A Level Further Mathematics 9231 Paper 22, May/June 2023

$\mid \begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\mid =(5\times 9-8\times 6)-2(4\times 9-7\times 6)+3(4\times 8-7\times 5)$

$=-3+12-9=0$

Uses all three equations to reduce to one equation with two unknowns.

$x+2y+3z=1$, $4x+5y+6z=1,⟹z=\frac{1}{3}(1-x-2y)⟹y=-2x-1$, $7x+8y+9z=1$

The three planes form a sheaf.

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}z}{\mathrm{d}x}-1$

$\frac{\mathrm{d}z}{\mathrm{d}x}-1=\frac{1+3z}{3z-1}$

$\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{1+3z}{3z-1}+1=\frac{6z}{3z-1}$

$\int (\frac{1}{2}-\frac{1}{6}{z}^{-1})\mathrm{d}z=\int 1\mathrm{d}x$

$\frac{1}{2}z-\frac{1}{6}\ln z=x+C⟹-\frac{1}{2}x+\frac{1}{2}y-\frac{1}{6}\ln (x+y)=C$

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

$\frac{1}{6}\ln (x+y)+\frac{1}{2}(x-y)-\frac{1}{2}=0$

$z+z^{-1}=2\cos \theta$

Expands and groups $(z+z^{-1}{)}^4=(z^4+z^{-4})+4(z^2+z^{-2})+6$

$(z+z^{-1}{)}^4=(z^4+z^{-4})+4(z^2+z^{-2})+6$

$(2\cos \theta {)}^4=2\cos 4\theta +4(2\cos 2\theta )+6$

${\cos }^4\theta =\frac{1}{8}(\cos 4\theta +4\cos 2\theta +3)$

Applies substitution gets to $\int {\cos }^4\theta \mathrm{d}\theta$

${\int }_0^{\frac{1}{2}}(1-x^2{)}^{\frac{3}{2}}\mathrm{d}x={\int }_0^{\frac{1}{6}\pi }{\cos }^4\theta \mathrm{d}\theta =\frac{1}{8}{[\frac{1}{4}\sin 4\theta +2\sin 2\theta +3\theta ]}_0^{\frac{1}{6}\pi }$

$\frac{1}{16}(\frac{9}{4}\sqrt{3}+\pi )$

Uses the product rule to differentiate $\frac{\mathrm{d}}{\mathrm{d}x}(x(1+x^5{)}^n)$ (or integrates by parts).

$\frac{\mathrm{d}}{\mathrm{d}x}(x(1+x^5{)}^n)=5n{x}^5(1+x^5{)}^{n-1}+(1+x^5{)}^n$

$5n(1+x^5-1)(1+x^5{)}^{n-1}+(1+x^5{)}^n$

${[x(1+x^5{)}^n]}_0^1=5n{I}_n-5n{I}_{n-1}+I_n$

$2^n=(5n+1)I_n-5n{I}_{n-1}⟹(5n+1)I_n=2^n+5n{I}_{n-1}$

$I_1={[x+\frac{1}{6}{x}^6]}_0^1=\frac{7}{6}$ or $I_0=1$

Applies reduction formula $11I_2=2^2+10I_1$

$I_2=\frac{47}{33}$

$16I_3=2^3+15I_2⟹I_3=\frac{323}{176}$

$\mid \begin{array}{ccc}18-\lambda & 5& -11\\ 8& 6-\lambda & -4\\ 32& 10& -20-\lambda \end{array}\mid =0$

$(18-\lambda )\mid \begin{array}{cc}6-\lambda & -4\\ 10& -20-\lambda \end{array}\mid -5\mid \begin{array}{cc}8& -4\\ 32& -20-\lambda \end{array}\mid -11\mid \begin{array}{cc}8& 6-\lambda \\ 32& 10\end{array}\mid$

${\lambda }^3-4{\lambda }^2-20\lambda +48=0$

$(\lambda -2)(\lambda +4)(\lambda -6)=0$ leading to $\lambda =2,-4,6$

Uses vector product (or equations) to find corresponding eigenvectors. Must attempt to solve equations for M1.

$\lambda =2:\left(\begin{array}{c}1\\ -1\\ 1\end{array}\right)$

$\lambda =-4:\left(\begin{array}{c}1\\ 0\\ 2\end{array}\right)$

$\lambda =6:\left(\begin{array}{c}1\\ 2\\ 2\end{array}\right)$

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

$\mathbf{P}=\left(\begin{array}{ccc}1& 1& 1\\ -1& 0& 2\\ 1& 2& 2\end{array}\right)$ and $\mathbf{D}=\left(\begin{array}{ccc}32& 0& 0\\ 0& -1024& 0\\ 0& 0& 7776\end{array}\right)$

$m^2-12m+36=0$

$x={\mathrm{e}}^{6t}(At+B)$

$x=p\sin t+q\cos t⟹x^{\mathrm{\prime }}=p\cos t-q\sin t⟹x^{\mathrm{\prime }\mathrm{\prime }}=-p\sin t-q\cos t$

Correct derivatives of particular integral

$-p+12q+36p=37-q-12p+36q=0$

$p=\frac{35}{37}q=\frac{12}{37}$

$x={\mathrm{e}}^{6t}(At+B)+\frac{35}{37}\sin t+\frac{12}{37}\cos t$

$x^{\mathrm{\prime }}=A{\mathrm{e}}^{6t}+6{\mathrm{e}}^{6t}(At+B)+\frac{35}{37}\cos t-\frac{12}{37}\sin t$

$B=-\frac{12}{37}A+6B+\frac{35}{37}=0$

$A=1$

$x={\mathrm{e}}^{6t}(t-\frac{12}{37})+\frac{35}{37}\sin t+\frac{12}{37}\cos t$

Applies given substitution.

