+5=6^{k+1}-1$. (method)

Then $u_{k+1}=6(6^k-1)+5=6^{k+1}-1$. (accuracy)

So, it is also true for $n=k+1$. Hence, by induction, $u_n=6^n-1$ is true for all positive integers.

$\frac{u_{2n}}{u_n}=\frac{6^{2n}-1}{6^n-1}=\frac{(6^n-1)(6^n+1)}{6^n-1}$

$=6^n+1$

A rotation, followed by a shear.

$\mathbf{M}=\left(\begin{array}{cc}\cos \theta +2\sin \theta & 2\cos \theta -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$

$\left(\begin{array}{cc}\cos \theta +2\sin \theta & 2\cos \theta -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}x\cos \theta +2x\sin \theta -y\sin \theta +2y\cos \theta \\ x\sin \theta +y\cos \theta \end{array}\right)$

$x\sin \theta +y\cos \theta =y\text{ AND }x\cos \theta +2x\sin \theta -y\sin \theta +2y\cos \theta =x$

$\Rightarrow x\cos \theta +2x\sin \theta +\frac{x\sin \theta }{1-\cos \theta }(2\cos \theta -\sin \theta )=x$

$(\cos \theta +2\sin \theta )(1-\cos \theta )+2\sin \theta \cos \theta -{\sin }^2\theta =1-\cos \theta$

$\cos \theta -{\cos }^2\theta +2\sin \theta -{\sin }^2\theta =1-\cos \theta \Rightarrow \sin \theta +\cos \theta =1$

$\theta =\frac{1}{2}\pi$

Spiral shape on $[0,2\pi ]$, with correct orientation, starting at pole and with one other point of intersection with the initial line.

$r$ strictly increasing, correct gradient at points of intersection with the initial line.

$\frac{1}{2}{\int }_0^{2\pi }{\theta }^2{\text{e}}^{\frac{1}{4}\theta }\text{d}\theta$

$=\frac{1}{2}{\left[4{\theta }^2{\text{e}}^{\frac{1}{4}\theta }\right]}_0^{2\pi }-\frac{1}{2}{\int }_0^{2\pi }8\theta {\text{e}}^{\frac{1}{4}\theta }\text{d}\theta$ (method)

$=\frac{1}{2}{\left[4{\theta }^2{\text{e}}^{\frac{1}{4}\theta }\right]}_0^{2\pi }-\frac{1}{2}{\int }_0^{2\pi }8\theta {\text{e}}^{\frac{1}{4}\theta }\text{d}\theta$ (accuracy)

$=\frac{1}{2}{\left[4{\theta }^2{\text{e}}^{\frac{1}{4}\theta }\right]}_0^{2\pi }-4{\left[4\theta {\text{e}}^{\frac{1}{4}\theta }\right]}_0^{2\pi }+16{\int }_0^{2\pi }{\text{e}}^{\frac{1}{4}\theta }\text{d}\theta ={[2{\theta }^2{\text{e}}^{\frac{1}{4}\theta }-16\theta {\text{e}}^{\frac{1}{4}\theta }+64{\text{e}}^{\frac{1}{4}\theta }]}_0^{2\pi }$ (method)

$=\frac{1}{2}{\left[4{\theta }^2{\text{e}}^{\frac{1}{4}\theta }\right]}_0^{2\pi }-4{\left[4\theta {\text{e}}^{\frac{1}{4}\theta }\right]}_0^{2\pi }+16{\int }_0^{2\pi }{\text{e}}^{\frac{1}{4}\theta }\text{d}\theta ={[2{\theta }^2{\text{e}}^{\frac{1}{4}\theta }-16\theta {\text{e}}^{\frac{1}{4}\theta }+64{\text{e}}^{\frac{1}{4}\theta }]}_0^{2\pi }$ (accuracy)

$=(8{\pi }^2-32\pi +64){\text{e}}^{\frac{1}{2}\pi }-64$

$y=\theta {\text{e}}^{\frac{1}{8}\theta }\sin \theta$

$\frac{\text{d}y}{\text{d}\theta }=\theta {\text{e}}^{\frac{1}{8}\theta }\cos \theta +\sin \theta (\frac{1}{8}\theta {\text{e}}^{\frac{1}{8}\theta }+{\text{e}}^{\frac{1}{8}\theta })$ (method)

$\frac{\text{d}y}{\text{d}\theta }=\theta {\text{e}}^{\frac{1}{8}\theta }\cos \theta +\sin \theta (\frac{1}{8}\theta {\text{e}}^{\frac{1}{8}\theta }+{\text{e}}^{\frac{1}{8}\theta })$ (accuracy)

${\text{e}}^{\frac{1}{8}\theta }\mathrm{\ne }0\Rightarrow \theta \cos \theta +(\frac{1}{8}\theta +1)\sin \theta =0$

