A Level Mathematics 9709 Paper 32, May/June 2026

Commence division and reach partial quotient of the form $2x^2\pm kx$ or $4x^3-4x^2+7x+2=(2x-1)(A{x}^2+Bx+C)+D$ with $A=2$ and $B=\pm k$

Obtain quotient $2x^2-x+3$ Or: state the factor as above and then at least $A=2$ and $B=-1,C=3$

Obtain remainder 5 Or: $D=5$

Use law of logarithm of a product or quotient (Allow later if seen applied to correct terms) OR Use a correct law of indices to obtain the same base on both sides

At least one use of the correct power law to a correct term OR Obtain a correct equation in any form with a single index term involving $x$

Obtain a correct linear equation in any form, e.g. $(5-2x)\ln 2=(x+1)\ln 4-\ln 3$ OR Correctly take logs of an equation in any form with a single index term to obtain a linear equation

Obtain $(x=)\frac{\ln 24}{\ln 16}$

$r=6$

$\theta =1$

Enlargement [scale] factor 2

Clockwise rotation of 1 [radians] or rotation of -1 [radians] (direction implied) or [anticlockwise] rotation of $(2\pi -1)$ [radians]

Square $x+\mathrm{i}y$ and equate real and imaginary parts to 3 and $-6\sqrt{2}$ respectively

Obtain equations $x^2-y^2=3$ and $2xy=-6\sqrt{2}$

Eliminate one variable and find an equation in the other

Obtain $x^4-3x^2-18=0$ or $y^4+3y^2-18=0$ or three-term equivalents

Obtain answers $\pm (\sqrt{6}-\sqrt{3}\mathrm{i})$ only

State $\frac{\mathrm{d}x}{\mathrm{d}\theta }=2-2{\sec }^{2}2\theta$

Use chain rule to obtain $\frac{\mathrm{d}}{\mathrm{d}\theta }\ln (\cos 2\theta )=\frac{\frac{\mathrm{d}}{\mathrm{d}\theta }(\cos 2\theta )}{\cos 2\theta }$

Obtain $\frac{\mathrm{d}y}{\mathrm{d}\theta }=6-\frac{4\sin 2\theta }{\cos 2\theta }$

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

Use ${\sec }^{2}2\theta =1+{\tan }^{2}2\theta$ and $\frac{\sin 2\theta }{\cos 2\theta }=\tan 2\theta$ to write $\frac{\mathrm{d}y}{\mathrm{d}x}$ or $\frac{\mathrm{d}x}{\mathrm{d}\theta }$ and $\frac{\mathrm{d}y}{\mathrm{d}\theta }$ in terms of $\tan 2\theta$

Obtain $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{6-4\tan 2\theta }{-2{\tan }^{2}2\theta }$

Set their $\frac{\mathrm{d}y}{\mathrm{d}x}=-1$, rearrange to a three-term quadratic in $\tan 2\theta$

Solve their three-term quadratic to obtain a value of $\theta$ or $2\theta$

Obtain $\theta =\frac{1}{8}\pi$ only within the range

Obtain $x=\frac{1}{4}\pi -1$ and $y=\frac{3}{4}\pi +2\ln \frac{\sqrt{2}}{2}$ only or exact equivalents

Use the correct product rule

Differentiate ${\tan }^{-1}3x$ to obtain $\frac{p}{1+q{x}^2}$

Obtain $\frac{3x}{1+9x^2}+{\tan }^{-1}3x$

Set $x=\frac{1}{3}$ and obtain $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{4}\pi +\frac{1}{2}$ only

Commence integration by parts and reach $A{x}^2{\tan }^{-1}3x\pm \int \frac{B{x}^2}{D+C{x}^2}\mathrm{d}x$

Obtain $\frac{1}{2}{x}^2{\tan }^{-1}3x-\frac{3}{2}\int \frac{x^2}{1+9x^2}\mathrm{d}x$

Attempt to reduce their $\frac{B{x}^2}{D+C{x}^2}$ to $E+\frac{F}{D+C{x}^2}$

Complete integration and obtain $\frac{1}{2}{x}^2{\tan }^{-1}3x-\frac{1}{6}x+\frac{1}{18}{\tan }^{-1}3x$

Substitute limits correctly in an expression of the form $a{x}^2{\tan }^{-1}3x+bx+c{\tan }^{-1}3x$

Obtain answer $\frac{1}{36}\pi -\frac{1}{18}$

Use the correct product or quotient rule

Obtain the correct derivative in any form, e.g. $\frac{(2-\cos 3x)(3\cos 3x)-(3+\sin 3x)(3\sin 3x)}{(2-\cos 3x{)}^2}$

Equate their derivative to $-\frac{3}{2}$, and obtain a three-term quadratic in $\sin 3a$ or $\sin 3x$

