A Level Further Mathematics 9231 Paper 12, May/June 2026

$(4r-3)(12r+3)=48r^2-24r-9$

Substitutes formulae for $\sum r^2$ and $\sum r$.

$=16n^3+12n^2-13n$

Finds partial fractions. Accept equivalent form.

Write enough correct sections of terms so that all that are left in the answer appear plus one pair that cancel.

All necessary complete terms to demonstrate the cancellation. Either two at the beginning and one at the end or one at the beginning and two at the end.

$=\frac{1}{12}-\frac{1}{12(4n+1)}$

${\sum }_{r=1}^{\mathrm{\infty }}\frac{1}{(4r-3)(12r+3)}=\frac{1}{12}$

Checks base case: $3^6+92+89=910$ is divisible by 91.

States inductive hypothesis, making it clear it is an assumption. Assume that $3^{6k}+{92}^k+89$ is divisible by 91 for some positive integer $k$.

Then $3^{6(k+1)}+{92}^{k+1}+89=(728+1)3^{6k}+(91+1){92}^k+89$ or considers difference between consecutive terms.

Separates $3^{6k}+{92}^k+89$ or evaluates difference to $91(8\times 3^{6k}+{92}^k)$.

Connects the divisibility with the hypothesis.

Hence, [by induction,] $3^{6n}+{92}^n+89$ is divisible by 91 for every positive integer $n$. Must have previous three marks and state the result in full.

Stretch parallel to the $x$-axis, scale factor $k$: $\left(\begin{array}{cc}k& 0\\ 0& 1\end{array}\right)$.

Shear, $x$-axis fixed, with $(0,1)$ mapped to $(2,1)$: $\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$.

Correct order: $\mathbf{M}=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\left(\begin{array}{cc}k& 0\\ 0& 1\end{array}\right)$.

$\mathbf{M}=\left(\begin{array}{cc}1\times k+2\times 0& 1\times 0+2\times 1\\ 0\times k+1\times 0& 0\times 0+1\times 1\end{array}\right)=\left(\begin{array}{cc}k& 2\\ 0& 1\end{array}\right)$

Transforms $\left(\begin{array}{c}x\\ y\end{array}\right)$ to $\left(\begin{array}{c}kx+2y\\ y\end{array}\right)$. Must see matrix applied to both coordinates.

Uses $\mathbf{M}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}x\\ y\end{array}\right)$. Must be considering a fixed point.

$y=\frac{1}{2}(1-k)x$

Uses $\det \mathbf{M}=k$ with P and S the correct way round.

$k=\pm 2$

Correct substitution seen used in equation: $y=x^{-2}⟹x=y^{-\frac{1}{2}}$, $5y^{-\frac{3}{2}}-2y^{-\frac{1}{2}}+1=0$.

Squares to eliminate fractional powers: $y^3=(2y-5{)}^2$.

$y^3-4y^2+20y-25=0$

${\alpha }^{-2}+{\beta }^{-2}+{\gamma }^{-2}=4{\alpha }^{-2}{\beta }^{-2}+{\beta }^{-2}{\gamma }^{-2}+{\gamma }^{-2}{\alpha }^{-2}=20$

${\alpha }^{-4}+{\beta }^{-4}+{\gamma }^{-4}=({\alpha }^{-2}+{\beta }^{-2}+{\gamma }^{-2}{)}^2-2({\alpha }^{-2}{\beta }^{-2}+{\beta }^{-2}{\gamma }^{-2}+{\gamma }^{-2}{\alpha }^{-2})$

$=-24$

Using ${\alpha }^{-6}-4{\alpha }^{-4}+20{\alpha }^{-2}-25=0$. Must use $S_0=3$.

$S_{-6}=-101$

Using ${\alpha }^{-8}-4{\alpha }^{-6}+20{\alpha }^{-4}-25{\alpha }^{-2}=0$.

$S_{-8}=176$

Finds direction of one line to another: $\left(\begin{array}{c}-3\\ -2\\ 4\end{array}\right)-\left(\begin{array}{c}-2\\ 4\\ 4\end{array}\right)=\left(\begin{array}{c}-1\\ -6\\ 0\end{array}\right)$.

Find common perpendicular by taking cross product of the two correct direction vectors.

Correct cross product: $\left(\begin{array}{c}-20\\ -5\\ 20\end{array}\right)\sim \left(\begin{array}{c}-4\\ -1\\ 4\end{array}\right)$.

