A Level Mathematics 9709 Paper 11, May/June 2023

For use of $\tan \theta =\frac{\sin \theta }{\cos \theta }$. BOD if $\theta$ missing.

WWW Factorise, not divide by $\sin \theta$ or $\tan \theta$. May see $\tan \theta (4\cos \theta +1)[=0]$ or $\sin \theta (4+\sec \theta )[=0]$.

$\theta ={104.5}^{\circ }$

$16+96x+216x^2$

$1-10x+40x^2$

$(16\times 40)-(10\times 96)+(1\times 216)$

$-104$

{Stretch} {factor 2} {in $y$-direction}

{Translation} $\left(\begin{array}{c}-6\\ 0\end{array}\right)$

$\frac{1}{2}\times 8^2\times \theta =\frac{16\pi }{3}\Rightarrow \theta =\frac{\pi }{6}$

$\text{Arc length}=8\times \text{their }\frac{\pi }{6}[=4.1887\dots ]$

$[BC=]2\times 8\sin (\frac{1}{2}\times \text{their }\frac{\pi }{6})[=4.1411\dots ]$

$\text{Perimeter}=8.33$

$kx-k=-\frac{1}{2x}\Rightarrow 2k{x}^2-2kx+1[=0]$ $\text{OR quadratic in }y:x=\frac{y+k}{k}\Rightarrow y=-\frac{1}{2\left(\frac{y+k}{k}\right)}\Rightarrow 2y^2+2ky+k=0$

$b^2-4ac[=0]\Rightarrow ([-2k{]}^2-4(2k)(1)[=0]$ $\text{or }4k^2-8k[=0\Rightarrow 4k(k-2)=0]$ $\text{OR using equation in }y:(2k{)}^2-4(2)(k)=0$

$k=2\text{ only}$

$4x^2-4x+1=0[\Rightarrow (2x-1{)}^2=0]\Rightarrow x=\frac{1}{2}$

$y=2\times \frac{1}{2}-2=-1$

$2(2p-6)=p+\frac{p^2}{6}\Rightarrow \frac{p^2}{6}-3p+12[=0]$ $\text{OR }(2p-6)-\frac{p^2}{6}=p-(2p-6)\Rightarrow \frac{p^2}{6}-3p+12[=0]$ $\text{OR }\frac{1}{6}{d}^2+d[=0]$

$p^2-18p+72[=0]\Rightarrow (p-6)(p-12)[=0]\text{ or }\frac{18\pm \sqrt{(-18{)}^2-4(1)(72)}}{2}$ $\text{OR }d(\frac{1}{6}d+1)[=0]\Rightarrow d=-6$

$p=12\text{ only}$

$\text{For GP }r=\left[\frac{2p-6}{\frac{p^2}{6}}\right]=[\frac{18}{24}=\frac{3}{4}]$

$\text{Sum to infinity}=\frac{24}{1-\frac{3}{4}}=96$

$[\text{Greatest}=]5$

$[\text{Least}=]-1$

One complete cycle starting and finishing at $y=2$. Maximum and minimum in correct quadrants. Shape and curvature approximately correct.

Maximum and minimum (indicated on $y$-axis with numbers or lines, or labelled on graph). FT *their* greatest and least values.

1

$1+\frac{2a}{7-a}=\frac{5}{2}[\Rightarrow \frac{2a}{7-a}=\frac{3}{2}\Rightarrow 7a=21]\Rightarrow a=\dots$ $\text{OR }1+\frac{2a}{7-a}=\frac{5}{2}[\Rightarrow (7-a)+2a=\frac{5}{2}(7-a)[\Rightarrow 7a=21]]\Rightarrow a=\dots$

$a=3$

$\text{f}(5)=1+\frac{2(\text{their }3)}{5-\text{their }3}=4[\Rightarrow 4b-2=4]\Rightarrow b=\dots$ $\text{OR gf}(5)=b(1+\frac{2(\text{their }3)}{5-\text{their }3})-2[\Rightarrow 4b-2=4]\Rightarrow b=\dots$

$b=\frac{3}{2}$

$x>1$

$\text{EITHER }x-1=\frac{6}{y-3}[\Rightarrow (y-3)(x-1)=6]$ $\text{OR }x=1+\frac{6}{y-3}\Rightarrow x(y-3)=(y-3)+6$

$y-3=\frac{6}{x-1}\text{ or }y(x-1)=3x+3$

$[{\text{f}}^{-1}(x)]=3+\frac{6}{x-1}$

$\frac{\text{d}V}{\text{d}h}=\frac{4}{3}\times 3(25+h{)}^2[=4900\text{ when }h=10]$

$\frac{\text{d}V}{\text{d}h}\times \frac{\text{d}h}{\text{d}t}=\frac{\text{d}V}{\text{d}t}\Rightarrow \text{their }\mathrm{"}4(25+10{)}^2\mathrm{"}\times \frac{\text{d}h}{\text{d}t}=500\Rightarrow \frac{\text{d}h}{\text{d}t}=\left[\frac{500}{4900}\right]$

