A Level Mathematics 9709 Paper 33, Oct/Nov 2024

Show an arc of a circle, centre the origin and radius 2. Only need 2 on Re(z) or 2i on Im(z) or $r=2$ to show correct radius

Show an arc centre the origin for $0⩽\arg z⩽\frac{1}{4}\pi$ with any radius

Show an arc of a circle, centre the origin and radius 4. Only need 4 on Re(z) or 4i on Im(z) or $r=4$ to show correct radius

Show an arc centre the origin for $0⩽\arg z⩽\frac{1}{2}\pi$ with any radius

State or imply the equation $x=\sqrt{\frac{4}{5-2x}}$ and square the equation

Rearrange this with at least one intermediate step in the form $2x^3-5x^2+4=0$

Use the iterative process correctly at least once

Obtain final answer 1.28

Show sufficient iterations to at least 4 dp to justify 1.28 to 2 dp or show that there is a sign change in the interval (1.275, 1.285)

State or imply that $\ln P=\ln a+kt$ or $\ln P=\ln a+k(\ln \mathrm{e})t$ $\ln P=\frac{1}{20}t+3$ B0 until associated with $a$ and /or $k$

State $k=\frac{1}{20}$, not from $\frac{\mathrm{d}P}{\mathrm{d}t}=k$

$\ln a=3\Rightarrow a=20$ to 2 sf

Form a correct equation in $t$ using $a$ and $k$, or their $a$ and $k$ where $a$ will cancel (or are both numerical)

Obtain $t=14$ [hours]

Substitute $z=x+\mathrm{i}y$ and obtain a horizontal equation Do not allow if this would lead to an equation containing $xy$ terms which do not cancel

Use ${\mathrm{i}}^2=-1$ anywhere

Obtain e.g. $5x+5(y-3)\mathrm{i}=(2x+9y+27)+\mathrm{i}(2y+6-9x)$ or e.g. $3x^2+3y^2-12y-63+(9x^2+9y^2-30x+54y+81)\mathrm{i}=0$

Obtain simultaneous equations by equating real and imaginary parts

Obtain [$z=$] $3-2\mathrm{i}$ only

Rewrite ${\cos }^4\theta$ as ${\left(\frac{1+\cos 2\theta }{2}\right)}^2$ or ${\sin }^4\theta$ as ${\left(\frac{1-\cos 2\theta }{2}\right)}^2$ or $4{\sin }^2\theta {\cos }^2\theta$ as ${\sin }^{2}2\theta$

Obtain ${\left(\frac{1+\cos 2\theta }{2}\right)}^2-{\left(\frac{1-\cos 2\theta }{2}\right)}^2-{\sin }^{2}2\theta$

Expand to $\frac{1}{4}+\frac{1}{2}\cos 2\theta +\frac{1}{4}{\cos }^{2}2\theta -(\frac{1}{4}-\frac{1}{2}\cos 2\theta +\frac{1}{4}{\cos }^{2}2\theta )-(1-{\cos }^{2}2\theta )$ and simplify to obtain ${\cos }^{2}2\theta +\cos 2\theta -1$

State a quadratic equation in $\cos 2\alpha$ and solve for $\alpha$ $({\cos }^{2}2\alpha +\cos 2\alpha -1=0)$

Obtain $\alpha ={25.9}^{\circ }$ or $\alpha ={154.1}^{\circ }$

Obtain $\alpha ={25.9}^{\circ }$ and $\alpha ={154.1}^{\circ }$ and no others in range

$P(2,1,-3)$

Use the correct method to find the scalar product of the direction vectors

Divide the scalar product by the product of the moduli to obtain $\pm \cos \theta$ using consistent vectors throughout

Obtain $\cos \theta =\frac{8}{5\sqrt{6}}$

Any two of $\mathrm{\mid }\overrightarrow{PA}\mathrm{\mid }=2\sqrt{5}$ $\mathrm{\mid }\overrightarrow{PB}\mathrm{\mid }=\sqrt{30}$ or $\mathrm{\mid }\overrightarrow{AB}\mathrm{\mid }=\sqrt{82}$ seen

$\text{Area}=\frac{1}{2}\times 2\sqrt{5}\times \sqrt{30}\times \sqrt{1-\frac{64}{150}}$

$=\sqrt{86}$

Obtain $\frac{\mathrm{d}x}{\mathrm{d}t}=6\cos 2t$

Obtain $\frac{\mathrm{d}y}{\mathrm{d}t}={\sec }^{2}t-{\mathrm{cosec}}^{2}t$

Use $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}÷\frac{\mathrm{d}x}{\mathrm{d}t}$