Uses formula for shortest distance using their common perpendicular and direction vector. Scalar product and magnitudes must be evaluated.

Correct answer: $\frac{10}{\sqrt{33}}$. Accept 1.74.

Finds the normal to ${\mathrm{\Pi }}_1$ by taking cross product of two correct direction vectors.

Correct normal to ${\mathrm{\Pi }}_1$: $\left(\begin{array}{c}22\\ 15\\ -3\end{array}\right)$.

Finds the normal to ${\mathrm{\Pi }}_2$ by taking cross product of two correct direction vectors.

Correct normal to ${\mathrm{\Pi }}_2$: $\left(\begin{array}{c}-22\\ 17\\ -8\end{array}\right)$.

Uses dot product of their normal vectors. Scalar product and magnitudes must be evaluated.

$180-\theta ={74.7}^{\circ }$

$x=-1$

Oblique asymptote - must show attempt to find constant term: $y=\frac{(x+1)(x-2)+4}{4x+4}$.

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

Sets $\frac{\mathrm{d}y}{\mathrm{d}x}=0$: $4x^2+8x-12=0$.

A1 for $x$-coordinates: $x=1,-3$.

A1 for $y$-coordinates correctly linked to $x$-coordinates: $(1,\frac{1}{4}),(-3,-\frac{7}{4})$.

Axes and asymptotes labelled. Follow through their two asymptotes from (a) for this mark only.

Left branch correct and also all approaches to the asymptotes.

Right branch correct including a smooth minimum in the first quadrant.

Must be fully correct, including vertical asymptote.

Finds critical points, award M1 for each case. Must be using equations.

All four correct, exact answers: $x=-5,-2$ and $x=\frac{1}{2}(9-\sqrt{105}),x=\frac{1}{2}(9+\sqrt{105})$.

$-5<x<-2\frac{1}{2}(9-\sqrt{105})<x<\frac{1}{2}(9+\sqrt{105})$

Uses $y=r\sin \theta$.

Differentiates using product rule.

or quotient rule. Must be using correct function.

Puts derivative = 0 and rearranges.

AG. Intermediate steps must be shown.

Shows sign change. Values calculated to at least one decimal place: $\cos (0.5)-\cdots =0.235$ and $\cos (0.6)-\cdots =-0.379$.

Initial line and pole clear. Curve in first quadrant through $(\sqrt{a},0)$ and $O$. Condone pole not labelled but must be clearly identifiable. $(\sqrt{a},0)$ not needed.

Correct shape (gradient at end points and approximately symmetrical). Maximum point should lie between 40% and 60% of width of the arc.

Using correct formula and correct limits: $\frac{1}{2}{\int }_0^{\frac{1}{4}\pi }{r}^2\mathrm{d}\theta =\frac{a}{2}{\int }_0^{\frac{1}{4}\pi }\frac{1}{1+\tan 2\theta }\mathrm{d}\theta$.

Applies given substitution: $v=\tan 2\theta ⟹\frac{\mathrm{d}v}{\mathrm{d}\theta }=2{\sec }^{2}2\theta$. Condone $\frac{\mathrm{d}v}{\mathrm{d}\theta }={\sec }^{2}2\theta$.

Clearly applies ${\sec }^{2}2\theta =1+{\tan }^{2}2\theta$. Allow ${\sec }^{2}2\theta =1+v^2$ written explicitly.

Some working must be shown, including calculations supporting change of limits. AG.

Partial fractions of correct form: $\frac{1}{(1+v)(1+v^2)}=\frac{1}{2}(\frac{1}{v+1}-\frac{v-1}{v^2+1})$.

and separates $\frac{-v+1}{v^2+1}$: $=\frac{1}{2}(\frac{1}{v+1}-\frac{v}{v^2+1}+\frac{1}{v^2+1})$.

Applies all three formulae from MF19 to a three-term expression from above. Follow through their constants from line above.

Correct integration: $\frac{a}{8}{[\ln (v+1)-\frac{1}{2}\ln (v^2+1)+{\tan }^{-1}v]}_0^{\mathrm{\infty }}$.

The four previous marks must have been awarded and logarithms must be combined for this mark to be awarded: $\frac{a}{8}{[\ln \left(\frac{v+1}{\sqrt{v^2+1}}\right)+{\tan }^{-1}v]}_0^{\mathrm{\infty }}=\frac{1}{16}\pi a$.