October/November 2025 Paper 11 Worked Answers (A-Level Maths 9709 AS)

23 questions · 75 marks · 110 minutes

Question papers and mark schemes are copyright Cambridge International. We do not reproduce them: the worked answers here are written by The Practice Book. Have the paper open alongside. Get the official paper from Cambridge

1. Question 1

   4 marksDiscriminant
   Step 1: For two distinct real roots, the discriminant must be positive: $b^2 - 4ac > 0$. Step 2: Here $a = 4k$, $b = k + 12$, $c = 4$. So $(k+12)^2 - 4(4k)(4) > 0$. Step 3: Expand: $k^2 + 24k + 144 - 64k > 0$, giving $k^2 - 40k + 144 > 0$. Step 4: Factorise: $(k - 4)(k - 36) > 0$. Step 5: Since the coefficient of $k^2$ is positive, the quadratic is positive outside the roots: $k < 4$ or $k > 36$.
   Method:

   Apply $b^2 - 4ac > 0$, expand and simplify to get a quadratic in $k$, then solve the inequality.
   Examiner tips

   • Remember: two distinct real roots means discriminant is strictly greater than zero
   • For a quadratic inequality $(k-a)(k-b) > 0$ with $a < b$, the solution is $k < a$ or $k > b$

2. Question 2a

   3 marksGeometric Progressions
   Step 1: The second term is $a\cos\alpha = 6$ and the fifth term is $a\cos^4\alpha = \frac{6}{27} = \frac{2}{9}$. Step 2: Divide the fifth term by the second term: $\cos^3\alpha = \frac{2/9}{6} = \frac{1}{27}$. Step 3: $\cos\alpha = \left(\frac{1}{27}\right)^{1/3} = \frac{1}{3}$. Step 4: $\alpha = \arccos\left(\frac{1}{3}\right) = 1.23$ (3 s.f.).
   Method:

   Set up two equations, divide to find $\cos\alpha$, then use $\arccos$ to find $\alpha$.
   Examiner tips

   • Dividing terms of a GP eliminates the first term and isolates the common ratio

3. Question 2b

   2 marksSum to Infinity
   Step 1: The sum to infinity of a GP with $|r| < 1$ is $S_\infty = \dfrac{a}{1 - r}$. Step 2: $S_\infty = \dfrac{18}{1 - 1/3} = \dfrac{18}{2/3} = 18 \times \dfrac{3}{2} = 27$.
   Method:

   Substitute $a$ and $r$ into the sum to infinity formula.
   Examiner tips

   • Always check $|r| < 1$ before applying the sum to infinity formula

4. Question 3

   5 marksBinomial Expansion
   Step 1: From $(px+2)^5$, the $x^4$ term is $\binom{5}{1}(2)(px)^4 = 10p^4x^4$. Step 2: From $\left(x^3 + \frac{p}{x}\right)^4$, the $x^4$ term is $\binom{4}{2}(x^3)^2\left(\frac{p}{x}\right)^2 = 6p^2x^4$. Step 3: Combined coefficient of $x^4$: $10p^4 - 6p^2 = 136$. Step 4: Let $u = p^2$: $10u^2 - 6u - 136 = 0$, i.e. $5u^2 - 3u - 68 = 0$. Step 5: Factorise: $(5u + 17)(u - 4) = 0$. Since $u = p^2 > 0$, we have $u = 4$, so $p = 2$.
   Method:

   Find the $x^4$ coefficient from each binomial expansion, combine, substitute $u = p^2$, and solve the resulting quadratic.
   Examiner tips

   • In mixed binomial problems, find the required term from each expansion separately
   • A substitution like $u = p^2$ can turn a quartic into a quadratic

