A Level Mathematics 9709 Paper 33, Oct/Nov 2023

State or imply non-modular inequality $-0.5<2^{x+1}-2<0.5$, can be in two separate statements, or $(2^{x+1}-2{)}^2<{0.5}^2$ or corresponding pair of linear equations $0.5=2^{x+1}-2$ and $-0.5=2^{x+1}-2$ or quadratic equation $(2^{x+1}-2{)}^2={0.5}^2$

Use correct method for solving an equation or inequality of the form $2^{x+1}=a$ or $2^x=b$ where $a,b>0$

Obtain critical values $x=0.322$ and $-0.415$ or awrt $x=0.32$ and $-0.42$ or exact equivalents

State final answer $-0.415<x<0.322$ or $(-0.415,0.322)$

Show a circle centre $(2,0)$

Show the relevant part of a circle with radius 1

Show the point representing $1-2\mathrm{i}$

Show the perpendicular bisector of the line joining $1-2\mathrm{i}$ and the origin. Perpendicular to $OP$ by eye and at midpoint of $OP$ by eye sufficient. Must reach midpoint of $OP$ and if extended will cut $BE$.

Shade the correct region. Dependent on all previous marks, except in case 3 below, and the perpendicular must cut axes between $CF$ and $BE$, but not actually through $C$ or $F$ and not through $B$ or $E$ Scale can be implied by dashes 1 Scale only on y-axis and $2OA=OC$ B1, B1FT, B1, B1FT, B1 2 Scale only on x-axis and $2OB=OE$ B1, B1FT, B1, B1FT, B1 3 No scale on either axis, but $2OA=OC$ then $2OB=OE$ B0, B1FT, B0, B1FT, B1

$2(-2{)}^3+a(-2{)}^2+b(-2)+6=-38$ Allow errors \begin{array}{r} 2x^2 + (a-4)x + b - 2a + 8 \\ x + 2 \enclose{longdiv}{2x^3 + ax^2 + bx + 6} \\ \underline{2x^3 + 4x^2\phantom{+bx+6}} \\ (a-4)x^2 + bx\phantom{+6} \\ \underline{(a-4)x^2 + (2a-8)x\phantom{+6}} \\ (b-2a+8)x + 6 \\ \underline{(b-2a+8)x + 2b - 4a + 16} \\ 4a - 2b - 10 \end{array}

Obtain a correct evaluated equation, e.g. $-16+4a-2b+6=-38$ or $4a-2b=-28$

$2{\left(\frac{1}{2}\right)}^3+a{\left(\frac{1}{2}\right)}^2+b\left(\frac{1}{2}\right)+6=\frac{19}{2}$ Allow errors \begin{array}{r} x^2 + \frac{a+1}{2}x + \frac{b}{2} + \frac{a}{4} + \frac{1}{4} \\ 2x - 1 \enclose{longdiv}{2x^3 + ax^2 + bx + 6} \\ \underline{2x^3 - x^2\phantom{+bx+6}} \\ (a+1)x^2 + bx\phantom{+6} \\ \underline{(a+1)x^2 - \left(\frac{a}{2} + \frac{1}{2}\right)x\phantom{+6}} \\ \left(b + \frac{a}{2} + \frac{1}{2}\right)x + 6 \\ \underline{\left(b + \frac{a}{2} + \frac{1}{2}\right)x - \left(\frac{b}{2} + \frac{a}{4} + \frac{1}{4}\right)} \\ 6 + \frac{b}{2} + \frac{a}{4} + \frac{1}{4} \end{array}

Obtain a correct evaluated equation, e.g. $\frac{1}{4}+\frac{a}{4}+\frac{b}{2}+6=\frac{19}{2}$ or $\frac{a}{4}+\frac{b}{2}=\frac{13}{4}$

Obtain $a=-3$ and $b=8$

$\frac{2\pm \sqrt{(-2{)}^2-4(3+\mathrm{i})(3-\mathrm{i})}}{2(3+\mathrm{i})}$

Use ${\mathrm{i}}^2=-1$ in $(3+\mathrm{i})(3-\mathrm{i})$

Obtain one of the answers $w=\frac{2+6\mathrm{i}}{6+2\mathrm{i}}$ or $w=\frac{2-6\mathrm{i}}{6+2\mathrm{i}}$

Show intention to multiply numerator and denominator by conjugate of their denominator.

