A Level Further Mathematics 9231 Paper 22, May/June 2025

Modulus and argument of $z^3$ (does not need to be in exponential form).

Finds one root. Cube root must be taken for M1.

$z_1=3\times 2^{\frac{1}{6}}{\text{e}}^{-\frac{1}{12}\text{i}\pi }$

$z_2=3\times 2^{\frac{1}{6}}{\text{e}}^{\frac{7}{12}\text{i}\pi }$

$z_3=3\times 2^{\frac{1}{6}}{\text{e}}^{-\frac{9}{12}\text{i}\pi }$

Applies integration by parts. Appropriate choice of parts for M1.

$I_n={[(1-x{)}^n\cosh x]}_0^1+{\int }_0^{1}n(1-x{)}^{n-1}\cosh x\text{ d}x$

Applies integration by parts again. Appropriate choice of parts for M1.

$I_n=-1+n(n-1)I_{n-2}$

Finds $I_0$.

Uses reduction formula from (a).

$I_2=2\cosh 1-3$

Use of $z-z^{-1}=2\text{i}\sin \theta$

Expands and groups. No clear grouping is M1 A0.

$(z-z^{-1}{)}^5=(z^5-z^{-5})-5(z^3-z^{-3})+10(z-z^{-1})$

Substitutes $z^n-z^{-n}=2\text{i}\sin n\theta$ on RHS.

$2^5\text{i}{\sin }^5\theta =2\text{i}\sin 5\theta -5(2\text{i}\sin 3\theta )+10(2\text{i}\sin \theta )$

Cancelation of i justified.

Forms sum of the areas of the rectangles.

Compares with integral. Wrong comparison (missing/wrong inequality or wrong limits) is A0.

Correct integration. Attempt at integration by parts is M0 A0.

${\int }_1^n\frac{1}{\sqrt{x}}{\text{e}}^{\sqrt{x}}\text{ d}x={\left[2{\text{e}}^{\sqrt{x}}\right]}_1^n$

AG, must have obtained all previous marks.

Forms the sum of the areas of appropriate rectangles.

Compares with integral.

Adds term from $r=1$.

${\sum }_{r=1}^n\frac{1}{\sqrt{r}}{\text{e}}^{\sqrt{r}}>{\left[2{\text{e}}^{\sqrt{x}}\right]}_1^n+\text{e}=2{\text{e}}^{\sqrt{n}}-\text{e}$

Auxiliary equation. M1 for correct type (complex) of roots.

Complimentary function. Allow 'x =' missing. Using x instead of t is A0.

Particular integral and its derivatives.

Substitutes and equates coefficients.

$k=\frac{1}{9}$

General solution. FT on CF and their value of k. Must have 'x ='.

Using product rule.

Substitutes initial conditions and forms simultaneous equations.

$A=-\frac{1}{9},B=\frac{1}{3\sqrt{15}}$

Must have 'x ='.

Writes in exponential from.

Writes over common denominator.

$={\text{sech}}^{2}u$

Differentiates implicitly.

$-\tanh u\text{sech}u\frac{\text{d}u}{\text{d}t}=1$

Uses identity from (a), AG.

$\frac{\text{d}u}{\text{d}t}=-\frac{1}{t\sqrt{1-t^2}}$

$\frac{\text{d}y}{\text{d}t}={\text{sech}}^{-1}t-\frac{1}{\sqrt{1-t^2}}$

$\frac{\text{d}x}{\text{d}t}=\frac{1}{1-t^2}$

Uses chain rule, AG.

$\frac{\text{d}y}{\text{d}x}=-\sqrt{1-t^2}+(1-t^2){\text{sech}}^{-1}t$

Applies product rule.

$\frac{\text{d}}{\text{d}t}\left(\frac{\text{d}y}{\text{d}x}\right)=\frac{t}{\sqrt{1-t^2}}-\frac{1-t^2}{t\sqrt{1-t^2}}-2t{\text{sech}}^{-1}t$

Uses chain rule.

$\frac{{\text{d}}^{2}y}{\text{d}{x}^2}=t\sqrt{1-t^2}-\frac{(1-t^2{)}^{\frac{3}{2}}}{t}-2t(1-t^2){\text{sech}}^{-1}t$

Finds integrating factor using $\int \frac{f^{\mathrm{\prime }}(x)}{f(x)}\text{ d}x=\ln \mathrm{\mid }f(x)\mathrm{\mid }$.

$\int \frac{x+5}{x^2+10x+61}\text{ d}x=\frac{1}{2}\ln (x^2+10x+61)$

${\text{e}}^{-\frac{1}{2}\ln (x^2+10x+61)}=\frac{1}{\sqrt{x^2+10x+61}}$

Correct form on LHS using their integrating factor.

Completes the square.

Integrates RHS. Attempting to integrate by parts is M0.

$\int \frac{1}{\sqrt{(x+5{)}^2+6^2}}\text{ d}x={\sinh }^{-1}\left(\frac{x+5}{6}\right)+C$

Substitutes initial conditions.

$C=-{\sinh }^{-1}\left(\frac{8}{6}\right)=-\ln (\frac{8}{6}+\frac{5}{3})=-\ln 3$

$\frac{y}{\sqrt{x^2+10x+61}}={\sinh }^{-1}\left(\frac{x+5}{6}\right)-{\sinh }^{-1}\left(\frac{4}{3}\right)$

Multiples by ${\mathbf{A}}^2$ on LHS. Trying to invert an eigenvector or missing eigenvector is M0.

$⟹{\mathbf{A}}^3\mathbf{e}=\lambda \mathbf{A}(\lambda \mathbf{e})={\lambda }^2(\mathbf{A}\mathbf{e})={\lambda }^3\mathbf{e}$

Forms $\det (\mathbf{A}-\lambda \mathbf{I})=0$. Do not accept checking each given value is an eigenvalue. Do not accept row operations on $\mathbf{A}$.

AG.

Uses vector product (or equations) of rows of $\mathbf{A}-\lambda \mathbf{I}$ to find corresponding eigenvectors.

Eigenvector for $\lambda =-1$: $\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$

Eigenvector for $\lambda =1$: $\left(\begin{array}{c}5\\ 2\\ 1\end{array}\right)$

Eigenvector for $\lambda =5$: $\left(\begin{array}{c}2\\ 0\\ 3\end{array}\right)$

Forms matrix $\mathbf{P}$ and matrix $\mathbf{D}$.

$\mathbf{P}=\left(\begin{array}{ccc}1& 5& 2\\ 0& 2& 0\\ 0& 1& 3\end{array}\right)$ and $\mathbf{D}=\left(\begin{array}{ccc}-3& 0& 0\\ 0& -1& 0\\ 0& 0& 3\end{array}\right)$

Expands $(\mathbf{A}-2\mathbf{I}{)}^3={\mathbf{A}}^3-6{\mathbf{A}}^2+12\mathbf{A}-8\mathbf{I}$ (or finds characteristic equation of $\mathbf{A}-2\mathbf{I}$).

Substitutes ${\mathbf{A}}^3=5{\mathbf{A}}^2+\mathbf{A}-5\mathbf{I}$ (or substitutes $\mathbf{A}-2\mathbf{I}$ and makes $(\mathbf{A}-2\mathbf{I}{)}^3$ the subject).

$=-{\mathbf{A}}^2+13\mathbf{A}-13\mathbf{I}$