5. Question 4a

   3 marksCompleting the Square
   Step 1: Rearrange: $1 - 6x - x^2 = -(x^2 + 6x) + 1$. Step 2: Complete the square inside the bracket: $x^2 + 6x = (x+3)^2 - 9$. Step 3: Substitute back: $-((x+3)^2 - 9) + 1 = -(x+3)^2 + 9 + 1 = 10 - (x+3)^2$. Step 4: Therefore $a = 10$ and $b = 3$.
   Method:

   Factor out the negative, complete the square inside the bracket, then simplify.
   Examiner tips

   • When the coefficient of $x^2$ is negative, factor it out before completing the square

6. Question 4b

   3 marksGraph Transformations
   Step 1: Start with $y = x^2$. Reflect in the $x$-axis: $y = -x^2$. Step 2: Now apply a translation $\begin{pmatrix} m \\ n \end{pmatrix}$: replace $x$ by $x - m$ and add $n$, giving $y = -(x-m)^2 + n$. Step 3: Compare with $y = 10 - (x+3)^2 = -(x+3)^2 + 10$. So $m = -3$ and $n = 10$. Step 4: $m + n = -3 + 10 = 7$.
   Method:

   Reflect in the $x$-axis first, then match the completed square form to find the translation vector.
   Examiner tips

   • When $y = -(x+3)^2 + 10$, the $+3$ inside means translate left (negative $m$)

7. Question 5a

   3 marksTrigonometric Identities
   Step 1: Write $\tan^4\theta = \frac{\sin^4\theta}{\cos^4\theta}$, so $\tan^4\theta - 1 = \frac{\sin^4\theta - \cos^4\theta}{\cos^4\theta}$. Step 2: The key step is recognising that $\sin^4\theta - \cos^4\theta$ is a difference of two squares: $(\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta)$. Step 3: Since $\sin^2\theta + \cos^2\theta = 1$, this simplifies to $\sin^2\theta - \cos^2\theta = (1 - \cos^2\theta) - \cos^2\theta = 1 - 2\cos^2\theta$. Step 4: Therefore $\tan^4\theta - 1 = \frac{1 - 2\cos^2\theta}{\cos^4\theta}$.
   Method:

   Convert to sines and cosines, factorise numerator as difference of squares, simplify.
   Examiner tips

   • $a^4 - b^4 = (a^2 - b^2)(a^2 + b^2)$ is a key factorisation to recognise

8. Question 5b

   4 marksSolving Trigonometric Equations
   Step 1: First derive the identity: $\tan^4\theta - 1 = \frac{\sin^4\theta - \cos^4\theta}{\cos^4\theta} = \frac{(\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta)}{\cos^4\theta} = \frac{\sin^2\theta - \cos^2\theta}{\cos^4\theta}$. Step 2: So $\cos^2\theta \times \frac{\sin^2\theta - \cos^2\theta}{\cos^4\theta} = \frac{\sin^2\theta - \cos^2\theta}{\cos^2\theta} = \tan^2\theta - 1 = 3$. Step 3: $\tan^2\theta = 4$, so $\tan\theta = \pm 2$. In $0° < \theta < 180°$: $\theta = \arctan(2)$ and $\theta = 180° - \arctan(2)$. That gives $2$ solutions.
   Method:

   Apply the identity, simplify to find $\cos^2\theta$, then find all solutions in the range.
   Examiner tips

   • When $\cos^2\theta = k$, remember $\cos\theta = \pm\sqrt{k}$, giving solutions in different quadrants

9. Question 6a

   1 marksDomain and Range
   Step 1: The vertex of $(x+3)^2 - 12$ is at $x = -3$, which is outside the domain $x \geq 0$. Step 2: Since the domain starts at $x = 0$ and the parabola opens upward, the minimum value on the domain occurs at $x = 0$. Step 3: $f(0) = (0+3)^2 - 12 = 9 - 12 = -3$. Step 4: Therefore the range is $f(x) \geq -3$.
   Method:

   Check that the vertex is outside the domain, then evaluate at the domain endpoint.
   Examiner tips

