A Level Further Mathematics 9231 Paper 21, Oct/Nov 2025

Evaluates determinant.

$=k^2-7$

$k=\pm \sqrt{7}$

Applies chain rule.

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1+\sinh t}{\cosh t}$

Finds $\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)$.

$=\frac{1-\sinh t}{{\cosh }^{2}t}$

Applies chain rule, AG.

$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{1-\sinh t}{{\cosh }^{3}t}$

Finds Maclaurin's series. M1 for $y(0)$, $y^{\mathrm{\prime }}(0)$ and $y^{\mathrm{\prime }\mathrm{\prime }}(0)$ evaluated.

$=1+x+\frac{1}{2}{x}^2$

Uses the product rule to differentiate.

$\frac{\mathrm{d}}{\mathrm{d}x}(x(1+x^2{)}^n)=2n{x}^2(1+x^2{)}^{n-1}+(1+x^2{)}^n$

Uses $x^2=x^2+1-1$.

Integrates both sides using the limits given.

Substitutes limits and rearranges. AG.

$I_{-1}={\left[{\tan }^{-1}x\right]}_0^1=\frac{1}{4}\pi$

Applies reduction formula with $n=-1$.

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

$I_{-2}=\frac{1}{4}+\frac{1}{8}\pi$

Auxiliary equation. Correct type of roots for M1.

Complementary function. Allow ‘ $y=$ ’ missing.

Particular integral and its derivatives.

Substitutes and equates coefficients.

$q=1r=-9$

General solution. FT on CF. Must have ‘ $y=$ ’.

Differentiates using the product rule, CF must be of the correct form.

Uses initial conditions.

$A=9,B=2$

Must have ‘ $y=$ ’.

Forms the sum of the areas of the rectangles. M1 for correct width and at least three complete terms including first and last.

Correct expression for the sum of the areas of the rectangles.

Applies formulae from MF19.

$=\frac{n^2(n+1{)}^2}{12n^4}+\frac{n(n+1)}{2n^2}$

AG.

Forms the sum of the areas of appropriate rectangles.

Correct expression for the sum of areas of rectangles.

Applies formulae from MF19. Accept $L_n=U_n-\frac{4}{3n}$.

At least one line of simplification after subbing in formula (powers of $n$ simplified).

Simplifies $U_n-L_n$ to $\frac{c}{n}$.

AG.

$a,2a,-3a$

${\mathbf{P}}^2=\left(\begin{array}{ccc}{a}^2& 3a& -2a-1\\ 0& 4a^2& a\\ 0& 0& 9a^2\end{array}\right)$

States that $\mathbf{P}$ satisfies its characteristic equation.

Multiplies through by ${\mathbf{P}}^{-1}$.

${\mathbf{P}}^{-1}=\frac{1}{6a^3}\left(\begin{array}{ccc}6a^2& -3a& 2a+1\\ 0& 3a^2& -a\\ 0& 0& -2a^2\end{array}\right)$

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

Applies $\mathbf{A}=\mathbf{P}\mathbf{D}{\mathbf{P}}^{-1}$.

Multiplies two adjacent matrices. For M1, P must correct.

Correct intermediate product of two matrices.

$\left(\begin{array}{ccc}1& \frac{1}{2a}& -\frac{4a+1}{6a^2}\\ 0& 2& -\frac{1}{3a}\\ 0& 0& 3\end{array}\right)$ OE.

Applies ${\tanh }^{-1}x=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)$.

$\frac{\mathrm{d}}{\mathrm{d}x}({\tanh }^{-1}x)=\frac{1}{2}\left(\frac{1-x}{1+x}\right)\frac{(1-x)(1)-(1+x)(-1)}{(1-x{)}^2}$

$=\frac{1}{1-x^2}$ AG.

Divides through by $x$.

Finds integrating factor.

$x^{-1}$

Correct form on LHS and RHS.

Integrates correct function on RHS.

$x^{-1}y=x{\tanh }^{-1}x+\frac{1}{2}\ln (1-x^2)+C$ (‘ $+C$ ’ not required).

‘ $+C$ ’ not required.

Substitutes initial conditions.

$x^{-1}y=x{\tanh }^{-1}x+\frac{1}{2}\ln (1-x^2)-\frac{1}{4}\ln \frac{27}{16}$

$\frac{z^{n+1}-z}{z-1}$

$\frac{z^{n+1}-z}{z-1}=\frac{(\frac{1}{2}{)}^{n+1}\cos (n+1)\theta -\frac{1}{2}\cos \theta +\mathrm{i}((\frac{1}{2}{)}^{n+1}\sin (n+1)\theta -\frac{1}{2}\sin \theta )}{\frac{1}{2}\cos \theta -1+\mathrm{i}\frac{1}{2}\sin \theta }$

Multiplies numerator and denominator by complex conjugate of $z-1$.

Takes imaginary part.

Correct expression for the imaginary part.

Uses $\sin (A-B)=\sin A\cos B-\cos A\sin B$ in numerator, AG.

$\frac{(\frac{1}{2}{)}^{n+2}\sin n\theta -(\frac{1}{2}{)}^{n+1}\sin (n+1)\theta +\frac{1}{2}\sin \theta }{\frac{5}{4}-\cos \theta }$

Differentiates the result from (b).

Applies quotient rule correctly.

Correct derivative expression.

$n{\left(\frac{1}{2}\right)}^n\to 0$ as $n\to \mathrm{\infty }$.

$\frac{\frac{5}{8}\cos \theta -\frac{1}{2}}{(\frac{5}{4}-\cos \theta {)}^2}$