A Level Further Mathematics 9231 Paper 22, May/June 2026

Finds first derivative. Accept $\frac{1}{x+1+\sqrt{(x+1{)}^2+1}}(1+\frac{2x+2}{2\sqrt{(x+1{)}^2+1}})$ or implicit $y^{\mathrm{\prime }}\cosh y=1$.

Finds second derivative. Allow not simplified or implicit $y^{\mathrm{\prime }\mathrm{\prime }}\cosh y+(y^{\mathrm{\prime }}{)}^2\sinh y=0$.

${\sinh }^{-1}1=\ln (1+\sqrt{2})$

Evaluates their derivatives at zero only. Must have attempted both derivatives.

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

Applies integration by parts. Allow limits missing. M1 for correct trigonometric functions and powers of $(1+t)$.

$I_n={[-\frac{1}{\pi }(1+t{)}^n\cos \pi t]}_0^1+\frac{n}{\pi }{\int }_0^1(1+t{)}^{n-1}\cos \pi t\mathrm{d}t$

Applies integration by parts again. M1 for correct trigonometric functions and powers of $(1+t)$.

${\pi }^2{I}_n=\pi (2^n+1)-n(n-1)I_{n-2}$

Finds $I_0$.

Uses reduction formula from (a).

$I_2=\frac{5}{\pi }-\frac{4}{{\pi }^3}$

$u=\cosh x⟹\mathrm{d}u=\sinh x\mathrm{d}x$

Applies correct identity ${\sinh }^{2}u={\cosh }^{2}u-1$ to find integral in terms of $u$.

$\int (u^2-1)\mathrm{d}u$

$=\frac{1}{3}{\cosh }^{3}x-\cosh x(+C)$

Finds integrating factor.

Integrating factor $=\sinh x$.

Correct form on LHS and attempt to integrate RHS.

Integrates RHS using their part (a). For M1, their answer to (a) must be of the correct form $a{\cosh }^{3}x+b\cosh x$ with $a,b$ non-zero.

$y\sinh x=\frac{1}{3}{\cosh }^{3}x-\cosh x+C$

Substitutes initial conditions. M0 if expression incorrectly manipulated before substitution. E.g. not all terms divided by $\sinh x$.

$y=\mathrm{cosech}x(\frac{1}{3}{\cosh }^{3}x-\cosh x+\frac{10}{81})$

Differentiates $(x-y{)}^5$ correctly.

Differentiates $\ln (xy)$ correctly.

Substitutes $(-1,-2)$ into $5(x-y{)}^4(1-y^{\mathrm{\prime }})+y^{\mathrm{\prime }}{y}^{-1}+x^{-1}=0$, AG. CWO. Might see $y^{\mathrm{\prime }}(y^{-1}-5(x-y{)}^4)=-5(x-y{)}^4-x^{-1}$.

Differentiates $5(x-y{)}^4(1-y^{\mathrm{\prime }})$. (1st mark)

Differentiates $5(x-y{)}^4(1-y^{\mathrm{\prime }})$. (2nd mark)

Differentiates $y^{\mathrm{\prime }}{y}^{-1}+x^{-1}$. Might see $\frac{x{y}^{\mathrm{\prime }}+y}{xy}$ differentiated using product or quotient rules. E.g. $\frac{xy(x{y}^{\mathrm{\prime }\mathrm{\prime }}+2y^{\mathrm{\prime }})-(x{y}^{\mathrm{\prime }}+y{)}^2}{(xy{)}^2}$ or $-(xy{)}^{-2}(x{y}^{\mathrm{\prime }}+y{)}^2+(xy{)}^{-1}(x{y}^{\mathrm{\prime }\mathrm{\prime }}+2y^{\mathrm{\prime }})$.

Substitutes $(-1,-2)$ and $y^{\mathrm{\prime }}=\frac{8}{11}$.

$-\frac{11}{2}{y}^{\mathrm{\prime }\mathrm{\prime }}+\frac{43}{121}=0⟹y^{\mathrm{\prime }\mathrm{\prime }}=\frac{86}{1331}$

Forms the sum of the areas of the rectangles. M1 for sum of $n$ complete terms.

