A Level Mathematics 9709 Paper 21, Oct/Nov 2025

Use identity to express integrand in the form $k_1+k_2\cos 2x$

Obtain correct $3-3\cos 2x$

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

Attempt solution of quadratic equation in ${\mathrm{e}}^{2x}$ to obtain at least one value of ${\mathrm{e}}^{2x}$

Obtain $({\mathrm{e}}^{2x}-12)({\mathrm{e}}^{2x}+4)=0$ OE, and hence at least ${\mathrm{e}}^{2x}=12$

Obtain $x=\frac{1}{2}\ln 12$ or 1.24 and no other solution

Solve $2x-3=5x+2$ to obtain $-\frac{5}{3}$

Attempt solution of linear equation where $2x$ and $5x$ have different signs

Obtain $\frac{1}{7}$

State or imply $\cos \theta =-\frac{3}{5}$

Attempt correct process for finding third quadrant angle

Obtain 4.07

Attempt to express equation in terms of $\tan \theta$ only

Obtain $\frac{1}{\tan \theta }\times \frac{\tan \theta +\tan 45}{1-\tan \theta \tan 45}=7$

Simplify to obtain equation $7{\tan }^2\theta -6\tan \theta +1=0$

Attempt solution of three-term quadratic equation in $\tan \theta$ to obtain at least one value of $\theta$

Obtain $\tan \theta =\frac{1}{7}(3\pm \sqrt{2})$ or equivalent and hence 12.8 and 32.2, and no others in the given range

Attempt solution of $8{\mathrm{e}}^{-\frac{1}{2}x}-1=0$ with use of a relevant logarithm property

Confirm $x=6\ln 2$

Integrate $8{\mathrm{e}}^{-\frac{1}{2}x}-1$ to obtain the form $k{\mathrm{e}}^{-\frac{1}{2}x}-x$

Obtain correct $-16{\mathrm{e}}^{-\frac{1}{2}x}-x$

Apply limits 0 and $6\ln 2$ to obtain $-16{\mathrm{e}}^{-\frac{1}{2}\times 6\ln 2}-6\ln 2+16$

Attempt area of triangle minus area under curve

Obtain $\frac{1}{2}\times 6\ln 2\times 7-(14-6\ln 2)$ or equivalent, and hence $27\ln 2-14$

Attempt use of chain rule for differentiation of y or simplification of y and use of the product rule.

Obtain $\frac{\mathrm{d}x}{\mathrm{d}\theta }={\sec }^2\theta$ and $\frac{\mathrm{d}y}{\mathrm{d}\theta }=\cos \theta -6{\sin }^2\theta \cos \theta$

Attempt expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $\cos \theta$ only using their $\frac{\mathrm{d}x}{\mathrm{d}\theta }$ and $\frac{\mathrm{d}y}{\mathrm{d}\theta }$

Obtain $6{\cos }^5\theta -5{\cos }^3\theta$

Attempt solution of $y=0$ to find value of $\theta$

Obtain ${\sin }^2\theta =\frac{1}{2}$ and hence $\theta =\frac{1}{4}\pi$

Obtain gradient of curve at the point is $-\frac{1}{\sqrt{2}}$ or exact equivalent

Attempt equation of normal using negative reciprocal for their gradient

Obtain $y=\sqrt{2}x-\sqrt{2}$ or exact equivalent

Substitute $x=-2$, equate to zero and attempt solution

Obtain $32-8k+4k-34+18=0$ or equivalent and hence $k=4$

State $\alpha =-2$

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

Obtain $2x^3+4x+9$

Rearrange $2x^3+4x+9=0$ to confirm $x=\sqrt[3]{-2x-4.5}$

Consider sign of $x-\sqrt[3]{-2x-4.5}$ or equivalent for $-1.4$ and $-1.0$

Obtain $-\mathrm{0.2...}$ and $\mathrm{0.3...}$ or equivalents and justify conclusion

Use iterative process correctly at least once

Obtain final answer $-1.26$

Show sufficient iterations to 5 sf to justify answer or show sign change in interval $[-1.265,-1.255]$

Differentiate to obtain the form $k_1{\mathrm{e}}^{1-2x}(3x-1{)}^{\frac{1}{2}}+k_2{\mathrm{e}}^{1-2x}(3x-1{)}^{-\frac{1}{2}}$

Obtain correct first term $-8{\mathrm{e}}^{1-2x}(3x-1{)}^{\frac{1}{2}}$

Obtain correct second term $+6{\mathrm{e}}^{1-2x}(3x-1{)}^{-\frac{1}{2}}$

Equate first derivative to zero and attempt solution as far as $3x-1=k$

Obtain $3x-1=\frac{3}{4}$ OE, and hence $x=\frac{7}{12}$

Obtain $y=2\sqrt{3}{\mathrm{e}}^{-\frac{1}{6}}$