A Level Mathematics 9709 Paper 13, Oct/Nov 2024

$a+3d=15$ and $a+7d=25$

Finding $a$ and $d$ $[d=\frac{5}{2},a=\frac{15}{2}]$

$u_{30}=\frac{15}{2}+29\times \frac{5}{2}=80$

$\cos \left(\frac{\pi }{6}\right)+\tan 2x+\frac{\sqrt{3}}{2}=0⟹\tan 2x=-\sqrt{3}$

$⟹2x=-\frac{\pi }{3}⟹x=-\frac{\pi }{6}$

$x^3:\left(\begin{array}{c}5\\ 3\end{array}\right)3^2(-ax{)}^3[-10\times 3^2\times a^3]$ or $x^4:\left(\begin{array}{c}5\\ 4\end{array}\right)3(-ax{)}^4[5\times 3\times a^4]$

$x^3:-90a^3$

$x^4:15a^4$

Coefficient of $x^4$ is $a\times their-90a^3+7\times their15a^4[=15a^4]$

$15a^4=240$

$⟹a^4=16⟹a=2$

Let $x={\sin }^2\theta$ $(2x+7)(2x-1)=0$ or $(2{\sin }^2\theta +7)(2{\sin }^2\theta -1)$

$⟹{\sin }^2\theta =\frac{1}{2}⟹\sin \theta =[\pm ]\frac{1}{\sqrt{2}}$

Any two correct values from $\theta ={45}^{\circ },{135}^{\circ },{225}^{\circ },{315}^{\circ }$

All correct and no others within the range from $\theta ={45}^{\circ },{135}^{\circ },{225}^{\circ },{315}^{\circ }$

Reflection

[in] $y$-axis

Translation or shift $\left(\begin{array}{c}-1\\ 0\end{array}\right)$

Stretch, factor 2, parallel to $y$-axis

Correct order and three correctly named transformations only

$a=2$

$b=-1$ and $c=-1$

$\frac{10(1-r^8)}{1-r}=\frac{17}{16}\frac{10(1-r^4)}{1-r}[a\frac{(1-r^8)}{(1-r)}=\frac{17}{16}\times a\frac{(1-r^4)}{(1-r)}]$

Simplifying to $16r^8-17r^4+1[=0]$ (or equivalent form)

$[(16r^4-1)(r^4-1)=0]⟹r=\pm \frac{1}{2}$

$S_{\mathrm{\infty }}=\frac{10}{1-([\pm ]\frac{1}{2})}$

$S_{\mathrm{\infty }}=20$ and $\frac{20}{3}$

Area of sector $BOF=\frac{1}{2}\times {20}^2\times (2\pi -2.4)[=776.63\dots ]$

Length $BD=DF=2\times 20\sin 0.6$ or $\sqrt{{20}^2+{20}^2-2\times 20\times 20\cos 1.2}[=22.58\dots ]$

Area of two semicircles $=\pi \times (20\sin 0.6{)}^2[=400.64\dots ]$

Area of triangles $=2\times \frac{1}{2}\times 20\times 20\sin 1.2[=372.81\dots ]$

Total area $=1550[{\text{cm}}^2]$

$\frac{1}{2}\pi r^2=50\pi ⟹r=10$

$⟹\theta =\frac{\pi }{3}$

Arc length of sector $BOF=20\times (2\pi -their\frac{2\pi }{3})$

Total perimeter $=20\times (2\pi -their\frac{2\pi }{3})+2\pi \times their10$

$\frac{140\pi }{3}$ or $46\frac{2}{3}\pi$

$a=-2$

$b=2$

$2$ or $k=2$ or $k\ge 2$

$3(x-2{)}^2+14-12=y⟹(x-2{)}^2=\frac{y-2}{3}$

$x=[\pm ]\sqrt{\frac{y-2}{3}}+2$

${\text{f}}^{-1}(x)=\sqrt{\frac{x-2}{3}}+2$

Finding ${\text{f}}^{-1}(29)[=5]$

Finding ${\text{f}}^{-1}(their5)$

$x=3$

$y=x^3-3x+3$ and $y=2x^3-4x^2+3⟹x^3-4x^2+3x[=0]$

$[x](x-1)(x-3)[=0]$

$x=0,1$ and $3$ {$x=0$ may be seen in the working}

Attempt at integration of both functions. Can be before or after subtraction of the functions or integrals

$=\pm (\frac{x^4}{4}-\frac{4x^3}{3}+\frac{3x^2}{2})$ or $\pm \{(\frac{x^4}{4}-\frac{3}{2}{x}^2+3x)-(\frac{2}{4}{x}^4-\frac{4}{3}{x}^3+3x)\}$

Correctly apply limits their 1 and 3 in the integrated expressions

$=\frac{8}{3}$

Gradient of $AB=\frac{-5-3}{8-4}[=-2]$

Midpoint $AB=(\frac{8+4}{2},\frac{-5+3}{2})[(6,-1)]$

Gradient of normal $=-\frac{1}{-2}[=\frac{1}{2}]$ and an attempt to find the required equation

Equation of perpendicular bisector is $y+1=\frac{1}{2}(x-6)$, so $y=\frac{1}{2}x-4$

Using the centre as $(a,\frac{1}{2}a-4)$

$(4-a{)}^2+(3-0.5a+4{)}^2=100$

$1.25a^2-15a-35[=0]⟹a^2-12a-28[=0]$ (or $b^2+2b-15[=0]$)

$[(a-14)(a+2)=0]⟹a=14,a=-2$

$⟹(x-14{)}^2+(y-3{)}^2=100$ and $(x+2{)}^2+(y+5{)}^2=100$

$\frac{\text{d}y}{\text{d}x}=\frac{1}{2}k{x}^{-\frac{1}{2}}-8x$

$\frac{{\text{d}}^{2}y}{\text{d}{x}^2}=-\frac{1}{4}k{x}^{-\frac{3}{2}}-8$

Setting their $\frac{\text{d}y}{\text{d}x}$ to zero and solving, providing their only error(s) are incorrect coefficients

$x=\frac{1}{4}$ only

$y=\frac{11}{4}$

$\frac{{\text{d}}^{2}y}{\text{d}{x}^2}=-\frac{1}{2}{x}^{-\frac{3}{2}}-8$ which is negative, so maximum

When $x=1$, attempting to find $y=k-2$ and gradient $=\frac{1}{2}k-8$

Equation of tangent is $y-k+2=(\frac{1}{2}k-8)(x-1)$

When $x=\frac{1}{4}$, attempting to find $y=\frac{1}{2}k+1.75$ and gradient $=k-2$

Equation of tangent is $y-\frac{1}{2}k-1.75=(k-2)(x-0.25)$

Meet at $(\frac{1}{2}k-8)(0.6-1)+k-2=(k-2)(0.6-0.25)+\frac{1}{2}k+1.75$ Equate two tangent equations and substitute $x=0.6$

$k=3$