A Level Mathematics 9709 Paper 12, May/June 2025

{Stretch} {factor 2} {‘parallel to y-axis’ or ‘in y-direction’ or ‘vertically’}

{Translation} $\left(\begin{array}{c}\{0\}\\ \{-14\}\end{array}\right)$ or $\{-14\}$ {‘parallel to the y-axis’ or ‘in the y-direction’ or ‘vertically’.}

$2\left(\frac{-y-4}{2}\right)y+5y^2=24$ or $2\left(\frac{24-5y^2}{2y}\right)+y+4[=0]$ or $2x(-4-2x)+5(-4-2x{)}^2=24$

$4y^2-4y=24$ or $16x^2+72x+56[=0]$

$\Rightarrow y=-2,y=3$ or $x=-1,x=-\frac{7}{2}$

$(-1,-2),(-\frac{7}{2},3)$

$\left(\begin{array}{c}5\\ 3\end{array}\right)$ or $\left(\begin{array}{c}5\\ 2\end{array}\right)$ or $10\times (p{x}^2{)}^{5-3}\times {\left(\frac{4}{p}x\right)}^3$

$10p^2\times \frac{64}{p^3}$

$\frac{\text{Their integer }k}{p}[x^7]=1280[x^7]$

$p=\frac{1}{2}$

$[\frac{\mathrm{d}y}{\mathrm{d}x}=]\frac{3}{2}a{x}^{\frac{1}{2}}-12$

$[\frac{\mathrm{d}y}{\mathrm{d}t}=\frac{\mathrm{d}y}{\mathrm{d}x}\times \frac{\mathrm{d}x}{\mathrm{d}t}=](\frac{3}{2}a\times 9^{\frac{1}{2}}-12)\times 5$

$[\frac{\mathrm{d}y}{\mathrm{d}t}=]5(\frac{9}{2}a-12)$ or $\frac{45}{2}a-60$ or $22.5a-60$ or $\frac{45a-120}{2}$ or $\frac{15}{2}(3a-8)$

$\frac{3}{2}a\times {\left(\frac{1}{4}\right)}^{\frac{1}{2}}-12=0$

$\Rightarrow a=16$

[Greatest] 7

[least] −1

Correct curve with two complete cycles, starting at (0,7), levelling off at 0 and 2pi

3 [solutions]

$6-3x=3\Rightarrow x\text{-coordinate of point of intersection}=1$

${[\frac{18}{5}(5x+4{)}^{\frac{1}{2}}]}_0^1$ (first B mark for $k(5x+4{)}^{\frac{1}{2}}$)

${[\frac{18}{5}(5x+4{)}^{\frac{1}{2}}]}_0^1$ (second B mark for correct $\frac{18}{5}$)

$\text{Area}=\frac{18}{5}\times 3-\frac{18}{5}\times 2[=\frac{18}{5}]$

$\text{Line crosses }x\text{-axis at }x=2\text{, so triangle area}=\frac{1}{2}\times 3\times 1$

$\text{Total area}=\frac{51}{10}$

For appropriate use of $\tan \theta =\frac{\sin \theta }{\cos \theta }$ or $\tan \theta \cos \theta =\sin \theta$ at least once

For appropriate use of ${\sin }^2\theta +{\cos }^2\theta =1$ or $1+{\tan }^2\theta ={\sec }^2\theta$

Fully correct proof

$\frac{\tan \theta +7}{{\tan }^2\theta -3}=\frac{5}{\tan \theta }\Rightarrow \tan \theta (\tan \theta +7)=5({\tan }^2\theta -3)$

$\Rightarrow 4{\tan }^2\theta -7\tan \theta -15=0\Rightarrow (4\tan \theta +5)(\tan \theta -3)=0$

$\tan \theta =-1.25$ and $\tan \theta =3$

$\theta ={71.6}^{\circ }$ and $\theta ={128.7}^{\circ }$

$(2,-2)$

$(12,-2)$

$(7,-4)$

$\tan \left(\frac{1}{2}\theta \right)=\frac{5}{2}[\Rightarrow \theta =2{\tan }^{-1}\frac{5}{2}]$ or $\sin \left(\frac{\theta }{2}\right)=\frac{5}{\sqrt{29}}$ or $\cos \left(\frac{\theta }{2}\right)=\frac{2}{\sqrt{29}}$ or $\theta =\pi -2\times {\tan }^{-1}\frac{2}{5}$ or $\pi -2\times 0.381$ or ${10}^2=29+29-2\times \sqrt{29}\times \sqrt{29}\cos \theta [\Rightarrow \cos \theta =\frac{29+29-100}{2\times \sqrt{29}\times \sqrt{29}}=\frac{-21}{29}]$

