October/November 2025 Paper 31 Worked Answers (A-Level Maths 9709 A2)

16 questions · 75 marks · 110 minutes

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1. Question 1

   4 marksDefinite Integration by Parts
   Step 1: Use integration by parts with $u = \ln 3x$, $dv = dx$, so $du = \dfrac{1}{x} dx$ and $v = x$. Then $\int \ln 3x \, dx = x \ln 3x - \int 1 \, dx = x \ln 3x - x$. Step 2: Apply the upper limit $x = 2$: $2 \ln 6 - 2$. Step 3: Apply the lower limit $x = 1$: $1 \cdot \ln 3 - 1 = \ln 3 - 1$. Step 4: Subtract: $(2 \ln 6 - 2) - (\ln 3 - 1) = 2 \ln 6 - \ln 3 - 1$. Step 5: Simplify: $2 \ln 6 = \ln 36$, so $\ln 36 - \ln 3 - 1 = \ln 12 - 1$. Hence $a = -1$ and $b = 12$.
   Method:

   Integrate by parts, evaluate at the limits, and combine logs into a single $\ln$.
   Examiner tips

   • Use $\ln A - \ln B = \ln(A/B)$ to simplify the final form
   • $\dfrac{d}{dx} \ln 3x = \dfrac{1}{x}$ (the 3 cancels)

2. Question 2a

   3 marksReducing a Logarithmic Equation to a Quadratic
   Step 1: Write $2 = \log_4 16$ and rearrange: $\log_4(2x+1) + \log_4 16 = \log_4(3x-1)^2$. Step 2: Combine: $\log_4 [16(2x+1)] = \log_4 (3x-1)^2$, so $16(2x+1) = (3x-1)^2$. Step 3: Expand: $32x + 16 = 9x^2 - 6x + 1$. Step 4: Rearrange: $9x^2 - 6x + 1 - 32x - 16 = 0 \Rightarrow 9x^2 - 38x - 15 = 0$.
   Method:

   Express the constant as a logarithm, combine logs into a single log on each side, equate the arguments, and expand.
   Examiner tips

   • Always rewrite constants as logs before combining
   • Cross-multiply by exponentiating both sides if needed

3. Question 2b

   2 marksSolving Logarithmic Equation
   Step 1: Apply the quadratic formula: $x = \dfrac{38 \pm \sqrt{1444 + 540}}{18} = \dfrac{38 \pm \sqrt{1984}}{18}$. Step 2: $\sqrt{1984} = \sqrt{64 \cdot 31} = 8\sqrt{31} \approx 44.54$, so $x \approx \dfrac{38 + 44.54}{18} \approx 4.59$ or $x \approx \dfrac{38 - 44.54}{18} \approx -0.36$. Step 3: Domain check: $\log_4(3x-1)$ requires $3x - 1 > 0$, i.e. $x > \dfrac{1}{3}$. The negative solution $-0.36$ fails this. Hence only $x \approx 4.59$ is valid.
   Method:

   Apply the quadratic formula, then reject any root outside the domain of the original logarithms.
   Examiner tips

   • Always check the original log domains for each candidate root
   • $\sqrt{1984} = 8\sqrt{31}$ may simplify nicely

4. Question 3a

   4 marksR-Formula (Cosine Form)
   Step 1: Expand $3\sqrt{2} \sin(x + 45^\circ) = 3\sqrt{2}\big[\sin x \cos 45^\circ + \cos x \sin 45^\circ\big] = 3\sqrt{2} \cdot \dfrac{\sqrt{2}}{2} (\sin x + \cos x) = 3 \sin x + 3 \cos x$. Step 2: Add $\cos x$: $3 \sin x + 4 \cos x$. Step 3: Match to $R \cos(x - \alpha) = R \cos\alpha \cos x + R \sin\alpha \sin x$. So $R \cos\alpha = 4$ and $R \sin\alpha = 3$. Step 4: $R^2 = 16 + 9 = 25$, so $R = 5$. $\tan\alpha = \dfrac{3}{4}$, so $\alpha = \tan^{-1}(0.75) \approx 36.87^\circ \approx 36.9^\circ$.
   Method:

