A Level Mathematics 9709 Paper 33, May/June 2024

Use law of the logarithm of product or quotient on each side

Use law of logarithm of a power involving $x$ on ONE side, e.g. $\ln 8^3+(-)6x\ln 8$ or $(3-6x)\ln 8$ or $(9-18x)\ln 2$ or $\ln 4-2x\ln 5$

Obtain a correct linear equation in $x$, e.g. $(3-6x)\ln 8=(9-18x)\ln 2=\ln 4-2x\ln 5$

Obtain answer $x=0.524$

Use correct product rule $\cos 2x$ may be $1-2{\sin }^{2}x$ or ...

Obtain correct derivative $2{\text{e}}^{2x}\sin 2x+2{\text{e}}^{2x}\cos 2x$

Equate derivative of the form $a{\text{e}}^{2x}\sin 2x+{\text{e}}^{2x}b\cos 2x$ to 0 and solve for $2x$ or $x$ using a correct method Note may have substituted for $\sin 2x$ and/or $\cos 2x$

Obtain $x=\frac{3}{8}\pi$ only or exact equivalent

Obtain $y=\frac{1}{2}\sqrt{2}{\text{e}}^{\frac{3}{4}\pi }$ only or exact simplified equivalent

Square $x+\text{i}y$ obtaining three terms when simplified and equate real and imaginary parts to 24 and $-7$ respectively

Obtain equations $x^2-y^2=24$ and $2xy=-7$

Eliminate one variable by correct method and find a horizontal equation in the other

Obtain $4x^4-96x^2-49=0$ or $4y^4+96y^2-49=0$ or 3-term equivalents

Obtain answers $\frac{7\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\text{i}$ and $-\frac{7\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\text{i}$ or exact equivalents and no others

State or imply that $\ln k+\ln y=cx$ or $\ln y=cx+\ln \frac{1}{k}$ etc.

Carry out a completely correct method for finding $\ln k$ or $c$

Obtain value $c=0.80$

Obtain value $k=6.5$

State or imply the form $A+\frac{B}{x-1}+\frac{C}{2x+1}$

Use a correct method for finding a constant

Obtain one of $A=3,B=2$ and $C=-3$

Obtain a second value

Obtain a third value

Show a circle centre $(4,3)$ Allow dashes for coordinates on axes

Show a circle with radius 2. Can be implied by at least two of the points $(2,3),(6,3),(4,1)$ and $(4,5)$ being correct

Point representing $(2,1)$

Show a half-line at their $(2,1)$ at an angle of $\frac{1}{3}\pi$, cutting top of circle between $x=3$ and $x=5$

Shade the correct region Needs correct half-line or "correct" full line extending into the third quadrant AND correct circle

Carry out a correct method for finding the greatest value of $\arg z$ in the correct region in (a)

Obtain answer 1.06, or 1.05 or 1.055 or 1.056 or ${60.4}^{\circ }$ or ${60.5}^{\circ }$

Show $8\times (-7{)}^3+54\times (-7{)}^2-17\times (-7)-21=0$ This is sufficient if no errors seen. [$-2744+2646+119-21=0$] Or complete division of $8x^3+54x^2-17x-21$ by $x+7$ to get quotient $8x^2-2x-3$ and remainder of 0 Or state $(x+7)(8x^2-2x-3)$ is sufficient Factors must be stated again in (b) to collect marks there

Commence division and reach partial quotient of the form $8x^2\pm 2x$ or $8x^3+54x^2-17x-21=(x+7)(A{x}^2+Bx+C)$ and reach $A=8$ and $B=\pm 2$ or $C=-3$

Obtain quotient $8x^2-2x-3$ with no errors seen Stating $(x+7)(8x^2-2x-3)$ is sufficient

Solve quadratic from (b) to obtain a value for $\theta ={\cos }^{-1}(-\frac{1}{2})$ or ${\cos }^{-1}\left(\frac{3}{4}\right)$