   • Always check whether the vertex lies inside or outside the given domain

10. Question 6b

    2 marksInverse Functions
    Step 1: Let $y = (x+3)^2 - 12$. Rearrange: $y + 12 = (x+3)^2$. Step 2: $x + 3 = \sqrt{y + 12}$ (take the positive root since $x \geq 0$ implies $x + 3 \geq 3 > 0$). Step 3: $x = -3 + \sqrt{y + 12}$. Step 4: Therefore $f^{-1}(x) = -3 + \sqrt{x + 12}$.
    Method:

    Rearrange $y = (x+3)^2 - 12$ to make $x$ the subject, using the positive root.
    Examiner tips

    • The domain restriction tells you which root to take — $x \geq 0$ means $x + 3 > 0$

11. Question 6c

    4 marksComposite Functions
    Step 1: $gf(x) = g(f(x)) = 2((x+3)^2 - 12) - 5 = 2(x+3)^2 - 24 - 5 = 2(x+3)^2 - 29$. Step 2: Set equal to $43$: $2(x+3)^2 - 29 = 43$, so $2(x+3)^2 = 72$, giving $(x+3)^2 = 36$. Step 3: $x + 3 = 6$ (taking positive root since $x \geq 0$), so $x = 3$. Step 4: Reject $x + 3 = -6$ giving $x = -9$ as it is outside the domain $x \geq 0$.
    Method:

    Form $gf(x)$, set equal to the given value, solve the quadratic, reject the solution outside the domain.
    Examiner tips

    • $gf(x)$ means apply $f$ first, then $g$
    • Always check domain restrictions when rejecting solutions

12. Question 7a

    2 marksArc Length and Sector Area
    Step 1: Area of segment $= \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta$ where $\theta = \frac{3\pi}{4}$. Step 2: $= \frac{1}{2}r^2\left(\frac{3\pi}{4} - \sin\frac{3\pi}{4}\right) = \frac{1}{2}r^2\left(\frac{3\pi}{4} - \frac{\sqrt{2}}{2}\right)$. Step 3: $= \frac{1}{2}r^2(2.3562 - 0.7071) = \frac{1}{2}r^2(1.6491) = 0.825r^2$. Step 4: Therefore $k = 0.825$ (3 s.f.).
    Method:

    Apply the segment area formula $\frac{1}{2}r^2(\theta - \sin\theta)$ and evaluate numerically.
    Examiner tips

    • Segment area = sector area $-$ triangle area: $\frac{1}{2}r^2(\theta - \sin\theta)$

13. Question 7b_i

    3 marksConnected Rates of Change
    Step 1: $A = 0.614r^2$, so $\dfrac{\mathrm{d}A}{\mathrm{d}r} = 2 \times 0.614r = 1.228r$. Step 2: By the chain rule: $\dfrac{\mathrm{d}A}{\mathrm{d}t} = \dfrac{\mathrm{d}A}{\mathrm{d}r} \times \dfrac{\mathrm{d}r}{\mathrm{d}t} = 1.228 \times 15 \times 0.5 = 9.21$ cm$^2$/s.
    Method:

    Differentiate $A$ with respect to $r$, then multiply by the given $\frac{\mathrm{d}r}{\mathrm{d}t}$.
    Examiner tips

    • Connected rates of change always use the chain rule to link rates

14. Question 7b_ii

    3 marksConnected Rates of Change
    Step 1: Arc length $l = r\theta = \frac{2\pi}{3}r$. Step 2: $\dfrac{\mathrm{d}l}{\mathrm{d}r} = \frac{2\pi}{3}$. Step 3: $\dfrac{\mathrm{d}l}{\mathrm{d}t} = \dfrac{\mathrm{d}l}{\mathrm{d}r} \times \dfrac{\mathrm{d}r}{\mathrm{d}t} = \frac{2\pi}{3} \times 0.3 = \frac{2\pi}{10} = \frac{\pi}{5}$ cm/s.
    Method:

