A Level Mathematics 9709 Paper 11, May/June 2026

Correct procedure to find $r^2$ or $r$.

$a=200$ and $r=\frac{1}{2}$

Use of formula for sum to infinity for their $a$ and $r$ provided that $\mathrm{\mid }r\mathrm{\mid }<1$.

$S_{\mathrm{\infty }}=400$

B1 for any two.

B1 for all three.

$(3+x^2{)}^6=729+1458x^2$

For using their values to find coefficient from $(a{x}^2+bx+c)(d+e{x}^2)\Rightarrow ad+ce$.

30618

B1 for $x-5$, or subtracting 2 or dividing by $4^3$.

Second B1 for another correct procedure.

Final B1 for fully correct OE, e.g. $y=\frac{1}{64}(x-5{)}^3-2$.

Use of ${\sin }^2\theta +{\cos }^2\theta =1$ to eliminate ${\cos }^2\theta$.

Solve their three-term quadratic in ${\sin }^2\theta$ using suitable method, e.g. factorisation, quadratic formula, completing the square.

$\sin \theta =\pm \frac{1}{\sqrt{2}}$

A1 for any correct pair.

A1 for all four and no other solutions in range.

B1 for $\frac{1}{2}(x^3-3x-2{)}^{-\frac{1}{2}}$

B1 for the rest.

For attempt at gradient by substituting $x=6$ into their $\frac{\mathrm{d}y}{\mathrm{d}x}$.

For use of product of gradients $=-1$ using their $\frac{\mathrm{d}y}{\mathrm{d}x}$.

Equation is $4x+13y-167=0$

OE Correct method for forming quadratic equation. Allow one sign error.

$p=1$ and $p=6$

For finding $d$ for one progression.

For finding $a$ for one progression.

For both.

Difference = 155

Correct use of cosine rule.

$=\frac{2}{3}\pi$ or ${120}^{\circ }$

Use of $r\theta$ with their angle. Using their numeric angle in radians.

Total perimeter $=(\frac{4}{3}\pi +1+\sqrt{7})r$

Sector area: Using their numeric angle.

Triangle area: Using their numeric angle.

For subtraction of their areas.

$=(\frac{4}{3}\pi -\frac{\sqrt{3}}{2})r^2$

For expressing as a three-term quadratic. All terms need not be on the same side.

$x=1$ and $x=3$

Subtracting two volumes. Condone subtraction the wrong way round.

Integration of $(7-2x{)}^2$: $\{49x-14x^2+\frac{4}{3}{x}^3\}$ or $\{-\frac{1}{6}(7-2x{)}^3\}$. Allow unsimplified.

Integration coefficient: Allow unsimplified.

Integration term: Allow unsimplified.

Use of their limits in an attempt at integration. Condone one sign error or missing brackets.

Volume $=\frac{32}{3}\pi$

$y^3=x\pm 8a$ or $x^3=y\pm 8a$

$[{\mathrm{g}}^{-1}(x)]=\sqrt[3]{x+8a}$

$\mathrm{f}(-1)=-4$

For $\mathrm{g}({\text{their }}^{\mathrm{\prime }}\mathrm{f}(-1{)}^{\mathrm{\prime }})$ and forming quadratic equation with all terms on one side.

Solving their quadratic using suitable method, e.g. factorisation, quadratic formula, completing the square. M0 if roots are not real.

$a=2$

For either $\mathrm{g}\mathrm{f}(x)=(x-3{)}^3-16$ or $\mathrm{f}\mathrm{g}(x)=(x^3-16)-3$

For forming equation with $a$ substituted with a real or $8a$ eliminated. $\mathrm{g}\mathrm{f}$ and $\mathrm{f}\mathrm{g}$ must be the correct way round.

For multiplying out and simplifying. All terms to one side. Allow one sign error only.

For use of $b^2-4ac$ correct for their $a,b,c$ from a three-term quadratic and including $k$.

$k>-3.75$ or $k>-\frac{15}{4}$ or $k>-\frac{135}{36}$

At $x=2$, $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=12-6[>0]\Rightarrow \text{minimum}$

$\left[\frac{\mathrm{d}y}{\mathrm{d}x}\right]=x^3-6x[+c]$

For substituting $\frac{\mathrm{d}y}{\mathrm{d}x}=0$ at $x=2$.

$y=\frac{1}{4}{x}^4-3x^2+4x+k$

For substituting $(2,3)$ into their “$y=\dots$”

$y=\frac{1}{4}{x}^4-3x^2+4x+3$

For using their $k$ in the given relationship or factorising otherwise.

Solving quadratic using suitable method, e.g. quadratic formula.

$x=-1\pm \sqrt{3}$

At $x=-1+\sqrt{3}$ $\left[\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}\right]=12-6\sqrt{3}-6\text{ or }-4.39\dots [<0]\Rightarrow \text{maximum}$

At $x=-1-\sqrt{3}$ $\left[\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}\right]=6\sqrt{3}+12-6\text{ or }16.4\dots [>0]\Rightarrow \text{minimum}$

For either normal gradient.

For either tangent gradient. Must be clear that they are finding tangent gradient.

Equation of tangent at $(6,16)$ is $y-16=\frac{1}{3}(x-6)[y=\frac{1}{3}x+14]$

Equation of tangent at $(14,8)$ is $y-8=3(x-14)[y=3x-34]$

Equating their two tangents.

$x=18$

For either. Coordinates of $P$ or $R$ must be clear or indication of where the lines cross the x- or y-axes.

For correct attempt at area, e.g. trapezium – triangle, integration.

$=360-54-\frac{200}{3}=\frac{718}{3}$ or $239\frac{1}{3}$