Obtain one answer, e.g. $\theta ={120}^{\circ }$

Obtain three further answers, e.g. $\theta ={240}^{\circ },{41.4}^{\circ }$ and ${318.6}^{\circ }$ (condone ${319}^{\circ }$) and no others in the interval

State $R=\sqrt{12}$ or exact equivalent

Use trig formula to find $\alpha$

Obtain $\alpha =\frac{1}{6}\pi$

Express integral in the form $A\int {\sec }^2(2x+\dots )\text{d}x$ or $A\int {\sec }^2(2x-\dots )\text{d}x$

Integrate and reach $B\tan (2x+\dots )$ or $B\tan (2x-\dots )$

Obtain $\frac{1}{8}\tan (2x+\dots )$

Use limits of $x=0$ and $x=\frac{1}{12}\pi$ in the correct order in expression of form $B\tan (2x\pm \dots )$ so $B\tan (\frac{\pi }{6}+\dots )-B\tan (\dots )$ or $B\tan (\frac{\pi }{6}-\dots )-B\tan (-\dots )$

Obtain answer $\frac{1}{12}\sqrt{3}$ or $\frac{1}{4\sqrt{3}}$ or $\frac{1}{\sqrt{48}}$ or single term exact equivalent

Obtain $\frac{\text{d}V}{\text{d}t}=[\pm ]\frac{k}{t}$ or $\frac{\text{d}V}{\text{d}t}=[\pm ]\frac{1}{kt}$

Obtain $\frac{\text{d}V}{\text{d}x}=20x-3x^2$

Correct use of chain rule involving $k$

Obtain $\frac{\text{d}x}{\text{d}t}=[\pm ]\frac{k}{t(20x-3x^2)}$ or equivalent,

Use $t=\frac{1}{10}$, $x=\frac{1}{2}$ and $\frac{\text{d}x}{\text{d}t}=-\frac{20}{37}$ to obtain given answer which must be stated $\frac{\text{d}x}{\text{d}t}=-\frac{20}{37}$ needed to score final A1

Separate variables correctly & integrate at least one side correctly

Obtain terms $10x^2-x^3$

Obtain term $\ln t$ with ‘correct’ coefficient from their separation of variables, for example $a\ln t$ for $\frac{a}{t}$.

Use $t=\frac{1}{10}$, $x=\frac{1}{2}$ to evaluate a constant or as limits in a solution containing terms of the form $x^2,x^3$ and $\ln t$ (or $\ln 2t$)

Obtain correct answer in any form, for example $10x^2-x^3=-\frac{\ln t}{2}+\frac{19}{8}+\frac{\ln 0.1}{2}$

Obtain answer $t=\frac{1}{10}{\text{e}}^{\frac{2x^3-20x^2+\frac{19}{4}}{\mathrm{.}\mathrm{.}\mathrm{.}}}$ or equivalent

Carry out correct process for evaluating the scalar product of direction vectors

Using the correct process for the moduli, divide the scalar product by the product of the moduli and equate to $\pm \frac{\sqrt{2}}{2}$, or equate the scalar product to the product of the moduli and $\pm \frac{\sqrt{2}}{2}$

State a correct equation in any form, e.g. $\frac{5+2a}{3\sqrt{25+a^2}}=[\pm ]\frac{\sqrt{2}}{2}$ Allow unsimplified as in guidance

Form a quadratic equation in $a$ with 3 or more terms all on one side and solve for $a$. DM1 depends on BOTH *M1

Obtain $a=5$ and $a=35$

Express general point of at least one line correctly in component form, i.e. $\left(\begin{array}{c}1+3\lambda \\ 1+4\lambda \\ 2a+a\lambda \end{array}\right)$ or $\left(\begin{array}{c}-3-\mu \\ -1+2\mu \\ 4+2\mu \end{array}\right)$

Equate at least two pairs of corresponding components and solve for $\lambda$ or $\mu$ or $a$

Obtain $\lambda =-1$ or $\mu =-1$

Obtain $a=2$

Obtain position vector of the point of intersection is $-2\mathbf{i}-3\mathbf{j}+2\mathbf{k}$ Two different answers for point of intersection scores A0 even if one is correct