A Level Mathematics 9709 Paper 11, May/June 2024

$a=-2$ (or in expression)

$b=-27$ (or in expression)

${(x^2-2)}^2=9$ leading to $x^2-2=\pm 3$

$x^2=-1$ or $x^2=5$

$x=\pm \sqrt{5}$

{Stretch}{factor 3}{ in $y$-direction}

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

$-3\sin x$

$-2$

$20\times 27\times a^3[=160]$

$[a]=\frac{2}{3}$

Coefficient of $x^2$ is $15\times 81\times {\left(\text{their }\frac{2}{3}\right)}^2[=540]$

$160\times 1-2\times \text{their }540$

$=-920$

$[k]=4.00063$

$[\text{Gradient }AE]=6.3566$

Suggests that $[{\mathrm{f}}^{\mathrm{\prime }}(2)]=6.25$

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

$=\frac{-\cos x(1+\cos x)}{1+\cos x}$

$=-\cos x$

$-\frac{1}{2}\cos x=\frac{1}{4}\Rightarrow x={\cos }^{-1}(-\frac{1}{2})$

$x={120}^{\circ }$ or $x={240}^{\circ }$

$x={240}^{\circ }$ or $x={120}^{\circ }$ (second solution)

For curve in correct quadrant.

Fully correct including line $y=x$. Horizontal asymptote closer to $x$ axis than vertical asymptote is to $y$ axis.

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

$y^2=\frac{2}{(x-4)}$ leading to $y=[\pm ]\sqrt{\frac{2}{x-4}}$ or $x=[\pm ]\sqrt{\frac{2}{y-4}}$

$[{\mathrm{f}}^{-1}(x)]=-\sqrt{\frac{2}{x-4}}$

$[x]=-2$

Because ${\mathrm{f}}^{-1}$ is always negative and $\mathrm{f}$ is always positive or curves do not intersect

Angle $\theta =\frac{\pi }{2}-{\cos }^{-1}\frac{10}{15}$ or ${\sin }^{-1}\frac{10}{15}=0.7297$

$BC=\sqrt{{15}^2-{10}^2}[=11.18\dots \text{ or }5\sqrt{5}]$

$\mathrm{Arc}AB=15\times 0.7297[=10.9455]$

$\text{Perimeter}=\text{their }BC+\text{their arc }AB+25+5\pi$

$\text{Perimeter}=62.8$

$\text{Area sector }AOB=\frac{1}{2}\times {15}^2\times 0.7297[=82.09]$

$\text{Area}=\frac{1}{2}\times 10\times \text{their }BC+\text{their sector }AOB+\frac{\pi }{4}\times {10}^2$

$\text{Area}=217$

$2(4p-1)=25+13-p$

$p=\frac{40}{9}$

$d=(4\times \frac{40}{9}-1)-25[=-\frac{74}{9}]$

${10}^{\text{th }}\text{ term }=25+9d=-49$

$(4q-1{)}^2=25(13-q)\Rightarrow 16q^2+17q-324[=0]$

$(q-4)(16q+81)[=0]$ leading to $[\Rightarrow q=4]$

$[r=]\frac{3}{5}$

$\text{Sum to infinity}=\frac{25}{1-\frac{3}{5}}=\frac{125}{2}$

$\text{Volume of cylinder}=\pi \times 1^2\times \frac{7}{5}=\frac{7}{5}\pi$

$\text{Volume under curve}=[\pi ]{\int }_1^{2.4}\frac{1}{(5x-4{)}^{\frac{2}{3}}}\mathrm{d}x$

$\left\{\frac{3}{5}\right\}$ in integrated expression

$\{(5x-4{)}^{\frac{1}{3}}\}$ in integrated expression

$=[\pi ]\left[\frac{3}{5}(8^{\frac{1}{3}}-1)\right][=\frac{3}{5}\pi ]$

$\text{Volume}=\frac{7}{5}\pi -\frac{3}{5}\pi =\frac{4}{5}\pi$

$(x-3{)}^2+y^2=18y=mx-9$ leading to $(x-3{)}^2+(mx-9{)}^2=18$

$x^2-6x+9+m^2{x}^2-18mx+81=18$ leading to $(m^2+1)x^2-(6+18m)x+72[=0]$

$(6+18m{)}^2-4(m^2+1)\times 72[=0]$

$36m^2+216m-252[=0][\text{leading to }{m}^2+6m-7=0]$

$m=1$ or $m=-7$

$m=1$ leading to $2x^2-24x+72=0$ leading to $x=6$

$m=-7$ leading to $50x^2+120x+72=0$ leading to $x=-\frac{6}{5}$

$(6,-3),(-\frac{6}{5},-\frac{3}{5})$

$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{12}{x^4}+\frac{3}{x^2}$

$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{12}{x^4}+\frac{3}{x^2}=0$ leading to $3x^4-12x^2=0$ or $-12+3x^2=0$

$3x^2(x^2-4)=0$ leading to $x=\pm 2$ only

$-2<x<0$ and $0<x<2$ or $(-2,0)$ and $(0,2)$ or $-2<x<2$ and $x\mathrm{\ne }0$ (first mark)

$-2<x<0$ and $0<x<2$ or $(-2,0)$ and $(0,2)$ or $-2<x<2$ and $x\mathrm{\ne }0$ (second mark)

$[\text{At }x=1]y=3\text{ and }m\tan =-9$

$m\text{ norm }=-\frac{1}{-9}=\frac{1}{9}$

$\text{Equation of normal is }y-3=\frac{1}{9}(x-1)[\text{leading to }y=\frac{1}{9}x+\frac{26}{9}]$

$\text{At }x=-1,y=1,m=-9$

$\text{Equation of tangent is }y-1=-9(x+1)[\text{leading to }y=-9x-8]$

$\text{Meet when }\frac{1}{9}x+\frac{26}{9}=-9x-8[\text{leading to }x=-1.19512\dots ,-\frac{49}{41}]$

$\text{Area}=\frac{1}{2}\times \text{their }1.19512\dots \times \text{their}(\frac{26}{9}+8)$

$6.51$