A Level Further Mathematics 9231 Paper 23, May/June 2025

Finds first derivative.

Finds second derivative.

Evaluates derivatives at zero. Must have attempted second derivative using the product rule.

Forms Maclaurin expansion.

Correct expansion: ${\mathrm{e}}^{(x+2{)}^{-1}}={\mathrm{e}}^{\frac{1}{2}}-\frac{1}{4}{\mathrm{e}}^{\frac{1}{2}}x+\frac{5}{32}{\mathrm{e}}^{\frac{1}{2}}{x}^2$

States exponential forms for $\tanh t$ and $\mathrm{sech}t$.

Writes in a single fraction over common denominator, AG.

Completes proof correctly to reach 1.

$\frac{\mathrm{d}x}{\mathrm{d}t}=\tanh t$

Applies $1+{\sinh }^{2}t={\cosh }^{2}t$.

$\frac{\mathrm{d}y}{\mathrm{d}t}=\mathrm{sech}t$

Substitutes their derivatives into a correct formula for arc length eg. $\int \sqrt{{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)}^2+{\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)}^2}\mathrm{d}t$

$=1$

Differentiates $y^2$ correctly.

Applies chain rule to differentiate ${\sinh }^{-1}(xy)$.

Differentiates ${\sinh }^{-1}(xy)$ correctly: $-3\frac{x{y}^{\mathrm{\prime }}+y}{\sqrt{1+x^2{y}^2}}$.

Substitutes $(4,\frac{1}{3})$ and shows at least one intermediate step, AG.

Differentiates $18y{y}^{\mathrm{\prime }}$.

Applies quotient/product rule to differentiate $-3\frac{x{y}^{\mathrm{\prime }}+y}{\sqrt{1+x^2{y}^2}}$.

Correct derivative of $-3\frac{x{y}^{\mathrm{\prime }}+y}{\sqrt{1+x^2{y}^2}}$.

Substitutes $(4,\frac{1}{3})$ and $y^{\mathrm{\prime }}=-\frac{1}{2}$.

$-\frac{6}{5}{y}^{\mathrm{\prime }\mathrm{\prime }}+\frac{87}{10}=0\Rightarrow y^{\mathrm{\prime }\mathrm{\prime }}=\frac{29}{4}$

Auxiliary equation. M1 for correct type of roots.

Complementary function. Allow "$x=$" missing. Using $x$ instead of $t$ is A0.

Particular integral and its derivatives.

Substitutes and equates coefficients.

$p=-1$, $q=-\frac{3}{2}$, $r=-\frac{5}{4}$

General solution. FT on CF and their values of $p$, $q$ and $r$. Must have "$x=$".

Differentiates.

Uses initial conditions.

$A=\frac{4}{3},B=-\frac{1}{12}$

Must have "$x=$". $x=\frac{4}{3}{\mathrm{e}}^t-\frac{1}{12}{\mathrm{e}}^{-2t}-t^2-\frac{3}{2}t-\frac{5}{4}$

Uses binomial theorem.

$\cos 5\theta =c^5-10s^2{c}^3+5s^{4}c$

Uses $s^2=1-c^2$.

$=16c^5-20c^3+5c$

Divides simplified numerator and denominator by $c^5$.

AG.

Relates with equation in part (a). Can be implied by M1 mark.

Solves $\cos 5\theta =\frac{\sqrt{3}}{2}$.

Gives one correct solution.

Gives four other solutions. Allow different values of $q$ as long as all five solutions are found. -1.35, -1.15, 1.01, 2.46, 4.81

Forms sum of the areas of the rectangles.

Correct sum expression.

Evaluates integral.

CAO: $\frac{1}{4}\pi$

AG. Clear and convincing comparison with the integral.

Forms the sum of the areas of appropriate rectangles. M0 for more than $n$ rectangles.

Correct sum of rectangles.

Compares with integral. Correctly adjust for $0^{\text{th}}$ and last rectangles. (Allow sign errors only for M1). Limits must match area covered by chosen rectangles.

Correct final inequality: ${\sum }_{r=1}^n\frac{n}{n^2+r^2}\ge \frac{1}{4}\pi -\frac{1}{2n}$

$\frac{1}{4}\pi$

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

$\ln (x^2+6x+5)$

${\mathrm{e}}^{-\ln (x^2+6x+5)}=\frac{1}{x^2+6x+5}$

Correct form on LHS.

Partial fractions. Or completing the square.

Integrates RHS. (RHS must be of the correct form.)

$\ln \left(\frac{x+1}{x+5}\right)+C$

Substitutes initial conditions.

Accept $y=(x^2+6x+5)\ln \left(\frac{5x+5}{x+5}\right)$.

Sets determinant equal to zero and forms 3 term quadratic equation.

$8a^2-3a-\frac{9}{2}=0$

$a=\frac{3}{16}(1\pm \sqrt{17})$

Finds characteristic equation.

Using C-H Theorem and replaces $\lambda$ with $\mathbf{A}$.

Multiplies by ${\mathbf{B}}^2={\mathbf{A}}^{-2}$

CAO: ${\mathbf{B}}^2=\frac{7}{9}\mathbf{I}-\frac{2}{9}\mathbf{A}$

Upper diagonal matrix or characteristic equation.

Uses vector product (or equations) to find corresponding eigenvectors.

Eigenvector for $\lambda =\frac{3}{2}$.

Eigenvector for $\lambda =3$.

Eigenvector for $\lambda =-1$.

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

Or correctly matched permutations of columns. Column of zeros or repeated column in $\mathbf{P}$, or repeated eigenvalues in $\mathbf{D}$ is M0.