A Level Further Mathematics 9231 Paper 13, May/June 2025

$\left(\begin{array}{cc}14& 0\\ 0& 1\end{array}\right)$

$\left(\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)$

$\mathbf{M}=\left(\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)\left(\begin{array}{cc}14& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}7& -\frac{\sqrt{3}}{2}\\ 7\sqrt{3}& \frac{1}{2}\end{array}\right)$

$2\mathbf{M}=\left(\begin{array}{cc}14& -\sqrt{3}\\ 14\sqrt{3}& 1\end{array}\right)$.

$\left(\begin{array}{cc}7& -\frac{\sqrt{3}}{2}\\ 7\sqrt{3}& \frac{1}{2}\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}7x-\frac{\sqrt{3}}{2}y\\ 7\sqrt{3}x+\frac{1}{2}y\end{array}\right)$

$7\sqrt{3}x+\frac{1}{2}mx=m(7x-\frac{\sqrt{3}}{2}mx)$ or $7\sqrt{3}+\frac{1}{2}m=7m-\frac{\sqrt{3}}{2}{m}^2$

$\sqrt{3}{m}^2-13m+14\sqrt{3}=0$

$m=2\sqrt{3}m=\frac{7}{\sqrt{3}}$

$y=2\sqrt{3}xy=\frac{7}{\sqrt{3}}x$

Method for finding ${\mathbf{M}}^{-1}$

${\mathbf{M}}^{-1}=\frac{1}{14}\left(\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -7\sqrt{3}& 7\end{array}\right)=\left(\begin{array}{cc}\frac{1}{28}& \frac{\sqrt{3}}{28}\\ -\frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)$

$2025+47-2=2070=45\times 46$ is divisible by 46.

Assume that ${2025}^k+{47}^k-2$ is divisible by 46 for some positive integer $k$.

Then ${2025}^{k+1}+{47}^{k+1}-2=(2024+1){2025}^k+(46+1){47}^k-2$

Separating ${2025}^k+{47}^k-2$ correctly or considering difference.

is divisible by 46 because $2024\times {2025}^k+46\times {47}^k=46(44\times {2025}^k)+46\times {47}^k$ is divisible by 46

Hence, [by induction], true for every positive integer $n$.

$0={\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2+2(7)$

${\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2=-14$

$S_4+7S_2+3S_1+88=0$

$S_4=10$

$(r+{\alpha }^2{)}^2=r^2+2{\alpha }^{2}r+{\alpha }^4$

${\sum }_{r=1}^{10}((r+{\alpha }^2{)}^2+(r+{\beta }^2{)}^2+(r+{\gamma }^2{)}^2+(r+{\delta }^2{)}^2)$

$={\sum }_{r=1}^{10}(4r^2-28r+10)$

$=4(\frac{10}{6}(10+1)(2(10)+1))-14(10)(10+1)+10(10)$

$=100$

$w_{r+1}-w_r=(r+1)(r+2)\dots (r+10)-r(r+1)(r+2)\dots (r+9)$

$(r+1)(r+2)\dots (r+9)(r+10-r)=10(r+1)(r+2)\dots (r+9)$

$\frac{1}{10}(w_2-w_1)+\frac{1}{10}(w_3-w_2)+\cdots +\frac{1}{10}(w_{n+1}-w_n)$

Writes at least three terms, including last.

$=\frac{1}{10}(w_{n+1}-w_1)=\frac{1}{10}(n+1)(n+2)\dots (n+10)-\frac{10!}{10}=\frac{1}{10}(n+1)(n+2)\dots (n+10)-9!$

$v_1+v_2+\cdots +v_n=x^{w_2}-x^{w_1}+x^{w_3}-x^{w_2}+\cdots +x^{w_{n+1}}-x^{w_n}=x^{w_{n+1}}-x^{w_1}$

For $-1<x<1$, sum to infinity is $-x^{w_1}=-x^{10!}$

For $x=\pm 1$, sum to infinity is 0.

Finds vector perpendicular to $\mathrm{\Pi }$.

$\left(\begin{array}{c}6\\ -1\\ 8\end{array}\right)$

$6(2)-(3)+8(-2)=-7$

$6x-y+8z=-7$

$\overrightarrow{OF}=\overrightarrow{OP}+\overrightarrow{PF}=\left(\begin{array}{c}4\\ 2\\ 9\end{array}\right)+t\left(\begin{array}{c}6\\ -1\\ 8\end{array}\right)$

$=\left(\begin{array}{c}4+6t\\ 2-t\\ 9+8t\end{array}\right)$

$6(4+6t)-(2-t)+8(9+8t)=-7⟹101t+94=-7$

$t=-1⟹\overrightarrow{OF}=\left(\begin{array}{c}-2\\ 3\\ 1\end{array}\right)$

Uses dot product of $3\mathbf{i}+5\mathbf{j}-\mathbf{k}$ and their normal in a correct formula.

$\cos \alpha =\frac{5}{\sqrt{35}\sqrt{101}}$

Acute angle between $l$ and $\mathrm{\Pi }$ is $90-\alpha ={4.8}^{\circ }$

$x=-a$

$y=\frac{(x+a)(x-a)+a^2+a}{x+a}=x-a+\frac{a^2+a}{x+a}$

$y=x-a$

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

Sets correct derivative to zero or ACF of quadratic equation $x^2+2ax-a=0$

$x=-a\pm \sqrt{a^2+a}$

Axes shown and asymptotes labelled.

Lower branch correct.

Upper branch with $(0,1)$ stated or 1 marked on vertical axis and no others.

Correct sketch of $y=\mathrm{\mid }f(x)\mathrm{\mid }$.

Finds critical points from $\frac{x^2+a}{x+a}=a⟹x^2-ax+a-a^2=0$

$\frac{x^2+a}{x+a}=-a⟹x^2+ax+a+a^2=0$

$a^2-4(a-a^2)=5a^2-4a>0$

$a>\frac{4}{5}$.

Finds $\frac{\mathrm{d}r}{\mathrm{d}\theta }$ and sets equal to 0.

${\cos }^2\theta -\sin \theta =0$

$1-{\sin }^2\theta -\sin \theta =0$

$\sin \theta =\frac{1}{2}(-1+\sqrt{5})$

$(1.208,0.666)$

Uses $x=r\cos \theta$. Finds $\frac{\mathrm{d}x}{\mathrm{d}\theta }$ and sets equal to 0.

$-3\sin \theta +{\cos }^2\theta =0$

$-3\sin \theta +1-{\sin }^2\theta =0$

$\sin \theta =\frac{1}{2}(-3+\sqrt{13})$

$(1.136,0.308)$

Closed loop passing through $O$ and one other point on the initial line. Does not go into 2nd and 3rd quadrants. More in the 1st than 4th quadrant.

$-\frac{1}{2}\pi \le \theta \le 0.666$ correct. Tangent needs to have positive gradient at the intersection $(1,0)$.

$0.666\le \theta \le \frac{1}{2}\pi$ correct. Graph should approach the line $\theta =\frac{\pi }{2}$.

$\frac{1}{2}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{e}^{\sin \theta }\cos \theta \mathrm{d}\theta$

$=\frac{1}{2}{\left[e^{\sin \theta }\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$

$=\frac{1}{2}(e-e^{-1})$