A Level Further Mathematics 9231 Paper 22, Oct/Nov 2024

${\int }_6^7\frac{1}{\sqrt{(x-5{)}^2-1^2}}\mathrm{d}x={[{\cosh }^{-1}(x-5)]}_6^7$ (method)

Applies formula.

$={\cosh }^{-1}(2)-{\cosh }^{-1}(1)$

$=\ln (2+\sqrt{3})$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(4y^2\right)=8y{y}^{\mathrm{\prime }}$

$\frac{\mathrm{d}}{\mathrm{d}x}(4\ln (xy))=4\frac{x{y}^{\mathrm{\prime }}+y}{xy}=4y^{\mathrm{\prime }}{y}^{-1}+4x^{-1}$

$8\left(\frac{1}{2}\right)y^{\mathrm{\prime }}+4(2)y^{\mathrm{\prime }}+4\left(\frac{1}{2}\right)=0\Rightarrow y^{\mathrm{\prime }}=-\frac{1}{6}$

Differentiates $8y^{\mathrm{\prime }}y+4y^{\mathrm{\prime }}{y}^{-1}+4x^{-1}=0$: for $8(y^{\mathrm{\prime }}{)}^2+8y^{\mathrm{\prime }\mathrm{\prime }}y$ (or in Solution 2, for $x^{-1}{(2y+y^{-1})}^{-2}(2y^{\mathrm{\prime }}-y^{-2}{y}^{\mathrm{\prime }})$)

for other terms (or in Solution 2, for $x^{-2}{(2y+y^{-1})}^{-1}$)

Substitutes $(2,\frac{1}{2})$ (and $y^{\mathrm{\prime }}=-\frac{1}{6}$)

$12y^{\mathrm{\prime }\mathrm{\prime }}-\frac{11}{9}=0\Rightarrow y^{\mathrm{\prime }\mathrm{\prime }}=\frac{11}{108}$

$\dot{x}={\mathrm{e}}^{2t}-t^2$

$\dot{y}=2{\mathrm{e}}^t+2(t-1){\mathrm{e}}^t=2t{\mathrm{e}}^t$

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

$={({\mathrm{e}}^{2t}+t^2)}^2$

${\int }_0^1{\mathrm{e}}^{2t}+t^2\mathrm{d}t$

${[\frac{1}{2}{\mathrm{e}}^{2t}+\frac{1}{3}{t}^3]}_0^1$

$=\frac{1}{2}{\mathrm{e}}^2-\frac{1}{6}$

Expands and takes real part for $\cos 6\theta$

${\cos }^6\theta -15{\cos }^4\theta {\sin }^2\theta +15{\cos }^2\theta {\sin }^4\theta -{\sin }^6\theta$

Takes imaginary part for $\sin 6\theta$

$6{\cos }^5\theta \sin \theta -20{\cos }^3\theta {\sin }^3\theta +6\cos \theta {\sin }^5\theta$

Divides numerator and denominator by ${\sin }^6\theta$.

$\cot 6\theta =\frac{{\cot }^6\theta -15{\cot }^4\theta +15{\cot }^2\theta -1}{6{\cot }^5\theta -20{\cot }^3\theta +6\cot \theta }$

$x=\cot \theta ,\cot 6\theta =1$

$6\theta =\frac{1}{4}\pi +k\pi$

$\cot \left(\frac{1}{24}\pi \right)$

$\cot \left(\frac{5}{24}\pi \right),\cot \left(\frac{3}{8}\pi \right),\cot \left(\frac{13}{24}\pi \right),\cot \left(\frac{17}{24}\pi \right),\cot \left(\frac{7}{8}\pi \right)$

$3m^2+2m+1=0\Rightarrow m=-\frac{1}{3}\pm \frac{\sqrt{2}}{3}\mathrm{i}$

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

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

$p=14p+q=06p+2q+r=0$

$q=-4r=2$

$y={\mathrm{e}}^{-\frac{1}{3}x}(A\cos \frac{\sqrt{2}}{3}x+B\sin \frac{\sqrt{2}}{3}x)+x^2-4x+2$