Express as a single fraction With $\frac{\mathrm{d}y}{\mathrm{d}t}$ correctly simplified as a single fraction in terms of $\sin t$ and $\cos t$ Allow with $6\cos 2t$ expressed as $\frac{1}{6\cos 2t}$ outside bracket

Obtain $(\frac{-4\cos 2t}{6\cos 2t\times {\sin }^{2}2t}=)\frac{-2}{3{\sin }^{2}2t}$ from full and correct working Numerator and denominator must have identical terms before cancelling, both $\cos 2t$ or both $+$ and $-({\sin }^{2}t-{\cos }^{2}t)$

Gradient of normal $=\frac{3}{2}$

Use correct method to find the equation of the normal

Obtain $2y-3x+5=0$

State or imply the form $\frac{A}{a-2x}+\frac{B}{3a+x}$ and use a correct method to find a constant

Obtain $A=2a$ or $B=a$

Obtain $A=2a$ and $B=a$

Use a correct method to obtain the first two terms of the expansion of $(a-2x{)}^{-1}$ or ${(1-\frac{2x}{a})}^{-1}$ or $(3a+x{)}^{-1}$ or ${(1+\frac{x}{3a})}^{-1}$

Obtain $+2(1+\frac{2x}{a}+\frac{4x^2}{a^2}+\dots )$

Obtain $+\frac{1}{3}(1-\frac{x}{3a}+\frac{x^2}{9a^2}+\dots )$

Obtain $\frac{7}{3}+\frac{35x}{9a}+\frac{217x^2}{27a^2}$

$\mathrm{\mid }x\mathrm{\mid }<\frac{a}{2}$

Divide to obtain quotient $x^2+k$

Obtain quotient $x^2-4$

Obtain remainder 32

$\frac{1}{3}{x}^3-4x$

Obtain $p{\tan }^{-1}qx$ where $q=2$ or $q=\frac{1}{2}$

Obtain $16{\tan }^{-1}\frac{1}{2}x$

Use limits correctly in an expression containing $p{\tan }^{-1}qx$ where $q=2$ or $q=\frac{1}{2}$ and $r{x}^3+sx$

Obtain $\frac{4}{3}(\pi +4)$ from full and correct working

State that $\frac{\mathrm{d}V}{\mathrm{d}t}=50000-600h$

[Use $V=40000h$ to] obtain $\frac{\mathrm{d}V}{\mathrm{d}h}=40000$ and use this and their $\frac{\mathrm{d}V}{\mathrm{d}t}$ in the correct chain rule to obtain $\frac{\mathrm{d}h}{\mathrm{d}t}$ or [Use $V=40000h$ to] obtain $\frac{\mathrm{d}V}{\mathrm{d}t}=40000\frac{\mathrm{d}h}{\mathrm{d}t}$ and equate to their $\frac{\mathrm{d}V}{\mathrm{d}t}$

Obtain $200\frac{\mathrm{d}h}{\mathrm{d}t}=250-3h$ from full and correct working

Separate variables correctly and integrate one side correctly

Obtain $-\frac{1}{3}\ln \mathrm{\mid }250-3h\mathrm{\mid }=\frac{t}{200}(+C)$

Use $t=0,h=50$ in an expression containing $\ln (250-3h)$ or $\ln \mathrm{\mid }250-3h\mathrm{\mid }$ to find the constant of integration.

Obtain $C=-\frac{1}{3}\ln 100$

$t=150$

Use of correct product rule and correct chain rule

Obtain $\frac{\mathrm{d}y}{\mathrm{d}x}=2\cos x\sqrt{2+\cos x}-\frac{2{\sin }^{2}x}{2\sqrt{2+\cos x}}$

Equate the derivative to zero and obtain a horizontal 3 term quadratic equation or 4 term quartic equation in $\cos a$ If M0 earlier then needs that expression to be such that arrive at 3 term quadratic or 4 term quartic equation in $\cos x$ without further trig errors. The only error in the form of the differential allowed is for $(2+\cos x{)}^{-\frac{1}{2}}$ to be $(2+\cos x{)}^{+\frac{1}{2}}$ or $(2+\cos x{)}^{-\frac{3}{2}}$

Solve for $\cos a$

Obtain $a=4.93$

State or imply $\mathrm{d}u=-\sin x\mathrm{d}x$

Substitute throughout for $u$ and $\mathrm{d}u$

Obtain $-\int 2\sqrt{u}\mathrm{d}u$

Integrate to obtain $k{u}^{\frac{3}{2}}(+C)$

Use correct limits correctly in an expression of the form $k{u}^{\frac{3}{2}}$ or $k(2+\cos x{)}^{\frac{3}{2}}$

Obtain $\frac{4}{3}(3\sqrt{3}-1)$ or $4\sqrt{3}-\frac{4}{3}$ or $\frac{4}{3}\sqrt{27}-\frac{4}{3}$