October/November 2025 Paper 13 Worked Answers (A-Level Maths 9709 AS)

23 questions · 75 marks · 110 minutes

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1. Question 1a

   3 marksBinomial Expansion
   Step 1: Use the binomial expansion: the general term is $\binom{6}{r}(2)^{6-r}\left(-\dfrac{x}{2}\right)^r$. Step 2: For the $x^2$ term, set $r = 2$: $\binom{6}{2}(2)^4\left(-\dfrac{1}{2}\right)^2 = 15 \times 16 \times \dfrac{1}{4} = 60$. Step 3: The coefficient of $x^2$ is $60$.
   Method:

   Apply the binomial expansion formula, pick out the $r=2$ term, and simplify.
   Examiner tips

   • Even powers of $(-1/2)$ are positive
   • Remember to include both the binomial coefficient and the powers of each part

2. Question 1b

   2 marksBinomial Expansion with Polynomial
   Step 1: Identify the three products that give $x^3$ terms: $3 \times (-20x^3)$, $(-x) \times 60x^2$, and $2x^3 \times 64$. Step 2: Compute each: $3(-20) = -60$, $(-1)(60) = -60$, $2(64) = 128$. Step 3: Sum: $-60 - 60 + 128 = 8$.
   Method:

   Identify all pairs of terms from each factor whose powers sum to 3, compute each product, and add.
   Examiner tips

   • Systematically identify all pairs of terms whose $x$-powers sum to the target power

3. Question 2a

   2 marksInverse Trigonometric Equations
   Step 1: Apply $\tan$ to both sides: $3x - 1 = \tan\left(-\dfrac{\pi}{4}\right) = -1$. Step 2: Solve: $3x = -1 + 1 = 0$, so $x = 0$.
   Method:

   Apply $\tan$ to both sides, use exact value of $\tan(-\pi/4)$, solve for $x$.
   Examiner tips

   • Remember that $\tan(-\pi/4) = -1$ and $\tan(\pi/4) = 1$

4. Question 2b

   4 marksTrigonometric Equations
   Step 1: Replace $\cos^2\theta$ with $1 - \sin^2\theta$: $3(1 - \sin^2\theta) = 2\sin\theta + 2$. Step 2: Rearrange: $3\sin^2\theta + 2\sin\theta - 1 = 0$. Step 3: Factorise: $(3\sin\theta - 1)(\sin\theta + 1) = 0$, so $\sin\theta = \dfrac{1}{3}$ or $\sin\theta = -1$. Step 4: For $\sin\theta = \dfrac{1}{3}$ in $[0, 2\pi]$: two solutions ($\theta \approx 0.3398$ and $\theta \approx \pi - 0.3398 \approx 2.802$). Step 5: For $\sin\theta = -1$: one solution ($\theta = \dfrac{3\pi}{2}$). Step 6: Total: $3$ solutions.
   Method:

   Substitute $\cos^2\theta = 1 - \sin^2\theta$, solve the quadratic, count solutions for each root in the range.
   Examiner tips

   • Always use $\cos^2\theta = 1 - \sin^2\theta$ when the equation mixes $\cos^2$ and $\sin$
   • $\sin\theta = k$ with $|k| < 1$ gives two solutions in $[0, 2\pi]$; $\sin\theta = -1$ gives exactly one

5. Question 3a

   1 marksDifferentiation from First Principles
   Step 1: Compute $8^{2/3} = (8^{1/3})^2 = 2^2 = 4$. Step 2: Compute $(8 - 2)^2 = 6^2 = 36$. Step 3: $f(8) = \dfrac{1}{2} \times 4 \times 36 = 72$.
   Method:

   Evaluate $8^{2/3} = 4$ and $(8-2)^2 = 36$, then multiply by $1/2$.
   Examiner tips

   • Break the calculation into parts: evaluate $x^{2/3}$ separately from $(x-2)^2$

6. Question 3b

   1 marksGradient of Chord
   Step 1: Gradient $= \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{12.719 - 12}{5.1 - 5} = \dfrac{0.719}{0.1} = 7.19$. Step 2: Correct to 1 decimal place: $7.2$.
   Method:

   Compute the difference in $y$-coordinates divided by the difference in $x$-coordinates.
   Examiner tips

   • The gradient of a chord is simply $\dfrac{\text{change in } y}{\text{change in } x}$

7. Question 3c

   1 marksDifferentiation from First Principles
   Step 1: As the second point gets closer to $A$ (from $D$ to $C$ to $B$), the chord gradient approaches $30$: $30.389 \to 30.153 \to 30.030$. Step 2: In the limit, the gradient of the tangent at $A$ is $f'(8) = 30$.
   Method:

