A Level Mathematics 9709 Paper 11, May/June 2025

For expressing as a 3-term quadratic. Terms need not all be on the same side.

For both. Allow AWRT 0.667.

A1 for any correct angle from a correct value of $\sin \theta$

A1 for all 4 and no others in the interval $-{180}^{\circ }<\theta <{180}^{\circ }$.

Substitute $x=4$ into $\frac{\text{d}y}{\text{d}x}$. $-\frac{11}{2}=4(2\times 4-5{)}^3-9\times 4^{\frac{1}{2}}$ is M0 unless they reach $-\frac{1}{90}$.

AWRT $-0.0111$.

$\{\frac{1}{2}(2x-5{)}^4\}$

$\{-6x^{\frac{3}{2}}\}$

Sub $x=4,y=-\frac{11}{2}$ into an integrated expression and attempt to find $c$.

Condone $c=2$ as final answer if 'y = …' seen previously. Fractions must be simplified. Accept f(x) in place of y.

For first and second, or first and third equations.

For expressing as a 3-term quadratic from two expressions with sign errors only. Terms need not all be on one side.

CAO Allow -0.75.

CAO Ignore $a=2$. May not find $a$, but instead use expression for $a$ in terms of $r$ (e.g. $a=\frac{18}{r^2}$) and then find $a$ in part (b).

AWRT. Ignore other values.

FT on values of $a$ and $r$, provided $-1<r<1$.

FT on values of $a$ and $r$, provided $-1<r<0$. Condone extra answers.

Attempt to integrate both terms and subtract areas. Accept subtraction either way round.

B1 for first integral (e.g. $\frac{1}{2}{x}^2-16x$).

B1 for second integral (e.g. $2x^{\frac{5}{2}}-10x^2$).

Limits either way round. Minimum -7.5 - 384. E.g. $((\frac{1}{2} - 16) - (2 - 10)) - ((128 - 256) - (2048 - 2560)) = (15.5 + 8) - (-128 + 512) = -7.5 - 384 = -391.5 Or (-2 + 10.5 - 16) - (-2048 + 2688 - 256) = -7.5 - 384 = -391.5 Or $-504+\frac{225}{2}=-391.5$

CAO

B2 for all correct. B1 for any two correct. May be in a list. Ignore terms with higher powers.

OE B2 for all correct. B1 for any two correct. May be in a list. Ignore terms with higher powers.

3 terms FT their values.

Set their 3-term coefficient to 93 and attempt to solve by factorising or other accepted method for solving their 3-term quadratic.

$p=-\frac{9}{4}$ and $p=\frac{1}{4}$

Substitutes $k=2,p=11$ and eliminates $y$ or $x$. Note: $\frac{2x^2+2}{2x}=11x+3$.

Simplifies to a 3-term quadratic. Terms need not all be on one side.

A1 for either both $x$-values correct or for both coordinates of one point correct.

A1 for all coordinates correct.

Substitute and reduce to 3-term quadratic. Terms need not all be on one side. Allow $2x^2-4k{x}^2-3kx+2[=0]$.

Correct quadratic. All terms to one side. Allow $2x^2-4k{x}^2-3kx+2[=0]$.

Use of $b^2-4ac$. Must be correct for their $a,b,c$. $a$ term must have two components.

Attempt to solve a 3-term quadratic in $k$ by factorising or other accepted method for solving their 3-term quadratic.

$-4<k<\frac{4}{9}$

OE Accept unsimplified.

OE For evaluating their $\frac{\text{d}y}{\text{d}x}$ or $\frac{\text{d}x}{\text{d}y}$.

For correct use of chain rule with $\pm 5$ and their $\frac{\text{d}y}{\text{d}x}$ (may be algebraic). Condone missing brackets.

Or decreasing at a rate of $\frac{4}{11}$. AWRT -0.364.

Equating their 2-term $\frac{\text{d}y}{\text{d}x}$ to zero.

AWRT $\pm 1.22$.

A0 A1 if one point correct. AWRT 4.00.

For differentiation. At least one correct term needed.

No need for reason. WWW on $x$-values.

Centre of circle is $(3,-5)$ (seen or implied).

3-term quadratic. Terms need not all be on one side. $(3,-5)$ must be correct if it is used.

$y=1$ and $y=-11$ (no method needed for solving quadratic).

Gradient of line $PC$ or gradient of line $QC$ (at least one correct).

Gradient of tangent at $P$ is $\frac{5}{6}$ or gradient of tangent at $Q$ is $-\frac{5}{6}$ (at least one correct).

Distance from point of intersection to line $x=-2$ is $6\times \frac{6}{5}[=\frac{36}{5}]$. Note $y=\frac{5}{6}x+\frac{8}{3}$ and $y=-\frac{5}{6}x-\frac{38}{3}$.

Area = $\frac{1}{2}\times 12\times their\frac{36}{5}$

$\frac{432}{10}$ or $\frac{216}{5}$ or $43.2$

$\frac{1}{2}{r}^2\alpha =8\alpha \Rightarrow r=4$

$\frac{1}{2}{r}^2\sin \alpha =4\Rightarrow \alpha =\frac{\pi }{6}$

Using their $r$ and their $\alpha$. Condone $4-\frac{1}{2}\times 4^2\times \frac{\pi }{6}$. Allow use of $8\times their\alpha -4$

Fraction must be simplified. Allow $\frac{4\pi -12}{3}$.

Or $\alpha =2{\sin }^{-1}\left(\frac{\sqrt{2}}{4}\right)$ or $2{\sin }^{-1}\left(\frac{1}{2\sqrt{2}}\right)$. Using $\alpha =\frac{\pi }{6}$ or $r=4$ implies 0/4.

Or $\alpha =41.4\Rightarrow \frac{1}{2}{r}^2\sin (their41.4)[=4]$

Accept $r^2=12.1$ AWRT.

Or $\frac{1}{2}\times {3.48}^2\times \frac{\pi }{180}\times 41.4-4=0.371$ AWRT.

{Stretch}{factor 3} {parallel to y-axis/in y-direction/vertically}

Translation $\left(\begin{array}{c}-2\\ -5\end{array}\right)$

Transformations correct and in the correct order.

For line or curve in correct quadrants only.

Must not pass through (0, 0). Fully correct including the line $y=x$. No label needed. Approximate reflection of $y=\text{f}(x)$. Curve must not come back on itself.

Allow x/y swap.

Must be in terms of $x$. Not 'x = …'.

Following their ${\text{g}}^{-1}(x)={\left(\frac{x+a}{b}\right)}^2-c$ where $a,b,c$ are non-zero. Not $x⩾-2$. Not $>-2$. Accept other notations, e.g. $[-2,\mathrm{\infty })$.

AWRT 3.44.

Minimum acceptable: 'The range of ${\text{g}}^{-1}$ is not within the domain of h'.