   Expand, simplify to $a \sin x + b \cos x$, match to $R \cos(x - \alpha)$, then compute $R = \sqrt{a^2+b^2}$ and $\alpha = \tan^{-1}(a/b)$.
   Examiner tips

   • For $R \cos(x - \alpha)$: $\tan\alpha = \dfrac{\text{coefficient of }\sin x}{\text{coefficient of }\cos x}$
   • Always expand the compound angle before matching

5. Question 3b

   4 marksSolving a Trig Equation Using R-Formula
   Step 1: Replace $x$ with $3\theta$: $5 \cos(3\theta - 36.9^\circ) = -4$, i.e. $\cos(3\theta - 36.9^\circ) = -\dfrac{4}{5}$. Step 2: Reference angle: $\cos^{-1}(0.8) \approx 36.87^\circ$. The general solutions of $\cos\phi = -0.8$ are $\phi = 180^\circ - 36.87^\circ = 143.13^\circ$ and $\phi = 180^\circ + 36.87^\circ = 216.87^\circ$ (mod $360^\circ$). Step 3: For $0 < \theta < 180^\circ$: $-36.9^\circ < 3\theta - 36.9^\circ < 503.1^\circ$. Solutions in this range: $143.13^\circ$ and $216.87^\circ$. Step 4: Solve: $3\theta = 143.13 + 36.9 = 180^\circ$, giving $\theta = 60^\circ$. And $3\theta = 216.87 + 36.9 = 253.77^\circ$, giving $\theta \approx 84.6^\circ$.
   Method:

   Use the harmonic form, find both general solutions for the cosine equation in the extended range, then divide by 3 and adjust for $36.9^\circ$.
   Examiner tips

   • $\cos\phi = c$ has solutions $\phi = \cos^{-1} c$ and $\phi = -\cos^{-1} c$ (mod $360^\circ$)
   • Always expand the search range to cover the multiplied argument $3\theta$

6. Question 5b

   3 marksLeast Argument in a Region of the Argand Plane
   Step 1: The line $\operatorname{Re}(z) = 2$ intersects the circle $(x-1)^2 + (y+2)^2 = 4$ at $x = 2$, giving $(y+2)^2 = 3$, so $y = -2 + \sqrt{3}$ or $y = -2 - \sqrt{3}$. Step 2: The candidate boundary points where the region's $\arg z$ is extreme are $z_1 = 2 + (-2 + \sqrt{3})i \approx 2 - 0.268i$ and $z_2 = 2 + (-2 - \sqrt{3})i \approx 2 - 3.732i$. Step 3: $\arg z_1 = \arctan(-0.268/2) \approx -7.6^\circ$ and $\arg z_2 = \arctan(-3.732/2) \approx \arctan(-1.866) \approx -61.8^\circ$. Step 4: The least (most negative) value is $\arg z = -61.8^\circ$ at $z = 2 + (-2 - \sqrt 3)i$.
   Method:

   Find the intersection of the bounding line $x = 2$ with the circle, evaluate $\arg z$ at each intersection, and take the minimum.
   Examiner tips

   • The least $\arg z$ in the lower half-plane is the most negative angle
   • Solve the boundary intersection to find candidate points before computing $\arg$

7. Question 6

   5 marksQuadratic Equation with Complex Coefficients
   Step 1: Apply the quadratic formula: $w = \dfrac{-4 \pm \sqrt{16 - 4(2+i)(2-i)}}{2(2+i)}$. Step 2: $(2+i)(2-i) = 4 - i^2 = 5$, so discriminant $= 16 - 20 = -4$, hence $\sqrt{-4} = \pm 2i$. Step 3: $w = \dfrac{-4 \pm 2i}{4 + 2i} = \dfrac{-2 \pm i}{2 + i}$. Step 4: Rationalise by multiplying numerator and denominator by $2 - i$. Denominator becomes $(2+i)(2-i) = 5$. Step 5: With the $+$ sign: numerator $(-2+i)(2-i) = -4 + 2i + 2i - i^2 = -4 + 4i + 1 = -3 + 4i$. So $w_1 = \dfrac{-3 + 4i}{5} = -\dfrac{3}{5} + \dfrac{4}{5} i$. Step 6: With the $-$ sign: numerator $(-2-i)(2-i) = -4 + 2i - 2i + i^2 = -4 - 1 = -5$. So $w_2 = \dfrac{-5}{5} = -1$.
   Method:

