A Level Mathematics 9709 Paper 32, May/June 2025

Obtain a 3-term quadratic in ${\text{e}}^x$

Obtain e.g. $3{\text{e}}^{2x}-12{\text{e}}^x-2=0$ or 3-term equivalent

Solve a 3-term quadratic to obtain a value for $x$ or ${\text{e}}^x$

Obtain root $\frac{6+\sqrt{42}}{3}$ or 4.16…

Obtain answer 1.426 only

Find the first two terms of the expansion of $(1-2x{)}^{-\frac{3}{2}}$

Obtain correct third term $\frac{-\frac{3}{2}(-\frac{3}{2}-1)}{2!}(-2x{)}^2$ or $\frac{-\frac{3}{2}(-\frac{3}{2}-1)}{2!}(2x{)}^2$

Multiply their 3 term expansion $a+bx+c{x}^2$ by $(6-x)$ obtaining all necessary terms

$6+17x+42x^2$

$\mathrm{\mid }x\mathrm{\mid }<\frac{1}{2}$ or $-\frac{1}{2}<x<\frac{1}{2}$ or (-0.5, 0.5) or ]-0.5, 0.5[

Show a circle centre $(0,3)$

Show a circle with radius 2

Show the point representing $(1,2)$

Show correct half-lines from $(1,2)$, one at an angle of $\frac{1}{4}\pi$ and the other at an angle of $\frac{3}{4}\pi$

Shade the correct region

Use correct trigonometric formulae to form an equation in $\tan x$ only, or an equation in terms of $\sin x$ and $\cos x$ only e.g. $\frac{3}{\tan x}-4\left(\frac{1-{\tan }^{2}x}{2\tan x}\right)=3$ $\frac{3}{\tan x}-\frac{4}{\frac{2\tan x}{1-{\tan }^{2}x}}=3$ $3\frac{\cos x}{\sin x}-4\frac{{\cos }^{2}x-{\sin }^{2}x}{2\sin x\cos x}=3$ $3\cot x-4\left(\frac{{\cot }^{2}x-1}{2\cot x}\right)=3$

Obtain a correct horizontal equation, in $\tan x$ or in $\cos x$ and $\sin x$, in any form

Reduce equation to a 3-term quadratic Allow if they square both sides and obtain a quartic in $\sin x$ or $\cos x$

Solve a 3-term quadratic to obtain a value for $x$ The quadratic could be arrived at from a correct cubic Not available if they have an incorrect cubic with an incorrect quadratic factor.

Obtain one answer, e.g. $(x=){45}^{\circ }$

Obtain a second answer, e.g. AWRT $(x=){26.6}^{\circ }$ and no other in the given interval, e.g. $x=0$

Square $x+\text{i}y$ and equate real and imaginary parts to –1 and $-4\sqrt{5}$ respectively

Obtain equations $x^2-y^2=-1$ and $2xy=-4\sqrt{5}$ from their expansion

Working from two equations in two unknowns, eliminate one variable and find an equation in the other

Obtain $x^4+x^2-20=0$ or $y^4-y^2-20=0$

Obtain answers $\pm (2-\sqrt{5}\text{i})$ and no others

Sketch a relevant graph for $0<x<\pi$ For $y=\mathrm{\mid }x-2\mathrm{\mid }$ graph should be symmetrical and have correct intercepts on the axes For $y=2\sin \frac{1}{2}x$, graph should pass through the origin, have correct curvature and $\max y=2$ when $x=\pi$

Sketch second relevant graph and confirm root. The vertex of the modulus graph in roughly correct position relative to $\pi$ and/or $\frac{1}{2}\pi$ If no mark on the graph, check to see if they have written something below the graph.

Calculate the values of a relevant expression or pair of expressions at $x=1$ and $x=1.5$ e.g. $f(x)=\mathrm{\mid }x-2\mathrm{\mid }-2\sin \frac{1}{2}x$ $f(1)=0.0411\dots >0$ $f(1.5)=-0.863\dots <0$ e.g. $f(x)=(x-2{)}^2-4{\sin }^2(\frac{1}{2}x)\Rightarrow f(1)=\mathrm{0.0806...},f(1.5)=-\mathrm{1.60...}$

Complete the argument correctly with correct calculated values

Use the iterative process correctly at least once starting at 1.03 (get as far as 1.0281)

Obtain final answer 1.02

Show sufficient iterations to 4 d.p. to justify 1.02 to 2 d.p. or show there is a sign change in the interval (1.015, 1.025) (or a smaller interval containing the root)

