A Level Mathematics 9709 Paper 31, Oct/Nov 2025

Commence integration by parts and reach $Ax\ln 3x+\int B\text{ d}x$

Obtain $x\ln 3x-x$

Substitute limits correctly in an expression of the form $Ax\ln 3x+Bx$

Obtain $-1+\ln 12$

State or imply $2={\log }_4{4}^2$

Use log power, product or quotient law e.g. $2{\log }_4(3x-1)={\log }_4(3x-1{)}^2$

Obtain a correct equation in any form, free of logs

Solve 3-term quadratic

Obtain answer 4.59 only

Obtain $3\sin x+4\cos x$

State $R=5$

Use trig formula to find $\alpha$ e.g. $\alpha ={\tan }^{-1}\left(\frac{3}{4}\right)$

Obtain $\alpha ={36.9}^{\circ }$ (or awrt ${36.9}^{\circ }$)

Evaluate ${\cos }^{-1}(-\frac{4}{R})$

Carry out a correct method to find a value of $\theta$ in the given interval

Obtain answer, e.g. $60(.0{)}^{\circ }$

Obtain a second answer, e.g. ${84.6}^{\circ }$ and no other in the interval

Use correct product rule: $p\sin 4x{\text{e}}^{\sin 2x}+q\cos 2x\cos 4x{\text{e}}^{\sin 2x}$

Obtain derivative $-4\sin 4x{\text{e}}^{\sin 2x}+2\cos 2x\cos 4x{\text{e}}^{\sin 2x}$

Equate derivative to zero and obtain an equation in one trig function

Obtain $2{\sin }^{2}2x+4\sin 2x-1=0$

Obtain $x=0.113$ and no other root

Obtain $\text{Re }z\ge 2$

Obtain answer of the form $\mathrm{\mid }z-a\mathrm{\mid }\le b$

Obtain answer $\mathrm{\mid }z-1+2\text{i}\mathrm{\mid }\le 2$

Identify the correct point

Carry out a correct method for finding the least value of $\arg z$

Obtain answer $-{61.8}^{\circ }$ or ${298.2}^{\circ }$ or $-{1.08}^{\text{c}}$ or $5.2(0{)}^{\text{c}}$

Use quadratic formula to solve for $w$

Use ${\text{i}}^2=-1$

Obtain one of the answers $w=\frac{-4-2\text{i}}{4+2\text{i}}$ or $w=\frac{-4+2\text{i}}{4+2\text{i}}$

Multiply numerator and denominator by conjugate of *their* denominator

Obtain final answers $-\frac{3}{5}+\frac{4}{5}\text{i}$ and $-1$ $[+0\text{i}]$

State $\frac{\text{d}x}{\text{d}t}=2t-\frac{2}{2t+1}$

Use correct quotient rule or correct product rule

Obtain $\frac{\text{d}y}{\text{d}t}=\frac{(2t+1).1-t(2)}{(2t+1{)}^2}$ oe

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

Obtain answer $\frac{1}{2(2t+1)(2t-1)(t+1)}$

Separate variables correctly

Obtain term $A{\text{e}}^{-2y}$

Obtain term $B\ln (x^2+1)$

Obtain $-\frac{1}{2}{\text{e}}^{-2y}=\frac{k}{2}\ln (x^2+1)(+C)$

Use $y=0$ when $x=0$ to evaluate $c$ or as limits in a solution containing terms of the form ${\text{e}}^{-2y}$ and $\ln (x^2+1)$

Use $y=-\frac{1}{2}$ when $x=1$ to evaluate $k$ or as limits in a solution containing terms of the form ${\text{e}}^{-2y}$ and $\ln (x^2+1)$

Obtain $c=-\frac{1}{2}$ and $k=\frac{1-\text{e}}{\ln 2}$

Obtain final answer $y=-\frac{1}{2}\ln (2\text{e}-1)$

Sketch a relevant graph, e.g. $y=\sec 2x$ May be sketching $\cos 2x$ and $-{\text{e}}^{-x}$ For $y=\sec 2x$ correct shape, asymptote in correct position. Cuts at $(0,1)$ For $-{\text{e}}^x$ correct shape, cuts at $(0,1)$ For $y=\cos 2x$ correct shape, correct max/min, cuts axis at $\frac{\pi }{4}$ For $-{\text{e}}^{-x}$ correct shape, cuts at $(0,-1)$

Sketch a second relevant graph, e.g. $y=-{\text{e}}^x$ and justify the given statement

Calculate the values of a relevant expression or pair of expressions at $x=0.9$ and $x=1$

Complete the argument correctly with correct calculated values

State the equation $x=\frac{1}{2}{\cos }^{-1}(-{\text{e}}^{-x})$ and rearrange to obtain $\sec 2x=-{\text{e}}^x$ Or rearrange $\sec 2x=-{\text{e}}^x$ as $x=\frac{1}{2}{\cos }^{-1}(-{\text{e}}^{-x})$ and state $x_{n+1}=\frac{1}{2}{\cos }^{-1}(-{\text{e}}^{-x_n})$

Use the iterative process correctly at least once

Obtain final answer 0.978

Show sufficient iterations to at least 5 d.p. to justify 0.978 to 3 d.p. or show there is a sign change in the interval $(0.9775,0.9785)$

State or imply the form $A+\frac{B}{3+x}+\frac{Cx+D}{2+x^2}$

Use a correct method for finding a constant

Obtain one of $A=1$, $B=-4$, $C=1$ and $D=-3$

Obtain a second value

Obtain a third value

Obtain all four correct values

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

Obtain correct unsimplified expansion up to the term in $x^2$ of first partial fraction

Obtain correct unsimplified expansion up to the term in $x^2$ of second partial fraction

Multiply $(Cx+D)$ by the expansion of ${(1+\frac{x^2}{2})}^{-1}$ up to the term in $x^2$ where $CD\mathrm{\ne }0$

Obtain final answer $-\frac{11}{6}+\frac{17}{18}x+\frac{65}{108}{x}^2$

Carry out a correct method for finding a vector equation for $m$

Obtain $\mathbf{r}=\left(\begin{array}{c}1\\ 5\\ 3\end{array}\right)+\lambda \left(\begin{array}{c}1\\ 1\\ 2\end{array}\right)$

Express general point of the second line in component form, i.e. $(1+2\mu ,-3-2\mu ,1+3\mu )$

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

Obtain $\lambda =-4$ or $\mu =-2$

Obtain position vector of point of intersection is $\left(\begin{array}{c}-3\\ 1\\ -5\end{array}\right)$

Find $\overrightarrow{CP}$ for a general point $P$ on $m$, e.g. $\left(\begin{array}{c}0\\ 8\\ 2\end{array}\right)+\lambda \left(\begin{array}{c}1\\ 1\\ 2\end{array}\right)$

Calculate scalar product of $\overrightarrow{CP}$ and a direction vector for $m$ and equate the result to zero

Obtain $\lambda =-2$

Obtain answer $\left(\begin{array}{c}-1\\ 3\\ -1\end{array}\right)$