A Level Mathematics 9709 Paper 22, Oct/Nov 2024

Apply logarithms correctly and apply power law at least once

Obtain $y=\frac{7\ln 6}{8\ln 5}x$ or equivalent perhaps involving decimals

Obtain $k=0.974$ or state $y=0.974x$

Differentiate to obtain form $k\sin 3x\cos 3x$ or $k\sin 6x$

Obtain correct $24\sin 3x\cos 3x$ or $12\sin 6x$

Substitute $x=\frac{1}{4}\pi$ to obtain $-12$

Express integrand in form $k_1+k_2\cos 6x$

Obtain $2-2\cos 6x$

Integrate to obtain $2x-\frac{1}{3}\sin 6x$

Use product rule to differentiate $6{\mathrm{e}}^{-x}{y}^2$

Obtain $-6{\mathrm{e}}^{-x}{y}^2+12{\mathrm{e}}^{-x}y\frac{\mathrm{d}y}{\mathrm{d}x}$

Obtain $+2{\mathrm{e}}^{2x}-12\frac{\mathrm{d}y}{\mathrm{d}x}$

Rearrange correctly to obtain $\frac{\mathrm{d}y}{\mathrm{d}x}=\dots$

Substitute for $x$ and $y$ to obtain an exact final answer

Obtain $\frac{5}{2}$ or 2.5

Show increasing curve above $x$-axis for $y=1+{\mathrm{e}}^{2x}$

Show V-shaped graph with vertex on positive $x$-axis and only one point of intersection with first curve

State or clearly imply $1+{\mathrm{e}}^{2x}=4-x$

Arrange to confirm $x=\frac{1}{2}\ln (3-x)$

Use iterative process correctly at least once

Obtain final answer 0.465

Show sufficient iterations to justify answer or show a sign change in the interval $[0.4645,0.4655]$

Substitute $x=-2$ and equate to zero

Substitute $x=2$ and equate to 24

Obtain $-6a+4b+8=0$ and $6a+4b-16=0$

Obtain $a=2$ and $b=1$

Divide by $x+2$ at least as far as the $x$ term

Obtain $(x+2)(2x^2-3x+4)$

Conclude with reference to root $-2$, discriminant is $-23$ and no further root

State $\mathrm{cosec}\theta =-4$

Attempt to find at least one value of $\theta$ from $\sin \theta =\pm \frac{1}{4}$

Obtain $-14.5$ only and no others in the range

Use $y$-values $\sqrt[3]{7}$, $\sqrt[3]{12}$ and $\sqrt[3]{27}$, or decimal equivalents

Use correct formula, or equivalent, with $h=1$

Obtain 4.75

Integrate to obtain form $k\ln (2x+5)$

Obtain correct $\frac{27}{2}\ln (2x+5)$

Apply limits 0 and 2 to obtain $\frac{27}{2}\ln \frac{9}{5}$ or exact equivalent of required form and no extra terms

Obtain 3.19

State under-estimate …

… because (a) is over-estimate due to tops of trapezia being above curve

Expand to obtain $2{\sin }^2\theta +2\sqrt{3}\sin \theta \cos \theta$

Attempt to express in terms of $\sin 2\theta$ and $\cos 2\theta$

Obtain $1-\cos 2\theta +\sqrt{3}\sin 2\theta$

Obtain $R=2$

Use appropriate trigonometry to find $\alpha$

Obtain $1+2\sin (2\theta -30)$

State $\sin (2\theta -30)=-\frac{6}{10}$

Attempt complete method to find smallest positive value

Obtain 123.4 only