A Level Mathematics 9709 Paper 12, May/June 2024

$240\left[x^2\right]$ or $80a^2\left[x^2\right]$

$240=12\times 80a^2$

$0.5$

Stretch factor 4 in $y$-direction/parallel to the $y$ axis/vertically.

Translation $\left(\begin{array}{c}3\\ 0\end{array}\right)$ or 3 parallel to the $x$ axis or in the $x$ direction, allow horizontally. $\left(\begin{array}{c}0\\ -8\end{array}\right)$ or $-8$ parallel to the $y$ axis or in the $y$ direction, allow vertically.

Two translations, one in each direction, and a stretch only.

Correct order of operations. The stretch which must be in the $y$ direction must come before the translation in the $y$ direction.

$7\frac{\sin \theta }{\cos \theta }÷\cos \theta +12[=0][\text{leading to }7\frac{\sin \theta }{\cos \theta }+12\cos \theta =0]$

$7\sin \theta +12(1-{\sin }^2\theta )[=0]$

$\Rightarrow 12{\sin }^2\theta -7\sin \theta -12=0$

$[12{\sin }^2\theta -7\sin \theta -12=0\text{ leading to}](4\sin \theta +3)(3\sin \theta -4)$

$\sin \theta =-\frac{3}{4}\left[\text{or }\frac{4}{3}\right]$

$[\theta =]{228.6}^{\circ },{311.4}^{\circ }$

$[{\text{f}}^{-1}(x)=](x+1{)}^2$

$0<\text{g}(x)⩽\frac{1}{2}$ or $\text{g}(x)>0$ and $\text{g}(x)⩽\frac{1}{2}$ or $(0,\frac{1}{2}]$

${\text{g}}^{-1}$ does not exist because it is one to many or ${\text{g}}^{-1}$ does not exist because it is not one to one. Or ${\text{g}}^{-1}$ does not exist because g is not one to one or ${\text{g}}^{-1}$ does not exist because g is many to one or ${\text{g}}^{-1}$ does not exist because g fails the horizontal line test.

$\text{f}\left(\frac{25}{16}\right)=\frac{1}{4}$

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

$[(\sqrt{x}-1{)}^2+2=4\text{ leading to}]\sqrt{x}-1=\sqrt{2}\text{ leading to }x=(1\pm \sqrt{2}{)}^2$ Or $[x-2\sqrt{x}+1+2=4\text{ leading to}]x-2\sqrt{x}-1=0\text{ leading to }x=(1\pm \sqrt{2}{)}^2$ Or $[x-1=2\sqrt{x}\text{ leading to}]x^2-6x+1=0\text{ leading to }x=\frac{6\pm \sqrt{36-4}}{2}$

$3+2\sqrt{2}$

$[d=]\sin \theta -\tan \theta$

$[d=]\frac{\sqrt{2}}{2}-1$

$[S_{40}=]\frac{40}{2}(2\times \tan \theta +39(\sin \theta -\tan \theta ))$

$390\sqrt{2}-740$ or $\frac{780}{\sqrt{2}}-740$

$r=\frac{\sin \theta }{\tan \theta }[=\cos \theta ]$

$S_{\mathrm{\infty }}=\frac{\tan \theta }{1-\cos \theta }$ or $\frac{\sin \theta }{\cos \theta -{\cos }^2\theta }$ or $\frac{{\tan }^2\theta }{\tan \theta -\sin \theta }$

$a=\sqrt{3}(\mathrm{1.73..})\text{ and }r=\frac{1}{2}$

$[S_{10}=]\sqrt{3}\left(\frac{1-{\left(\frac{1}{2}\right)}^{10}}{1-\frac{1}{2}}\right)$

$=3.46$

$\frac{\mathrm{d}y}{\mathrm{d}x}=2-\frac{1}{2}\times 8x^{-\frac{1}{2}}$

$2-4x^{-\frac{1}{2}}=0$

[$A$ is] $(4,-8)$ or $x=4,y=-8$

[$B$ is] $(16,0)$ or $x=16,y=0$

$[\pm ]\frac{2x^2}{2}-\frac{8}{\frac{3}{2}}{x}^{\frac{3}{2}}[+C]$

$[\pm ]\frac{x^2-32x}{3}$ or $\frac{(2x-32{)}^2}{12}[+C]$

Attempt to integrate, defined by at least one correct power in each expression, and then subtract.