   Observe the chord gradients converging to 30 as the interval shrinks, conclude $f'(8) = 30$.
   Examiner tips

   • The derivative is the limit of the chord gradient as the second point approaches the first

8. Question 4a

   2 marksArithmetic Progressions
   Step 1: For an AP, the common difference is constant: $k - 24 = (k - 7) - k = -7$. Step 2: So $k - 24 = -7$, giving $k = 17$ and $d = -7$. Step 3: The 20th term: $a + 19d = 24 + 19(-7) = 24 - 133 = -109$.
   Method:

   Equate consecutive differences to find $d$, then use the $n$th term formula.
   Examiner tips

   • In an AP, the difference between consecutive terms is constant
   • The $n$th term is $a + (n-1)d$, not $a + nd$

9. Question 4b

   4 marksGeometric Progressions - Sum to Infinity
   Step 1: For a GP, $\dfrac{k}{12} = \dfrac{k-3}{k}$, so $k^2 = 12(k - 3) = 12k - 36$. Step 2: $k^2 - 12k + 36 = 0$, so $(k - 6)^2 = 0$, giving $k = 6$. Step 3: Common ratio $r = \dfrac{6}{12} = \dfrac{1}{2}$. Since $|r| < 1$, the sum to infinity exists. Step 4: $S_\infty = \dfrac{a}{1 - r} = \dfrac{12}{1 - 1/2} = \dfrac{12}{1/2} = 24$.
   Method:

   Equate ratios of consecutive terms, solve the quadratic for $k$, find $r$, then apply $S_\infty = a/(1-r)$.
   Examiner tips

   • When consecutive terms form a GP, equate the ratio of consecutive terms
   • A repeated root means there is only one valid value of $k$

10. Question 6b

    3 marksGraph Transformations
    Step 1: Write $f(x) = 5(x - 4)^3 + 7$ in the form $a(x + b)^3 + c$. Step 2: Compare: $x - 4 = x + (-4)$, so $b = -4$. Step 3: Also $a = 5$ and $c = 7$.
    Method:

    Compare the given expression with $a(x+b)^3 + c$ and read off the values.
    Examiner tips

    • Be careful with signs: $x - 4 = x + (-4)$ so $b = -4$

11. Question 7

    4 marksInverse Functions
    Step 1: Let $y = \dfrac{2}{4x - 3} + \dfrac{1}{2}$. Rearrange: $y - \dfrac{1}{2} = \dfrac{2}{4x - 3}$. Step 2: $4x - 3 = \dfrac{2}{y - 1/2} = \dfrac{2}{(2y - 1)/2} = \dfrac{4}{2y - 1}$. Step 3: $4x = \dfrac{4}{2y - 1} + 3$, so $x = \dfrac{1}{2y - 1} + \dfrac{3}{4}$. Step 4: Swap $x$ and $y$: $g^{-1}(x) = \dfrac{1}{2x - 1} + \dfrac{3}{4}$.
    Method:

    Rearrange $y = g(x)$ to make $x$ the subject, swap variables to get $g^{-1}(x)$.
    Examiner tips

    • When finding an inverse, isolate the term containing $x$ first
    • Check your answer by verifying $g(g^{-1}(x)) = x$

12. Question 8a

    3 marksEquation of a Circle
    Step 1: Centre $= \left(\dfrac{5 + (-3)}{2}, \dfrac{-1 + 5}{2}\right) = (1, 2)$. Step 2: Radius $=$ distance from centre to either endpoint $= \sqrt{(5 - 1)^2 + (-1 - 2)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
    Method:

    Find the midpoint for the centre, compute the distance from centre to an endpoint for the radius.
    Examiner tips

    • The centre of a circle is the midpoint of any diameter
    • The radius is half the diameter, not the full length

13. Question 8b

    5 marksTangent to a Circle
    Step 1: The normal to the tangent has gradient $2$ (negative reciprocal of $-1/2$). Step 2: Line through centre $(3, 1)$ with gradient $2$: $y - 1 = 2(x - 3)$, so $y = 2x - 5$. Step 3: Substitute into the circle equation: $(x - 3)^2 + (2x - 5 - 1)^2 = 20$. Step 4: $(x - 3)^2 + 4(x - 3)^2 = 20$, so $5(x - 3)^2 = 20$, $(x - 3)^2 = 4$. Step 5: $x = 3 \pm 2$, giving $x = 1$ or $x = 5$.
    Method:

    Find the normal gradient, write the line through the centre, substitute into the circle equation and solve.
    Examiner tips

    • The tangent point lies on both the circle and the normal line through the centre
    • Use the negative reciprocal relationship between tangent and normal gradients

