A Level Further Mathematics 9231 Paper 11, May/June 2026

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

$48(\frac{1}{6}n(n+1)(2n+1))-24(\frac{1}{2}n(n+1))-9n$

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

$\frac{1}{(4r-3)(12r+3)}=\frac{1}{12}(\frac{1}{4r-3}-\frac{1}{4r+1})$

${\sum }_{r=1}^n\frac{1}{(4r-3)(12r+3)}=\frac{1}{12}(1-\frac{1}{5})+\frac{1}{12}(\frac{1}{5}-\frac{1}{9})+\frac{1}{12}(\frac{1}{9}-\frac{1}{13})$

$+\cdots +\frac{1}{12}(\frac{1}{4n-3}-\frac{1}{4n+1})$

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

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

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 $3^{6(k+1)}+{92}^{k+1}+89-(3^{6k}+{92}^k+89)$.

Separates $3^{6k}+{92}^k+89$: $=728\times 3^{6k}+91\times {92}^k+(3^{6k}+{92}^k+89)$ or $=(728)3^{6k}+(91){92}^k=91(8\times 3^{6k}+{92}^k)$.

Connects the divisibility with the hypothesis.

Hence, [by induction,] $3^{6n}+{92}^n+89$ is divisible by 91 for every positive integer $n$.

$\left(\begin{array}{cc}k& 0\\ 0& 1\end{array}\right)$

$\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$

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

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

$\left(\begin{array}{cc}k& 2\\ 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}kx+2y\\ y\end{array}\right)$

$kx+2y=x$

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

Area of P = $\mathrm{\mid }\det \mathbf{M}\mathrm{\mid }\times \text{Area of S}⟹1=\frac{1}{2}\mathrm{\mid }k\mathrm{\mid }$

$k=\pm 2$

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

Squares to eliminate fractional powers: $y^3=(2y-5{)}^2$ or squares to eliminate odd powers of $x$: $(5x^2-2{)}^2{x}^2=1$.

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

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

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

$=-24$

$S_{-6}=4S_{-4}-20S_{-2}+75$

$S_{-6}=-101$

$S_{-8}=4S_{-6}-20S_{-4}+25S_{-2}$

$S_{-8}=176$

Finds direction of one line to another: $\left(\begin{array}{c}-3\\ -2\\ 4\end{array}\right)-\left(\begin{array}{c}-2\\ 4\\ 4\end{array}\right)=\left(\begin{array}{c}-1\\ -6\\ 0\end{array}\right)$ or (M1 for Alt: Vector joining P on one line to Q on the other line, perpendicular to one line direction).

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

Correct perpendicular vector $\left(\begin{array}{c}-20\\ -5\\ 20\end{array}\right)\sim \left(\begin{array}{c}-4\\ -1\\ 4\end{array}\right)$ or Solves equations from using perpendicular to both lines to find $\lambda$ and $\mu$: $\lambda =\frac{101}{165},\mu =\frac{139}{165}$.

Uses formula for shortest distance using their common perpendicular and direction vector: $\frac{1}{\sqrt{33}}\mid \left(\begin{array}{c}-1\\ -6\\ 0\end{array}\right)\cdot \left(\begin{array}{c}-4\\ -1\\ 4\end{array}\right)\mid =\frac{10}{\sqrt{33}}$ or Finds length of PQ: $\sqrt{{(-\frac{40}{33})}^2+{(-\frac{10}{33})}^2+{\left(\frac{40}{33}\right)}^2}$.

$=\frac{10}{\sqrt{33}}$

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

$\left(\begin{array}{c}22\\ 15\\ -3\end{array}\right)$ Correct.

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

$\left(\begin{array}{c}-22\\ 17\\ -8\end{array}\right)$ Correct.

Uses dot product of their normal vectors. Scalar product and magnitudes must be evaluated: $\left(\begin{array}{c}22\\ 15\\ -3\end{array}\right)\cdot \left(\begin{array}{c}-22\\ 17\\ -8\end{array}\right)=\sqrt{718}\sqrt{837}\cos \theta ⟹\cos \theta =\frac{-205}{\sqrt{718}\sqrt{837}}$

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

$x=-1$

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

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

Sets $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4x^2+8x-12}{(4x+4{)}^2}=0⟹4x^2+8x-12=0$

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.

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: $\frac{x^2-x+2}{4x+4}=-2⟹x^2+7x+10=0$ and $\frac{x^2-x+2}{4x+4}=2⟹x^2-9x-6=0$. Must be using equations.

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

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

Uses $y=r\sin \theta$

Differentiates using product rule or quotient rule.

Correct derivative expression. Must be using correct function.

Puts derivative $=0$ and rearranges.

$\cos \theta -\frac{{\sec }^{2}2\theta \sin \theta }{1+\tan 2\theta }=\cos \theta -\frac{\sin \theta }{\cos 2\theta (\sin 2\theta +\cos 2\theta )}=0$

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

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.

$\frac{1}{2}{\int }_0^{\frac{1}{4}\pi }{r}^2\mathrm{d}\theta =\frac{a}{2}{\int }_0^{\frac{1}{4}\pi }\frac{1}{1+\tan 2\theta }\mathrm{d}\theta$

$v=\tan 2\theta ⟹\frac{\mathrm{d}v}{\mathrm{d}\theta }=2{\sec }^{2}2\theta$

$\frac{a}{2}{\int }_0^{\frac{1}{4}\pi }\frac{1}{1+\tan 2\theta }\mathrm{d}\theta =\frac{a}{4}{\int }_0^{\mathrm{\infty }}\frac{1}{1+v}\frac{1}{{\sec }^{2}2\theta }\mathrm{d}v$

$=\frac{a}{4}{\int }_0^{\mathrm{\infty }}\frac{1}{1+v}\frac{1}{1+{\tan }^{2}2\theta }\mathrm{d}v=\frac{a}{4}{\int }_0^{\mathrm{\infty }}\frac{1}{(1+v)(1+v^2)}\mathrm{d}v$

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

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

Applies all three formulae from MF19 to a three-term expression from above: $\frac{a}{8}{[\ln (v+1)-\frac{1}{2}\ln (v^2+1)+{\tan }^{-1}v]}_0^{\mathrm{\infty }}$

Correct integration.

$\frac{a}{8}{[\ln \left(\frac{v+1}{\sqrt{v^2+1}}\right)+{\tan }^{-1}v]}_0^{\mathrm{\infty }}=\frac{1}{16}\pi a$