A Level Mathematics 9709 Paper 13, May/June 2023

{Translation} $\left(\begin{array}{c}\{0\}\\ \{-2\}\end{array}\right)$

{Stretch}{[scale] factor 2} {parallel to $x$-axis}

$x^2-6x+c>2$ leading to $(x-3{)}^2-9+c>2$

$(x-3{)}^2-9+c>2$

$c>11-(x-3{)}^2$ and $(x-3{)}^2\ge 0$

$c>11$

$x^5+10x^3+40x+\frac{80}{x}+\frac{80}{x^3}+\frac{32}{x^5}$ or $x^5+10x^3+40x+80x^{-1}+80x^{-3}+32x^{-5}$

$their40\times a+(their\text{ coefficient of }{x}^{-1})\times b=0$

$(their\text{ coefficient of }{x}^{-1})\times a+(their\text{ coefficient of }{x}^{-3})\times b=80$

$a=2$

$b=-1$

$3{\sin }^{2}x-3{\sin }^{2}x{\cos }^{2}x-4{\cos }^{2}x[=0]$

$3(1-{\cos }^{2}x)-3(1-{\cos }^{2}x){\cos }^{2}x-4{\cos }^{2}x[=0]$

$3{\cos }^{4}x-10{\cos }^{2}x+3=0$ or $-3{\cos }^{4}x+10{\cos }^{2}x-3=0$

$(3{\cos }^{2}x-1)({\cos }^{2}x-3)[=0]$

$\cos x=[\pm ]\frac{1}{\sqrt{3}}$

${54.7}^{\circ },$

${125.3}^{\circ }$

$(x-1{)}^2+(x-9+4{)}^2=40$

$x^2-6x-7[=0]$ leading to $(x+1)(x-7)[=0]$

$(-1,-10)$ or $x=-1,y=-10$

$(7,-2)$ or $x=7,y=-2$

[C is mid-point =] $(\frac{their{x}_1+their{x}_2}{2},\frac{their{y}_1+their{y}_2}{2})$

$\text{Radius}=\sqrt{(theirx-their3{)}^2+(theiry-their(-6){)}^2}$ OR $their\sqrt{((7-(-1){)}^2+(-2-(-10){)}^2)}\mathrm{/}2$

$(x-3{)}^2+(y+6{)}^2=32$

$\frac{1\mathrm{/}2r^2\theta }{r\theta }=\frac{76.8}{9.6}$ or $\frac{1}{2}\left(\frac{{9.6}^2}{{\theta }^2}\right)\theta =76.8$

$r=16$

$\theta =0.6$

$\mathrm{\Delta }OAB=1\mathrm{/}2\times their{16}^2\times \sin their0.6$

[Area = $76.8-72.27=$] $4.53$

$AB=2\times 16\times \sin 0.3$ OR $A{B}^2={16}^2+{16}^2-2\times {16}^2\cos 0.6$

$\text{Perimeter}=9.6+9.46=19.1$

$[y]<2$ OR $[\text{f}(x)]<2$

$y=2-\frac{5}{x+2}$ leading to $y(x+2)=2(x+2)-5$ leading to $xy+2y=2x-1$

$2y+1=2x-xy$ leading to $2y+1=x(2-y)$

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

Domain is $x<2$

$\text{fg}(x)=2-\frac{5}{x+3+2}$

$=\frac{2(x+5)-5}{x+5}$ or $\frac{2(x+5)}{x+5}-\frac{5}{x+5}$

$=\frac{2x+5}{x+5}$

$r=\frac{a}{a+2}$

$\frac{a}{1-\frac{a}{a+2}}=264$

$\frac{a(a+2)}{a+2-a}=264$ leading to $\frac{a(a+2)}{2}=264$ leading to $a^2+2a-528[=0]$

$(a-22)(a+24)[=0]$

$a=22\text{ (only)}$

$d=\frac{6^2}{6+2}-6=-\frac{3}{2}$

$\left\{\frac{n}{2}[12+(n-1)\left(\frac{-3}{2}\right)]\right\}[<]-480$

$[3](n^2-9n-640)[>0]$

$[n=]\frac{9\pm \sqrt{81+2560}}{2}$

$31\text{ only}$

$\{x\}$

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

$\text{Sub }x=0,y=3\text{ leading to }3=0+1+c$

$y=x+(x-1{)}^{-2}+2$ or $\text{f}(x)=x+(x-1{)}^{-2}+2$

[Gradient of tangent =] ${\text{f}}^{\mathrm{\prime }}(0)=3$

Equation of tangent is $y-3=their\text{ gradient at }x=0(x-0)$

Intersection given by $3x+3=x+(x-1{)}^{-2}+2$

$2x+1=\frac{1}{(x-1{)}^2}\to (2x+1)(x-1{)}^2-1=0$ or solve equation before given form reached and show solution ($x=3\mathrm{/}2$) satisfies given result

Substitute $x=\frac{3}{2}$ leading to $(2x+1)(x-1{)}^2-1$ leading to $4\times 1\mathrm{/}4-1=0$. Hence $x=\frac{3}{2}$ If shown in (b) must be referenced here (in part (c))

When $x=\frac{3}{2}y=7\frac{1}{2}$

$\{9\}$

$\{-\frac{3}{2}(2x+1{)}^{1\mathrm{/}2}\times 2\}$

$9-3(2x+1{)}^{1\mathrm{/}2}=0$ leading to $2x+1=9$

Max point = (4, 9)

When $x=1\frac{1}{2}$, shows substitution or $\frac{\text{d}y}{\text{d}x}=3$

Gradient of $AB$ is $\frac{5\frac{1}{2}-3\frac{1}{2}}{1\frac{1}{2}-7\frac{1}{2}}[=-\frac{1}{3}]$

$-\frac{1}{3}\times 3=-1$. [Hence $AB$ is the normal]

$\left\{\frac{9x^2}{2}\right\}$

$\left\{\frac{-(2x+1{)}^{\frac{5}{2}}}{\frac{5}{2}\times 2}\right\}$

$\{\frac{9}{2}\cdot {7.5}^2-\frac{1}{5}(2\times 7.5+1{)}^{2.5}\}-\{\frac{9}{2}\cdot {1.5}^2-\frac{1}{5}(2\times 1.5+1{)}^{2.5}\}$ or $(\frac{9}{2}\times \frac{225}{4}-\frac{1024}{5})-(\frac{81}{8}-\frac{32}{5})$ or $\frac{1933}{40}-\frac{149}{40}$ or $48.325-3.725$

$\frac{1}{2}(5\frac{1}{2}+3\frac{1}{2})\times 6$ or ${\int }_{\frac{3}{2}}^{\frac{15}{2}}(-\frac{1}{3}x+6)\text{d}x=(-\frac{1}{6}{\left(\frac{15}{2}\right)}^2+6\times \frac{15}{2})-(-\frac{1}{6}{\left(\frac{3}{2}\right)}^2+6\times \frac{3}{2})$ or $\frac{285}{8}-\frac{69}{8}[=27]$

[Shaded area = $44.6-27=$] $17.6$