A Level Mathematics 9709 Paper 22, May/June 2026

Expand integrand to obtain $1+2{\text{e}}^{-x}+{\text{e}}^{-2x}$

Integrate to obtain at least two of the terms of the form $x+k_1{\text{e}}^{-x}+k_2{\text{e}}^{-2x}$

Obtain correct $x-2{\text{e}}^{-x}-\frac{1}{2}{\text{e}}^{-2x}$

Apply limits correctly to obtain $a-\frac{3}{2}{\text{e}}^{-2a}-\frac{1}{2}{\text{e}}^{-4a}+2{\text{e}}^{-a}$

Use correct identities to produce a correct equation in one trigonometric function

Obtain ${\sin }^2\theta =\frac{1}{3}$ or ${\cos }^2\theta =\frac{2}{3}$ or ${\tan }^2\theta =\frac{1}{2}$ or ${\cot }^2\theta =2$ or ${\mathrm{cosec}}^2\theta =3$

Obtain at least two of the correct answers $\pm 35.3$, $\pm 144.7$

Obtain all four correct answers and no others between $-{180}^{\circ }$ and ${180}^{\circ }$

State or imply equation $\ln y=\ln k+(x+2)\ln a$

Equate gradient of line to $\ln a$ and attempt to find value of $a$

Obtain $\ln a=0.588$ and hence $a=1.8$

Substitute appropriate values to find $\ln k$

Obtain $\ln k=1.162$ and hence $k=3.2$

Attempt use of correct quotient rule

Obtain $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{{\text{e}}^{3x}+4{\text{e}}^{2x}}{({\text{e}}^x+2{)}^2}$

Substitute $x=0$ to find value of derivative at $P$

Obtain $-\frac{9}{5}$ for gradient of normal

Use $x=0,y=\frac{1}{3}$ and their negative reciprocal of derivative to find equation

Obtain $27x+15y-5=0$ or equivalent of requested form

Solve $4x-7=2x+3$ to obtain $x=5$

Attempt solution of linear equation where $4x$ and $2x$ have different signs

Obtain $x=\frac{2}{3}$ or $x=\frac{4}{6}$

Conclude $\frac{2}{3}⩽x⩽5$ or $x⩽5$ and $x⩾\frac{2}{3}$

Attempt solution for $N$ of $\ln (\frac{1}{2}N)$ equated to either critical value from 5(a)

Obtain $\ln (\frac{1}{2}N)=\frac{2}{3}$ and hence smallest integer is 4

Obtain $\ln (\frac{1}{2}N)=5$ and hence largest integer is 296

Attempt solution for $a$ of $\mathrm{f}(-2)=0$

Obtain $16a-32-12a+42-18=0$ and hence $a=2$

Attempt division of quartic with their $a$ to obtain quotient at least as far as the $x$ term

Obtain $2x^3-6x-9$

Rearrange $2x^3-6x-9=0$ to confirm $x=\sqrt[3]{3x+4.5}$

Use iteration process correctly at least once

Obtain final answer 2.238

Show sufficient iterates to 6 significant figures to justify answer or show a sign change in the interval $[2.2375,2.2385]$

Attempt to find solution of $8\cos 2x-\frac{1}{{\cos }^{2}2x}=0$

Obtain $\cos 2x=\frac{1}{2}$ and hence $x=\pm \frac{1}{6}\pi$

Recognise $\frac{1}{{\cos }^{2}2x}$ as ${\sec }^{2}2x$

Integrate to obtain the form $k_1\sin 2x+k_2\tan 2x$

Obtain correct $4\sin 2x-\frac{1}{2}\tan 2x$

Apply correct method for their limits to find exact value

Obtain $3\sqrt{3}$

Obtain $\frac{\mathrm{d}x}{\mathrm{d}t}=6-2\cos 2t$

Obtain $\frac{\mathrm{d}y}{\mathrm{d}t}=-6\sin 2t+5\cos t$

Attempt value of $\frac{\mathrm{d}y}{\mathrm{d}x}$ for $t=0$

Obtain $\frac{5}{4}$

Equate $\frac{\mathrm{d}y}{\mathrm{d}x}$ or $\frac{\mathrm{d}y}{\mathrm{d}t}$ to zero and use identity $\sin 2t=2\sin t\cos t$

Obtain $\cos t(5-12\sin t)=0$ or equivalent and hence at least $\sin t=\frac{5}{12}$

Attempt to express $y$ in terms of $\sin t$ and/or $\cos t$ only

Obtain $3-6{\sin }^{2}t+5\sin t$ or $3({\cos }^{2}t-{\sin }^{2}t)+5\sin t$ and $\cos t=\frac{\sqrt{119}}{12}$ / $3(2{\cos }^{2}t-1)+5\sin t$ and $\cos t=\frac{\sqrt{119}}{12}$

Substitute exact value of $\sin t$ and $\cos t$ (if appropriate) to obtain $\frac{97}{24}$