A Level Further Mathematics 9231 Paper 23, May/June 2024

Completes the square.

Applies formula.

${\left[{\sin }^{-1}\left(\frac{x-2}{\sqrt{3}}\right)\right]}_2^{\frac{7}{2}}$

Uses limits.

Answer must be in radians: $\frac{1}{3}\pi$

Differentiates $x$ and $y$ with respect to $t$.

Applies ${\cosh }^{2}t+{\sinh }^{2}t=\cosh 2t$. May use exponential form or ${(1+{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2)}^{\frac{1}{2}}\mathrm{d}x={(1+\frac{{\cosh }^{2}t}{{\sinh }^{2}t})}^{\frac{1}{2}}\sinh t\mathrm{d}t$.

${\dot{x}}^2+{\dot{y}}^2={\sinh }^{2}t+{\cosh }^{2}t=\cosh 2t$

AG: $s={\int }_0^{\frac{3}{5}}\sqrt{\cosh 2t}\mathrm{d}t$

Finds first derivative.

Finds second derivative

Applies Maclaurin's series.

Evaluates integral ${[t+\frac{1}{3}{t}^3]}_0^{\frac{3}{5}}$

CWO: $\frac{84}{125}=0.672$

Differentiates $y^3$ correctly.

Differentiates $xy$ correctly.

Substitutes $(-2,-1)$ into $3x^2+2x{y}^{\mathrm{\prime }}+2y+24y^2{y}^{\mathrm{\prime }}=0$, AG. Must see intermediary step between substitution and the final answer.

Differentiates $3x^2+2x{y}^{\mathrm{\prime }}$.

Differentiates $2y+24y^2{y}^{\mathrm{\prime }}$: for $2y^{\mathrm{\prime }}+24y^2{y}^{\mathrm{\prime }\mathrm{\prime }}$.

for $48y(y^{\mathrm{\prime }}{)}^2$. (Allow alternative using quotient rule.)

Substitutes $(-2,-1)$ into their expression (must have derivatives $y^{\mathrm{\prime }}$ and $y^{\mathrm{\prime }\mathrm{\prime }}$).

OE: $y^{\mathrm{\prime }\mathrm{\prime }}=\frac{13}{10}$

Forms sum of the areas of the rectangles. Need to see this sum written explicitly, may be in summation form, cannot be embedded in working.

Evaluates integral, correct limits for M1.

${[-x^{-1}]}_2^N=\frac{1}{2}-\frac{1}{N}$

Compares with integral correctly (using >).

AG: ${\sum }_{r=1}^N{r}^{-2}>\frac{3}{2}-\frac{1}{N}+\frac{1}{N^2}$

Forms the sum of the areas of appropriate rectangles. Allow addition of one extra rectangle or shift one left or right if limits on the integral are adjusted accordingly.

Compares with integral, correct limits and correct inequality symbol (<).

${\sum }_{r=1}^N{r}^{-2}<\frac{7}{4}-\frac{1}{N}$

Needs to clear (using > or words) is that value is lower bound.

Needs to clear (using < or words) is that value is upper bound. Must have scored final A mark in part (b). FT on fully correct alternative to part 4(b).

Auxiliary equation.

Complementary function. Allow ‘x = ’ missing.

Particular integral and its derivatives. For M1, both derivatives must be present and of the correct form (only allow sign errors).

Derivatives of particular integral correct.

Substitutes and equates coefficients.

$p=12q=-5$

Must have ‘x =’: $x={\mathrm{e}}^{-5t}(At+B)+12\sin t-5\cos t$

Considering PI only.

A1 for $R$.

A1 for $\varphi$. Condone ${22.6}^{\circ }$. Withhold final A1 mark if working with incorrect PI. CWO.

Uses sum of geometric series.

Divides numerator and denominator by $z^2$, AG.

Simplifies denominator.

Applies de Moivre’s theorem to numerator. (Used at least once for M1.) Accept $z^2={\cos }^2\theta +2\mathrm{i}\sin \theta \cos \theta -{\sin }^2\theta$.

Numerator correctly simplified with de Moivre's theorem.

Equates imaginary parts.

AG. Must see evidence of eliminating i from the denominator.

Applies product rule and $\frac{\mathrm{d}}{\mathrm{d}x}\left({\cosh }^{-1}\frac{x}{3}\right)=\frac{1}{\sqrt{x^2-9}}$

OE: $\frac{x^2}{2\sqrt{x^2-9}}+\frac{1}{2}\sqrt{x^2-9}-\frac{9}{2\sqrt{x^2-9}}$

Puts over common denominator, AG.

Divides through by $x$.

Finds integrating factor.

Integrating factor $=x^{-1}$.

Correct form on LHS, $\frac{\mathrm{d}}{\mathrm{d}x}(Iy)$ for their integrating factor $I$.

Integrates RHS. For M1, RHS must be of the form $c\sqrt{x^2-9}$, where $c$ is a non-zero constant.

$x^{-1}y=\frac{x}{2}\sqrt{x^2-9}-\frac{9}{2}{\cosh }^{-1}\frac{x}{3}+C$

Substitutes initial conditions.

Divides through by their integrating factor.

Accept $y=x(\frac{x}{2}\sqrt{x^2-9}-\frac{9}{2}\ln (\frac{x}{3}+\frac{1}{3}\sqrt{x^2-9})+\frac{1}{3})$.

Substitutes point into $x+2y+3z$.

$x+2y+3z=13$

$x+2y+3z=-9$

Forms vector equation of line.

AG. CWO.

Uses $\det (\mathbf{A}-\lambda \mathbf{I})$ (attempt an expansion).

Expands determinant, expansion of brackets shown, AG. CWO. (Must have ‘= 0’ here or above.)

$\lambda =0,\frac{7}{2},-\frac{1}{2}$

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

Finds eigenvector for $\lambda =0$.

Finds eigenvector for $\lambda =\frac{7}{2}$.

Finds eigenvector for $\lambda =-\frac{1}{2}$.

Matching their eigenvectors correctly with their eigenvalues. M0 if a column of zeros appears in $\mathbf{P}$.

Correct $\mathbf{P}$ and $\mathbf{D}$, or correctly matched permutations of columns. Do not accept $\mathbf{D}={\left(\begin{array}{ccc}0& 0& 0\\ 0& \frac{7}{2}& 0\\ 0& 0& -\frac{1}{2}\end{array}\right)}^n$.