Obtain final answers $\frac{3}{5}+\frac{4}{5}\mathrm{i}$ and $-\mathrm{i}$ Accept $0.6+0.8\mathrm{i}$ and $0-\mathrm{i}$

Use correct product or quotient rule

Obtain correct derivative in any form, e.g. $\frac{6x(1-x^2){\mathrm{e}}^{3x^2-1}+2x{\mathrm{e}}^{3x^2-1}}{(1-x^2{)}^2}$

Equate derivative (or its numerator) to zero and solve for $x$

Obtain the point $(0,{\mathrm{e}}^{-1})$ or exact equivalent

Obtain the point $(\frac{2\sqrt{3}}{3},-3{\mathrm{e}}^3)$ or exact equivalent

Obtain the point $(-\frac{2\sqrt{3}}{3},-3{\mathrm{e}}^3)$ or exact equivalent

Use correct Pythagoras ${\cot }^2\theta ={\mathrm{cosec}}^2\theta -1$ or ${\cot }^2\theta =1\mathrm{/}{\sin }^2\theta -1$ or ${\cot }^2\theta ={\cos }^2\theta \mathrm{/}{\sin }^2\theta$ and then ${\cos }^2\theta =1-{\sin }^2\theta$, together with double angle formula $\cos 2\theta =1-2{\sin }^2\theta$, to obtain an equation in $\sin \theta$ or $\sin \theta$ and ${\mathrm{cosec}}^2\theta$

Obtain a correct equation in $\sin \theta$ in any form

Reduce to the given answer of $4{\sin }^4\theta +3{\sin }^2\theta -1=0$ correctly

Solve the given quadratic to obtain a value for $\theta$

Obtain answer, e.g. $\theta ={30}^{\circ }$

Obtain three further answers, e.g. $\theta ={150}^{\circ }$, ${210}^{\circ }$ and ${330}^{\circ }$ and no others in the interval

State or imply $2y\frac{\mathrm{d}y}{\mathrm{d}x}$ as the derivative of $y^2$

Equate derivative of LHS to zero and solve for $\frac{\mathrm{d}y}{\mathrm{d}x}$

Obtain the given answer

Equate numerator to zero and solve for $x$

Obtain $x=0$ and $x=-2$ only

Substitute their $x$, [$x=0$ or $x=-2$] in curve equation to obtain quadratic equation in $y$ equal to 0

Obtain $y=1$ and $y=-4$ [when $x=0$]

Obtain $y=0$ and $y=-3$ [when $x=-2$]

Separate variables correctly and reach $a{\sec }^{2}3y$ or $b{\mathrm{e}}^{-4x}$

Obtain term $-\frac{1}{4}{\mathrm{e}}^{-4x}$

Obtain only a term of the form $a\tan 3y$

Obtain term $\frac{1}{3}\tan 3y$

Use $x=2,y=0$ to evaluate a constant or as limits in a solution containing terms of the form $a\tan by$ and $c{\mathrm{e}}^{\pm 4x}$

Obtain correct answer in any form

Obtain final answer $y=\frac{1}{3}{\tan }^{-1}(\frac{3}{4}{\mathrm{e}}^{-8}-\frac{3}{4}{\mathrm{e}}^{-4x})$

