A Level Mathematics 9709 Paper 21, May/June 2024

Differentiate to obtain $2{\sec }^{2}x-5\cos x$

Equate first derivative to zero and solve correctly at least as far as $\cos x=\dots$

Obtain $\cos x=\sqrt[3]{0.4}$ and hence $x=0.742$

Attempt use of product rule for differentiating $x^2\ln y$

Obtain $2x\ln y+\frac{x^2}{y}\frac{\mathrm{d}y}{\mathrm{d}x}$

Obtain $+2y\frac{\mathrm{d}y}{\mathrm{d}x}$

Substitute $x=2,y=1$ in equation equal to 0 involving at least one $\frac{\mathrm{d}y}{\mathrm{d}x}$ and solve for $\frac{\mathrm{d}y}{\mathrm{d}x}$

Obtain $-\frac{2}{3}$

Draw V-shaped graph with vertex on positive $x$-axis in the first quadrant.

Draw correct graph of $y=5-x$ correctly positioned with respect to modulus graph.

Solve $3x-8=5-x$ to obtain $\frac{13}{4}$

Solve linear equation or inequality with signs of $3x$ and $x$ the same

Obtain $\frac{3}{2}$

Conclude $\frac{3}{2}<x<\frac{13}{4}$ or $x>\frac{3}{2}$ and $x<\frac{13}{4}$

Attempt value of $N$ (maybe non-integer at this stage) for ${\mathrm{e}}^{0.1N}<\text{their }\frac{13}{4}$

Conclude with single integer 11

State $3\tan 2\theta =\frac{6\tan \theta }{1-{\tan }^2\theta }$

State $\tan (\theta +{45}^{\circ })=\frac{\tan \theta +1}{1-\tan \theta }$

Attempt to express left-hand side in terms of $\tan \theta$ as a single fraction

Confirm $\frac{{\tan }^2\theta +8\tan \theta +1}{1-{\tan }^2\theta }$

Solve quadratic equation $5{\tan }^2\theta +8\tan \theta -3=0$ to obtain at least one value of $\theta$

Obtain 17.4

Obtain 117.6

Differentiate using quotient rule

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

Equate first derivative to zero and arrange as far as $x=\dots$

Confirm $x=\frac{1}{6}+\frac{1}{2}{\mathrm{e}}^{-2x}$

Consider sign of $x-\frac{1}{6}-\frac{1}{2}{\mathrm{e}}^{-2x}$

Obtain $-0.06\dots$ (– 0.064959…) and $0.08\dots$ (0.0800…) or equivalents and justify conclusion

Use iterative process correctly at least once

Obtain final answer 0.394

Show sufficient iterations to 5 sf to justify answer or show sign change in interval [0.3935, 0.3945]

Use $y$-values $(0)$, $\sqrt{\sin \frac{1}{6}\pi +{\sin }^2\frac{1}{6}\pi }$, $\sqrt{\sin \frac{1}{3}\pi +{\sin }^2\frac{1}{3}\pi }$ or decimal equivalents

Use correct formula, or equivalent, with $h=\frac{1}{12}\pi$

Obtain 0.39

Use $(\pi )\int (\sin 2x+{\sin }^{2}2x)\mathrm{d}x$

Express integrand in the form $k_1\sin 2x+k_2+k_3\cos 4x$

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

Integrate to obtain $-\frac{1}{2}\cos 2x+\frac{1}{2}x-\frac{1}{8}\sin 4x$

Apply limits correctly to obtain exact value

Obtain volume = $\frac{1}{4}\pi +\frac{1}{12}{\pi }^2-\frac{1}{16}\pi \sqrt{3}$

Carry out division at least as far as $3x^2+n_1$

Obtain quotient $3x^2+4$

Confirm remainder is $k-8$

Integrate to obtain at least a term in $x^3$ and term of form $n_2\ln (3x+2)$

Obtain $x^3+4x+\frac{1}{3}(k-8)\ln (3x+2)$

Apply limits correctly to expression with three terms

Obtain $a=235$

Equate logarithm term to $\ln 64$ and apply appropriate logarithm properties

Obtain $k=17$