A Level Further Mathematics 9231 Paper 21, Oct/Nov 2023

Evaluates determinant. Can expand along any row e.g. $-(36-36)+(-126+126)+k(84-84)$. If using row operations, they must show an inconsistent system for M1.

$=14(-9-6k)+4(-9+21k)+6(6+21)=0$

Two parallel planes, not identical.

Other plane not parallel.

Finds modulus and argument of $4+4\mathrm{i}\sqrt{3}$.

Method for finding one root.

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

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

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

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

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

Evaluates derivatives at $x=0$.

Uses logarithmic form of ${\tanh }^{-1}$.

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

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

Auxiliary equation.

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

Particular integral and its derivatives: $y=p{x}^2+qx+r\Rightarrow y^{\mathrm{\prime }}=2px+q\Rightarrow y^{\mathrm{\prime }\mathrm{\prime }}=2p$

Substitutes and equates coefficients: $3p=273q+4p=02p+2q+3r=0$

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

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

Differentiates. Must use product rule: $y^{\mathrm{\prime }}={\mathrm{e}}^{-x}(-\sqrt{2}A\sin \sqrt{2}x+\sqrt{2}B\cos \sqrt{2}x)-{\mathrm{e}}^{-x}(A\cos \sqrt{2}x+B\sin \sqrt{2}x)+18x-12$

Uses initial conditions: $A+2=2\sqrt{2}B-A-12=-8$

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

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

Differentiates $x$ and $y$ with respect to $t$.

Forms ${\dot{x}}^2+{\dot{y}}^2$.

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

Applies correct formula for arc length: ${\int }_0^3\sqrt{{\dot{x}}^2+{\dot{y}}^2}\mathrm{d}t$.

Answer must be simplified to $4\sqrt{3}$.

Finds first derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$.

Differentiates $\frac{\mathrm{d}y}{\mathrm{d}x}$ with respect to $t$.

Applies chain rule: $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\times \frac{\mathrm{d}t}{\mathrm{d}x}$.

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

$0<t<1$

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

Expands product of exponentials.

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

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

Applies identities to find integral in terms of $u$.

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

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

Method to find integrating factor ${\mathrm{e}}^{\int \tanh x\mathrm{d}x}$.

Integrating factor $=\cosh x$.

Correct form on LHS and attempt to integrate RHS: $\frac{\mathrm{d}}{\mathrm{d}x}(y\cosh x)={\sinh }^{2}2x\cosh x$.

Integrates RHS using their part (b).

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

Substitutes initial conditions to find $C=4$.

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

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

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

Eigenvector for $\lambda =-6$: $\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$

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

Eigenvector for $\lambda =8$: $\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right)$

Forms $\mathbf{P}$ and $\mathbf{D}$ matching their (non-zero) eigenvectors to their eigenvalues.

$$\mathbf{P}=\left(\begin{array}{ccc}1& 1& 2\\ 0& 2& 1\\ 0& 0& 2\end{array}\right)\text{ and }\mathbf{D}=\left(\begin{array}{ccc}-\frac{1}{6}& 0& 0\\ 0& -\frac{1}{2}& 0\\ 0& 0& \frac{1}{8}\end{array}\right)$$ (or correctly matched permutations of columns)

Substitutes $\mathbf{A}$ into characteristic equation: ${\mathbf{A}}^3-52\mathbf{A}-96\mathbf{I}=0$.

Multiplies through by ${\mathbf{A}}^{-1}$: $96{\mathbf{A}}^{-1}={\mathbf{A}}^2-52\mathbf{I}$.

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

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

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

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

Multiplies numerator and denominator by complex conjugate.

Takes real part. $\cos (n-1)\theta =\cos n\theta \cos \theta +\sin n\theta \sin \theta$.

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

Factorises.

Divides through by denominator.

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

Forms sum of areas of rectangles given in the diagram.

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

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

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

Forms sum of areas of rectangles.

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

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