A Level Further Mathematics 9231 Paper 13, Oct/Nov 2023

$r^2+2r+1-r^2=2r+1$

$2{\sum }_{r=1}^{n}r+n=(n+1{)}^2-1^2$

Uses method of differences and sums both sides.

$\Rightarrow 2{\sum }_{r=1}^{n}r=n^2+n=n(n+1)$

${\sum }_{r=1}^n(r+a)={\sum }_{r=1}^{n}r+an$

$\frac{1}{2}n(n+1)+an=n$

$a=\frac{1}{2}(1-n)$

$1=\frac{1-2x+x^2}{(1-x{)}^2}=\frac{(1-x{)}^2}{(1-x{)}^2}$ so ${\text{H}}_1$ is true.

Assume that ${\sum }_{r=1}^{k}r{x}^{r-1}=\frac{1-(k+1)x^k+k{x}^{k+1}}{(1-x{)}^2}$

${\sum }_{r=1}^{k+1}r{x}^{r-1}=\frac{1-(k+1)x^k+k{x}^{k+1}}{(1-x{)}^2}+(k+1)x^k$

$\frac{1-(k+1)x^k+k{x}^{k+1}+(k+1)x^k(1-2x+x^2)}{(1-x{)}^2}$

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

So ${\text{H}}_{k+1}$ is true. By induction, ${\text{H}}_n$ is true for all positive integers $n$.

$b=-(\alpha +\beta +\gamma +\delta )=-3$

$5=(-3{)}^2-2(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta )$

Uses formula for sum of squares.

$c=2$

$6=\frac{\alpha \beta \gamma +\beta \gamma \delta +\gamma \delta \alpha +\delta \alpha \beta }{\alpha \beta \gamma \delta }=\frac{-d}{-2}$

$d=12$

${\alpha }^4+{\beta }^4+{\gamma }^4+{\delta }^4=3(-27)-2(5)-12(3)+2(4)$

$-119$

$\left(\begin{array}{c}2\\ -2\\ 3\end{array}\right)-\left(\begin{array}{c}-2\\ -3\\ -5\end{array}\right)=\left(\begin{array}{c}4\\ 1\\ 8\end{array}\right)$

$\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -4& 3& 5\\ 2& -3& 1\end{array}\mid =\left(\begin{array}{c}18\\ 14\\ 6\end{array}\right)\sim \left(\begin{array}{c}9\\ 7\\ 3\end{array}\right)$

Find common perpendicular.

$\frac{1}{\sqrt{139}}\mid \begin{array}{c}\left(\begin{array}{c}4\\ 1\\ 8\end{array}\right)\cdot \left(\begin{array}{c}9\\ 7\\ 3\end{array}\right)\end{array}\mid =\frac{67}{\sqrt{139}}(=5.68)$

Uses formula for shortest distance.

$\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 0& 1\\ -4& 3& 5\end{array}\mid =\left(\begin{array}{c}-3\\ -9\\ 3\end{array}\right)\sim \left(\begin{array}{c}1\\ 3\\ -1\end{array}\right)$

Finds vector perpendicular to the plane.

$1(-1)+3(-3)-1(-4)=-6\Rightarrow x+3y-z=-6$

Uses point in the plane.

$\mid \begin{array}{cc}1& 3\\ 2& 5\end{array}\mid -k\mid \begin{array}{cc}2& 3\\ 3& 5\end{array}\mid +3\mid \begin{array}{cc}2& 1\\ 3& 2\end{array}\mid =0\Rightarrow -1-k+3=0\Rightarrow k=2$

Sets determinant of A equal to zero.

$\left(\begin{array}{ccc}-2& -1& 1\\ 1& 1& 3\end{array}\right)\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\\ 3& 2& 5\end{array}\right)\left(\begin{array}{cc}0& -2\\ -1& 3\\ 0& 0\end{array}\right)=\left(\begin{array}{ccc}-2& -1& 1\\ 1& 1& 3\end{array}\right)\left(\begin{array}{cc}-2& 4\\ -1& -1\\ -2& 0\end{array}\right)$

$\left(\begin{array}{cc}3& -7\\ -9& 3\end{array}\right)$

Completing matrix multiplication, AG.

