A Level Further Mathematics 9231 Paper 21, 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.

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

Finds second root. FT on modulus. Withhold one answer mark for omission of i.

Finds third root. FT on modulus. Withhold one answer mark for omission of i.

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

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

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

AG, withhold for no evidence of substitution or incorrect use of limits.

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.

$(z-z^{-1}{)}^5=(z^5-z^{-5})-5(z^3-z^{-3})+10(z-z^{-1})$. No clear grouping is M1 A0.

Substitutes $z^n-z^{-n}=2\text{i}\sin n\theta$ on RHS. Omission of i is M0.

$2^5\text{i}{\sin }^5\theta =2\text{i}\sin 5\theta -5(2\text{i}\sin 3\theta )+10(2\text{i}\sin \theta )$. FT on signs of coefficients only. Substituting $z^n-z^{-n}=\text{i}\sin n\theta$ on RHS is M1 A0.

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.

Withold this mark if working with LHS and RHS simultaneously.

Differentiates implicitly (or differentiates ${\cosh }^{-1}\left(\frac{1}{t}\right)$ / logarithmic form).

Implicit derivative correct.

Uses identity from (a) or relevant hyperbolic identity / applies chain rule.

Simplifies to AG.

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

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

Uses chain rule.

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

Applies product rule. Might see $\frac{2t^2-1}{t\sqrt{1-t^2}}-2t{\text{sech}}^{-1}t$.

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

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.

${\sinh }^{-1}\left(\frac{x+5}{6}\right)+C$

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.

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

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

AG.

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

Finds first eigenvector.

Finds second eigenvector.

Finds third eigenvector.

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

Correct $\mathbf{P}$ and $\mathbf{D}$ (or correctly matched permutations of columns).

Expands $(\mathbf{A}-2\mathbf{I}{)}^3$ / Finds characteristic equation of $\mathbf{A}-2\mathbf{I}$.

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

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