    Differentiate the arc length formula with respect to $r$, then multiply by $\frac{\mathrm{d}r}{\mathrm{d}t}$.
    Examiner tips

    • Since $\theta$ is constant, differentiating $l = r\theta$ gives a constant times $\frac{\mathrm{d}r}{\mathrm{d}t}$

15. Question 8a

    3 marksArea Under a Curve
    Step 1: Area under the curve: $\int_0^9 \frac{1}{2}\sqrt{x}\,\mathrm{d}x = \frac{1}{2} \times \frac{2}{3}\left[x^{3/2}\right]_0^9 = \frac{1}{3}(27) = 9$. Step 2: Area of the rectangle (under the line $y = \frac{3}{2}$ from $x = 0$ to $x = 9$): $\frac{3}{2} \times 9 = \frac{27}{2}$. Step 3: Shaded area = rectangle area $-$ area under curve $= \frac{27}{2} - 9 = \frac{27}{2} - \frac{18}{2} = \frac{9}{2}$.
    Method:

    Integrate the curve from 0 to 9, find the rectangle area, and subtract.
    Examiner tips

    • When the shaded region is between a curve and a horizontal line, use rectangle area minus integral

16. Question 8b

    4 marksVolume of Revolution
    Step 1: Rotating about the $y$-axis: $V = \pi\int_0^{3/2} x^2\,\mathrm{d}y = \pi\int_0^{3/2} (4y^2)^2\,\mathrm{d}y = \pi\int_0^{3/2} 16y^4\,\mathrm{d}y$. Step 2: $= \pi\left[\frac{16y^5}{5}\right]_0^{3/2} = \frac{16\pi}{5} \times \left(\frac{3}{2}\right)^5 = \frac{16\pi}{5} \times \frac{243}{32} = \frac{243\pi}{10}$.
    Method:

    Express $x = 4y^2$, set up $\pi\int_0^{3/2}(4y^2)^2\,\mathrm{d}y$, integrate and evaluate.
    Examiner tips

    • For rotation about the $y$-axis, always express $x$ as a function of $y$ first

17. Question 9a

    4 marksArithmetic Progressions
    Step 1: $S_2 = 2(2) + \frac{2(1)}{2}d = 4 + d$, so $S_2 - 1 = 3 + d$. Step 2: $S_4 = 4(2) + \frac{4(3)}{2}d = 8 + 6d$. Step 3: $S_9 = 9(2) + \frac{9(8)}{2}d = 18 + 36d$. Step 4: For three terms to form an AP: $2(8 + 6d) = (3 + d) + (18 + 36d)$. Step 5: $16 + 12d = 21 + 37d$, so $25d = -5$, giving $d = -\frac{1}{5}$.
    Method:

    Express each sum in terms of $d$, apply the condition for three terms to form an AP, and solve.
    Examiner tips

    • Three terms $a$, $b$, $c$ form an AP if and only if $2b = a + c$

18. Question 9b

    4 marksArithmetic Progressions
    Step 1: First AP: 15th term $= 2 + 14 \times (-\frac{1}{5}) = 2 - \frac{14}{5} = \frac{10-14}{5} = -\frac{4}{5}$. Step 2: Second AP: 15th term $= \frac{14}{5} + 14 \times 4 = \frac{14}{5} + 56 = \frac{14 + 280}{5} = \frac{294}{5}$. Step 3: Difference $= \frac{294}{5} - \left(-\frac{4}{5}\right) = \frac{298}{5} = 59.6$.
    Method:

    Find the 15th term of each AP using the formula, then subtract.
    Examiner tips

