A Level Further Mathematics 9231 Paper 21, May/June 2024

Finds modulus and argument of $-108\sqrt{3}+108\mathrm{i}$

Finds one root. For M1 must have divided their argument of $-108\sqrt{3}+108\mathrm{i}$ by 3.

$z_1=6\cos \frac{5\pi }{18}+\mathrm{i}6\sin \frac{5\pi }{18}$ (FT on their modulus)

Finds other two roots (method).

$z_2=6\cos \frac{17\pi }{18}+\mathrm{i}6\sin \frac{17\pi }{18},z_3=6\cos \frac{29\pi }{18}+\mathrm{i}6\sin \frac{29\pi }{18}$

Changes ${\mathrm{e}}^{1+x^2}$ or ${\mathrm{e}}^{1-x}$ so that the formula given in the list of formulae (MF19) can be applied.

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

Sums power series.

${\mathrm{e}}^{1+x^2}+{\mathrm{e}}^{1-x}=2\mathrm{e}-\mathrm{e}x+\frac{3}{2}\mathrm{e}{x}^2+\dots$

$\frac{\mathrm{d}y}{\mathrm{d}t}={\cos }^{-1}t-\frac{t}{\sqrt{1-t^2}},\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{1}{\sqrt{1-t^2}}$

Uses chain rule

$\frac{\mathrm{d}y}{\mathrm{d}x}=-t+\sqrt{1-t^2}{\cos }^{-1}t$ (AG)

Applies product rule.

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

Uses chain rule.

$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=-2\sqrt{1-t^2}-t{\cos }^{-1}t$

Separates into correct structure.

Uses integration by parts correctly.

Uses $1-{\mathrm{sech}}^{2}x={\tanh }^{2}x$.

${\left(\frac{3}{5}\right)}^{n-2}\left(\frac{4}{5}\right)+(n-2)(I_{n-2}-I_n)$

$(1+n-2)I_n={\left(\frac{3}{5}\right)}^{n-2}\left(\frac{4}{5}\right)+(n-2)I_{n-2}\Rightarrow (n-1)I_n={\left(\frac{3}{5}\right)}^{n-2}\left(\frac{4}{5}\right)+(n-2)I_{n-2}$ (AG)

$I_2=\frac{4}{5}$

Applies reduction formula with $n=4$.

$I_4=\frac{236}{375}=0.629$

Forms the sum of the areas of the rectangles, may be in summation form. M1 is for correct number of rectangles.

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

Applies formulae from MF19. Substituted in correctly once for M1.

$=\frac{2}{n^2}{\sum }_{r=1}^{n}r-\frac{1}{n^3}{\sum }_{r=1}^n{r}^2=\frac{1}{n^2}n(n+1)-\frac{1}{6n^3}n(n+1)(2n+1)$ (Summations must have (correct) limits for A1)

$=1+\frac{1}{n}-\frac{1}{6}(1+\frac{1}{n})(2+\frac{1}{n})=(1+\frac{1}{n})(\frac{2}{3}-\frac{1}{6n})=\frac{2}{3}+\frac{1}{2n}-\frac{1}{6n^2}$ (AG)

Forms the sum of the areas of appropriate rectangles. M1 is for seeing correct height of last rectangle.

Correct expression for the sum of the areas of rectangles.

Applies formulae from MF19, substitutes correct limit. Substituted in correctly once for M1.

$(1-\frac{1}{n})(\frac{2}{3}+\frac{1}{6n})=\frac{2}{3}-\frac{1}{2n}-\frac{1}{6n^2}$

Expresses their $U_n-L_n$ in terms of $\frac{1}{n}$. For M1 they must taking the limit of an expression that tends to a constant. (Must be fully expanded if they do not take the limit at any stage.)

$\frac{1}{n}\to 0\text{ as }n\to \mathrm{\infty }\text{ or }\frac{2}{3}-\frac{2}{3}=0$ (AG, CWO)

Substitutes sinh and cosh in terms of exponentials.

$({\mathrm{e}}^x{)}^{\frac{1}{2}}={\mathrm{e}}^{\frac{1}{2}x}$ (AG)

Auxiliary equation.

Complimentary function. Allow '$y=$' missing.

Particular integral and its derivatives.

Substitutes and equates coefficients. PI must be correct form ($a{\mathrm{e}}^{bx}$).

$k=\frac{4}{3}$

Must have '$y=$'. FT on their CF.

$\frac{\mathrm{d}y}{\mathrm{d}x}={\mathrm{e}}^{-\frac{1}{2}x}(-\frac{\sqrt{11}}{2}A\sin \frac{\sqrt{11}}{2}x+\frac{\sqrt{11}}{2}B\cos \frac{\sqrt{11}}{2}x)-\frac{1}{2}{\mathrm{e}}^{-\frac{1}{2}x}(A\cos \frac{\sqrt{11}}{2}x+B\sin \frac{\sqrt{11}}{2}x)+\frac{2}{3}{\mathrm{e}}^{\frac{1}{2}x}$

Substitutes initial conditions and forms simultaneous equations. For M1, CF must be correct form

$A=-\frac{1}{3},B=\frac{1}{\sqrt{11}}$

Must have '$y=$'. $y={\mathrm{e}}^{-\frac{1}{2}x}(-\frac{1}{3}\cos \frac{\sqrt{11}}{2}x+\frac{1}{\sqrt{11}}\sin \frac{\sqrt{11}}{2}x)+\frac{4}{3}{\mathrm{e}}^{\frac{1}{2}x}$

Applies substitution.

$\sqrt{1+x^2}[+C]$

Divides through by $x$.

Finds integrating factor (method).

${\mathrm{e}}^{-\int x^{-1}\mathrm{d}x}={\mathrm{e}}^{-\ln x}=x^{-1}$

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

Integrates RHS. RHS must be of the form $c{\sinh }^{-1}x$.

$x^{-1}y=x{\sinh }^{-1}x-\int \frac{x}{\sqrt{1+x^2}}\mathrm{d}x$

$x^{-1}y=x{\sinh }^{-1}x-\sqrt{1+x^2}+C$

Substitutes initial conditions.

Divides through by their integrating factor.

$y=x^2{\sinh }^{-1}x-x\sqrt{1+x^2}-x\ln (1+\sqrt{2})+x(1+\sqrt{2})$

Finds determinant or solves equations.

$a\mathrm{\ne }-3$

If $a\mathrm{\ne }-3$ then system has unique solution (so consistent).

Eliminates variable using all 3 equations. Or states there are two distinct equations and three unknowns. Or gives accurate geometrical description with $a=-3$.

So the system has infinitely many solutions (so consistent).

Eigenvalues of $\mathbf{A}$ are 6, $-1$ and 4.

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

$\lambda =6:\left(\begin{array}{c}14\\ 4\\ 17\end{array}\right)$

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

$\lambda =4:\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$

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

Thus $\mathbf{P}=\left(\begin{array}{ccc}14& 0& 0\\ 4& -1& 0\\ 17& 1& 1\end{array}\right)$ and $\mathbf{D}=\left(\begin{array}{ccc}11664& 0& 0\\ 0& 100& 0\\ 0& 0& 6400\end{array}\right)$

Characteristic equation.

Substitutes for $\mathbf{A}$ and makes $14\mathbf{A}+24\mathbf{I}$ the subject.

Squares and factorises (method).

$(14\mathbf{A}+24\mathbf{I}{)}^2={(-{\mathbf{A}}^3+9{\mathbf{A}}^2)}^2={\mathbf{A}}^4(\mathbf{A}-9\mathbf{I}{)}^2$ (CWO)