A Level Mathematics 9709 Paper 31, May/June 2024

State correct unsimplified first two terms of the expansion of $(1-2x{)}^{\frac{1}{2}}$, e.g. $1+\frac{1}{2}(-2x)$

State correct unsimplified term in $x^2$, e.g. $\frac{\frac{1}{2}(\frac{1}{2}-1)(-2x{)}^2}{2!}$

Obtain sufficient terms of the product of $(3+x)$ and the expansion up to the term in $x^2$

Obtain final answer $3-2x-\frac{5}{2}{x}^2$

Use law of logarithm of a product (or quotient) on correct terms

Use correct method to eliminate logarithm

Obtain a correct quadratic in $x$, e.g. $x^2-5x-{\text{e}}^7=0$ (allow decimals)

Obtain answer $x=35.71$ only

State or imply that $y\ln a=\ln b+\ln x$

Carry out a completely correct method for finding $\ln a$ or $\ln b$

Obtain value $a=7$

Obtain value $b=5$

State or imply $r=2$

State or imply $\theta =-\frac{2}{3}\pi$

State or imply $r=\frac{5}{2}$

State or imply $\theta =\frac{5}{6}\pi$

Use correct quotient (or product) rule

Obtain correct derivative $\frac{{\text{e}}^{\sin x}{\cos }^{3}x-(-2{\text{e}}^{\sin x}\sin x\cos x)}{({\cos }^{2}x{)}^2}$ or equivalent

Equate numerator to zero

Obtain equation in one unknown

Solve a 3 term quadratic in $\sin x$ to find a value for $x$

Obtain a correct solution to the quadratic equation, e.g. ${3.57}^{\circ }$

Obtain a further correct solution, e.g. $x={5.86}^{\circ }$ and no others in the interval

Sketch a relevant graph, e.g. $y={\text{e}}^x-3$ Correct shape, correct vertical intercept

Sketch a second relevant graph, e.g. $y=\text{cosec}\frac{x}{2}$ (correct shape, minimum above the axis) and justify the given statement. Need to mark intersection with a dot, a cross, or say root at points of intersection, or equivalent

Calculate the values of a relevant expression or pair of expressions at $x=1$ and $x=2$

Complete the argument correctly with correct calculated values

State $x=\ln (\text{cosec}\frac{1}{2}x+3)$ and rearrange to the given equation $\text{cosec}\frac{x}{2}={\text{e}}^x-3$

Use the iterative formula correctly at least twice

Obtain final answer 1.50

Show sufficient iterations to 4 dp to justify 1.50 to 4 dp or show there is a sign change in the interval (1.495, 1.505)

4

Show a circle centre $(3,-2)$

Show a circle with radius 2 FT centre not at the origin

Show the point representing $(-3,4)$ or the midpoint $(0,1)$

Show the perpendicular bisector of the line joining $(-3,4)$ and centre of the circle FT is on the position of $(-3,4)$ and centre of the circle

Carry out a correct method for finding the least value of $\mathrm{\mid }z-w\mathrm{\mid }$

Obtain answer $\sqrt{18}-2$ or $3\sqrt{2}-2$

State or imply $\text{d}u=-\cos x\text{d}x$

Use $\sin 2x=2\sin x\cos x$ and write the integral in terms of $u$

Obtain $\pm 2\int \frac{(1-u)}{\sqrt{u}}\text{d}u$ or equivalent

Integrate correctly to obtain $a{u}^{\frac{1}{2}}+b{u}^{\frac{3}{2}}$

Obtain correct $-4u^{\frac{1}{2}}+\frac{4}{3}{u}^{\frac{3}{2}}$

Correctly use limits $u=2$ and $0$ in an expression of the form $a{u}^{\frac{1}{2}}+b{u}^{\frac{3}{2}}$ OR limits $x=\frac{3}{2}\pi$ and $\frac{1}{2}\pi$ in an expression of the form $a(1-\sin x{)}^{\frac{1}{2}}+b(1-\sin x{)}^{\frac{3}{2}}$

Obtain $\frac{8}{3}-\frac{4}{3}\sqrt{2}$

Carry out correct process for evaluating the scalar product of direction vectors, equate the result to zero and obtain given value of $a=4$

Express general point of at least one line correctly in component form, i.e. $(1+2\lambda ,-2-\lambda ,3+4\lambda )$ or $(-1+3\mu ,-1-2\mu ,-1-2\mu )$

Equate at least two pairs of corresponding components and solve for $\lambda$ or for $\mu$

Obtain $\lambda =-1$ or $\mu =0$

Obtain position vector of point of intersection is $-\mathbf{i}-\mathbf{j}-\mathbf{k}$

Equate one component of $l_1$ to matching component of $A$ and solve to find $\lambda$

Use $\lambda =-3$ in equation of $l_1$ and show this gives position vector of $A$

Method to find position vector of $B$

Obtain position vector of $B$ is $3\mathbf{i}-3\mathbf{j}+7\mathbf{k}$

Obtain $2={\sec }^{2}y\frac{\text{d}y}{\text{d}x}$ or equivalent

Use ${\sec }^{2}y=1+{\tan }^{2}y$

Replace $\tan y$ with $2x$ and rearrange to obtain given answer $\frac{\text{d}y}{\text{d}x}=\frac{2}{1+4x^2}$

Integrate by parts and reach $a{x}^2{\tan }^{-1}2x+b\int \frac{x^2}{1+4x^2}\text{d}x$

Obtain $\frac{1}{2}{x}^2{\tan }^{-1}2x-\int \frac{x^2}{1+4x^2}\text{d}x$

Reduce integral to expression of the form $\int (m+\frac{n}{1+4x^2})\text{d}x$

Complete integration and reach $p{x}^2{\tan }^{-1}2x+qx+r{\tan }^{-1}2x$

Obtain $\frac{1}{2}{x}^2{\tan }^{-1}2x-\frac{1}{4}x+\frac{1}{8}{\tan }^{-1}2x$

Use limits of $x=\frac{1}{2}$ and $x=\frac{1}{2}\sqrt{3}$ in the correct order, having integrated twice

Obtain answer $\frac{5}{48}\pi -\frac{1}{8}\sqrt{3}+\frac{1}{8}$ or exact equivalent

State or imply equation of the form $\frac{\text{d}x}{\text{d}t}=kx(300-x)$ and use $\frac{\text{d}x}{\text{d}t}=0.2$ and $x=1$

Obtain $k=\frac{1}{1495}$ and rearrange to the given answer

Separate variables correctly

Correct integration of $t$ term

State or imply partial fractions of the form $\frac{A}{x}+\frac{B}{300-x}$

Correct method to find $A$ or $B$

Obtain terms $\frac{1495}{300}\ln x-\frac{1495}{300}\ln (300-x)$

Use $t=0$, $x=1$ to evaluate a constant or as limits in a solution containing terms of the form $\ln x$, $\ln (300-x)$ and $t$.

Obtain correct answer in any form

Use law of logarithms twice to obtain an expression for $t$

Obtain final answer $t=\frac{299}{60}\ln \frac{299x}{300-x}$ or equivalent single logarithm