14. Question 10a_i

    3 marksCompleting the Square
    Step 1: Factor out the coefficient of $x^2$: $3(x^2 + 4x) + 7$. Step 2: Complete the square inside the bracket: $3(x^2 + 4x + 4 - 4) + 7 = 3((x + 2)^2 - 4) + 7$. Step 3: Expand: $3(x + 2)^2 - 12 + 7 = 3(x + 2)^2 - 5$. Step 4: So $a = 3$, $b = 2$, $c = -5$.
    Method:

    Factor out the coefficient of $x^2$, complete the square, then simplify.
    Examiner tips

    • When the coefficient of $x^2$ is not 1, factor it out first before completing the square

15. Question 10a_ii

    1 marksDomain and Range
    Step 1: Since $(x + 2)^2 \ge 0$ for all real $x$, the minimum value of $3(x+2)^2$ is $0$. Step 2: Therefore the minimum value of $f(x) = 3(x+2)^2 - 5$ is $0 - 5 = -5$, achieved when $x = -2$. Step 3: Range: $f(x) \ge -5$.
    Method:

    Read off the minimum value from the completed square form.
    Examiner tips

    • The minimum of $a(x+b)^2 + c$ with $a > 0$ is $c$, and it is attained

16. Question 10b

    5 marksDiscriminant and Quadratic Roots
    Step 1: By Vieta's formulas for $px^2 + 4x - 8 = 0$: sum of roots $= -4/p$ and product of roots $= -8/p$. Step 2: Sum: $4m + (-6m) = -2m = -4/p$, so $m = 2/p$. Step 3: Product: $(4m)(-6m) = -24m^2 = -8/p$, so $24m^2 = 8/p$, giving $3m^2 = 1/p$. Step 4: Substitute $m = 2/p$: $3 \times 4/p^2 = 1/p$, so $12/p^2 = 1/p$, giving $p = 12$. Step 5: Then $m = 2/12 = 1/6$.
    Method:

    Use Vieta's formulas to form two equations, eliminate $m$, solve for $p$, then find $m$.
    Examiner tips

    • Vieta's formulas relate roots to coefficients without solving the quadratic
    • Be careful with signs: sum $= -b/a$, not $b/a$

17. Question 11a

    6 marksTangents and Normals
    Step 1: Differentiate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-24}{(3x - 8)^2} + \dfrac{6}{(x - 1)^2}$. Step 2: At $x = 3$: $\dfrac{-24}{(9 - 8)^2} + \dfrac{6}{(3 - 1)^2} = \dfrac{-24}{1} + \dfrac{6}{4} = -24 + \dfrac{3}{2} = -\dfrac{45}{2}$.
    Method:

    Differentiate using the chain rule, evaluate at $x=3$ to get the gradient, form the tangent equation, solve simultaneously with $y = -8x$.
    Examiner tips

    • When differentiating $k/(ax+b)$, rewrite as $k(ax+b)^{-1}$ and use the chain rule
    • Be careful with the chain rule: the derivative of $(3x-8)^{-1}$ includes a factor of 3

18. Question 11b_i

    3 marksStationary Points
    Step 1: Set $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$: $\dfrac{-24}{(3x-8)^2} + \dfrac{6}{(x-1)^2} = 0$. Step 2: Rearrange: $\dfrac{6}{(x-1)^2} = \dfrac{24}{(3x-8)^2}$, so $\dfrac{(3x-8)^2}{(x-1)^2} = 4$. Step 3: Take square roots: $\dfrac{3x - 8}{x - 1} = \pm 2$. Step 4: Case $+2$: $3x - 8 = 2x - 2$, so $x = 6$. Step 5: Case $-2$: $3x - 8 = -2x + 2$, so $5x = 10$, $x = 2$. Step 6: $p + q = 2 + 6 = 8$.
    Method:

    Set derivative to zero, rearrange to a ratio of squares, take both $\pm$ square roots, solve each case.
    Examiner tips

    • When you have a ratio of squares equal to a constant, take both positive and negative square roots

19. Question 11b_ii

    4 marksNature of Stationary Points
    Step 1: Find $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$. Differentiate $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -24(3x-8)^{-2} + 6(x-1)^{-2}$. Step 2: $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{144}{(3x-8)^3} - \dfrac{12}{(x-1)^3}$. Step 3: At $x = 2$: $\dfrac{144}{(6-8)^3} - \dfrac{12}{(2-1)^3} = \dfrac{144}{-8} - \dfrac{12}{1} = -18 - 12 = -30$. Step 4: Since $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -30 < 0$, the stationary point at $x = 2$ is a maximum.
    Method:

    Differentiate again to get the second derivative, evaluate at each stationary point, use the sign to determine nature.
    Examiner tips

    • Be careful with cube powers of negative numbers: $(-2)^3 = -8$
    • State clearly whether the second derivative is positive or negative and what this implies