$\int \frac{x}{\sqrt{x^2-1}}\mathrm{d}x=\frac{1}{2}\int \frac{1}{\sqrt{u}}\mathrm{d}u$

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

${\cosh }^{-1}r=\ln (r+\sqrt{r^2-1})$

${\cosh }^{-1}2+{\cosh }^{-1}3+\dots +{\cosh }^{-1}N$

$>{\int }_1^N{\cosh }^{-1}x\mathrm{d}x$

${\int }_1^N{\cosh }^{-1}x\mathrm{d}x={\left[x{\cosh }^{-1}x\right]}_1^N-{\int }_1^N\frac{x}{\sqrt{x^2-1}}\mathrm{d}x$

${\sum }_{r=2}^N\ln (r+\sqrt{r^2-1})>N\ln (N+\sqrt{N^2-1})-\sqrt{N^2-1}$

Compares with integral with correct limits.

$({\cosh }^{-1}1)+{\cosh }^{-1}2+\dots +{\cosh }^{-1}(N-1)<{\int }_1^N{\cosh }^{-1}x\mathrm{d}x(\text{or }<{\int }_2^N{\cosh }^{-1}x\mathrm{d}x)$

${\sum }_{r=2}^N\ln (r+\sqrt{r^2-1})<(N+1)\ln (N+\sqrt{N^2-1})-\sqrt{N^2-1}$ or ${\sum }_{r=2}^N\ln (r+\sqrt{r^2-1})<(N+1)\ln (N+\sqrt{N^2-1})-\sqrt{N^2-1}-2\ln (2+\sqrt{3})+\sqrt{3}$

$\mathrm{sech}t=\frac{2}{{\mathrm{e}}^t+{\mathrm{e}}^{-t}}\tanh t=\frac{{\mathrm{e}}^t-{\mathrm{e}}^{-t}}{{\mathrm{e}}^t+{\mathrm{e}}^{-t}}$

Expands, gets to $\frac{{\mathrm{e}}^{2t}+2+{\mathrm{e}}^{-2t}-4}{({\mathrm{e}}^t+{\mathrm{e}}^{-t}{)}^2}$

$1-{\left(\frac{2}{{\mathrm{e}}^t+{\mathrm{e}}^{-t}}\right)}^2=\frac{({\mathrm{e}}^t+{\mathrm{e}}^{-t}{)}^2-4}{({\mathrm{e}}^t+{\mathrm{e}}^{-t}{)}^2}=\frac{{\mathrm{e}}^{2t}+2+{\mathrm{e}}^{-2t}-4}{({\mathrm{e}}^t+{\mathrm{e}}^{-t}{)}^2}=\frac{({\mathrm{e}}^t-{\mathrm{e}}^{-t}{)}^2}{({\mathrm{e}}^t+{\mathrm{e}}^{-t}{)}^2}$

$\frac{\mathrm{d}y}{\mathrm{d}t}=4{\tanh }^{3}t{\mathrm{sech}}^{2}t$

M1 sensible attempt at derivative of $x$.

$\frac{\mathrm{d}x}{\mathrm{d}t}=\tanh t{\mathrm{sech}}^{2}t-\tanh t(=\tanh t({\mathrm{sech}}^{2}t-1)=-{\tanh }^{3}t)$

Applies chain rule, must substitute their $\frac{\mathrm{d}y}{\mathrm{d}t}$ and $\frac{\mathrm{d}x}{\mathrm{d}t}$

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{4{\tanh }^{3}t{\mathrm{sech}}^{2}t}{-{\tanh }^{3}t}=-4{\mathrm{sech}}^{2}t$

Finds $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}$. For M1 $\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=c{\mathrm{sech}}^{2}t\tanh t$, AEF.

$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{8{\mathrm{sech}}^{2}t\tanh t}{-{\tanh }^{3}t}=-8\frac{{\mathrm{sech}}^{2}t}{{\tanh }^{2}t}$

Sets equal to $-\frac{9}{2}$, uses identity from (a) or equivalent.

$-8\frac{{\mathrm{sech}}^{2}t}{{\tanh }^{2}t}=-\frac{9}{2}⟹8(1-{\tanh }^{2}t)=\frac{9}{2}{\tanh }^{2}t⟹{\tanh }^{2}t=\frac{16}{25}$

Correct $x$-coordinate: $x=\frac{8}{25}+\ln \frac{3}{5}$

Correct $y$-coordinate: $y=\frac{881}{625}$