A Level Further Mathematics 9231 Paper 22, Oct/Nov 2025

Forms length of $C$ correctly (integral setup).

$s=\frac{5}{3}{\int }_0^{\pi }{\text{e}}^{\frac{4}{3}\theta }\text{d}\theta$

Integrates an expression of the form $a{\text{e}}^{b\theta }$.

$=\frac{5}{4}({\text{e}}^{\frac{4}{3}\pi }-1)$

$z^4=8-8\text{i}\sqrt{3}=16{\text{e}}^{-\text{i}\frac{1}{3}\pi }$

Method for finding one root using their $r$ and their $\frac{\theta }{4}$.

$z_1=2{\text{e}}^{-\text{i}\frac{1}{12}\pi }$

$z_2=2{\text{e}}^{\text{i}\frac{5}{12}\pi }$

$z_3=2{\text{e}}^{-\text{i}\frac{7}{12}\pi },z_4=2{\text{e}}^{\text{i}\frac{11}{12}\pi }$

Differentiates $x^{2}y$ correctly: $\frac{\text{d}}{\text{d}x}(x^{2}y)=x^2{y}^{\mathrm{\prime }}+2xy$

Differentiates $(x+y{)}^3$ correctly: $\frac{\text{d}}{\text{d}x}((x+y{)}^3)=3(x+y{)}^2(1+y^{\mathrm{\prime }})$

$y^{\mathrm{\prime }}-4+3(1+y^{\mathrm{\prime }})=0\Rightarrow y^{\mathrm{\prime }}=\frac{1}{4}$

Differentiates $x^2{y}^{\mathrm{\prime }}$ to give $x^2{y}^{\mathrm{\prime }\mathrm{\prime }}+2x{y}^{\mathrm{\prime }}$.

Differentiates $2xy$ to give $2x{y}^{\mathrm{\prime }}+2y$.

Differentiates $3(x+y{)}^2{y}^{\mathrm{\prime }}$ term involving $y^{\mathrm{\prime }\mathrm{\prime }}$: $3(x+y{)}^2{y}^{\mathrm{\prime }\mathrm{\prime }}$.

Differentiates remaining product: $+6(1+y^{\mathrm{\prime }}{)}^2(x+y)$.

Substitutes $(-1,2)$ and $y^{\mathrm{\prime }}=\frac{1}{4}$ into their expression with at least a $y^{\mathrm{\prime }\mathrm{\prime }}$ in their expression.

$y^{\mathrm{\prime }\mathrm{\prime }}=-\frac{99}{32}$

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

Expands $(z+z^{-1}{)}^6$.

Groups $(z^6+z^{-6})+6(z^4+z^{-4})+15(z^2+z^{-2})+20$.

Substitutes $z^n+z^{-n}=2\cos n\theta$.

${\cos }^6\theta =\frac{1}{32}\cos 6\theta +\frac{3}{16}\cos 4\theta +\frac{15}{32}\cos 2\theta +\frac{5}{16}$

Applies (a). Must have the $\frac{1}{2}$ and the correct limits for this mark.

$=\frac{1}{2}{[\frac{1}{192}\sin 6\theta +\frac{3}{64}\sin 4\theta +\frac{15}{64}\sin 2\theta +\frac{5}{16}\theta ]}_0^{\pi }$

$=\frac{5}{32}\pi$

Auxiliary equation: $m^2+4m+3=0\Rightarrow m=-1,-3$.

Complementary function: $[y=]A{\text{e}}^{-x}+B{\text{e}}^{-3x}$.

Particular integral form $y=p\sin x+q\cos x$.

Derivatives: $y^{\mathrm{\prime }}=p\cos x-q\sin x\Rightarrow y^{\mathrm{\prime }\mathrm{\prime }}=-p\sin x-q\cos x$.

Substitutes their particular integral and their derivatives of the correct form and equates coefficients: $-p-4q+3p=0-q+4p+3q=5$.

$p=1q=\frac{1}{2}$

$y=A{\text{e}}^{-x}+B{\text{e}}^{-3x}+\sin x+\frac{1}{2}\cos x$

${\text{e}}^{-x}\to 0$ and for setting up for Pythagoras and applying compound angle formula.

