A Level Mathematics 9709 Paper 33, Oct/Nov 2025

State or imply non-modular inequality $(3x+2{)}^2<3^2(2x-1{)}^2$, or pair of linear equations $(3x+2)=\pm 3(2x-1)$

Make reasonable attempt at solving a 3-term quadratic, or solve two linear equations for $x$

Obtain critical values $x=\frac{1}{9}$ and $x=\frac{5}{3}$

State final answer $x<\frac{1}{9},x>\frac{5}{3}$ only

Commence division and reach partial quotient of the form $3x^3\pm 3x^2$ or $3x^4-2x^2\equiv (x+1)(A{x}^3+B{x}^2+Cx+D)+Ex+F$, and reach $A=3$ and $B=\pm 3$

Obtain quotient $3x^3-3x^2+x-1$ Do not ISW

Obtain remainder of 1

Use log quotient law, e.g. $\ln \frac{3}{5^x}=\ln 3-\ln 5^x$ or log product law, e.g. $\ln 5^x+\ln (2^{3x-4})$

Use log power law, e.g. $\ln 5^x=x\ln 5$ or $\ln (2^{3x-4})=(3x-4)\ln 2$

Obtain correct expression for $x$ in any exact form, e.g. $\frac{\ln 3+4\ln 2}{3\ln 2+\ln 5}$

Obtain final answer $\frac{\ln 48}{\ln 40}$

Show a circle centre $(0,-2)$

Show a circle with radius 3

Show the point representing $(2,-4)$ or the midpoint $(1,-3)$

Show the perpendicular bisector of the line joining $(2,-4)$ and $(0,-2)$ or the perpendicular bisector of the line joining $(2,-4)$ and centre of their circle or the perpendicular bisector going through $(1,-3)$

Shade the correct region

Use correct double angle formula to express $\cos 4x$ in terms of ${\sin }^{2}2x$

Use correct double angle formula to express $\cos 4x$ in terms of single angles

Obtain a correct expression in $\sin x$ in any form

Obtain the given answer $8{\sin }^{4}x-6{\sin }^{2}x$

Obtain answers $0^{\circ }$, ${180}^{\circ }$ and $-{180}^{\circ }$

Carry out a correct method to find a value of $x$ in the given interval for $8{\sin }^{2}x-6=0$

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

Obtain remaining answers, e.g. $-{60}^{\circ }$, ${120}^{\circ }$ and $-{120}^{\circ }$ and no other in the interval

Commence integration by parts and reach $A{x}^2\cos 2x-\int Bx\cos 2x\mathrm{d}x$

Obtain $-\frac{1}{2}{x}^2\cos 2x+\int x\cos 2x\mathrm{d}x$

Complete integration by parts and reach $A{x}^2\cos 2x+Bx\sin 2x+C\cos 2x$

Obtain $-\frac{1}{2}{x}^2\cos 2x+\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x$

Substitute limits correctly in an expression of the form $A{x}^2\cos 2x+Bx\sin 2x+C\cos 2x$ and evaluate the trig expressions

Obtain answer $-\frac{1}{144}{\pi }^2+\frac{\sqrt{3}}{24}\pi -\frac{1}{8}$ or exact three-term equivalent

Attempt to remove fraction by multiplying throughout by $2-\mathrm{i}$ or multiply fraction by $\frac{2+\mathrm{i}}{2+\mathrm{i}}$

Substitute $z=x+\mathrm{i}y$ and $z^{\ast }=x-\mathrm{i}y$ throughout the equation

Use ${\mathrm{i}}^2=-1$ correctly at least once and equate real and imaginary parts to zero

Obtain two correct equations e.g. $5x-2(x^2+y^2)+48=0$ and $5y+(x^2+y^2)-4=0$

Solve a quadratic and a linear equation for $x$ or for $y$

Obtain answers $2-5\mathrm{i}$ and $-\frac{16}{5}-\frac{12}{5}\mathrm{i}$ only

Use the correct product or quotient rule, e.g. ${\mathrm{e}}^{-5x}\frac{\mathrm{d}}{\mathrm{d}x}(\ln 5x)+\ln 5x\frac{\mathrm{d}}{\mathrm{d}x}({\mathrm{e}}^{-5x})$

Obtain the correct derivative in any form e.g. $\frac{1}{x}{\mathrm{e}}^{-5x}-5{\mathrm{e}}^{-5x}\ln 5x$

Obtain the given answer $\ln 5p=\frac{1}{5p}$ after full and correct working

Sketch an acceptable graph, e.g. $y=\ln 5x$ or $y=\frac{1}{5x}$

Sketch a second acceptable graph, e.g. $y=\frac{1}{5x}$ or $\ln 5x$, and justify the given statement by dot, cross or statement only one intersection.