A1 for at least 3 correct terms including first and last. May be written in summation form.

Applies formula for $\sum r^3$ from MF19.

$=\frac{11}{40}{(1+\frac{1}{n})}^2+\frac{1}{10}$

Forms the sum of the areas of appropriate rectangles. M1 for sum of $n$ complete terms.

A1 for at least 3 correct terms including first and last.

Applies formulae from MF19. Allow $U_n+\frac{1}{10n}-\frac{12}{10n}$.

$=\frac{11}{40}{(1-\frac{1}{n})}^2+\frac{1}{10}$

Their $U_n-L_n$ simplifies to $\frac{c}{n}$. Accept $\frac{11}{40}{(1+\frac{1}{n})}^2+\frac{1}{10}-\frac{11}{40}{(1-\frac{1}{n})}^2-\frac{1}{10}$.

$\frac{11}{10n}\to 0\text{ as }n\to \mathrm{\infty }$

Auxiliary equation. M1 for correct type of roots $a\pm b\mathrm{i}$.

Complementary function. Allow '$y=$' missing.

Particular integral and its derivatives.

Substitutes and equates coefficients.

$q=-1r=-5$

General solution. Must have '$y=$'. FT on CF.

Differentiates. General solution must be of the correct form. Must use product rule.

Uses initial conditions.

Both $A$ and $B$ correct ($A=5,B=15$).

$y={\mathrm{e}}^{-\frac{2}{5}x}(5\cos \frac{1}{5}x+15\sin \frac{1}{5}x)+x^2-x-5$

$\frac{z-z^{n+1}}{1-z}$ or $\frac{z^{n+1}-z}{z-1}$ or $\frac{z(z^n-1)}{z-1}$

States one root of $z^7=1,z\mathrm{\ne }1$.

Five other roots. No repeats or incorrect roots, $z=1$ must be omitted.

Applies sum to infinity and substitutes for $z$.

Expect $\frac{{\mathrm{e}}^{\mathrm{i}\theta }}{\mathrm{e}-{\mathrm{e}}^{\mathrm{i}\theta }}$ or $\frac{{\mathrm{e}}^{-1+\mathrm{i}\theta }}{1-{\mathrm{e}}^{-1+\mathrm{i}\theta }}$.

Rationalises denominator.

Applies ${\mathrm{e}}^{\mathrm{i}\theta }+{\mathrm{e}}^{-\mathrm{i}\theta }=2\cos \theta$ or ${\sin }^2\theta +{\cos }^2\theta =1$.

Correct simplified expression before taking real part.

Takes the real part.

$\mathrm{Re}\left\{{\sum }_{m=1}^{\mathrm{\infty }}{z}^m\right\}=\frac{\mathrm{e}\cos \theta -1}{{\mathrm{e}}^2-2\mathrm{e}\cos \theta +1}$

Finds determinant in terms of $a$ and $b$.

$4a+ab-b-2$

$a=\frac{b+2}{b+4}[\text{since }b\mathrm{\ne }-4]$

Finds characteristic equation.

Replaces $\lambda$ with $\mathbf{A}$ and multiplies through by $\mathbf{A}$.

Substitutes ${\mathbf{A}}^3=2{\mathbf{A}}^2+\mathbf{A}-2\mathbf{I}$.

${\mathbf{A}}^4=5{\mathbf{A}}^2-4\mathbf{I}$

Upper diagonal matrix or characteristic equation to state eigenvalues of $\mathbf{A}$ are $-1,1$ and $2$.

Uses vector product (or equations) to find corresponding eigenvectors. M0 if all three vectors are zero vectors.

First eigenvector correct.

Second eigenvector correct.

Third eigenvector correct.

Forms matrix $\mathbf{P}$ and matrix $\mathbf{D}$. A column of zeros or repeated column in $\mathbf{P}$ is M0.

$\mathbf{P}=\left(\begin{array}{ccc}1& 1& 1\\ 0& -1& -1\\ 0& 0& 1\end{array}\right)$ and $\mathbf{D}=\left(\begin{array}{ccc}-1& 0& 0\\ 0& 1& 0\\ 0& 0& 128\end{array}\right)$ (or correctly matched permutations of columns).