A Level Mathematics 9709 Paper 12, May/June 2023

$[y]=\frac{4}{-2}(x-3{)}^{-3+1}$ or $\frac{4}{-2(x-3{)}^2}[+c]$

$5=\frac{4}{-2}(4-3{)}^{-2}+c$ or $5=\frac{4}{-2(4-3{)}^2}+c$ leading to $c=$

$y=\frac{-2}{(x-3{)}^2}+7$ or $y=-2(x-3{)}^{-2}+7$

[Coefficient of $x^4=p=]15a^2$

[Coefficient of $x^2=q=]54a^2$

Equating their $p+\text{their }q$ to 276 leading to an equation in $a^2$ only

$a=\pm 2$

$4(x-3{)}^2$ seen or $a=4\text{ and }b=-3$

$-36+p$ or $p-36$ seen or $c=p-36$

$p-36>0$ leading to $p>36$ or ${24}^2-4\times 4p(0\Rightarrow p)36$ or $36<p$

$[8x^6+215x^3-27=0]$ leading to $(8x^3-1)(x^3+27)[=0]$ OR $\frac{-215\pm \sqrt{{215}^2-\mathrm{4.8.}-27}}{2.8}$ or $\frac{-215\pm \sqrt{47089}}{2.8}$

$\frac{1}{8},-27$

$\frac{1}{2}$ or $0.5,-3$

$\left\{\frac{10}{\frac{3}{2}}{x}^{\frac{3}{2}}\right\}$

$\left\{\frac{5}{2\times \frac{5}{2}}{x}^{\frac{5}{2}}\right\}$

$=(\text{their }\frac{20}{3}\times 8-32)[-0]$

[Area of shaded region = ] $\frac{64}{3},21\frac{1}{3}$ or $21.3[333\dots ]$

$\frac{1}{2}OA=x\cos \theta$ or $\frac{OA}{\sin (\pi -2\theta )}=\frac{x}{\sin \theta }$ or $O{A}^2=x^2+x^2-2x^2\cos (\pi -2\theta )$ or $x^2=r^2+x^2-2rx\cos \theta$ or other valid method.

$OA=2x\cos \theta$ leading to $\text{Arc length}=2x\theta \cos \theta$

$\text{Sector area}=\frac{1}{2}(2x\cos \theta {)}^2\times \theta$

$\text{Triangle area}=\frac{1}{2}\times 2x\cos \theta \times x\sin \theta$ OR $\frac{1}{2}{x}^2\sin (\pi -2\theta )$

$[\text{Area }APB=]\text{ Their sector area}-\text{their triangle area}$

$[\text{Area }APB=]\frac{1}{2}(2x\cos \theta {)}^2\times \theta -\frac{1}{2}{x}^2\sin (\pi -2\theta )$ $[=x^2(2\theta {\cos }^2\theta -\frac{1}{2}\sin 2\theta )\text{ or }{x}^2\cos \theta (2\theta \cos \theta -\sin \theta )]$

${\cos }^2\theta +2\sin \theta \cos \theta +{\sin }^2\theta =1$ leading to $2\sin \theta \cos \theta =0$ or $\sin 2\theta =0$

$[\theta =]0,\frac{\pi }{2},\pi$

$\cos 0+\sin 0=[1+0=]1$ and $\cos \frac{\pi }{2}+\sin \frac{\pi }{2}=[0+1=]1$

$\cos \pi +\sin \pi =[-1+0=]-1$ or $\mathrm{\ne }1$

$\frac{(\cos \theta -\sin \theta )\sin \theta +(\cos \theta +\sin \theta )(1-\cos \theta )}{(\cos \theta +\sin \theta )(\cos \theta -\sin \theta )}$

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

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

$\frac{\cos \theta +\sin \theta -1}{1-2{\sin }^2\theta }=2(\cos \theta +\sin \theta -1)$ leading to $1=2(1-2{\sin }^2\theta )$

$k{\sin }^2\theta =1\text{ or }3$ leading to $\sin \theta =[\pm ]\sqrt{\frac{1\text{ or }3}{k}}$ $[4{\sin }^2\theta =1\text{ leading to }\sin \theta =\pm \frac{1}{2}]$

Solutions $0,\frac{1}{6}\pi ,\frac{1}{2}\pi ,\frac{5}{6}\pi$

The line $y=x$ correctly drawn.

