A Level Further Mathematics 9231 Paper 12, Oct/Nov 2025

Substitutes formulae.

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

Factorises to $\frac{1}{4}n(n+1)(n-1)(n+2)$

Finds partial fractions. M mark is for an equation that would lead to values for A, B and C.

$\frac{r+3}{r^3-r}=\frac{2}{r-1}-\frac{3}{r}+\frac{1}{r+1}$

Shows five complete terms including first two and last two.

Correct list of terms showing cancellation structure

$\frac{3}{2}-\frac{2}{n}+\frac{1}{n+1}$

$\frac{3}{2}$

$b=-(\alpha +\beta +\gamma )=-2$

Uses formula for sum of squares: $3=(-2{)}^2-2c$

$c=\frac{1}{2}$

Uses original equation or formula for sum of cubes: $S_3+b{S}_2+c{S}_1+3d=0$

$S_3=-3b-2c-3d=5-3d$

Uses original equation: $S_4+b{S}_3+c{S}_2+d{S}_1=0$

$5+(-2)(5-3d)+(\frac{1}{2})(3)+2d=0⟹8d-\frac{7}{2}=0$

$d=\frac{7}{16}$

States base case: $u_1<5$ (given)

States inductive hypothesis: Assume that $u_k<5$ for some positive integer $k$.

Considers $5-u_{k+1}$, puts over common denominator: $5-\frac{6u_k+5}{u_k+2}=\frac{5u_k+10-6u_k-5}{u_k+2}$

$=\frac{5-u_k}{u_k+2}>0$.

Hence, by induction, $u_n<5$ for all positive integers $n$.

Considers $u_{n+1}-u_n$ or uses $5-u_{n+1}=\frac{5-u_n}{u_n+2}$.

$=\frac{-u_n^2+4u_n+5}{u_n+2}$

$=\frac{(5-u_n)(1+u_n)}{u_n+2}$. So $u_n<5⟹u_{n+1}-u_n>0$

Stretch

parallel to the x-axis, scale factor k.

Enlargement,

centre the origin, scale factor k.

Multiplies two matrices correctly.

$\left(\begin{array}{cc}0& m\\ -k& m\\ k& m\end{array}\right)$

$\left(\begin{array}{ccc}m& m& 2m\\ -2k+m& k+m& -k+2m\\ 2k+m& -k+m& k+2m\end{array}\right)$

Finds determinant.

$m\mid \begin{array}{cc}k+m& -k+2m\\ -k+m& k+2m\end{array}\mid -m\mid \begin{array}{cc}-2k+m& -k+2m\\ 2k+m& k+2m\end{array}\mid +2m\mid \begin{array}{cc}-2k+m& k+m\\ 2k+m& -k+m\end{array}\mid$

$=m(k+m)(k+2m)-m(-k+m)(-k+2m)-m(-2k+m)(k+2m)+m(2k+m)(-k+2m)+2m(-2k+m)(-k+m)-2m(2k+m)(k+m)=6m^{2}k+6m^{2}k-12m^{2}k=0$

Initial line drawn and correct shape.

Maximum distance of C from the pole is 1.

Uses $\frac{1}{2}\int r^2\mathrm{d}\theta$ with correct limits.

Integrates a correct function to get $-\frac{1}{2}[\ln \cos 2\theta ]$ condoning the lack of limits and factor of $\frac{1}{2}$

$-\frac{1}{4}[\ln \cos 2\theta {]}_0^{\frac{1}{8}\pi }$

$-\frac{1}{4}\ln \frac{\sqrt{2}}{2}=\frac{1}{8}\ln 2$

Uses relevant identity for $\tan 2\theta$.

For the M mark uses either $r=\sqrt{x^2+y^2}$ or $\tan \theta =\frac{y}{x}$.

Use both $r=\sqrt{x^2+y^2}$ and $\tan \theta =\frac{y}{x}$ correctly.

$(x^2+y^2)(x^2-y^2)=2xy⟹x^4-y^4=2xy⟹x^4-2xy-y^4=0$.

Area of triangle minus their (b): $\frac{1}{2}(\cos \frac{1}{8}\pi )(\sin \frac{1}{8}\pi )-\frac{1}{8}\ln 2$

$\frac{1}{8}\sqrt{2}-\frac{1}{8}\ln 2$

$\frac{1}{\sqrt{1^2+3^2+2^2}}$

$=\frac{1}{\sqrt{14}}$

Finds normal to OAB. Allow 1 error for the M mark to be awarded.

$\left(\begin{array}{c}1\\ 4\\ 2\end{array}\right)$

Takes dot product of normal vectors: $\left(\begin{array}{c}1\\ 4\\ 2\end{array}\right)\cdot \left(\begin{array}{c}1\\ 3\\ 2\end{array}\right)=\sqrt{21}\sqrt{14}\cos \alpha ⟹\cos \alpha =\frac{17}{\sqrt{21}\sqrt{14}}$

${7.5}^{\circ }$

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

$\overrightarrow{PQ}=\left(\begin{array}{c}2\mu -2\lambda \\ -1+\mu +\lambda \\ 2-3\mu +\lambda \end{array}\right)$

Uses that dot product of $\overrightarrow{PQ}$ with line direction is zero, or, alternatively, $\overrightarrow{PQ}$ is a multiple of the common perpendicular $\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$.

Deduces one equation: $6\mu -6\lambda =1$

Deduces second equation: $14\mu -6\lambda =7$

Solves for $\mu$ and substitutes into $\overrightarrow{OQ}$ and $\overrightarrow{PQ}$ or $\lambda =\frac{7}{12}⟹\overrightarrow{OP}=\frac{7}{12}\left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)$

$\mu =\frac{3}{4}⟹\overrightarrow{OQ}=\frac{1}{4}\left(\begin{array}{c}6\\ -1\\ -1\end{array}\right)$

$\mathbf{r}=\frac{1}{4}\left(\begin{array}{c}6\\ -1\\ -1\end{array}\right)+k\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$

Vertical asymptote: $x=-1$

Oblique asymptote: $y=\frac{x(x+1)+1}{x+1}$

$y=x$

Sets $\frac{\mathrm{d}y}{\mathrm{d}x}=0$: $\frac{x^2+2x}{(x+1{)}^2}=0$

$(0,1)$

$(-2,-3)$

Axes and asymptotes labelled.

Left branch correct with all approaches to the asymptotes correct.

Right branch correct with a smooth minimum point.

Symmetrical about $x=0$ and follow through from part (c) right branch

Correct shape of a curve and smooth at $(0,1)$ from a correct graph in part (c)

Finds critical point: $\frac{x^2+x+1}{x+1}=3⟹x^2-2x-2=0$

$x=1+\sqrt{3}$ (for selecting just the positive value of x)

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