A Level Mathematics 9709 Paper 12, Oct/Nov 2023

[Coefficient of $x^3$ from $(3+2ax{)}^5=$] $10\times 9\times 8a^3[=720a^3]$

[Coefficient of $x^2$ from $(2+ax{)}^6=$] $15\times 16\times a^2[=240a^2]$

$their(10\times 9\times 8a^3)=6\times their(15\times 16\times a^2)[\Rightarrow 720a^3=1440a^2]$

$a=2$

$[{\tan }^{-1}4x=](their-\frac{\pi }{6})\pm \frac{\pi }{6}[{\tan }^{-1}4x=\pm \frac{\pi }{3},\pm 1.047\text{ or }0]$

$[4x=-\sqrt{3},x=]-\frac{\sqrt{3}}{4}$

[Gradient of normal =] $\frac{-1}{\text{Their }\frac{11}{2}}[\frac{-1}{\frac{11}{2}}=-\frac{2}{11}]$

$\frac{y-8}{x-2}=-\frac{2}{11}$ or $11y+2x=92$ or $y=-\frac{2x}{11}+\frac{92}{11}$

$[y=]\{\frac{1}{2}{x}^2÷2\}$

$[y=]\{\frac{72}{x^3}÷-3\}[+c][\frac{x^2}{4}-\frac{24}{x^3}+c]$

$8=\frac{1}{4}\times 4-\frac{24}{8}+c$

$y=\left(\frac{1}{4}\text{ or }0.25\right)x^2-\frac{24}{x^3}+10$

$[\text{Arc length}=]2\times \frac{\pi }{3}\text{ or }\frac{60}{360}\times 2\pi \times 2$

$[\text{Perimeter}=]2\pi \text{ or }6.28$

$[\text{Area of one sector}=]\frac{1}{2}\times 2^2\times \frac{\pi }{3}\text{ or }\frac{60}{360}\times \pi \times 2^2[=\frac{2\pi }{3}\text{ or }2.09]$

$[\text{Area of triangle}=]\frac{1}{2}\times 2^2\times \sin \left(\frac{\pi }{3}\right)\text{ or other valid method}[=\sqrt{3}\text{ or }1.73]$

$[\text{Area of coin}=3\text{ segments}+\text{triangle}\Rightarrow ]3(\frac{2\pi }{3}-\sqrt{3})+\sqrt{3}[=2.82]$

$2\pi -2\sqrt{3}\text{ or }2(\pi -\sqrt{3})$

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

${\cos }^2\theta +{\sin }^2\theta =2\sin \theta$ leading to $\sin \theta =\left[\frac{1}{2}\right]$

$[\theta =]\frac{\pi }{6}\text{ or }0.524$

$a=\frac{1}{2}r=\sqrt{3}$

$S_{10}=\sin \left(their\frac{\pi }{6}\right)\left(\frac{1-(their\sqrt{3}{)}^{10}}{1-(their\sqrt{3})}\right)$

$[S_{10}=]\frac{121}{\sqrt{3}-1}$

$\frac{\text{d}}{\text{d}x}(x^2-8x+5)=0[2x-8=0]$ OR $y=(x-4{)}^2-11$ OR $x=\frac{-b}{2a}=\frac{\pm 8}{2}$

$x=4,y=-11$

$x=(theirx\text{ value from }a)+4[=8]$

$y=\{(theiry\text{ value from }a)\times 2\}+1[-21]$

$2(x^2-8x+5)\text{ or }2\{(x-4{)}^2-11\}$

For the $x$ translation, each $x$ becomes $(x-4)$

For the $y$ translation of $+1$

$y=2x^2-32x+107\text{ or }a=2,b=-32,c=107$

$(2x-1)(4x^2+2x-1)=8x^3+4x^2-2x-4x^2-2x+1=8x^3-4x+1$

For use of $\tan \theta =\frac{\sin \theta }{\cos \theta }$ in the numerator and denominator (or use of ${\sin }^2\theta +{\cos }^2\theta =1$ twice / use of $1+{\tan }^2\theta ={\sec }^2\theta$ twice)

For use of ${\sin }^2\theta +{\cos }^2\theta =1$ twice, in a correct expression, resulting in an expression in ${\cos }^2\theta$ (or dividing throughout by ${\cos }^2\theta$ / multiplying throughout by ${\cos }^2\theta$ to give the RHS)

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

$\frac{1}{1-2{\cos }^2\theta }=4\cos \theta$ leading to $1=4\cos \theta (1-2{\cos }^2\theta )[8{\cos }^3\theta -4\cos \theta +1=0]$

