A Level Mathematics 9709 Paper 33, May/June 2025

Symmetrical. In correct position. Lines intended to be straight. Must be in both first and second quadrants. Key coordinates must be correct.

Obtain critical value $\frac{7a}{2}$ from $x+5a=3x-2a$

Obtain critical value $-\frac{3a}{4}$ from $x+5a=2a-3x$

State final answer $-\frac{3a}{4}<x<\frac{7a}{2}$

State $z_1{z}_2=r_1{r}_2{\mathrm{e}}^{\mathrm{i}({\theta }_1+{\theta }_2)}$

State correct conjugate of $z_1{z}_2,z_1$ or $z_2$

Obtain given result from correct working

State $(z^{\ast }=)3{\mathrm{e}}^{-\frac{1}{4}\pi \mathrm{i}}$

Complete method to find both $b$ and $c$

Obtain $b=-3\sqrt{2},c=9$

State or imply $y+x\frac{\mathrm{d}y}{\mathrm{d}x}$ as the derivative of $xy$

State or imply $2y{\mathrm{e}}^{-x}\frac{\mathrm{d}y}{\mathrm{d}x}-y^2{\mathrm{e}}^{-x}$ as the derivative of $y^2{\mathrm{e}}^{-x}$

Equate attempted derivative to zero and solve for $\frac{\mathrm{d}y}{\mathrm{d}x}$

Obtain $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y^2-y{\mathrm{e}}^x}{x{\mathrm{e}}^x+2y}$ from correct working

Obtain one correct gradient

Obtain second correct gradient

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

Obtain one of $A=3$ and $B=-1$

Obtain the second value

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

Obtain the correct unsimplified expansions in terms of $a$, up to the term in $x^2$.

Obtain final answer $\frac{1}{2a}-\frac{11}{12a^2}x+\frac{23}{72a^3}{x}^2$

State $\mathrm{\mid }x\mathrm{\mid }<\frac{3}{2}a$

Express the left hand side in terms of $\sin \theta$ and $\cos \theta$

Combine to a single term and factorise the numerator

Use correct double angle formulae in numerator and denominator

Obtain $4\cot 2\theta \mathrm{cosec}2\theta$ from correct working

Use the identity to obtain an expression in one trigonometric function

Obtain ${\tan }^{2}2x=\frac{4}{5}$

Obtain one solution e.g. ${20.9}^{\circ }$

Obtain a second value e.g. ${69.1}^{\circ }$ and no extras in range

Use a correct method to form an equation for the line through $B$ and $C$

Obtain $\mathbf{r}=\mathbf{i}+3\mathbf{j}-2\mathbf{k}+\lambda (\mathbf{i}-4\mathbf{j}+5\mathbf{k})$

Find $\overrightarrow{AP}$ for a general point $P$ on $l$

Calculate the scalar product of $\overrightarrow{AP}$ (not $\overrightarrow{OP}$) and a direction vector for $l$ and equate the result to zero.

Obtain $\lambda =\frac{1}{3}$ or $\mu =-\frac{2}{3}$

Obtain $\frac{4}{3}\mathbf{i}+\frac{5}{3}\mathbf{j}-\frac{1}{3}\mathbf{k}$

Use a correct method to find _their_ position vector of $D$

Obtain $\frac{5}{3}\mathbf{i}+\frac{4}{3}\mathbf{j}-\frac{2}{3}\mathbf{k}$

Separate variables correctly

Use correct double angle formula to simplify integral in $y$

Obtain $\sin 2y$

Commence integration by parts and obtain $px\cos 3x+q\int \cos 3x\mathrm{d}x$

Obtain $-\frac{1}{3}x\cos 3x+\frac{1}{3}\int \cos 3x\mathrm{d}x$

Complete integration and obtain $-\frac{1}{3}x\cos 3x+\frac{1}{9}\sin 3x$

Use $y=\frac{1}{12}\pi$ when $x=\frac{1}{2}\pi$ in an expression with $\sin 2y,x\cos 3x$ and $\sin 3x$ to obtain the constant of integration

Obtain $\sin 2y=-\frac{1}{3}x\cos 3x+\frac{1}{9}\sin 3x+\frac{11}{18}$

Solve $\sin 2y=\frac{11}{18}$ to obtain one solution, e.g. 0.329

Obtain a second solution, e.g. 1.24, and no others in range

Use the correct product rule to differentiate

Obtain $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{2\sqrt{x}}\sin 2x+2\sqrt{x}\cos 2x$

Equate the derivative to zero and form an equation without surds

Obtain $\tan 2a=-4a$ from correct work

Calculate the values of a relevant expression or pair of expressions at $x=0.9$ and $x=0.95$

Complete the argument correctly with correct calculated values

State $a=\frac{1}{2}(\pi -{\tan }^{-1}4a)$ and rearrange to $\pi -2a={\tan }^{-1}4a$

State $\tan (\pi -2a)=4a$ and rearrange to $\tan 2a=-4a$

Use the iterative process correctly at least once

Obtain final answer 0.9183

Show sufficient iterations to at least 6 d.p. to justify 0.9183 to 4 d.p. or show there is a sign change in the interval $(0.91825,0.91835)$