A Level Mathematics 9709 Paper 22, Oct/Nov 2023

Substitute $x=-2$ and equate to 33

Obtain $-8a+16a+14-5=33$ and hence $a=3$

Attempt to express equation in terms of $\cos \theta$ and $\sin \theta$ only, using a correct identity and $\sec \theta =\frac{1}{\cos \theta }$

Obtain $\cos \theta \cos 60+\sin \theta \sin 60=4\cos \theta$

Either divide by $\cos \theta$ to find value of $\tan \theta$ or use another correct method to find $\sin \theta$ or $\cos \theta$

Obtain $\tan \theta =\frac{7}{\sqrt{3}}$ or $\tan \theta =4.04\dots$ or $\sin \theta =\frac{7}{\sqrt{52}}$ or $\sin \theta =0.970\dots$ or $\cos \theta =\frac{\sqrt{3}}{\sqrt{52}}$ or $\cos \theta =0.240\dots$

Obtain $-103.9$ and $76.1$

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

Substitute $x=2$ to obtain $-3{\mathrm{e}}^{-1}$ or $-\frac{3}{\mathrm{e}}$

Integrate to obtain $-12{\mathrm{e}}^{-\frac{1}{2}x}$

Use limits 0 and 2 correctly to an integral of the form $k{\mathrm{e}}^{-\frac{x}{2}}$, retaining exactness

Subtract from 12 to obtain final answer $12{\mathrm{e}}^{-1}$ or $\frac{12}{\mathrm{e}}$

Draw V-shaped graph with vertex on positive $x$-axis

Draw (more or less) correct graph of $y=9-2x$ with steeper negative gradient

Solve linear equation or inequality with signs of $x$ and $2x$ different

Obtain critical value 4

Conclude $x>4$ only

State or imply $(3x-10)\ln 2<\ln 500$

Obtain critical value 6.32

Obtain $x<6.32$

State 5 and 6 only

Carry out division at least as far as $3x^2+k_{1}x$

Obtain quotient $3x^2-4x-10$

Confirm given result of remainder is 6 with sufficient detail

Integrate to obtain at least $x^3$ and term of form $k_2\ln (2x+1)$

Obtain $x^3-2x^2-10x+3\ln (2x+1)$

Apply limits correctly to expression with four terms

Apply appropriate logarithm properties correctly to obtain the form $k_3\ln a$

Obtain $195+\ln 27$

Obtain $\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{6}{2t-3}$

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

Obtain $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(4\ln t+4)(2t-3)}{6}$

Attempt to find $t$ corresponding to point $A$ using a complete and correct method

Obtain $t=2$ and hence gradient is $\frac{2}{3}\ln 2+\frac{2}{3}$

Equate $\frac{\mathrm{d}y}{\mathrm{d}x}$ to 12 and attempt rearrangement to $2t-3=\frac{k}{4\ln t+4}$

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

Use iteration process correctly at least once

Obtain final answer 4.96

Show sufficient iterations to 5 sf to justify answer or show sign change in interval $[4.955,4.965]$

Attempt to express LHS in terms of $\sin x$ and $\cos x$ only

Obtain $2{\cos }^{2}x+6{\sin }^{2}x$

Use relevant identities to express in terms of $\cos 2x$ only

Confirm given result $4-2\cos 2x$ with sufficient detail

Substitute $x=\frac{1}{12}\pi$ and attempt exact value of $\frac{4-2\cos 2\left(\frac{\pi }{12}\right)}{\sin 2\left(\frac{\pi }{12}\right)}$

Obtain $2(4-2\cos \frac{1}{6}\pi )$ or equivalent and hence $8-2\sqrt{3}$

Differentiate to obtain $4\sin 2x$

Equate to 4 and obtain limits $\frac{1}{4}\pi$ and $\pi$

Integrate to obtain $4x-\sin 2x$

Use limits their $a$ and $b$ to find exact value for integral of form $k_{1}x+k_2\sin 2x$

Obtain $4\pi -\sin 2\pi -(\pi -\sin \frac{1}{2}\pi )$ and confirm given result $3\pi +1$ with sufficient detail