$\left(\begin{array}{cc}3& -7\\ -9& 3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}3x-7y\\ -9x+3y\end{array}\right)$

$-9x+3mx=m(3x-7mx)$

Uses $y=mx$ and $Y=mX$.

$-9+3m=3m-7m^2\Rightarrow 7m^2=9$

$y=\frac{3}{\sqrt{7}}x$ and $y=-\frac{3}{\sqrt{7}}x$

$\mathbf{D}=\left(\begin{array}{cc}\alpha & 0\\ 0& \alpha \end{array}\right)$

$\mathbf{E}=\left(\begin{array}{cc}\beta & 0\\ 0& 1\end{array}\right)$

$\mathbf{F}=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$

$\left(\begin{array}{cc}3& -7\\ -9& 3\end{array}\right)=\left(\begin{array}{cc}\alpha & 0\\ 0& \alpha \end{array}\right)-9\left(\begin{array}{cc}0& \beta \\ 1& 0\end{array}\right)$

$\mathbf{D}=\left(\begin{array}{cc}3& 0\\ 0& 3\end{array}\right)\mathbf{E}=\left(\begin{array}{cc}\frac{7}{9}& 0\\ 0& 1\end{array}\right)$

$x^2-x+\frac{1}{4}+y^2=\frac{1}{4}\Rightarrow r^2-r\cos \theta +\frac{1}{4}=\frac{1}{4}$

$r(r-\cos \theta )=0$

$[r\mathrm{\ne }0\Rightarrow ]r=\cos \theta$

$\sin 2\theta =\cos \theta \Rightarrow 2\sin \theta \cos \theta =\cos \theta$

$\cos \theta \mathrm{\ne }0\Rightarrow \sin \theta =\frac{1}{2}$

$(\frac{1}{2}\sqrt{3},\frac{1}{6}\pi )$

Initial line drawn and one curve correct.

Other curve correct.

Intersection marked in correct position and both curves labelled.

$\frac{1}{2}{\int }_0^{\frac{1}{6}\pi }{\sin }^{2}2\theta \mathrm{d}\theta +\frac{1}{2}{\int }_{\frac{1}{6}\pi }^{\frac{1}{2}\pi }{\cos }^2\theta \mathrm{d}\theta$

$\frac{1}{2}{\int }_0^{\frac{1}{6}\pi }{\sin }^{2}2\theta \mathrm{d}\theta =\frac{1}{4}{\int }_0^{\frac{1}{6}\pi }1-\cos 4\theta \mathrm{d}\theta$

$=\frac{1}{4}{[\theta -\frac{1}{4}\sin 4\theta ]}_0^{\frac{1}{6}\pi }$

$\frac{1}{2}{\int }_{\frac{1}{6}\pi }^{\frac{1}{2}\pi }{\cos }^2\theta \mathrm{d}\theta =\frac{1}{4}{\int }_{\frac{1}{6}\pi }^{\frac{1}{2}\pi }1+\cos 2\theta \mathrm{d}\theta$

$=\frac{1}{4}{[\theta +\frac{1}{2}\sin 2\theta ]}_{\frac{1}{6}\pi }^{\frac{1}{2}\pi }$

$\frac{1}{4}(\frac{1}{6}\pi -\frac{1}{8}\sqrt{3})+\frac{1}{4}(\frac{1}{2}\pi -\frac{1}{6}\pi -\frac{1}{4}\sqrt{3})=\frac{1}{8}(\pi -\frac{3}{4}\sqrt{3})$

$x=-1$, $x=2$

$y=1$

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

$x^2+8x-2=0$

$(-8.2,0.9)$

$(0.2,-0.9)$.

Axes and all three asymptotes.

Correct shape and position, crossing horizontal asymptote.

States $(0,-1)$ coordinates of intersection with axes, may be seen on diagram.

FT from sketch in (c)

All correct.

$\frac{x^2+2}{x^2-x-2}=1$ or $\frac{x^2+2}{x^2-x-2}=-1$ $x+4=0$ or $2x^2-x=0$

$x=-4$ or $x=0$, $x=\frac{1}{2}$

$-4<x<-1$, $0<x<\frac{1}{2}$, $x>2$