$\{({16}^2-\frac{8}{\frac{3}{2}}\cdot {16}^{\frac{3}{2}})-(4^2-\frac{8}{\frac{3}{2}}\cdot 4^{\frac{3}{2}})\}[-]\{\left(\frac{{16}^2-32\times 16}{3}\right)-\left(\frac{4^2-32\times 4}{3}\right)\}$

$\frac{32}{3}$, $10\frac{2}{3}$ or $10.7$

$(x-6{)}^2+(2a-x+a{)}^2=18$

$2x^2-12x-6ax+9a^2+36-18[=0]$

$(12+6a{)}^2-4\times 2\times (9a^2+18)[=0]$

$-36a^2+144a[+0=0]$

$a=0,a=4$

[Centre is] $(6,-4)$ or [Point of intersection is] $(9,-1)$

[Gradient of diameter] $=1$

$y+4=x-6$ or $y+1=x-9[\text{leading to }y=x-10]$

$\frac{XE}{0.4}=\sin \frac{\pi }{6}$ or $\frac{XE}{0.4}=\cos \frac{\pi }{3}[XE=0.2]$

Length $EF=2+2\times 0.2=2.4$

$[CX=]0.4\cos \frac{\pi }{6}$ or $0.4\sin \frac{\pi }{3}$ or $\sqrt{{0.4}^2-{0.2}^2}$

$[\text{Sector}=]\frac{1}{2}\times (0.4{)}^2\times \frac{\pi }{3}$

Either Area of *their* (rectangle + two triangles + two sectors) Or Area of *their* (trapezium + two sectors)

$0.930$

$[\text{Length }AD=]2+2r$

$[\text{Arc length}=]r\times \frac{\pi }{3}$

$[EF=]2+2r\sin \frac{\pi }{6}$ or $2+2r\cos \frac{\pi }{3}$ or $2+r$

$[4+3r+\frac{2\pi r}{3}=6\text{ leading to }]0.393$

$6(2x-3{)}^2-6x<0$ or $=0$

$24x^2-78x+54$ or $4x^2-13x+9$ or $(x-1)(4x-9)$ OR $6(2x-3{)}^2<6x\text{ leading to }(2x-3)<\sqrt{x}\text{ leading to }2x-\sqrt{x}-3$

$[x=]1,\frac{9}{4}$

$1<x<\frac{9}{4}$ or $x>1$ and $x<\frac{9}{4}$ or $(1,\frac{9}{4})$

$\{\frac{6}{3\times 2}(2x-3{)}^3\}$

$\{-\frac{6}{2}{x}^2\}$

$-1=(-1{)}^3-3\times 1^2+C$

$[\text{f}(x)=](2x-3{)}^3-3x^2+3$

$[x=]-2$

$\frac{\mathrm{d}y}{\mathrm{d}x}=k(5-2x{)}^{\frac{1}{2}}[=-2\times \frac{3}{2}(5-2x{)}^{\frac{1}{2}}]$

$[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}\times \frac{\mathrm{d}t}{\mathrm{d}x}\text{ leading to}]-9=\pm 5\times \frac{\mathrm{d}t}{\mathrm{d}x}$

$\frac{5}{9}$ or $0.556=$

$k(5-2x{)}^{\frac{1}{2}}=-3$

[$B$ is] $(2,6)$

Gradient $AB=m_1=\frac{32-6}{-2-2}$, gradient of perpendicular $=-\frac{1}{m_1}=\frac{4}{26}$

Mid point is $(\frac{2-2}{2},\frac{6+32}{2})=(0,19)$

$y-19=\frac{2}{13}(x-0)$

$2x-13y+247=0$ or $\pm$ integer multiples of this.