A Level Mathematics 9709 Paper 21, May/June 2026

Apply natural logarithm to both sides and use power law

Obtain $(x+y)\ln 3=(4x+1)\ln 2$

Attempt arrangement to $y=mx+c$ form with use of at least one logarithm law

Obtain $m=\frac{\ln \frac{16}{3}}{\ln 3}$ and $c=\frac{\ln 2}{\ln 3}$ as final answers

Either Solve $3x+7=4x-1$ to obtain $x=8$ OR State or imply non-modulus equation $(3x+7{)}^2=(4x-1{)}^2$

Attempt solution of linear equation where $3x$ and $4x$ have different signs OR Attempt complete solution of three-term quadratic equation

Obtain $x=-\frac{6}{7}$ OR Obtain $-\frac{6}{7}$ and $8$

Attempt solution of $\tan \theta =k$ for at least one of *their* values from part (a) to obtain at least one positive solution.

Obtain $\tan \theta =\frac{1}{8}$ and hence $\theta =0.124$

Obtain $\tan \theta =-\frac{7}{6}$ and hence $\theta =2.279$

Use $y$-values $3,\sqrt{10},\sqrt{17},6$ or decimal equivalents and no others

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

Obtain $\frac{1}{2}(3+2\sqrt{10}+2\sqrt{17}+6)$ or equivalent, and hence $11.8$

State over-estimate because tops of trapeziums are above the curve

Obtain $R=\sqrt{34}$

Use appropriate trigonometry to find $\alpha$

Obtain $\alpha =30.96$

Use correct method to find one value of $\theta$ using *their* answer to 4(a)

Obtain $69.94+30.96$ or equivalent and hence $100.9$

Use correct method to find second value of $\theta$ between $-{180}^{\circ }$ and ${180}^{\circ }$

Obtain $-69.94+30.96$ or equivalent and hence $-39.0$ and no others between $-{180}^{\circ }$ and ${180}^{\circ }$

Attempt division at least as far as $x^3+kx$ for the quotient

Obtain correct quotient $x^3+2x-7$

Confirm that remainder is 21

State $\alpha =-\frac{3}{2}$

Equate quotient to 0 and confirm $x=\sqrt[3]{7-2x}$

Consider sign of $x-\sqrt[3]{7-2x}$, or $y=x^3+2x-7$, or $y=2x^4+3x^3+4x^2-8x-21$ for $x=1.5$ and $x=1.6$ OR $1.5<\sqrt[3]{7-2(1.5)}$ leading to $1.5<1.587$ and $1.6>\sqrt[3]{7-2(1.6)}$ leading to $1.6>1.56$

Obtain $-\mathrm{0.08...}$ and $0.03\dots$ or $-0.625$ and $0.296$ or $-3.75$ and $1.83$ and justify conclusion OR Justify conclusion

Use iterative formula correctly at least once

Obtain final answer 1.569

Show sufficient iterations to 6 significant figures to justify answer or show sign change in interval $[1.5685,1.5695]$

Differentiate $y$ using correct quotient rule or equivalent

Obtain $\frac{9{\mathrm{e}}^t-{\mathrm{e}}^{3t}}{(9+{\mathrm{e}}^{2t}{)}^2}$

Attempt expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $t$

Obtain $\frac{9{\mathrm{e}}^t-{\mathrm{e}}^{3t}}{4{\mathrm{e}}^{2t}(9+{\mathrm{e}}^{2t}{)}^2}$

Equate $\frac{\mathrm{d}y}{\mathrm{d}x}$ or $\frac{\mathrm{d}y}{\mathrm{d}t}$ to zero and attempt at least the value of ${\mathrm{e}}^t$ or ${\mathrm{e}}^{2t}$

Obtain ${\mathrm{e}}^t=3$, $t=\frac{1}{2}\ln 9$, $t=\ln 3$ or $t=1.099$ or better (may be implied by correct coordinates)

Substitute to obtain the coordinates $(19,\frac{1}{6})$

Use value $x=33$ to find at least the value of ${\mathrm{e}}^t$ or ${\mathrm{e}}^{2t}$

Obtain ${\mathrm{e}}^t=4$, $t=\frac{1}{2}\ln 16$, $t=\ln 4$ or $t=1.386$ (may be implied by correct gradient)

Hence obtain gradient $-\frac{7}{10000}$ or $-0.0007$

Differentiate to obtain the form $k_1\sin 2x\cos 2x+k_2\sin 2x$ or $k_3\sin 4x+k_2\sin 2x$

Obtain correct $4\sin 2x\cos 2x+2\sin 2x$ or $2\sin 4x+2\sin 2x$

Equate first derivative to zero and attempt to find the values of $\alpha$ or $\beta$

Obtain $\cos 2x=-\frac{1}{2}$ and hence $\alpha =\frac{1}{3}\pi$ and $\beta =\frac{2}{3}\pi$

Attempt to express integrand in terms of $\cos 4x$ and $\cos 2x$

Obtain $\int (3+\frac{1}{2}-\frac{1}{2}\cos 4x-\cos 2x)\mathrm{d}x$

Integrate to obtain $\frac{7}{2}x-\frac{1}{8}\sin 4x-\frac{1}{2}\sin 2x$

Use their limits correctly to attempt exact value of area

Obtain $(\frac{7}{3}\pi -\frac{1}{16}\sqrt{3}+\frac{1}{4}\sqrt{3})-(\frac{7}{6}\pi +\frac{1}{16}\sqrt{3}-\frac{1}{4}\sqrt{3})$ and hence $\frac{7}{6}\pi +\frac{3}{8}\sqrt{3}$ or exact equivalent