A Level Further Mathematics 9231 Paper 23, Oct/Nov 2023

$\mid \begin{array}{ccc}14& -4& 6\\ 1& 1& k\\ -21& 6& -9\end{array}\mid =14\mid \begin{array}{cc}1& k\\ 6& -9\end{array}\mid +4\mid \begin{array}{cc}1& k\\ -21& -9\end{array}\mid +6\mid \begin{array}{cc}1& 1\\ -21& 6\end{array}\mid =14(-9-6k)+4(-9+21k)+6(6+21)$

= 0

Two parallel planes, not identical.

Other plane not parallel.

$(z+5\mathrm{i}{)}^3=4+4\mathrm{i}\sqrt{3}=8{\mathrm{e}}^{\frac{1}{3}\pi \mathrm{i}}$

$z_1=2(\cos \frac{1}{9}\pi +\mathrm{i}\sin \frac{1}{9}\pi )-5\mathrm{i}$

$=2\cos \frac{1}{9}\pi +\mathrm{i}(2\sin \frac{1}{9}\pi -5)$

$z_2=2\cos \frac{7}{9}\pi +\mathrm{i}(2\sin \frac{7}{9}\pi -5)$

$z_3=2\cos \frac{13}{9}\pi +\mathrm{i}(2\sin \frac{13}{9}\pi -5)$

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{1}{2}{\mathrm{e}}^x}{1-\frac{1}{4}{\mathrm{e}}^{2x}}$

$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{(1-\frac{1}{4}{\mathrm{e}}^{2x})\left(\frac{1}{2}{\mathrm{e}}^x\right)-\frac{1}{2}{\mathrm{e}}^x(-\frac{1}{2}{\mathrm{e}}^{2x})}{{(1-\frac{1}{4}{\mathrm{e}}^{2x})}^2}$

$f^{\mathrm{\prime }}(0)=\frac{2}{3}{f}^{\mathrm{\prime }\mathrm{\prime }}(0)=\frac{10}{9}$

$f(0)={\tanh }^{-1}\left(\frac{1}{2}\right)=\frac{1}{2}\ln (\frac{3}{2}\times 2)$

Applies $f(x)=f(0)+f^{\mathrm{\prime }}(0)x+\frac{1}{2!}{f}^{\mathrm{\prime }\mathrm{\prime }}(0)x^2$

$\frac{1}{2}\ln 3+\frac{2}{3}x+\frac{5}{9}{x}^2$

$m^2+2m+3=0$

$[y=]{\mathrm{e}}^{-x}(A\cos \sqrt{2}x+B\sin \sqrt{2}x)$

$y=p{x}^2+qx+r\Rightarrow y^{\mathrm{\prime }}=2px+q\Rightarrow y^{\mathrm{\prime }\mathrm{\prime }}=2p$

$3p=273q+4p=02p+2q+3r=0$

$p=9q=-12r=2$

$y={\mathrm{e}}^{-x}(A\cos \sqrt{2}x+B\sin \sqrt{2}x)+9x^2-12x+2$

$y^{\mathrm{\prime }}={\mathrm{e}}^{-x}(-\sqrt{2}A\sin \sqrt{2}x+\sqrt{2}B\cos \sqrt{2}x)-{\mathrm{e}}^{-x}(A\cos \sqrt{2}x+B\sin \sqrt{2}x)+18x-12$

$A+2=2\sqrt{2}B-A-12=-8$

$\Rightarrow A=0,B=2\sqrt{2}$

$y=2\sqrt{2}{\mathrm{e}}^{-x}\sin \sqrt{2}x+9x^2-12x+2$

$\dot{x}=t^{\frac{1}{2}}-t^{-\frac{1}{2}},\dot{y}=2$

${\dot{x}}^2+{\dot{y}}^2={(t^{\frac{1}{2}}-t^{-\frac{1}{2}})}^2+4=t+2+t^{-1}$

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

${\int }_0^3{t}^{\frac{1}{2}}+t^{-\frac{1}{2}}\mathrm{d}t={[\frac{2}{3}{t}^{\frac{3}{2}}+2t^{\frac{1}{2}}]}_0^3$

$=4\sqrt{3}$

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

$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{2}{t^{\frac{1}{2}}-t^{-\frac{1}{2}}}\right)=\frac{-2(\frac{1}{2}{t}^{-\frac{1}{2}}+\frac{1}{2}{t}^{-\frac{3}{2}})}{{(t^{\frac{1}{2}}-t^{-\frac{1}{2}})}^2}$

$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{2}{t^{\frac{1}{2}}-t^{-\frac{1}{2}}}\right)\times \frac{\mathrm{d}t}{\mathrm{d}x}=\frac{-2(\frac{1}{2}{t}^{-\frac{1}{2}}+\frac{1}{2}{t}^{-\frac{3}{2}})}{{(t^{\frac{1}{2}}-t^{-\frac{1}{2}})}^3}$

$=-\frac{t+1}{(t-1{)}^3}$

$0<t<1$

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

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

$=\sinh 2x$

$u=\sinh x\Rightarrow \mathrm{d}u=\cosh x\mathrm{d}x$

$\int {\sinh }^{2}2x\cosh x\mathrm{d}x=4\int {\sinh }^{2}x{\cosh }^{2}x\mathrm{d}u=4\int {\sinh }^{2}x({\sinh }^{2}x+1)\mathrm{d}u$

$=4\int u^2(u^2+1)\mathrm{d}u$

$=4(\frac{1}{5}{u}^5+\frac{1}{3}{u}^3)(+C)=4(\frac{1}{5}{\sinh }^{5}x+\frac{1}{3}{\sinh }^{3}x)(+C)$

${\mathrm{e}}^{\int \tanh x\mathrm{d}x}={\mathrm{e}}^{\ln \cosh x}$

$=\cosh x$

$\frac{\mathrm{d}}{\mathrm{d}x}(y\cosh x)={\sinh }^{2}2x\cosh x$

$y\cosh x=4(\frac{1}{5}{\sinh }^{5}x+\frac{1}{3}{\sinh }^{3}x)+C$

Integrates RHS using their part (b).