   Apply the quadratic formula, evaluate the discriminant using $i^2 = -1$, and rationalise the resulting complex fractions.
   Examiner tips

   • $(a+bi)(a-bi) = a^2 + b^2$ — useful for simplifying complex denominators
   • Even with complex coefficients, the quadratic formula still applies

8. Question 7

   5 marksParametric Differentiation Simplified
   Step 1: $\dfrac{dx}{dt} = 2t - \dfrac{2}{2t+1} = \dfrac{2t(2t+1) - 2}{2t+1} = \dfrac{4t^2 + 2t - 2}{2t+1}$. Step 2: Factor: $4t^2 + 2t - 2 = 2(2t^2 + t - 1) = 2(2t - 1)(t + 1)$. So $\dfrac{dx}{dt} = \dfrac{2(2t-1)(t+1)}{2t+1}$. Step 3: $\dfrac{dy}{dt}$ by the quotient rule: $\dfrac{(2t+1) \cdot 1 - t \cdot 2}{(2t+1)^2} = \dfrac{1}{(2t+1)^2}$. Step 4: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{1/(2t+1)^2}{2(2t-1)(t+1)/(2t+1)} = \dfrac{1}{(2t+1)^2} \cdot \dfrac{2t+1}{2(2t-1)(t+1)} = \dfrac{1}{2(2t+1)(2t-1)(t+1)}$.
   Method:

   Differentiate $x$ and $y$ with respect to $t$, factor each, then divide and simplify.
   Examiner tips

   • Always factor numerator and denominator before dividing
   • Use $2t^2 + t - 1 = (2t - 1)(t + 1)$ — recognise common factorisations

9. Question 8

   8 marksSeparable Differential Equation
   Step 1: Separate variables: $e^{-2y} dy = \dfrac{k x}{x^2 + 1} dx$. Step 2: Integrate: $-\dfrac{1}{2} e^{-2y} = \dfrac{k}{2} \ln(x^2 + 1) + C$. Step 3: Apply IC $y = 0$ at $x = 0$: $-\dfrac{1}{2} \cdot 1 = 0 + C$, so $C = -\dfrac{1}{2}$. Step 4: Apply IC $y = -\dfrac{1}{2}$ at $x = 1$: $-\dfrac{e}{2} = \dfrac{k \ln 2}{2} - \dfrac{1}{2}$, so $k \ln 2 = 1 - e$, giving $k = \dfrac{1 - e}{\ln 2}$. Step 5: At $x = \sqrt{3}$: $x^2 + 1 = 4$, $\ln 4 = 2 \ln 2$. So $-\dfrac{1}{2} e^{-2y} = \dfrac{1-e}{2 \ln 2} \cdot 2 \ln 2 - \dfrac{1}{2} = (1-e) - \dfrac{1}{2}$. Step 6: Multiply by $-2$: $e^{-2y} = 2(e - 1) + 1 = 2e - 1$. Therefore $-2y = \ln(2e - 1)$, so $y = -\dfrac{1}{2} \ln(2e - 1)$.
   Method:

   Separate variables, integrate both sides, apply both initial conditions to find the two constants, then evaluate at $x = \sqrt{3}$.
   Examiner tips

   • Apply both initial conditions in sequence to find the constants
   • Use $\ln 4 = 2 \ln 2$ to simplify