Fully correct (needs to reach $y=2\pi$ and $x$-axis and cross the line $y=x$ in the correct squares).

$y=3+2\sin \frac{1}{4}x$ leading to $\sin \frac{1}{4}x=\frac{y\pm 3}{2}$

$x=4{\sin }^{-1}\left(\frac{y-3}{2}\right)$ leading to $[{\text{f}}^{-1}(x)\text{ or }y=]4{\sin }^{-1}\left(\frac{x-3}{2}\right)$

Continuing given graph from $y$ intercept to $-2\pi$.

Yes it does have an inverse, because the graph is always increasing OR because it is one-one OR because it passes the horizontal line test OR it is not a many to one [function].

{Stretch} {factor 4} {in $x$-direction}

{Stretch} {factor 2} {in $y$-direction}

{Translation} $\left(\begin{array}{c}0\\ 3\end{array}\right)$

$[ar=16,\frac{a}{1-r}=100]$ leading to $a=\frac{16}{r}$ and $a=100(1-r)$

$100(1-r)r=16$ leading to $[100r^2-100r+16=0]$

$(5r-4)(5r-1)=0$ OR $\frac{25\pm \sqrt{{25}^2-\mathrm{4.25.4}}}{2.25}$ leading to $r=\left[\frac{4}{5}\text{ or }\frac{1}{5}\right]$

$a=20,a=80$

$r=\frac{4}{5},\frac{1}{5}$

$[u_n=]\text{their }20\times \text{their }{\left(\frac{4}{5}\right)}^{n-1}[v_n=]\text{their }80\times \text{their }{\left(\frac{1}{5}\right)}^{n-1}$

Correctly separating the numerator and denominator of their ${\left(\frac{4}{5}\right)}^{n-1}$ or dividing two nth terms of the correct format and simplifying their terms in $r$

Given result clearly shown ($u_n=4^{n-2}\times v_n$ or $\frac{u_n}{v_n}=4^{n-2}$)

$(x-a{)}^2+{(\frac{1}{2}x+6-3)}^2=20$ or using $x=2y-12$

$\frac{5}{4}{x}^2+(3-2a)x+a^2-11[=0]$

$(3-2a{)}^2-4\times \frac{5}{4}(a^2-11)[=0]$

Using $a=4$: $(3-8{)}^2-5(5)=0$ (or $-a^2-12a+64=0\Rightarrow (a-4)(a+16)=0\Rightarrow a=4$)

$a=-16$

Centre $(4,3)$ identified or used or the point P is $(2,7)$

$\therefore \text{gradient of normal}=-2$

Forming normal equation using their gradient (not 0.5) and their centre or P

$\frac{y-3}{(x-4)}=-2$ or $y-7=-2(x-2)$

Diameter: $y-3=\frac{1}{2}(x-4)$ or $2(x-4)+2(y-3)\frac{\text{d}y}{\text{d}x}=0$ (or Vector approach / unsimplified equation in $x$ only with $y=-2x+c$)

$(x-4{)}^2+{(\frac{1}{2}x+1-3)}^2=20$ (or $(4-4c{)}^2-20((c-3{)}^2-4)[=0]$)

$x=0\text{ or }8,y=1\text{ or }5[(0,1)\text{ and }(8,5)]$ (or $[4c^2-88c+84=0]$)

Equations are $y-1=-2x$ and $y-5=-2(x-8)$ (or $c=21\text{ and }c=1\text{ or }y=-2x+21\text{ and }y=-2x+1$)

$\frac{\text{d}y}{\text{d}x}=\{k\frac{1}{2}(4x+1{)}^{-\frac{1}{2}}\}\{\times 4\}\{-1\}$

$2k(4x+1{)}^{-\frac{1}{2}}-1=0$ leading to $(4x+1{)}^{\frac{1}{2}}=2k$ or $\frac{2k}{(4x+1{)}^{\frac{1}{2}}}=1$

$x=\frac{4k^2-1}{4}$

$2\times 10.5(4x+1{)}^{-\frac{1}{2}}-1=2$

$7=(4x+1{)}^{\frac{1}{2}}$ leading to $4x+1=49$ leading to $x=12$

$y=[10.5\sqrt{4x+1}-x+5=]66.5[\text{leading to }(12,66.5)]$

$y-66.5=-\frac{1}{2}(x-12)$