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

$[x\text{ or }\cos \theta =]\frac{1}{2}\text{ and }\frac{-2\pm \sqrt{4+16}}{8}\text{OR}0.31,-0.81\text{ AWRT}$

$[\theta =]{60}^{\circ },{72}^{\circ },{144}^{\circ }$

$y=(x+a{)}^2-a$ leading to $(x+a{)}^2=y\pm a$ OR $x=(y+a{)}^2-a$ leading to $y^2+2ay+a^2-a-x[=0]$

$x=[\pm ]\sqrt{y\pm a}\pm a$ OR $y=\frac{-2a\pm \sqrt{4a^2-4(a^2-a-x)}}{2}$

$[y\text{ or }{f}^{-1}(x)=]-\sqrt{x+a}-a$

$x⩾-a$

$y\text{ or }{f}^{-1}[(x)]⩽-a$

$[\text{gf}\left(\frac{7}{2}\right)=]2({(x+\frac{7}{2})}^2-\frac{7}{2})-1$ or $2x^2+4\left(\frac{7}{2}\right)x+2{\left(\frac{7}{2}\right)}^2-2\left(\frac{7}{2}\right)-1[=0]$

$[x=]-\frac{7}{2}\pm 2$ or $\frac{-14\pm \sqrt{{14}^2-4\times 2\times \frac{33}{2}}}{4}\left[\frac{-14\pm \sqrt{64}}{4}\right]$ or factorising

$[x=]-\frac{11}{2}$

Equating the two expressions in $x$ and then multiplying each term by $x^{\frac{1}{2}}$ or by their substitution for $x^{\frac{1}{2}}$

Solving their three-term quadratic

$x=1\text{ and }25$

$y=15\text{ and }12\frac{3}{5}$

Attempt to integrate, defined by at least one correct fractional power, and subtract – condone the wrong way round.

B1 for either { } (or one mark for each correct expression)

B1 for completely correct integration of their expression following through +/- sign errors from the subtraction.

Substitution of their positive limits from part (a) in their integrated expression (or in both integrated expressions), defined by at least one correct fractional power, and subtraction.

$[\text{Area}=]\frac{128}{3},42\frac{2}{3},42.7$

$\frac{\text{d}y}{\text{d}x}=\{\frac{5}{3}(4x-3{)}^{\frac{2}{3}}\}\{x4\}-\left\{\frac{20}{3}\right\}$

$[\frac{20}{3}(4x-3{)}^{\frac{2}{3}}-\frac{20}{3}=0]$ leading to $(4x-3{)}^2=k,k>0$ leading to $4x-3=\pm m$

$[4x-3=\pm 1][x=]\frac{1}{2},1$

$\frac{{\text{d}}^{2}y}{\text{d}{x}^2}=\frac{40}{9}(4x-3{)}^{-\frac{1}{3}}\times 4$

$[x=\frac{1}{2}]\frac{{\text{d}}^{2}y}{\text{d}{x}^2}<0\text{ so max and }[x=1]\frac{{\text{d}}^{2}y}{\text{d}{x}^2}>0\text{ so min}$

$x<\frac{1}{2},x>1$

$\left(their\frac{7-4}{p-6}\right)\left(their\frac{18-7}{14-p}\right)=-1$ OR Scalar product leading to $(14-p)(6-p)-33=0$ OR correct use of Pythagoras with A,B and C OR with the centre, B and one of the other two points

$p^2-20p+84=33$ leading to $p^2-20p+51[=0]$ or $p^2-20p=-51$ (or $2p^2-40p+102[=0]$)

$[2](p-3)(p-17)\text{ or }\frac{20\pm \sqrt{{20}^2-4\times 51}}{2}$

$p=3$

[Midpoint or Centre is] (10, 11)

$\frac{1}{2}\sqrt{(14-6{)}^2+(18-4{)}^2}$ or $(18-their11{)}^2+(14-their10{)}^2$ or $(their11-4{)}^2+(their10-6{)}^2[r^2=65\text{ or }r=\sqrt{65}]$

$(x-10{)}^2+(y-11{)}^2=65$ or $x^2+y^2-20x-22y+156=0$

$\frac{18-their11}{14-their10}\text{ or }\frac{their11-4}{their10-6}\text{ or }\frac{18-4}{14-6}[=\frac{7}{4}]$

$y-18=-\frac{1}{their\frac{7}{4}}(x-14)$

$4x+7y-182=0$