A Level Further Mathematics 9231 Paper 13, May/June 2024

Evaluates determinant, forms quadratic expression. (Allow 1 slip in the calculation of the determinant for M1 only)

$=-5k^2-2$

No real value of $k\Rightarrow$ Non-singular

Uses $\mathbf{A}{\mathbf{A}}^{-1}=\mathbf{I}$ or $\det ({\mathbf{A}}^{-1})=(\det \mathbf{A}{)}^{-1}$ to find equation in $k$. Could also use a minor determinant e.g. $-k\times 1-3\times 0=0$

$k=0$

$y=x^2+1$

Substitutes and obtains an equation not involving radicals.

$y^3-y^2+4y-5=0$

Uses ${\alpha }^2+{\beta }^2+{\gamma }^2=(\alpha +\beta +\gamma {)}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )$ on their answer for part 2(a).

$-7$

Uses their equation from part 2(a) (or in a valid formula).

$4$

[One-way] stretch, shear

Stretch followed by shear

Stretch parallel to the $x$-axis, scale factor 7

Shear, $x$-axis fixed, with $(0,1)$ mapped to $(2,1)$.

$\mathbf{M}=\left(\begin{array}{cc}7& 2\\ 0& 1\end{array}\right)$

$\left(\begin{array}{cc}7& 2\\ 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}7x+2y\\ y\end{array}\right)$

$mx=m(7x+2mx)$

$2m^2+6m=0$

$y=0\text{ and }y=-3x$

$\text{Area of }PQR=\mathrm{\mid }7\mathrm{\mid }\times \text{Area of }DEF$

$\text{Area of }DEF=5{\text{ cm}}^2$

$1^2=\frac{1}{6}(1)(2)(3)\text{ so }{\mathrm{H}}_1\text{ is true.}$

$\text{Assume that }{\sum }_{r=1}^k{r}^2=\frac{1}{6}k(k+1)(2k+1)\mathrm{.}$

${\sum }_{r=1}^{k+1}{r}^2=\frac{1}{6}k(k+1)(2k+1)+(k+1{)}^2$

$\frac{1}{6}(k+1)(2k^2+k+6k+6)=\frac{1}{6}(k+1)(2k^2+7k+6)=\frac{1}{6}(k+1)(k+2)(2k+3)$

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

$3^5-1^5+5^5-3^5+7^5-5^5\dots +(2n+1{)}^5-(2n-1{)}^5$

$(2n+1{)}^5-1$

Sums RHS and applies standard formulae.

$160S_n+\frac{40}{3}n(n+1)(2n+1)+2n$

Makes $S_n$ the subject and factorises.

$(2n+1)(16n^4+32n^3+\frac{32}{3}{n}^2-\frac{16}{3}n)=\frac{16}{3}n(2n+1)(3n^3+6n^2+2n-1)\Rightarrow S_n=\frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)$

$n^{-5}{S}_n=\frac{1}{30}(1+\frac{1}{n})(2+\frac{1}{n})(3+\frac{3}{n}-\frac{1}{n^2})$

$\frac{1}{5}$

$\left(\begin{array}{c}-3\\ 4\\ 0\end{array}\right)-\left(\begin{array}{c}1\\ 4\\ -1\end{array}\right)=\left(\begin{array}{c}-4\\ 0\\ 1\end{array}\right)$

Find common perpendicular.

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

Uses formula for shortest distance.

$\frac{21}{\sqrt{30}}(=3.83)$

$5(1)-2(4)-1(-1)=-2$

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

$-\mathbf{i}-\mathbf{j}+\mathbf{k}$

Find common perpendicular.

$x+2y+3z=6$

$x^2+3=0$ has no real roots.

$y=0$

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

Sets equal to 0 and simplifies to quadratic equation.

$(-3,-\frac{1}{6})$

$(1,\frac{1}{2})$

Axes and approach to asymptote.

Correct smooth shape and position.

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

Sections for $x⩾0$ and $x<-1$ correct

$-1⩽x<0$ correct.

$(-1,0),(0,\pm \frac{1}{\sqrt{3}})$

$(1,\pm \frac{1}{\sqrt{2}})$

One loop correct with initial line.

Second loop correct with correct form at the pole.

$\theta =\frac{1}{2}\pi$

Use of $\sin 2\theta =2\sin \theta \cos \theta$, and $x=r\cos \theta$ or $y=r\sin \theta$.

$r^5=2y{x}^2$

$(x^2+y^2{)}^{\frac{5}{2}}=2x^{2}y$

Applies $\frac{1}{2}\int r^2\mathrm{d}\theta$.

Attempt to integrate in a valid way. May apply $\sin (2\theta )=2\sin \theta \cos \theta$.

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

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

Differentiates with respect to $\theta$ (or differentiates [RHS] with respect to $\theta$).

$\frac{\mathrm{d}r}{\mathrm{d}\theta }=\frac{1}{2}(\sin 2\theta \cos \theta {)}^{-\frac{1}{2}}(2\cos 2\theta \cos \theta -\sin 2\theta \sin \theta )$

Correct use of relevant identities to express in terms of a single trig function.

Sets derivative equal to 0 and solves to find one of ${\sin }^2\theta =\frac{1}{3}$, ${\cos }^2\theta =\frac{2}{3}$, ${\tan }^2\theta =\frac{1}{2}$ or $\theta =0.6155$

$6{\cos }^3\theta -4\cos \theta =0\Rightarrow {\cos }^2\theta =\frac{2}{3}$

$r=\sqrt{\frac{4}{3\sqrt{3}}}=\sqrt{\frac{4\sqrt{3}}{9}}=0.877$