$y^{\mathrm{\prime }}={\mathrm{e}}^{-\frac{1}{3}x}(-\frac{\sqrt{2}}{3}A\sin \frac{\sqrt{2}}{3}x+\frac{\sqrt{2}}{3}B\cos \frac{\sqrt{2}}{3}x)-\frac{1}{3}{\mathrm{e}}^{-\frac{1}{3}x}(A\cos \frac{\sqrt{2}}{3}x+B\sin \frac{\sqrt{2}}{3}x)+2x-4$

$A+2=0\frac{\sqrt{2}}{3}B-\frac{1}{3}A-4=0$

$A=-2,B=5\sqrt{2}$

$y={\mathrm{e}}^{-\frac{1}{3}x}(-2\cos \frac{\sqrt{2}}{3}x+5\sqrt{2}\sin \frac{\sqrt{2}}{3}x)+x^2-4x+2$

Forms the sum of the areas of the rectangles (M1 for correct number of rectangles).

$\left(\frac{1}{n}\right){\mathrm{e}}^{1-\frac{1}{n}}+\left(\frac{1}{n}\right){\mathrm{e}}^{1-\frac{2}{n}}+\cdots +\left(\frac{1}{n}\right){\mathrm{e}}^{1-\frac{n-1}{n}}$

Applies sum of geometric progression

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

Forms the sum of the areas of appropriate rectangles (M1 for correct number rectangles).

$\left(\frac{1}{n}\right){\mathrm{e}}^{1-\frac{1}{n}}+\left(\frac{1}{n}\right){\mathrm{e}}^{1-\frac{2}{n}}+\cdots +\left(\frac{1}{n}\right){\mathrm{e}}^{1-\frac{n-1}{n}}+\left(\frac{1}{n}\right){\mathrm{e}}^{1-\frac{n}{n}}$

Applies sum of geometric progression.

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

$U_n-L_n=\frac{\mathrm{e}-1}{n}$

$\frac{\mathrm{e}-1}{n}\to 0\text{ as }n\to \mathrm{\infty }$

Substitutes power series for ${\mathrm{e}}^{-\frac{1}{z}}$

$=1-\frac{1}{2z}+\frac{1}{6z^2}+\dots$ (for enough terms)

${\lim }_{n\to \mathrm{\infty }}(U_n)=\mathrm{e}-1$

$\frac{\mathrm{d}}{\mathrm{d}x}(\ln (\tanh x))$

$=\frac{{\mathrm{sech}}^{2}x}{\tanh x}$

$=\frac{1}{\sinh x\cosh x}=\frac{2}{\sinh 2x}$

$\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{2y}{\sinh (2x)}=1$

$\frac{\mathrm{d}}{\mathrm{d}x}(y\tanh x)$

$=\tanh x$

Integrates RHS

$y\tanh x=\ln (\cosh x)+C$

$5\left(\frac{3}{5}\right)=\ln \frac{5}{4}+C$

$y\tanh x=\ln (\cosh x)+3-\ln \frac{5}{4}$

$\mid \begin{array}{ccc}-2-\lambda & 0& 0\\ 0& 7-\lambda & 9\\ 4& 1& 7-\lambda \end{array}\mid =0$

$(-2-\lambda )((7-\lambda {)}^2-9)=0\Rightarrow (-2-\lambda )(49-14\lambda +{\lambda }^2-9)$

$\Rightarrow {\lambda }^3-12{\lambda }^2+12\lambda +80=0$

$\lambda =-2,4,10$

${\mathbf{A}}^3-12{\mathbf{A}}^2+12\mathbf{A}+80\mathbf{I}=0$

$\Rightarrow {\mathbf{A}}^4-12{\mathbf{A}}^3+12{\mathbf{A}}^2+80\mathbf{A}=0$

Substitutes ${\mathbf{A}}^3$

$=132{\mathbf{A}}^2-224\mathbf{A}-960\mathbf{I}$

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

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

$\lambda =4$: $\left(\begin{array}{c}0\\ 3\\ -1\end{array}\right)$

$\lambda =10$: $\left(\begin{array}{c}0\\ 3\\ 1\end{array}\right)$

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

$\mathbf{P}=\left(\begin{array}{ccc}2& 0& 0\\ 1& 3& 3\\ -1& -1& 1\end{array}\right)\text{ and }\mathbf{D}=\left(\begin{array}{ccc}625& 0& 0\\ 0& 1& 0\\ 0& 0& 2401\end{array}\right)$