A Level Mathematics 9709 Paper 21, May/June 2023

Apply logarithms to both sides and apply power law correctly at least once

Obtain $x\ln 12=(2x+1)\ln 3$

Attempt solution of linear equation

Obtain 3.82

Use quotient rule (or equivalent) to find first derivative

Obtain $\frac{(1+2x)\frac{3}{x}-(2+3\ln x)2}{(1+2x{)}^2}$

Substitute $x=1$ and obtain $\frac{5}{9}$

Attempt equation of tangent through $(1,\frac{2}{3})$ with _their_ numerical gradient

Obtain $5x-9y+1=0$ or equivalent of required form

Integrate to obtain the form $k_1{\mathrm{e}}^{2x}+k_{2}x$

Obtain correct $\frac{3}{2}{\mathrm{e}}^{2x}-x$

Use limits correctly and attempt rearrangement at least as far as ${\mathrm{e}}^{2a}=\dots$

Confirm given result $a=\frac{1}{2}\ln (9+\frac{2}{3}a)$ with sufficient detail

Use iteration process correctly at least once

Obtain final answer 1.139

Show sufficient iterations to 6 sf to justify answer or show a sign change in interval [1.1385, 1.1395]

Substitute $x=3$, equate to zero and attempt solution

Obtain $k=-17$

Divide by $x-3$ at least as far as $2x^2+mx$

Obtain $2x^2+9x+10$

Obtain $(x-3)(2x+5)(x+2)$

Solve linear equation $4y-5=3$ to obtain $y=2$

Attempt solution of linear equation $4y-5=-3$ or equivalent

Obtain $y=\frac{1}{2}$

Use product rule to find $\frac{\mathrm{d}y}{\mathrm{d}t}$

Obtain $\frac{\mathrm{d}y}{\mathrm{d}t}=-5{\mathrm{e}}^{-t}\cos 2t-10{\mathrm{e}}^{-t}\sin 2t$

Obtain $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{-5{\mathrm{e}}^{-t}\cos 2t-10{\mathrm{e}}^{-t}\sin 2t}{8{\mathrm{e}}^{2t}}$

Equate $\frac{\mathrm{d}y}{\mathrm{d}x}$ to zero and simplify at least as far as $\tan 2t=\dots$

Obtain $\tan 2t=-\frac{1}{2}$

Obtain $t=-0.231\dots$

Substitute negative value of $t$ in expressions for $x$ and $y$

Obtain $x=2.52$ and $y=5.64$

Express $4{\cos }^{2}2x$ in the form $k_1\cos 4x+k_2$

Obtain correct $2\cos 4x+2$

State or imply $\frac{1}{{\cos }^{2}x}={\sec }^{2}x$

Integrate to obtain $k_3\sin 4x+k_{4}x+\tan x$

Obtain correctly $\frac{1}{2}\sin 4x+2x+\tan x$

Use limits correctly with correct values of $\sin \frac{4}{3}\pi$ and $\tan \frac{1}{3}\pi$ indicated

Confirm given result $\frac{3}{4}\sqrt{3}+\frac{1}{6}\pi -1$ with sufficient detail

State $R=25$

Use appropriate trigonometry to find $\alpha$

Obtain $\alpha =73.74$

Use correct method to find one value of $\theta$

Obtain 29.8 (or 117.7)

Use correct method to find second value of $\theta$ between 0 and 360

Obtain 117.7 (or 29.8)

State or imply expression is $\frac{150}{25\cos (\frac{1}{2}\beta -73.74)+50}$

Obtain $V=6$

Attempt complete method to find positive value from $\cos (\frac{1}{2}\beta -73.74)=-1$

Obtain 507.5