A Level Mathematics 9709 Paper 13, May/June 2024

Correct second term $30x$ in expansion of $(1+3x{)}^{10}$

Correct third term $+405x^2$

Multiply $(2-5x)$ by *their* $30x+405x^2$ to obtain two $x^2$ terms only

Coefficient is 660

State $(\frac{5}{3}\pi ,0)$ for point $A$

$x=\frac{19}{6}\pi$ for point $B$

$y=-k$ for point $B$

Solve at least as far as ${\sin }^{-1}3t=k\pi$ with correct value for ${\cos }^{-1}(\frac{1}{2}\sqrt{2})$

${\sin }^{-1}3t=\frac{1}{6}\pi$ and hence $t=\frac{1}{6}$

State $2r+r\theta =65$ and $\frac{1}{2}{r}^2\theta =225$

Form a 3-term quadratic or cubic in $r$ or $\theta$ or $r\theta$ from correct arc and sector formula

Solve *their* 3 term quadratic or cubic to obtain values of $r$ or $\theta$

$r=10$ and $\theta =4.5$ ignore $r=22.5$ and $\theta =\frac{8}{9}$, do not ignore $r=0$

Use correct formula for area of triangle with clear use of angle being $2\pi -\text{their }\theta$

48.9

Use identity $\tan \theta =\frac{\sin \theta }{\cos \theta }$

Use identity ${\cos }^2\theta =1-{\sin }^2\theta$

$\pm (5{\sin }^2\theta +7\sin \theta -6=0)$

Attempt solution of *their* 3 term equation and correct process to find at least 1 value of $\sin x$ or $\sin 2x$ or $\sin \theta$

$x=18.4$

$x=71.6$ or $(90-\text{their }18.4)$ or greater accuracy; and no other solutions for $0^{\circ }<x<{180}^{\circ }$

Differentiate to obtain $4x+\frac{1}{2}{x}^{-2}$

Equate first derivative to zero and solve $4x+\frac{K}{x^2}=0$ as far as $x^3=k$, $K$ and $k$ non-zero

$x=-\frac{1}{2}$ and $y=\frac{9}{2}$

Differentiate *their* first derivative, substitute *their* $x$ value. Substitution may be implied by a correct inequality or correct value,

conclude minimum

State increasing …

… with clear reference to first derivative always being positive [for $x>0$]

Integrate to obtain form $k(5x-3{)}^{-1}$

$4(5x-3{)}^{-1}$

Substitute $x=\frac{4}{5}$ and $y=-3$ to attempt value of $c$

$y=4(5x-3{)}^{-1}-7$ allow $\text{f}(x)$ or $f=4(5x-3{)}^{-1}-7$

Carry out stretch by replacing $x$ by $2x$ in *their* equation

Carry out translation by replacing $x$ by $x-2$ and $y$ by $y-10$

$y=\frac{4}{10x-23}+3$

$5(3+9d)=127.5$

$d=2.5$

Attempt to find either the first term or the last term in the set by considering $1.5+2.5(n-1)>25$ or $1.5+2.5(n-1)<100$ or equivalent equations

State or imply that 11th term or 26.5 is the first in the set

State or imply that 40th term or 99 is the last in the set

Either use $S_{40}-S_{10}$ Or use $\frac{1}{2}n(a+l)$ with correct results for *their* $d$ Or use $\frac{1}{2}n[2a+(n-1)d]$ with correct results for *their* $d$

Obtain 1882.5

Substitute $x=0$ and attempt solution of 3-term quadratic equation in $y$

$-5$ and 3

State or imply centre of circle is $(3,-1)$

Attempt gradient of $AC$ or $BC$

$-\frac{4}{3}$ or $\frac{4}{3}$

State or imply gradient of tangent is $\frac{3}{4}$ or $-\frac{3}{4}$

Either solve simultaneous equations (of 2 tangent equations) to find $x$-coordinate Or Substitute $y$-value of centre into either tangent equation

$x=-\frac{16}{3},y=-1$

Differentiate to obtain form $k{x}^2(2x^3+10{)}^{-\frac{1}{2}}$

$3x^2(2x^3+10{)}^{-\frac{1}{2}}$

Substitute $x=3$ in first derivative and evaluate to find gradient

Attempt equation of tangent at $(3,8)$ with numerical gradient

$[\pm ](27x-8y-17)=0$ or integer multiples

State or imply volume is $\pi \int (2x^3+10)\mathrm{d}x$

Integrate to obtain $k_1{x}^4+k_{2}x$ and evaluate using limits 1 and 3

$60\pi$

Substitute to obtain equation $9r^4+14r^2-8=0$

Attempt solution of quadratic equation in $r^2$ to obtain at least one value of $r$ or $r^2$

$r=\frac{2}{3}$ only

Substitute $a=2$ and *their* $r$ in correct formula and attempt to evaluate

5.998

Identify $a_2=\frac{4}{3}$ and common ratio as $\frac{8}{27}$.

Substitute *their* new $a$ and $r$ in correct formula for sum to infinity and evaluate

$\frac{36}{19}$

Express $\text{f}(x)$ as: $a-(x-3{)}^2$ or $a-(3-x{)}^2$ where $a=\pm 19$ or $\pm 1$

$19-(3-x{)}^2$ or $19-(x-3{)}^2$

$f(x)⩽19$ or $y⩽19$ with $⩽$, not $< or $-\mathrm{\infty }<\text{f}(x)⩽19$ or $-\mathrm{\infty }⩽\text{f}(x)⩽19$ or $(-\mathrm{\infty },19]$ or $[-\mathrm{\infty },19]$

${\text{g}}^{-1}(x)=\frac{1}{4}(x-k)$

${\text{g}}^{-1}\text{f}(x)=\frac{1}{4}(10+6x-x^2-k)=4x+k$

Simplify the quadratic equation obtained from ${\text{g}}^{-1}\text{f}(x)=\text{g}(x)$ provided $k$ is present and apply $b^2-4ac=0$ to this quadratic equation

Obtain $100-4(5k-10)=0$ and hence $k=7$

Use *their* $k$ to form and solve a quadratic in $x$

$(-5,-13)$ only