$[\theta =]2.38$

$[\text{Length }AC=\text{Length }BC=\sqrt{5^2+2^2}]=\sqrt{29}[=5.385\dots ]$

$[\text{Area of large sector}=]\frac{1}{2}\times \sqrt{29}\times \sqrt{29}\times (2\pi -\text{their }\theta )[=56.587\dots ]$ or $[\text{Area of small sector}=]\frac{1}{2}\times \sqrt{29}\times \sqrt{29}\times \text{their }\theta [=34.518\dots ]$

$\text{Area of triangle}=\frac{1}{2}\times \sqrt{29}\times \sqrt{29}\times \sin \theta$ or $\frac{1}{2}\times 10\times 2[=10]$

$[\text{Segment area}=]66.6$

$[\frac{\mathrm{d}y}{\mathrm{d}x}=]\frac{-24x^{-2}}{-2}[=\frac{12}{x^2}][+c]$

$\frac{12}{(-2{)}^2}+c=0[\Rightarrow c=-3]$

$[\frac{\mathrm{d}y}{\mathrm{d}x}=]\frac{12}{x^2}-3$ or $12x^{-2}-3$

$[\frac{12}{x^2}-3=0\Rightarrow x^2=4\Rightarrow x=\pm 2\Rightarrow ]x=2$

$[\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=]-\frac{24}{(2{)}^3}$ or $-3$ or $<0\Rightarrow$ maximum

$[y=]-12x^{-1}+\text{their }cx[=-\frac{12}{x}-3x][+d]$

$19=-\frac{12}{-2}-3\times (-2)+d[\Rightarrow d=7]$

$y=-\frac{12}{x}-3x+7$

$\frac{12}{x^2}-3=-\frac{9}{4}\Rightarrow \frac{12}{x^2}=\frac{3}{4}\Rightarrow x=\dots [\Rightarrow x^2=16\Rightarrow x=4,y=-8]$

$\text{Gradient of normal}=\frac{4}{9}$

$\frac{y+8}{(x-4)}=\frac{4}{9}$

$4x-9y-88=0$

$8k-k^2=k^2-4k$ or $2k^2=4k+8k$ or $4k+2(k^2-4k)=8k$

$[2k(k-6)=0\Rightarrow ]k=6[\text{or }k=0]$

$d[=8k-k^2\text{ or }2k\text{ or }{k}^2-4k]=12$ and $a[=4k]=24$

$S_{20}=\frac{20}{2}(2\times 24+19\times 12)$ or $=\frac{20}{2}(24+252)$

$S_{20}=2760$

$a{r}^3=36$ and $a{r}^5=6$ or $r^2=\frac{1}{6}$

$r=[\sqrt{\frac{6}{36}}=]\frac{1}{\sqrt{6}}\text{ or }\frac{\sqrt{6}}{6}$

$a=\frac{36}{{\left(\frac{1}{\sqrt{6}}\right)}^3}=[36\times {\sqrt{6}}^3,216\sqrt{6}\text{ or }\frac{1296}{\sqrt{6}}]$

$S_{\mathrm{\infty }}=\frac{216\sqrt{6}}{1-\frac{1}{\sqrt{6}}}$

$[S_{\mathrm{\infty }}=]\frac{1296}{\sqrt{6}-1}$ or $\frac{7776}{\sqrt{216}-6}$

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

$\{-2\}$

$y=(x+2{)}^2-2\Rightarrow y+2=(x+2{)}^2$

$x=[\pm ]\sqrt{y+2}-2$

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

$[\mathrm{g}\mathrm{f}(x)=]-\{\text{their completed square form}\}-4$ or $-(x^2+4x+2)-4$

$[\mathrm{g}\mathrm{f}(x)=]-(x+2{)}^2-2$

$x=[\pm ]\sqrt{-y-2}-2$

$[(\mathrm{g}\mathrm{f}{)}^{-1}(x)=]-\sqrt{-x-2}-2$