A Level Further Mathematics 9231 Paper 13, Oct/Nov 2024

$$\left(\begin{array}{cc}k& 0\\ 0& 1\end{array}\right)$$

$$\left(\begin{array}{cc}1& k\\ 0& 1\end{array}\right)$$

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

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

$$\left(\begin{array}{cc}k& k\\ 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}kx+ky\\ y\end{array}\right)$$

$$kx+ky=x$$

$$y=\frac{1-k}{k}x\text{ oe}$$

$${\mathbf{M}}^{-1}=\frac{1}{k}\left(\begin{array}{cc}1& -k\\ 0& k\end{array}\right)$$

$$\mathrm{\mid }k\mathrm{\mid }=3k^2$$

$$k\mathrm{\ne }0⟹k=\pm \frac{1}{3}$$

$$\frac{\mathrm{d}({\tan }^{-1}x)}{\mathrm{d}x}=\frac{1}{1+x^2}\text{ so true when }n=1.$$

$$\text{Assume that }\frac{{\mathrm{d}}^k}{\mathrm{d}{x}^k}({\tan }^{-1}x)={\mathrm{P}}_k(x)(1+x^2{)}^{-k},\text{ where }\mathrm{deg}{\mathrm{P}}_k(x)=k-1.$$

$$\frac{{\mathrm{d}}^{k+1}({\tan }^{-1}x)}{\mathrm{d}{x}^{k+1}}={\mathrm{P}}_k^{\mathrm{\prime }}(x)(1+x^2{)}^{-k}-2kx{\mathrm{P}}_k(x)(1+x^2{)}^{-k-1}$$ (differentiating using product rule)

$$\frac{{\mathrm{d}}^{k+1}({\tan }^{-1}x)}{\mathrm{d}{x}^{k+1}}={\mathrm{P}}_k^{\mathrm{\prime }}(x)(1+x^2{)}^{-k}-2kx{\mathrm{P}}_k(x)(1+x^2{)}^{-k-1}$$ (correct derivative)

$$=({\mathrm{P}}_k^{\mathrm{\prime }}(x)(1+x^2)-2kx{\mathrm{P}}_k(x))(1+x^2{)}^{-k-1}\text{ so }\mathrm{deg}{\mathrm{P}}_{k+1}(x)=k$$

$$\text{So true when }n=k+1.\text{ By induction, true for all positive integers }n\mathrm{.}$$

$$y=x^4$$

$$y+2y^{\frac{3}{4}}-1=0⟹16y^3=(1-y{)}^4$$

$$16y^3=1-4y+6y^2-4y^3+y^4$$

$$y^4-20y^3+6y^2-4y+1=0$$

$${\alpha }^4+{\beta }^4+{\gamma }^4+{\delta }^4=20$$

$$\alpha +\beta +\gamma +\delta =-2$$

$$x^5+2x^4-x=0⟹{\alpha }^5+{\beta }^5+{\gamma }^5+{\delta }^5=-2(20)+(-2)$$

$$-42$$

$${\alpha }^8+{\beta }^8+{\gamma }^8+{\delta }^8={20}^2-2(6)$$

$$388$$

$$\frac{5k}{(5r+k)(5r+5+k)}=k(\frac{1}{5r+k}-\frac{1}{5r+5+k})$$ (method for partial fractions)

$$\frac{5k}{(5r+k)(5r+5+k)}=k(\frac{1}{5r+k}-\frac{1}{5r+5+k})$$ (correct partial fractions)

$$\sum _{r=1}^n\frac{5k}{(5r+k)(5r+5+k)}=k(\frac{1}{5+k}-\frac{1}{10+k}+\frac{1}{10+k}-\frac{1}{15+k}+\cdots +\frac{1}{5n+k}-\frac{1}{5n+5+k})$$

$$=k(\frac{1}{5+k}-\frac{1}{5n+5+k})$$

$$\frac{k}{5+k}=\frac{1}{3}⟹3k=5+k$$

$$k=\frac{5}{2}$$

$$\sum _{r=n}^{n^2}\frac{5k}{(5r+k)(5r+5+k)}=\sum _{r=1}^{n^2}\frac{5k}{(5r+k)(5r+5+k)}-\sum _{r=1}^{n-1}\frac{5k}{(5r+k)(5r+5+k)}$$

$$=\frac{1}{2n+1}-\frac{1}{2n^2+3}$$

$$r^4=6r^2\sin \theta \cos \theta$$

$$r^2=3\sin 2\theta$$

Correct position and symmetrical about $\theta =\frac{1}{4}\pi$.