$\frac{\text{d}h}{\text{d}t}=0.102[{\text{cms}}^{-1}]$

$\frac{\text{d}V}{\text{d}t}=\frac{\text{d}V}{\text{d}h}\times \frac{\text{d}h}{\text{d}t}\Rightarrow 500=\text{their }\mathrm{"}4(25+h{)}^2\mathrm{"}\times 0.075$

$[(25+h{)}^2=\frac{5000}{3}]\Rightarrow h=[15.8248\dots ]$

$V=69900{\text{ cm}}^3$

$[\pi ]\int \frac{16}{(2x-1{)}^4}[\text{d}x]=[\pi ]\int 16(2x-1{)}^{-4}[\text{d}x]=[\pi ](-\frac{16}{3\times 2\times (2x-1{)}^3})$

$[\pi ](-\frac{16}{3\times 2\times (2x-1{)}^3})$

$[\pi ](-\frac{16}{6\times 8}+\frac{16}{6\times 1})=[[\pi ]\frac{112}{48}=[\pi ]\frac{7}{3}]$

$\text{Volume of cylinder }[=\pi \times 1^2\times \frac{1}{2}]=\frac{1}{2}\pi \text{ OR }[\pi ]{\int }_1^{1.5}1[\text{d}x]=\frac{1}{2}\pi$

$\text{Volume of revolution }[=\frac{7}{3}\pi -\frac{1}{2}\pi ]=\frac{11}{6}\pi$

$[\frac{\text{d}y}{\text{d}x}=]\{-8(2x-1{)}^{-3}\}\{\times 2\}$

$\text{At }B\text{ gradient}=-2$

$\text{Eqn of tangent }y-1=\text{their }\mathrm{"}-2\mathrm{"}(x-\frac{3}{2})$ $\text{OR Eqn of normal }y-1=\text{their }\mathrm{"}\frac{1}{2}\mathrm{"}(x-\frac{3}{2})$

$\text{Tangent crosses }x\text{-axis at }2\text{ or normal crosses }x\text{-axis at }-\frac{1}{2}$

$\text{Area}=\frac{5}{4}$

$6a^2-30a+6a=0[\Rightarrow 6a(a-4)=0]$

$a=4\text{ only}$

$\frac{{\text{d}}^{2}y}{\text{d}{x}^2}=12x-30\text{ or correct values of }\frac{\text{d}y}{\text{d}x}\text{ either side of }x=4$

$\text{At }x=4,\frac{{\text{d}}^{2}y}{\text{d}{x}^2}>0\therefore \text{minimum or }\frac{{\text{d}}^{2}y}{\text{d}{x}^2}=18\therefore \text{minimum}$ $\text{or concludes minimum from }\frac{\text{d}y}{\text{d}x}\text{ values}$

$[y=]\frac{6}{3}{x}^3-\frac{30}{2}{x}^2+6(\text{their }a)x[+c]$

$-15=2(\text{their }\mathrm{"}4\mathrm{"}{)}^3-15(\text{their }\mathrm{"}4\mathrm{"}{)}^2+6(\text{their }\mathrm{"}4\mathrm{"}{)}^2+c$

$y=2x^3-15x^2+24x+1$

$\frac{\text{d}y}{\text{d}x}=6x^2-30x+6(\text{their }\mathrm{"}4\mathrm{"})[=0]$ $\text{If correct, }[6](x-1)(x-4)[=0]\text{ or }\frac{30\pm \sqrt{(-30{)}^2-4(6)(24)}}{12}$

$\text{Coordinates }(1,12)$

$x^2+(y-2{)}^2=100$

$\text{Gradient of radius}=[\frac{10-2}{6-0}=\frac{4}{3}]\text{ or gradient of tangent}=-\frac{3}{4}$

$\text{Equation of tangent is }y-10=-\frac{3}{4}(x-6)[\Rightarrow y=-\frac{3}{4}x+\frac{29}{2}]$

$\text{Coordinates of centre of circle }Q\text{ are }(0,\text{their }\frac{29}{2})$

$\text{Equation of circle }Q\text{ is }{x}^2+{(y-\text{their }\frac{29}{2})}^2={\left(\frac{5\sqrt{5}}{2}\right)}^2[=\frac{125}{4}]$

$x^2+(11-2{)}^2=100\Rightarrow x^2=19\text{ and }{x}^2+{(11-\frac{29}{2})}^2=\frac{125}{4}\Rightarrow x^2=19$ $\text{OR e.g. }\frac{125}{4}-{(y-\frac{29}{2})}^2+(y-2{)}^2=100\Rightarrow 25y=275\Rightarrow y=11$

$x^2+{(-\frac{3}{4}x+\frac{29}{2}-\frac{29}{2})}^2=\frac{125}{4}[\Rightarrow \frac{25}{16}{x}^2=\frac{125}{4}\Rightarrow x^2=20]$ $\text{or }{y}^2-29y+199[=0]$

$x=\pm 2\sqrt{5}\text{ or }y=\frac{29\mp 3\sqrt{5}}{2}$

$[y=(-\frac{3}{4}\times \pm \sqrt{20})+\frac{29}{2}=]\frac{29\mp 3\sqrt{5}}{2}$