A Level Further Mathematics 9231 Paper 21, May/June 2023

Shows that determinant is zero or row operations to obtain e.g. $x+2y+3z=1$, $y+2z=1$,

Correct evaluation showing determinant is zero.

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

$z=\frac{1}{3}(1-x-2y)⟹y=-2x-1$ correctly derived using all three equations.

The three planes form a sheaf.

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

Substitutes and derives first order separable equation.

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

Separates variables and integrates both sides.

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

Substitutes initial conditions into their expression.

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

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

Expands and groups.

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

Substitutes $z^n+z^{-n}=2\cos n\theta$.

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

Applies substitution, gets to $\int {\cos }^4\theta \mathrm{d}\theta$, changes limits.

Integration correct: $\frac{1}{8}{[\frac{1}{4}\sin 4\theta +2\sin 2\theta +3\theta ]}_0^{\frac{1}{6}\pi }$.

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

Uses the product rule to differentiate (or integrates by parts in alternative method).

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

Uses $x^5=x^5+1-1$.

Integrates both sides using the limits given (or forms recursive formula). Requires previous method mark.

Substitutes limits and rearranges to $(5n+1)I_n=2^n+5n{I}_{n-1}$. AG.

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

Applies reduction formula.

$$11I_2=2^2+10I_1⟹I_2=\frac{47}{33}$$

$$16I_3=2^3+15I_2⟹I_3=\frac{323}{176}$$

Sets determinant equal to zero.

Expands determinant. Can use other rows/columns.

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

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

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

First eigenvector correct for $\lambda =2$.

Second eigenvector correct for $\lambda =-4$.

Third eigenvector correct for $\lambda =6$.

Constructs matrix $\mathbf{P}$ and matrix $\mathbf{D}$.

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

Auxiliary equation.

Complementary function. Allow "$x=$" missing.

Particular integral and its derivatives.

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

Substitutes and equates coefficients. Must have two unknowns.

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

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

Differentiates using product rule. Must have their PI differentiated.

Forms simultaneous equations using initial conditions.

$$B=-\frac{12}{37}A+6B+\frac{35}{37}=0⟹A=1$$

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

Applies given substitution.

$$\frac{1}{2}\int \frac{1}{\sqrt{u}}\mathrm{d}u$$

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

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

Forms sum of the areas of the rectangles.

Compares with integral with correct limits.

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

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

Compares with integral with correct limits.

Correct evaluation of integral comparison.

Adds $\ln (N+\sqrt{N^2-1})$ to both sides (or reaches the corresponding final expression).

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

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

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

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

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 $\frac{\mathrm{d}x}{\mathrm{d}t}$.

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

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.

$$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=-8\frac{{\mathrm{sech}}^{2}t}{{\tanh }^{2}t}$$

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

$${\tanh }^{2}t=\frac{16}{25}$$

$x=\frac{8}{25}+\ln \frac{3}{5}$

$y=\frac{881}{625}$