    • Be careful with the signs: subtracting a negative number adds

19. Question 10a

    4 marksIntersection of Lines and Circles
    Step 1: From the line: $y = 8 - 2x$. Substitute into the circle equation. Step 2: $x^2 + (8-2x)^2 + 4(8-2x) - 21 = 0$. Step 3: Expand: $x^2 + 64 - 32x + 4x^2 + 32 - 8x - 21 = 0 \Rightarrow 5x^2 - 40x + 75 = 0 \Rightarrow x^2 - 8x + 15 = 0$. Step 4: $(x-3)(x-5) = 0$, so $x = 3$ and $x = 5$. Sum $= 3 + 5 = 8$.
    Method:

    Rearrange the line for $y$, substitute into the circle, solve the quadratic.
    Examiner tips

    • Always expand and simplify carefully — sign errors are common in substitution

20. Question 10b

    3 marksEquation of a Circle
    Step 1: Note that $B(5,-2)$ and $C(0,-2)$ share the same $y$-coordinate, so $BC$ is horizontal with length $|5-0| = 5$. Step 2: The height from $A(3,2)$ to the line $y = -2$ is $|2 - (-2)| = 4$. Step 3: Area $= \frac{1}{2} \times 5 \times 4 = 10$.
    Method:

    Find the centre, then use base-height or the coordinate formula for the area.
    Examiner tips

    • Look for a horizontal or vertical base to simplify the calculation

21. Question 11a

    3 marksTangents and Normals
    Step 1: At $x = 4$: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{8}{16} - \dfrac{10}{25} = \dfrac{1}{2} - \dfrac{2}{5} = \dfrac{5-4}{10} = \dfrac{1}{10} = 0.1$. Step 2: The gradient of the normal is the negative reciprocal: $-\dfrac{1}{0.1} = -10$.
    Method:

    Substitute into the derivative, find the tangent gradient, take negative reciprocal.
    Examiner tips

    • The normal gradient is $-1/m$ where $m$ is the tangent gradient

22. Question 11b

    3 marksStationary Points
    Step 1: Differentiate $8x^{-2}$: $-16x^{-3}$. Differentiate $-10(2x-3)^{-2}$: $-10 \times (-2)(2x-3)^{-3} \times 2 = 40(2x-3)^{-3}$. Step 2: $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -16x^{-3} + 40(2x-3)^{-3}$. Step 3: At $x = 4$: $-\dfrac{16}{64} + \dfrac{40}{125} = -\dfrac{1}{4} + \dfrac{8}{25} = \dfrac{-25 + 32}{100} = \dfrac{7}{100}$.
    Method:

    Differentiate each term of $\frac{\mathrm{d}y}{\mathrm{d}x}$ using the power and chain rules, then evaluate at $x = 4$.
    Examiner tips

    • When differentiating $(ax+b)^n$, remember to multiply by the coefficient $a$ from the chain rule

23. Question 11c

    5 marksFinding the equation of a curve from its derivative
    Step 1: Integrate $\frac{8}{x^2} = 8x^{-2}$: $\int 8x^{-2}\,\mathrm{d}x = -8x^{-1} = -\frac{8}{x}$. Step 2: Integrate $-\frac{10}{(2x-3)^2} = -10(2x-3)^{-2}$: $\int -10(2x-3)^{-2}\,\mathrm{d}x = \frac{-10}{-1 \times 2}(2x-3)^{-1} = \frac{5}{2x-3}$. Step 3: $y = -\frac{8}{x} + \frac{5}{2x-3} + c$. Step 4: Substitute $(4, 3)$: $3 = -\frac{8}{4} + \frac{5}{5} + c = -2 + 1 + c$, so $c = 4$. Step 5: At $x = -1$: $q = -\frac{8}{-1} + \frac{5}{2(-1)-3} + 4 = 8 + \frac{5}{-5} + 4 = 8 - 1 + 4 = 11$.
    Method:

    Integrate term by term, find $c$ using the known point, then evaluate at $x = -1$.
    Examiner tips

    • When integrating $(ax+b)^n$, divide by $a \times (n+1)$
    • Always remember the constant of integration