A Level Mathematics 9709 Paper 21, Oct/Nov 2024

State or imply $2y\ln a=3x+k$ and conclude that gradient is $\frac{3}{2\ln a}$

Equate $\frac{3}{2\ln a}$ to gradient of line

Obtain $\frac{3}{2\ln a}=\frac{2.85}{2.9}$ or equivalent and hence obtain $a=4.6$ or $a={\text{e}}^{\frac{29}{19}}$

Substitute appropriate values to find value of $k$

Obtain $k=1.7$

Attempt solution of equation or inequality, where signs of $x$ and $4x$ are different

Obtain $\frac{4}{5}\dots$

... and finally no other value

Conclude $x<\frac{4}{5}$

Differentiate to obtain form $k\tan \frac{1}{2}x{\sec }^2\frac{1}{2}x$

Obtain correct $\tan \frac{1}{2}x{\sec }^2\frac{1}{2}x$

Substitute $\frac{2}{3}\pi$ to obtain $4\sqrt{3}$

Express integrand as ${\sec }^2\frac{1}{2}x-1+\sin x$

Integrate to obtain $k_1\tan \frac{1}{2}x-x+k_2\cos x$

Obtain correct $2\tan \frac{1}{2}x-x-\cos x$

Apply limits correctly to obtain $3-\frac{1}{2}\pi$ or exact equivalent

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

Obtain $a=4$

Divide by $x+2$ at least as far as $k_1{x}^2+k_{2}x$

Obtain $4x^2-12x+9$

Obtain $(x+2)(2x-3{)}^2$ or equivalent with integer coefficients only

Equate ${\sin }^2\theta$ to appropriate value from factorised form and attempt solution

Obtain 54.7

Obtain $-54.7$

Obtain integral of form $k\ln (2x+1)$

Obtain correct $5\ln (2x+1)$

Apply limits correctly and equate to 7

Apply appropriate logarithm property to reach at least $a^3=\dots$

Confirm $a=\sqrt[3]{0.5{\text{e}}^{1.4}(2a+1)-0.5}$

Use iterative process correctly at least once

Obtain final answer 2.18

Show sufficient iterations to 5 sf to justify answer or show a sign change in the interval $[2.175,2.185]$

Differentiate $x$ using quotient rule or correct equivalent

Obtain $\frac{({\text{e}}^{2t}+1)2{\text{e}}^{2t}-({\text{e}}^{2t}-2)2{\text{e}}^{2t}}{({\text{e}}^{2t}+1{)}^2}$ or equivalent

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

Obtain $\frac{1}{2}{\text{e}}^t({\text{e}}^{2t}+1{)}^2$ or (unsimplified) equivalent

Identify $t=\frac{1}{2}\ln 2$ at point where curve crosses $y$-axis

Substitute non-zero value of $t$ in their expression for $\frac{\text{d}y}{\text{d}x}$ and attempt simplification

Obtain $\frac{9}{2}\sqrt{2}$ or exact equivalent

State $(\cos \theta \cos 30-\sin \theta \sin 30)(\cos \theta \cos 60-\sin \theta \sin 60)$

Expand and use correct exact values

Obtain $\frac{1}{4}\sqrt{3}({\cos }^2\theta +{\sin }^2\theta )-\sin \theta \cos \theta$ or similarly simplified equivalent

Conclude $\frac{1}{4}\sqrt{3}-\frac{1}{2}\sin 2\theta$

Use identity to obtain value for $\sin 4\alpha$

Obtain $\sin 4\alpha =\frac{1}{2}\sqrt{3}-\frac{2}{5}$ or $0.466\dots$

Show correct process to obtain one value of $\alpha$

Obtain 6.9 and 38.1

Substitute $\theta =-10$ to obtain $\cos 20\cos 50=\frac{1}{4}\sqrt{3}-\frac{1}{2}\sin (-20)$

Substitute $\theta =10$ to obtain $\cos 40\cos 70=\frac{\sqrt{3}}{4}-\frac{1}{2}\sin 20$

Add and confirm $\frac{1}{2}\sqrt{3}$ with clear indication that $\sin (-20)=-\sin 20$