A Level Further Mathematics 9231 Paper 13, Oct/Nov 2025

Substitutes correct formulae from MF19 into expression

$$2n^2(n+1{)}^2+2n(n+1)(2n+1)+2n(n+1)+3n$$

$$=2n^4+8n^3+10n^2+7n$$

Puts over a common denominator and expands

$$\frac{1}{r^4}-\frac{1}{(r+1{)}^4}=\frac{(r+1{)}^4-r^4}{r^4(r+1{)}^4}=\frac{r^4+4r^3+6r^2+4r+1-r^4}{r^4(r+1{)}^4}=\frac{4r^3+6r^2+4r+1}{r^4(r+1{)}^4}$$

Uses method of differences

Shows three complete terms, including first and last: $$1-\frac{1}{2^4}+\frac{1}{2^4}-\frac{1}{3^4}+\cdots +\frac{1}{n^4}-\frac{1}{(n+1{)}^4}$$

$$=1-\frac{1}{(n+1{)}^4}$$

1

Shear, [$x$-axis fixed]

With $(0,1)$ mapped to $(\frac{3}{2},1)$.

Shear, [$y$-axis fixed],

with $(1,0)$ mapped to $(1,\frac{3}{2})$.

$$\mathbf{A}\mathbf{B}=\left(\begin{array}{cc}\frac{13}{4}& \frac{3}{2}\\ \frac{3}{2}& 1\end{array}\right)$$ or $$\det \mathbf{A}\mathbf{B}=\det \mathbf{A}\det \mathbf{B}$$

$$\det \mathbf{A}\mathbf{B}=1$$

$$\left(\begin{array}{cc}\frac{13}{4}& \frac{3}{2}\\ \frac{3}{2}& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}\frac{13}{4}x+\frac{3}{2}y\\ \frac{3}{2}x+y\end{array}\right)$$

Uses $Y=mX$

$$\frac{3}{2}x+mx=m(\frac{13}{4}x+\frac{3}{2}mx)$$

$$\frac{3}{2}+m=\frac{13}{4}m+\frac{3}{2}{m}^2⟹m^2+\frac{3}{2}m-1=0$$

$$y=-2x\text{ and }y=\frac{1}{2}x\mathrm{.}$$

$$\frac{\text{d}}{\text{d}x}(x\cos x)=\cos x-x\sin x=(-1{)}^1(x\sin x-(2(1)-1)\cos x)$$

Assume true for $n=k$, so $\frac{{\text{d}}^{2k-1}}{\text{d}{x}^{2k-1}}(x\cos x)=(-1{)}^k(x\sin x-(2k-1)\cos x)$

Differentiates once

$\frac{{\text{d}}^{2k}}{\text{d}{x}^{2k}}(x\cos x)=(-1{)}^k(x\cos x+2k\sin x)$

Differentiates again

$$\frac{{\text{d}}^{2k+1}}{\text{d}{x}^{2k+1}}(x\cos x)=(-1{)}^k(-x\sin x+\cos x+2k\cos x)=(-1{)}^{k+1}(x\sin x-(2k+1)\cos x)$$

So, it is also true for $n=k+1$. Hence, [by induction], $\frac{{\text{d}}^{2n-1}}{\text{d}{x}^{2n-1}}(x\cos x)=(-1{)}^n(x\sin x-(2n-1)\cos x)$ is true for all positive integers.

$$y=2x+1⟹x=\frac{1}{2}(y-1)$$

Expands substitution into the equation

$$\frac{1}{16}(y-1{)}^4+\frac{1}{8}(y-1{)}^3+\frac{1}{4}(y-1{)}^2+\frac{1}{2}(y-1)+1=0$$ $$\frac{1}{16}(y^4-4y^3+6y^2-4y+1)+\frac{1}{8}(y^3-3y^2+3y-1)+\frac{1}{4}(y^2-2y+1)+\frac{1}{2}(y-1)+1=0$$

$$\frac{1}{16}{y}^4-\frac{1}{8}{y}^3+\frac{1}{4}{y}^2+\frac{1}{8}y+\frac{11}{16}=0⟹y^4-2y^3+4y^2+2y+11=0$$

$$S_2=2^2-2(4)$$

$$S_2=-4$$

$$S_4-2S_3+4S_2+2S_1+44=0$$

$$S_4=-76$$

Finds perpendicular to ${\mathrm{\Pi }}_1$ via cross product

$$\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 0& 1& 2\\ 1& 3& 1\end{array}\mid =\left(\begin{array}{c}-5\\ 2\\ -1\end{array}\right)\sim \left(\begin{array}{c}5\\ -2\\ 1\end{array}\right)$$