Single correct loop.

$$\sqrt{3}$$

$$\frac{3}{2}{\int }_0^{\frac{1}{2}\pi }\sin 2\theta \mathrm{d}\theta =\frac{3}{2}{[-\frac{1}{2}\cos 2\theta ]}_0^{\frac{1}{2}\pi }$$

$$\frac{3}{2}$$

$$y=3^{\frac{1}{2}}{\sin }^{\frac{1}{2}}2\theta \sin \theta$$

Sets $\frac{\mathrm{d}y}{\mathrm{d}\theta }=0$

$${\sin }^{-\frac{1}{2}}2\theta \cos \theta +{\sin }^{-\frac{1}{2}}2\theta \cos 2\theta \sin \theta =0$$

$$\sin 2\theta \cos \theta +\cos 2\theta \sin \theta =0⟹\tan 2\theta =-\tan \theta ⟹\frac{2\tan \theta }{1-{\tan }^2\theta }=-\tan \theta$$

$$\theta =\frac{1}{3}\pi$$

$$\frac{3^{\frac{5}{4}}}{2^{\frac{3}{2}}}=1.40$$

$$x=\frac{1}{2},x=3$$

$$y=2$$

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(2x^2-7x+3)(8x+1)-(4x^2+x+1)(4x-7)}{(2x^2-7x+3{)}^2}$$

$$-3x^2+2x+1=0$$

$$(-\frac{1}{3},\frac{1}{5})$$

$$(1,-3)$$

Axes and asymptotes. Clear identification (label or clear intersection with axes at correct place).

$x>3$ correctly approaching asymptotes, not too truncated.

$\frac{1}{2}<x<3$ correct.

$x<\frac{1}{2}$ correct.

$$(0,\frac{1}{3})$$

FT from sketch in (c). At least two branches.

$$k>3$$

$$\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 1& 1\\ 1& -4& 2\end{array}\mid$$ (method for common perpendicular)

$$=\left(\begin{array}{c}6\\ -3\\ -9\end{array}\right)\sim \left(\begin{array}{c}-2\\ 1\\ 3\end{array}\right)$$

$$-2(1)+(3)+3(-2)=-5$$

$$2x-y-3z=5$$

$$\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& -4& 2\\ 1& 1& -2\end{array}\mid$$ (method for normal to ${\mathrm{\Pi }}_2$)

$$=\left(\begin{array}{c}6\\ 4\\ 5\end{array}\right)$$

$$\left(\begin{array}{c}2\\ -1\\ -3\end{array}\right)\cdot \left(\begin{array}{c}6\\ 4\\ 5\end{array}\right)=\sqrt{14}\sqrt{77}\cos \theta ⟹\cos \theta =\frac{-7}{\sqrt{14}\sqrt{77}}$$

$${77.7}^{\circ }$$

Finds $\overrightarrow{PQ}$ using general points on $l_1$ and $l_2$

$$\overrightarrow{PQ}=\left(\begin{array}{c}\mu -2\lambda \\ -5-4\mu -\lambda \\ 11+2\mu -\lambda \end{array}\right)$$

$$\left(\begin{array}{c}\mu -2\lambda \\ -5-4\mu -\lambda \\ 11+2\mu -\lambda \end{array}\right)\cdot \left(\begin{array}{c}2\\ 1\\ 1\end{array}\right)=0\text{ or }\left(\begin{array}{c}\mu -2\lambda \\ -5-4\mu -\lambda \\ 11+2\mu -\lambda \end{array}\right)=k\left(\begin{array}{c}6\\ -3\\ -9\end{array}\right)$$

$$-6\lambda +6=0$$

$$\left(\begin{array}{c}\mu -2\lambda \\ -5-4\mu -\lambda \\ 11+2\mu -\lambda \end{array}\right)\cdot \left(\begin{array}{c}1\\ -4\\ 2\end{array}\right)=0⟹21\mu +42=0$$

$$\lambda =1⟹\overrightarrow{OP}=\left(\begin{array}{c}3\\ 4\\ -1\end{array}\right)\text{ or }\mu =-2⟹\overrightarrow{OQ}=\left(\begin{array}{c}-1\\ 6\\ 5\end{array}\right)$$

$$\mathbf{r}=\left(\begin{array}{c}-1\\ 6\\ 5\end{array}\right)+t\left(\begin{array}{c}6\\ -3\\ -9\end{array}\right)$$