$4=C$

$y=4\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}x(\frac{1}{5}{\sinh }^{5}x+\frac{1}{3}{\sinh }^{3}x+1)$

$\lambda =-6,\lambda =-2,\lambda =8$

$\lambda =-6:\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 0& 4& 5\\ 0& 0& 2\end{array}\mid$

$=\left(\begin{array}{c}8\\ 0\\ 0\end{array}\right)\sim \left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$

$\lambda =-2:\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -4& 2& 13\\ 0& 0& 5\end{array}\mid =\left(\begin{array}{c}10\\ 20\\ 0\end{array}\right)\sim \left(\begin{array}{c}1\\ 2\\ 0\end{array}\right)$

$\lambda =8:\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -14& 2& 13\\ 0& -10& 5\end{array}\mid =\left(\begin{array}{c}140\\ 70\\ 140\end{array}\right)\sim \left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$

Eigenvectors matched to their eigenvalues

$\mathbf{P}=\left(\begin{array}{ccc}1& 1& 2\\ 0& 2& 1\\ 0& 0& 2\end{array}\right)$ and $\mathbf{D}=\left(\begin{array}{ccc}-\frac{1}{6}& 0& 0\\ 0& -\frac{1}{2}& 0\\ 0& 0& \frac{1}{8}\end{array}\right)$

${\mathbf{A}}^3-52\mathbf{A}-96\mathbf{I}=0$

$96{\mathbf{A}}^{-1}={\mathbf{A}}^2-52\mathbf{I}$

${\mathbf{A}}^2=\left(\begin{array}{ccc}36& -16& 36\\ 0& 4& 30\\ 0& 0& 64\end{array}\right)$

${\mathbf{A}}^{-1}=\left(\begin{array}{ccc}-\frac{1}{6}& -\frac{1}{6}& \frac{3}{8}\\ 0& -\frac{1}{2}& \frac{5}{16}\\ 0& 0& \frac{1}{8}\end{array}\right)$

$\frac{z^n-1}{z-1}$

$\frac{z^n-1}{z-1}=\frac{\cos n\theta -1+\mathrm{i}\sin n\theta }{\cos \theta -1+\mathrm{i}\sin \theta }$

$\frac{(\cos n\theta -1+\mathrm{i}\sin n\theta )(\cos \theta -1-\mathrm{i}\sin \theta )}{(\cos \theta -1+\mathrm{i}\sin \theta )(\cos \theta -1-\mathrm{i}\sin \theta )}$

$\mathrm{R}\mathrm{e}\left(\frac{z^n-1}{z-1}\right)=\frac{\cos n\theta \cos \theta +\sin n\theta \sin \theta -\cos n\theta -\cos \theta +1}{(\cos \theta -1{)}^2+{\sin }^2\theta }$

$=\frac{\cos n\theta \cos \theta +\sin n\theta \sin \theta -\cos n\theta }{2(1-\cos \theta )}+\frac{1}{2}$

$=\frac{\cos n\theta (\cos \theta -1)+\sin n\theta \sin \theta }{2(1-\cos \theta )}+\frac{1}{2}$

Divides through by denominator.

$=\frac{1}{2}(1-\cos n\theta +\frac{\sin n\theta \sin \theta }{1-\cos \theta })$

Forms sum of areas of rectangles given in the diagram.

${\int }_0^1\cos x\mathrm{d}x<\frac{1}{n}+\frac{1}{n}\cos \frac{1}{n}+\frac{1}{n}\cos \frac{2}{n}+\dots \frac{1}{n}\cos \frac{n-1}{n}$

Applies result from part (b) with $\theta =\frac{1}{n}$.

$=\frac{1}{n}(1+\cos \frac{1}{n}+\cos \frac{2}{n}+\dots +\cos \frac{n-1}{n})=\frac{1}{2n}(1-\cos \frac{n}{n}+\frac{\sin \frac{n}{n}\sin \frac{1}{n}}{1-\cos \frac{1}{n}})$

Forms sum of areas of rectangles.

${\int }_0^1\cos x\mathrm{d}x>\frac{1}{n}\cos \frac{1}{n}+\frac{1}{n}\cos \frac{2}{n}+\dots \frac{1}{n}\cos \frac{n}{n}$

$=\frac{1}{2n}(1-\cos 1+\frac{\sin 1\sin \frac{1}{n}}{1-\cos \frac{1}{n}})+\frac{1}{n}\cos 1-\frac{1}{n}=\frac{1}{2n}(\cos 1-1+\frac{\sin 1\sin \frac{1}{n}}{1-\cos \frac{1}{n}})$