Uses point on ${\mathrm{\Pi }}_1$: $$5(-3)-2(-1)+(-1)=-14$$

$$5x-2y+z=-14$$

Uses correct formula for perpendicular distance from point to ${\mathrm{\Pi }}_1$

$$\frac{1}{\sqrt{5^2+2^2+1^2}}\mid \left(\begin{array}{c}-2\\ -1\\ 1\end{array}\right)\cdot \left(\begin{array}{c}-5\\ 2\\ -1\end{array}\right)\mid$$

$$\frac{7}{\sqrt{30}}$$

States point common to both planes e.g. $\left(\begin{array}{c}0\\ 7\\ 0\end{array}\right)$.

Finds direction of line

$$\mid \begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 5& -2& 1\\ 3& 2& -1\end{array}\mid =\left(\begin{array}{c}0\\ 8\\ 16\end{array}\right)\sim \left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)$$

$$\mathbf{r}=\left(\begin{array}{c}0\\ 7\\ 0\end{array}\right)+\lambda \left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)$$

Initial line drawn, pole clear and one loop in the first quadrant.

Symmetrical with correct shape at extremities.

$$\theta =\frac{1}{6}\pi \mathrm{.}$$

Uses $\frac{1}{2}\int r^2\text{d}\theta$ with correct limits.

Applies double angle formula correctly and integrates.

$$=\frac{1}{4}{\int }_0^{\frac{1}{3}\pi }1-\cos 6\theta \text{d}\theta =\frac{1}{4}{[\theta -\frac{1}{6}\sin 6\theta ]}_0^{\frac{1}{3}\pi }$$

$$=\frac{1}{12}\pi$$

$$y=3{\sin }^2\theta -4{\sin }^4\theta$$

$$6\sin \theta \cos \theta -16{\sin }^3\theta \cos \theta =0$$

$$\sin \theta \cos \theta \mathrm{\ne }0⟹6-16{\sin }^2\theta =0$$

$${\sin }^2\theta =\frac{3}{8}$$

$$y=\frac{9}{16}\mathrm{.}$$

$$r^4=r\sin \theta (3r^2-4r^2{\sin }^2\theta )$$ $$r=\frac{3y}{r}-4\frac{y^3}{r^3}$$

Uses $r^2=x^2+y^2$

$$(x^2+y^2{)}^2=y(3x^2-y^2)$$

$$x=-\frac{3}{2}-\frac{1}{2}\sqrt{5},x=-\frac{3}{2}+\frac{1}{2}\sqrt{5}$$

$$y=0$$

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

$$x^2+4x+5=0$$

$$\text{Discriminant is }{4}^2-20=-4<0\text{ so no real roots.}$$

Axes and asymptotes labelled.

Correct shape and position.

States coordinates of intersections with axes, $(0,2)$ and $(-2,0)$, may be seen on diagram.

FT from sketch in part 7(c)

From a correct graph and with correct shape at $(-2,0)$.

Finds critical points for case $\frac{x+2}{x^2+3x+1}=2$

Finds critical points for case $\frac{x+2}{x^2+3x+1}=-2$

$$x=-\frac{5}{2},x=0\text{or}x=-\frac{7}{4}-\frac{1}{4}\sqrt{17},x=-\frac{7}{4}+\frac{1}{4}\sqrt{17}$$

$$-\frac{7}{4}-\frac{1}{4}\sqrt{17}<x<-\frac{5}{2},-\frac{7}{4}+\frac{1}{4}\sqrt{17}<x<0,x\mathrm{\ne }-\frac{3}{2}\pm \frac{1}{2}\sqrt{5}$$