$5\cos 5+(\frac{5}{8}+1)\sin 5=-0.1399$ and $5.05\cos 5.05+(\frac{5.05}{8}+1)\sin 5.05=0.1336$

$\overrightarrow{AB}=2\mathbf{j}+4\mathbf{k}\overrightarrow{AC}=\mathbf{i}-\mathbf{j}+\mathbf{k}$

$\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 0& 2& 4\\ 1& -1& 1\end{array}\mid =\left(\begin{array}{c}6\\ 4\\ -2\end{array}\right)\sim \left(\begin{array}{c}3\\ 2\\ -1\end{array}\right)$ (method)

$\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 0& 2& 4\\ 1& -1& 1\end{array}\mid =\left(\begin{array}{c}6\\ 4\\ -2\end{array}\right)\sim \left(\begin{array}{c}3\\ 2\\ -1\end{array}\right)$ (accuracy)

$3(1)+2(0)-(-2)=5\Rightarrow 3x+2y-z=5$ (method)

$3(1)+2(0)-(-2)=5\Rightarrow 3x+2y-z=5$ (accuracy)

$\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 0& 2& 4\\ 0& 0& t+2\end{array}\mid =\left(\begin{array}{c}2t+4\\ 0\\ 0\end{array}\right)\sim \left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$ (method)

$\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 0& 2& 4\\ 0& 0& t+2\end{array}\mid =\left(\begin{array}{c}2t+4\\ 0\\ 0\end{array}\right)\sim \left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$ (accuracy)

$\left(\begin{array}{c}3\\ 2\\ -1\end{array}\right)\cdot \left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)=\sqrt{14}\cos \theta$

$(t\mathrm{\ne }-2)\Rightarrow \theta ={36.7}^{\circ }$

$\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 0& 2& 4\\ -1& 1& t+1\end{array}\mid =\left(\begin{array}{c}2t-2\\ -4\\ 2\end{array}\right)\sim \left(\begin{array}{c}t-1\\ -2\\ 1\end{array}\right)$ (method)

$\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 0& 2& 4\\ -1& 1& t+1\end{array}\mid =\left(\begin{array}{c}2t-2\\ -4\\ 2\end{array}\right)\sim \left(\begin{array}{c}t-1\\ -2\\ 1\end{array}\right)$ (accuracy)

$\mid \frac{1}{\sqrt{t^2-2t+6}}\left(\begin{array}{c}1\\ -1\\ 1\end{array}\right)\cdot \left(\begin{array}{c}t-1\\ -2\\ 1\end{array}\right)\mid =\mid \frac{t+2}{\sqrt{t^2-2t+6}}\mid$ (method)

$\mid \frac{1}{\sqrt{t^2-2t+6}}\left(\begin{array}{c}1\\ -1\\ 1\end{array}\right)\cdot \left(\begin{array}{c}t-1\\ -2\\ 1\end{array}\right)\mid =\mid \frac{t+2}{\sqrt{t^2-2t+6}}\mid$ (accuracy)

$\frac{t+2}{\sqrt{t^2-2t+6}}=\sqrt{2}\Rightarrow (t+2{)}^2=2(t^2-2t+6)$

$t^2-8t+8=0\Rightarrow t=2(2+\sqrt{2})$

$t=2(2-\sqrt{2})$

$x=-\frac{1}{2},x=4$

$y=1$

$\frac{\text{d}y}{\text{d}x}=\frac{(2x^2-7x-4)(4x-5)-(2x^2-5x)(4x-7)}{(2x^2-7x-4{)}^2}$

$4x^2+16x-20=0$

$(-5,\frac{25}{27})$

$(1,\frac{1}{3})$

Axes and asymptotes labelled.

Correct shape and position.

$(0,0),(\frac{5}{2},0)$

Correct shape and position.

$\frac{2x^2-5x}{2x^2-7x-4}=\frac{1}{9}\text{ or }\frac{2x^2-5x}{2x^2-7x-4}=-\frac{1}{9}$ $16x^2-38x+4=0\text{ or }20x^2-52x-4=0$

$x=\frac{19}{16}+\frac{3}{16}\sqrt{33},x=\frac{19}{16}-\frac{3}{16}\sqrt{33}$

$x=\frac{13}{10}+\frac{3}{10}\sqrt{21},x=\frac{13}{10}-\frac{3}{10}\sqrt{21}$

$\frac{13}{10}-\frac{3}{10}\sqrt{21}<x<\frac{19}{16}-\frac{3}{16}\sqrt{33},\frac{19}{16}+\frac{3}{16}\sqrt{33}<x<\frac{13}{10}+\frac{3}{10}\sqrt{21}$