Obtain ${\sin }^{2}3a+6\sin 3a-3=0$ with full and correct working

Find a value of $a$ in the given interval

Obtain an answer, e.g. 0.161

Obtain all answers 0.161, 0.886, 2.26, 2.98 and no others in the interval (${9.2}^{\circ },{50.8}^{\circ },{129.2}^{\circ },{170.8}^{\circ }$)

Sketch $y=\mathrm{cosec}\frac{1}{2}x$ for $0\le x\le \pi$. The horizontal level of the minimum should not look as if it is on the $x$-axis

Sketch $y=\frac{1}{2}(x+3)$ for $0\le x\le \pi$ and justify the given statement by clear marking of the intersect or a statement or a dotted line down to the $x$-axis. Consistent vertical scale: e.g., $y$-intercept above the minimum for the trigonometric graph

Calculate the values of a relevant expression or pair of expressions at $x=1$ and $x=1.2$

Complete the argument correctly

Express $x_{n+1}=2{\sin }^{-1}\left(\frac{2}{x_n+3}\right)$ as $x=2{\sin }^{-1}\left(\frac{2}{x+3}\right)$ OR Rearrange $\mathrm{cosec}\frac{1}{2}x=\frac{1}{2}(x+3)$ to $x=2{\sin }^{-1}\left(\frac{2}{x+3}\right)$

Rearrange $x=2{\sin }^{-1}\left(\frac{2}{x+3}\right)$ to $\mathrm{cosec}\frac{1}{2}x=\frac{1}{2}(x+3)$ with full and correct working OR Express $x=2{\sin }^{-1}\left(\frac{2}{x+3}\right)$ as $x_{n+1}=2{\sin }^{-1}\left(\frac{2}{x_n+3}\right)$ after full and correct working

Use the iterative process correctly at least once

Obtain final answer 1.037

Show sufficient iterations to at least 5 dp to justify 1.037 to 3 dp or show there is a sign change in the interval (1.0365, 1.0375)

Use a correct method to form an equation for line $m$

Obtain $\mathbf{r}=\left(\begin{array}{c}3\\ 4\\ 1\end{array}\right)+\lambda \left(\begin{array}{c}1\\ -1\\ 0\end{array}\right)$

Express $l$ or $m$ in component form, e.g. $(6+\mu ,-1-\mu ,-3-2\mu )$ or $(3+{\lambda }_1,4-{\lambda }_1,1)$ or $(4+{\lambda }_2,3-{\lambda }_2,1)$

Equate pairs of components of general points on $l$ and their $m$ to obtain an answer for $\lambda$ or $\mu$, e.g. $\mu =-2$ or ${\lambda }_1=1$ or ${\lambda }_1=3$ if equations correct, or determine no value can be found for $\lambda$ or $\mu$ if using equations for $\mathbf{i}$ and $\mathbf{j}$

Determine that all three equations are not satisfied by showing that they cannot be solved or obtaining a contradiction

Justify lines are not parallel, e.g. $\left(\begin{array}{c}1\\ -1\\ -2\end{array}\right)\mathrm{\ne }d\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right)$, and conclude the lines are skew

Express $\overrightarrow{PA}$ or $\overrightarrow{AP}$ in component form, e.g. $\pm (-3-\mu ,5+\mu ,4+2\mu )$

Carry out correct process for evaluating the scalar product of their $\overrightarrow{PA}$ and their $\overrightarrow{BA}$

Complete the scalar product process using the moduli of their vectors and use $\cos {60}^{\circ }=\frac{1}{2}$ to form an equation involving a single parameter

Form a horizontal equation involving a single parameter after correct removal of square roots and correct processing of $\frac{1}{2}$

Solve their three-term quadratic and use a root to find a position vector for $P$

Obtain position vector $\left(\begin{array}{c}3\\ 2\\ 3\end{array}\right)$ only

State or imply the form $\frac{A}{x-1}+\frac{B}{2x-6}$

Use a correct method for finding a constant

Obtain $A=-1$ or $B=3$

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

Separate variables correctly – seen or implied by correct structure to the integration of both sides

Obtain term $-\ln (x-1)$

Obtain term $\frac{3}{2}\ln (2x-6)$

Commence integration by parts and reach $py\ln 2y\pm \int \frac{q}{y}y\mathrm{d}y$

Complete integration by parts to obtain $y\ln 2y-y$

Use $x=4,y=1$ to evaluate a constant of the correct form or as limits in a solution containing terms of the form $a\ln (x-1)$, $b\ln (2x-6)$, $cy\ln 2y$ and $fy$

Obtain correct answer in any form (with ${\mathrm{e}}^{\ln 2}$, and similar terms, simplified if seen)