10. Question 9d

    3 marksIterative Method to 3 Decimal Places
    Step 1: Compute $x_1 = \dfrac{1}{2} \cos^{-1}(-e^{-1}) = \dfrac{1}{2} \cos^{-1}(-0.36788) \approx \dfrac{1}{2}(1.94752) \approx 0.97376$. Step 2: $x_2 = \dfrac{1}{2} \cos^{-1}(-e^{-0.97376}) \approx \dfrac{1}{2} \cos^{-1}(-0.37768) \approx 0.97903$. Step 3: $x_3 \approx 0.97796$, $x_4 \approx 0.97813$, $x_5 \approx 0.97810$. The sequence stabilises around $0.978$ (3 d.p.).
    Method:

    Apply the iterative formula in radians from $x_0 = 1$, retaining at least 5 d.p., until two successive values agree to 4 d.p.
    Examiner tips

    • Always work in radians for inverse trig in this style of problem
    • Keep at least 5 d.p. through the iteration

11. Question 10a

    6 marksPartial Fractions with Cubic Numerator
    Step 1: Since the numerator and denominator have equal degree, perform long division. $(3+x)(2+x^2) = x^3 + 3x^2 + 2x + 6$. So $f(x) = 1 + \dfrac{(x^3 + 2x - 11) - (x^3 + 3x^2 + 2x + 6)}{(3+x)(2+x^2)} = 1 + \dfrac{-3x^2 - 17}{(3+x)(2+x^2)}$. Step 2: Set $\dfrac{-3x^2 - 17}{(3+x)(2+x^2)} = \dfrac{B}{3+x} + \dfrac{Cx + D}{2 + x^2}$. Step 3: Multiply through: $-3x^2 - 17 = B(2 + x^2) + (Cx + D)(3 + x)$. Step 4: Expand: $B(2+x^2) + (Cx + D)(3 + x) = 2B + B x^2 + 3 C x + C x^2 + 3D + D x = (B + C) x^2 + (3C + D) x + (2B + 3D)$. Step 5: Equate coefficients: $B + C = -3$, $3C + D = 0$, $2B + 3D = -17$. Step 6: From $3C + D = 0$: $D = -3C$. Substitute: $2B - 9C = -17$ and $B = -3 - C$. So $2(-3 - C) - 9C = -17 \Rightarrow -11C = -11 \Rightarrow C = 1$. Then $B = -4$, $D = -3$. Step 7: $f(x) = 1 - \dfrac{4}{3+x} + \dfrac{x - 3}{2+x^2}$.
    Method:

    Perform polynomial long division to extract the constant term, then decompose the proper fraction using a linear and an irreducible quadratic factor.
    Examiner tips

    • Compare degrees first: if numerator $\ge$ denominator, divide first
    • An irreducible quadratic factor needs $\dfrac{Cx + D}{...}$, not just $\dfrac{C}{...}$

12. Question 10b

    5 marksBinomial Expansion of Partial Fractions
    Step 1: $-\dfrac{4}{3+x} = -\dfrac{4}{3}\left(1 + \dfrac{x}{3}\right)^{-1} = -\dfrac{4}{3}\left(1 - \dfrac{x}{3} + \dfrac{x^2}{9} - \cdots\right) = -\dfrac{4}{3} + \dfrac{4x}{9} - \dfrac{4 x^2}{27}$. Step 2: $\dfrac{x-3}{2+x^2} = (x - 3) \cdot \dfrac{1}{2}\left(1 + \dfrac{x^2}{2}\right)^{-1} = \dfrac{x-3}{2}\left(1 - \dfrac{x^2}{2} + \cdots\right)$. Up to $x^2$: $\dfrac{x-3}{2} - \dfrac{(x-3) x^2}{4}$. The $x^2$ term comes only from $-\dfrac{-3 x^2}{4} = \dfrac{3 x^2}{4}$. So $-\dfrac{3}{2} + \dfrac{x}{2} + \dfrac{3 x^2}{4}$. Step 3: Constant: $1 - \dfrac{4}{3} - \dfrac{3}{2} = \dfrac{6 - 8 - 9}{6} = -\dfrac{11}{6}$. Step 4: $x$-coefficient: $\dfrac{4}{9} + \dfrac{1}{2} = \dfrac{8 + 9}{18} = \dfrac{17}{18}$. Step 5: $x^2$-coefficient: $-\dfrac{4}{27} + \dfrac{3}{4} = \dfrac{-16 + 81}{108} = \dfrac{65}{108}$.
    Method:

    Expand each component using the standard $(1+u)^{-1}$ binomial series, multiply where necessary, and add the like-power coefficients.
    Examiner tips

    • Always factor out the constant from each linear factor before expanding
    • Only keep terms up to $x^2$ — discard higher powers

13. Question 11a

    2 marksVector Equation of a Line Through Two Points
    Step 1: $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (0-1, 4-5, 1-3) = (-1, -1, -2)$. Step 2: A direction vector is $(-1, -1, -2)$ or equivalently $(1, 1, 2)$ (scalar multiple). Step 3: Vector equation: $\mathbf{r} = \overrightarrow{OA} + \lambda \cdot \text{direction} = (\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}) + \lambda(\mathbf{i} + \mathbf{j} + 2\mathbf{k})$.
    Method:

    Compute $\overrightarrow{AB}$ and write the line as $\mathbf{r} = \overrightarrow{OA} + \lambda \overrightarrow{AB}$ (or scalar multiple).
    Examiner tips

    • Either $A$ or $B$ can be used as the position vector
    • Direction $(1,1,2)$ and $(-1,-1,-2)$ are both valid

14. Question 11b

    4 marksIntersection of Two Lines in 3D
    Step 1: Direction of $CD$: $\overrightarrow{CD} = (3-1, -5-(-3), 4-1) = (2, -2, 3)$. Line $CD$: $\mathbf{r} = (1, -3, 1) + \mu(2, -2, 3)$. Step 2: Equate to $m$'s general point $(1+\lambda, 5+\lambda, 3+2\lambda)$: $1+\lambda = 1+2\mu$, $5+\lambda = -3-2\mu$, $3+2\lambda = 1+3\mu$. Step 3: From eqn 1: $\lambda = 2\mu$. Substitute into eqn 2: $5 + 2\mu = -3 - 2\mu \Rightarrow 4\mu = -8 \Rightarrow \mu = -2$, so $\lambda = -4$. Step 4: Verify eqn 3: $3 + 2(-4) = -5$ and $1 + 3(-2) = -5$. ✓ Step 5: Substitute $\lambda = -4$ into $m$: $(1-4, 5-4, 3-8) = (-3, 1, -5)$.
    Method:

    Form equations from equating components of the two parameterised lines, solve the system, and substitute the parameter back.
    Examiner tips

    • Always check the third equation — if it fails, the lines do not intersect (skew)
    • Either $\lambda$ or $\mu$ alone suffices to give the point

15. Question 11c

    4 marksFoot of Perpendicular from a Point to a Line
    Step 1: General point on $m$: $P = (1+\lambda, 5+\lambda, 3+2\lambda)$. Then $\overrightarrow{CP} = P - C = (\lambda, 8 + \lambda, 2 + 2\lambda)$. Step 2: Perpendicularity: $\overrightarrow{CP} \cdot (1, 1, 2) = 0$. Step 3: $\lambda \cdot 1 + (8 + \lambda) \cdot 1 + (2 + 2\lambda) \cdot 2 = \lambda + 8 + \lambda + 4 + 4\lambda = 6\lambda + 12 = 0 \Rightarrow \lambda = -2$. Step 4: Substitute $\lambda = -2$: $P = (1 - 2, 5 - 2, 3 - 4) = (-1, 3, -1)$.
    Method:

    Parameterise $P$ on the line, write $\overrightarrow{CP}$, set the scalar product with the line's direction to zero, solve for $\lambda$, and substitute.
    Examiner tips

    • Distinguish $\overrightarrow{CP}$ from $\overrightarrow{OP}$
    • Substitute $\lambda$ back into the line equation, not into $\overrightarrow{CP}$