State $R=25$

Use correct trig formula to find $\alpha$, e.g. $\alpha ={\tan }^{-1}\left(\frac{7}{24}\right)$

Obtain $\alpha =0.2838$

${\cos }^{-1}\left(\frac{24.5}{25}\right)$

Carry out a complete correct method to find a value of $x$ Sight of a correct equation using (a) is needed, e.g. $\cos (\frac{1}{3}x-0.2838)=0.98$ OE

Obtain answer $(x=)1.45$

Obtain $(x=)0.250$ or $(x=)0.251$, and no other in the interval

Separate variables correctly

Obtain term $\frac{1}{4}\ln (4x+3)$

Obtain term of the form $A\ln (\sin 2\theta )$

Obtain term $\frac{1}{2}\ln (\sin 2\theta )$

Use $x=0$ when $\theta =\frac{1}{12}\pi$ to evaluate a constant or as limits in a solution containing terms of the form $\ln (\sin 2\theta )$ and $\ln (4x+3)$ $c=\dots$ seen or implied

Obtain correct answer in any form with the trigonometry evaluated

Obtain final answer $x=\frac{12{\sin }^{2}2\theta -3}{4}$

Carry out a correct method for finding a vector equation for the line through $A$ and $B$

Obtain $\mathbf{r}=\left(\begin{array}{c}1\\ -4\\ 2\end{array}\right)+\lambda \left(\begin{array}{c}-3\\ 5\\ 1\end{array}\right)$

Obtain a direction vector for $\overrightarrow{AC}=\left(\begin{array}{c}1\\ 7\\ 3\end{array}\right)$

Carry out correct process for evaluating the scalar product of $\overrightarrow{AB}$ and $\overrightarrow{AC}$ or $\overrightarrow{BA}$ and $\overrightarrow{CA}$

Using the correct process for the moduli, divide their scalar product by the product of the moduli

Obtain $(\cos BAC=)\frac{35}{\sqrt{59}}$ or exact simplified equivalent

Use $\text{area}=\frac{1}{2}\mathrm{\mid }\overrightarrow{AB}\mathrm{\mid }\mathrm{\mid }\overrightarrow{AC}\mathrm{\mid }\sin BAC$ with their $\mathrm{\mid }\overrightarrow{AB}\mathrm{\mid }\mathrm{\mid }\overrightarrow{AC}\mathrm{\mid }$

Use $\sin x=\sqrt{1-{\cos }^{2}x}$ or an equivalent exact method with their $\cos x(<1)$ to obtain an exact value for $\sin x$ $\frac{1}{2}\sqrt{35}\sqrt{59}\sqrt{1-\frac{{35}^2}{35\times 59}}=\frac{1}{2}\sqrt{35}\sqrt{59}\frac{\sqrt{840}}{\sqrt{35}\sqrt{59}}$

Obtain answer $\sqrt{210}$ from correct working

Obtain quotient $\frac{1}{4}$

Obtain remainder $-\frac{1}{4}$

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

Obtain $\frac{1}{2}{x}^2{\tan }^{-1}2x-\int \frac{x^2}{1+4x^2}\text{d}x$

Integrate $\int \frac{x^2}{1+4x^2}\text{d}x$ to obtain an expression of the form $p{\tan }^{-1}2x+qx$

Complete integration and obtain $\frac{1}{2}{x}^2{\tan }^{-1}2x+\frac{1}{8}{\tan }^{-1}2x-\frac{1}{4}x$

Substitute limits correctly in an expression of the form $F{x}^2{\tan }^{-1}2x+G{\tan }^{-1}2x+Hx$

Obtain answer $\frac{1}{16}\pi -\frac{1}{8}$ or exact one- or two-term equivalent with trigonometry evaluated

Differentiate ${\cos }^{2}x$ to obtain $-2\sin x\cos x$

Use correct product rule

Obtain derivative $-10\sin 2x\sin x\cos x+10\cos 2x{\cos }^{2}x$

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

Obtain $3{\tan }^{2}x=1$, $4{\sin }^{2}x=1$ or $4{\cos }^{2}x=3$ or $\cos 3x=0$

Obtain $x=\frac{1}{6}\pi$ only

$\frac{\text{d}u}{\text{d}x}=-\sin x$

Reach an integral of the form $\int A{u}^3\text{d}u$

Obtain $-10\int u^3\text{d}u$

Substitute correct limits correctly in an expression of the form $C{u}^4$ or $C{\cos }^{4}x$

Obtain answer $\frac{15}{8}$ or 1.875