$R=\frac{\sqrt{5}}{2}$

$\varphi =\mathrm{0.464...}$

Applies substitution and ${\sinh }^{2}u+1={\cosh }^{2}u$.

Obtains $\frac{1}{2}\sqrt{2}\int 1\text{d}u$.

$=\frac{1}{2}\sqrt{2}{\sinh }^{-1}(x\sqrt{2})[+C]$

Forms sum of the areas of the rectangles (first line of working).

Simplifies to $\frac{1}{n}(\frac{n}{\sqrt{2\times 1^2+n^2}}+\frac{n}{\sqrt{2\times 2^2+n^2}}+\cdots +\frac{n}{\sqrt{2\times n^2+n^2}})$.

Compares with integral: $<{\int }_0^1\frac{1}{\sqrt{2x^2+1}}\text{d}x$.

Evaluates integral: ${\int }_0^1\frac{1}{\sqrt{2x^2+1}}\text{d}x={[\frac{1}{2}\sqrt{2}{\sinh }^{-1}(x\sqrt{2})]}_0^1=\frac{1}{2}\sqrt{2}\ln (\sqrt{2}+\sqrt{3})$.

$\frac{1}{n}{\sum }_{r=1}^n\frac{n}{\sqrt{2r^2+n^2}}={\sum }_{r=1}^n\frac{1}{\sqrt{2r^2+n^2}}<\frac{1}{2}\sqrt{2}\ln (\sqrt{2}+\sqrt{3})$

Forms the sum of the areas of appropriate rectangles.

Expresses sum correctly as $=\frac{1}{n}(1+\frac{n}{\sqrt{2\times 1^2+n^2}}+\cdots +\frac{n}{\sqrt{2(n-1{)}^2+n^2}})$.

Compares with integral: $⩾{\int }_0^1\frac{1}{\sqrt{2x^2+1}}\text{d}x=\frac{1}{2}\sqrt{2}\ln (\sqrt{2}+\sqrt{3})$.

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

$\frac{1}{2}\sqrt{2}\ln (\sqrt{2}+\sqrt{3})$

Applies product rule and formula from MF19 for the derivative of ${\sin }^{-1}x$.

Obtains $-\frac{1}{2}{x}^2(4-x^2{)}^{-\frac{1}{2}}+\frac{1}{2}(4-x^2{)}^{\frac{1}{2}}+2(4-x^2{)}^{-\frac{1}{2}}$.

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

Method for finding integrating factor.

Integrating factor $=\sqrt{2+x}$.

Sets up correct form on LHS.

Simplifies RHS: $\frac{\text{d}}{\text{d}x}(y\sqrt{2+x})=\sqrt{4-x^2}$.

Integrates RHS using (a).

$y\sqrt{2+x}=\frac{1}{2}x\sqrt{4-x^2}+2{\sin }^{-1}(\frac{1}{2}x)[+C]$

Substitutes initial conditions: $\frac{1}{2}\sqrt{3}=\frac{1}{2}\sqrt{3}+\frac{1}{3}\pi +C$.

$y\sqrt{2+x}=\frac{1}{2}x\sqrt{4-x^2}+2{\sin }^{-1}(\frac{1}{2}x)-\frac{1}{3}\pi$

Finds determinant equal and forms quadratic equation.

$k=-\frac{1}{2}$

$k=-1$

Eliminates a variable to obtain a system of two equations.

Derives contradiction clearly: $\frac{1}{2}x+z=3,x+2z=2$.

Triangular prism.

Eliminates variables to derive contradiction.

Derives contradiction: $\Rightarrow 1=-2$.

Two parallel planes. Other plane not parallel.

Parallel planes not identical.

Finds characteristic equation of $\mathbf{A}$.

$-{\lambda }^3+3{\lambda }^2-8\lambda =0$

Substitutes matrix and makes ${\mathbf{A}}^3$ the subject: ${\mathbf{A}}^3=3{\mathbf{A}}^2-8\mathbf{A}$.

Multiplies by $\mathbf{A}$ and substitutes for ${\mathbf{A}}^3$: ${\mathbf{A}}^4=3{\mathbf{A}}^3-8{\mathbf{A}}^2=3(3{\mathbf{A}}^2-8\mathbf{A})-8{\mathbf{A}}^2$.

${\mathbf{A}}^4={\mathbf{A}}^2-24\mathbf{A}$