Calculate the values of a relevant expression or pair of expressions at $p=0.2$ and $p=0.6$ Note can use, e.g., $p=0.3$ and $p=0.5$, or any smaller interval which works

Complete the argument correctly with correct calculated values

Use the iterative formula correctly at least twice

Obtain final answer $p=0.35$, $\text{Answer}=0.35$, or just 0.35 stated

Show sufficient iterations to 4 dp to justify 0.35 to 2 dp or show there is a sign change in the interval $(0.345,0.355)$

Obtain $\mathbf{r}=3\mathbf{i}+\mathbf{j}-6\mathbf{k}+\lambda (2\mathbf{i}+\mathbf{j}+4\mathbf{k})$

Carry out a correct method for finding a direction vector for $l_2,(3\mathbf{i}-2\mathbf{j}+\mathbf{k})\cdot (y\mathbf{j}+z\mathbf{k})=0$

Obtain $\mathbf{r}=-\mathbf{i}+3\mathbf{j}-6\mathbf{k}+\mu (\mathbf{j}+2\mathbf{k})$

Carry out correct process for evaluating the scalar product of the direction vectors of $l_1$ and $l_2$

Using the correct process for the moduli, divide the scalar product by the product of the moduli and state $\cos \theta =$ the result

Obtain answer ${28.6}^{\circ }$ or ${0.498}^{\text{c}}$

Equate components of general points on their $l_1$ and their $l_2$, provided these are both equations of lines and solve for $\lambda$ or for $\mu$

Obtain correct answer for $\lambda$ or $\mu$, e.g. $\lambda =-2,\mu =-4$

Obtain position vector of point of intersection is $-\mathbf{i}-\mathbf{j}-14\mathbf{k}$

Carry out a relevant method to find $A$ and $B$ such that $\frac{2}{1-9y^2}=\frac{A}{1+3y}+\frac{B}{1-3y}$

Obtain $A=1$ and $B=1$

Separate variables correctly and attempt integration of at least one side

Integrate $\frac{2}{1-9y^2}$ to obtain $\frac{1}{3}\ln (1+3y)-\frac{1}{3}\ln (1-3y)$

Obtain $r\tan 3x$

Obtain term $\frac{1}{3}\tan 3x$

Use $y=0$ when $x=\frac{1}{12}\pi$ to evaluate a constant or as limits in a solution of the form $p\ln (1+3y)+q\ln (1-3y)+r\tan 3x$ where $p,q,r\mathrm{\ne }0$

$\frac{1}{3}\ln (1+3y)-\frac{1}{3}\ln (1-3y)=\frac{1}{3}\tan 3x-\frac{1}{3}$ Obtain answer $y=\frac{1}{3}-\frac{2}{3(1+{\mathrm{e}}^{\tan 3x-1})}$ or $y=\frac{{\mathrm{e}}^{\tan 3x-1}-1}{3({\mathrm{e}}^{\tan 3x-1}+1)}$

Differentiate $\sqrt{3+2\tan x}$ to obtain ${\sec }^{2}x(3+2\tan x{)}^{-\frac{1}{2}}$

Differentiate ${\sec }^{2}x$ to obtain $2\sec x\sec x\tan x$

Use correct product (or quotient) rule

[Obtain derivative, if correct $2{\sec }^{2}x\tan x(3+2\tan x{)}^{\frac{1}{2}}+{\sec }^{4}x(3+2\tan x{)}^{-\frac{1}{2}}$] and equate derivative to zero and obtain an equation in one trig function

Obtain $5{\tan }^{2}x+6\tan x+1=0$

Obtain AWRT $x=-0.197$ only

$\frac{\mathrm{d}u}{\mathrm{d}x}=2{\sec }^{2}x$

Reach an integral of the form $\int A\sqrt{u}\mathrm{d}u$

Obtain $\int \frac{1}{2}{u}^{\frac{1}{2}}\mathrm{d}u$

Obtain $\frac{1}{3}{u}^{\frac{3}{2}}$

Substitute correct limits correctly in an expression of the form $B{u}^{\frac{3}{2}}$ or $B(3+2\tan x{)}^{\frac{3}{2}}$ and obtain $c(5^{\frac{3}{2}}-1^{\frac{3}{2}})$

Obtain answer $\frac{5\sqrt{5}}{3}-\frac{1}{3}$