State or imply the form $\frac{Ax+B}{2+3x^2}+\frac{C}{2-x}$

Use a correct method for finding a coefficient

Obtain one of $A=-2,B=3$ and $C=5$

Obtain a second value

Obtain the third value

Use a correct method to find the first two terms of the expansion $(2-x{)}^{-1}=2^{-1}+(-1)2^{-2}(-x)+[(-1)(-2)2^{-3}(-x{)}^2\mathrm{/}2!]$, ${(1+\frac{3x^2}{2})}^{-1}=1-\frac{3x^2}{2}$ or ${(1-\frac{x}{2})}^{-1}=1-\left(\frac{-x}{2}\right)$

$$\frac{Ax+B}{2}[1+(-1)\frac{3x^2}{2}\dots ]A=-2B=3$$ $$\frac{C}{2}[1+(-1)\left(\frac{-x}{2}\right)+\frac{(-1)(-2)}{2}{\left(\frac{-x}{2}\right)}^2+\frac{(-1)(-2)(-3)}{6}{\left(\frac{-x}{2}\right)}^3\dots ]C=5$$

$$=\frac{3-2x}{2}(1-\frac{3x^2}{2})+\frac{5}{2}(1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{8})$$ $$=(\frac{3}{2}+\frac{5}{2})+(-1+\frac{5}{4})x+(-\frac{9}{4}+\frac{5}{8})x^2+(\frac{3}{2}+\frac{5}{16})x^3$$

Multiply expansion of ${(1+\frac{3x^2}{2})}^{-1}$ (must reach $1\pm \frac{3x^2}{2}$) by $Ax+B$, where $AB\mathrm{\ne }0$, up to the term in $x^3$. Allow if used $Cx+D$ ($Ax+B$ miscopied).

Obtain final answer $4+\frac{1}{4}x-\frac{13}{8}{x}^2+\frac{29}{16}{x}^3$, or equivalent [If final answer has been multiplied throughout, e.g. by 16 then A0 at the end]

State answer $\mathrm{\mid }x\mathrm{\mid }<\sqrt{\frac{2}{3}}$ or $-\sqrt{\frac{2}{3}}<x<\sqrt{\frac{2}{3}}$ clear conclusion required

Use the product rule correctly on $y=x\cos 2x$

Obtain the correct derivative in any form

Obtain $y=-\frac{\pi }{2}$ and $\frac{\mathrm{d}y}{\mathrm{d}x}=-1$ when $x=\frac{\pi }{2}$

Obtain answer $x+y=0$

Integrate by parts and reach $ax\sin 2x+b\int \sin 2x\mathrm{d}x$

Obtain $\frac{1}{2}x\sin 2x-\frac{1}{2}\int \sin 2x\mathrm{d}x$

Complete integration and obtain $\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x$

Use limits of $x=0$ and $x=\frac{\pi }{4}$ in the correct order, having integrated twice to obtain $ax\sin 2x+c\cos 2x$

Obtain answer $\frac{\pi }{8}-\frac{1}{4}$ or exact simplified two term equivalent

Use correct process for modulus on direction vector of $l$, e.g. $\sqrt{(-1{)}^2+1^2+2^2}$

$[\pm ]\frac{1}{\sqrt{6}}(-\mathbf{i}+\mathbf{j}+2\mathbf{k})$

Use a correct method to form an equation for line $m$

Obtain $\mathbf{r}=-2\mathbf{i}+2\mathbf{j}-\mathbf{k}+{\mu }_1(-5\mathbf{i}+3\mathbf{j}-2\mathbf{k})$

Justify lines are not parallel

Express $l$ or $m$ in component form e.g. $(-2-5{\mu }_1,2+3{\mu }_1,-1-2{\mu }_1)$ or $(3-5{\mu }_2,-1+3{\mu }_2,1-2{\mu }_2)$ or $(1-\lambda ,-2+\lambda ,-3+2\lambda )$

Equate two pairs of components of general points on $l$ and their $m$ and solve simultaneously for $\lambda$ or for $\mu$

Obtain correct answer for $\lambda$ or $\mu$, e.g. $\lambda =\frac{11}{2},{\mu }_1=\frac{1}{2}$

Determine that all three equations are not satisfied and the lines fail to intersect and conclude the lines are skew. Conclusion needs to follow correct working