May/June 2025 Paper 31 Worked Answers (A-Level Maths 9709 A2)

16 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 1b

   2 marksModulus Inequalities
   Step 1: Square both sides (valid since both are non-negative): $(2x-1)^2 < (x+3)^2$. Step 2: Expand: $4x^2 - 4x + 1 < x^2 + 6x + 9$, giving $3x^2 - 10x - 8 < 0$. Step 3: Factorise: $(3x + 2)(x - 4) < 0$. Critical values are $x = -\dfrac{2}{3}$ and $x = 4$. Step 4: The quadratic opens upwards, so it is negative between its roots: $-\dfrac{2}{3} < x < 4$.
   Method:

   Square both sides to eliminate moduli, factorise the resulting quadratic, then read off the interval where it is negative.
   Examiner tips

   • Squaring is valid because $|a| < |b| \iff a^2 < b^2$
   • For an upward-opening quadratic, $(x-r_1)(x-r_2) < 0$ on the interval $(r_1, r_2)$

2. Question 2

   3 marksLogarithm Manipulation
   Step 1: Rewrite using log laws: $3 \ln a = \ln a^3$ and $\dfrac{1}{2}\ln(b+4) = \ln \sqrt{b+4}$. Step 2: The equation becomes $\ln \dfrac{a^3}{\sqrt{b+4}} = 2$, so $\dfrac{a^3}{\sqrt{b+4}} = e^2$. Step 3: Rearrange: $\sqrt{b+4} = \dfrac{a^3}{e^2}$. Step 4: Square both sides: $b + 4 = \dfrac{a^6}{e^4}$, hence $b = \dfrac{a^6}{e^4} - 4$.
   Method:

   Use the power law of logs to rewrite each term, combine into a single log, exponentiate to remove the log, and square to clear the square root.
   Examiner tips

   • Convert all coefficients of logs into powers using $k \ln x = \ln x^k$
   • Remember to square (not just remove the square root) when isolating the variable

3. Question 3

   6 marksComplex Numbers from Real and Modulus Conditions
   Step 1: Multiply numerator and denominator by the conjugate of the denominator $(x-5) - iy$. The numerator becomes $x(x-5) + y(y+5) + i\big[(y+5)(x-5) - xy\big]$. Step 2: For the quotient to be real, the imaginary part must vanish: $(y+5)(x-5) - xy = 0 \Rightarrow xy + 5x - 5y - 25 - xy = 0 \Rightarrow x - y = 5$. Step 3: Combine with $|z|^2 = x^2 + y^2 = 17$: substitute $y = x - 5$ to get $x^2 + (x-5)^2 = 17$, i.e. $2x^2 - 10x + 25 = 17$, so $x^2 - 5x + 4 = 0$. Step 4: Factorise: $(x-1)(x-4) = 0$, giving $x = 1$ or $x = 4$. The corresponding values are $z = 1 - 4i$ or $z = 4 - i$.
   Method:

   Substitute $z = x + iy$, multiply by the conjugate of the denominator, set the imaginary part of the resulting expression to zero, then combine with $|z|^2 = 17$ to solve for $x$ and $y$.
   Examiner tips

   • When a quotient must be real, set its imaginary part to zero
   • Always double-check candidate solutions against both conditions

4. Question 4

   6 marksTangent to a Parametric Curve
   Step 1: At $(e, 3)$, since $x = e^{\tan t} = e$, we have $\tan t = 1$ (so $t = \pi/4$). Step 2: Differentiate: $\dfrac{dx}{dt} = \sec^2 t \cdot e^{\tan t}$. At $\tan t = 1$: $\sec^2 t = 1 + \tan^2 t = 2$, so $\dfrac{dx}{dt} = 2 \cdot e = 2e$. Step 3: $\dfrac{dy}{dt} = 6 \tan t \cdot \sec^2 t = 6 \cdot 1 \cdot 2 = 12$. Step 4: Gradient at the point: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{12}{2e} = \dfrac{6}{e}$. Step 5: Tangent line: $y - 3 = \dfrac{6}{e}(x - e)$, which simplifies to $y = \dfrac{6}{e} x - 6 + 3 = \dfrac{6}{e} x - 3$.
   Method:

   Determine $t$ from the given point, differentiate $x$ and $y$ with respect to $t$, divide to get $\dfrac{dy}{dx}$, then write the tangent in point-gradient form.
   Examiner tips

   • Use $\sec^2 t = 1 + \tan^2 t$ to evaluate without finding $t$ explicitly
   • $e^{\tan t}$ differentiates by chain rule: $\sec^2 t \cdot e^{\tan t}$

5. Question 5

   5 marksRemainder Theorem with Parameters
   Step 1: Apply the remainder theorem with $x = -2a$: $f(-2a) = 3(-2a)^3 + pa(-2a)^2 + 7a^2(-2a) + qa^3 = -24a^3 + 4pa^3 - 14a^3 + qa^3 = (-38 + 4p + q)a^3 = -22a^3$. So $4p + q = 16$. Step 2: Apply with $x = a/3$: $f(a/3) = 3(a/3)^3 + pa(a/3)^2 + 7a^2(a/3) + qa^3 = \dfrac{a^3}{9} + \dfrac{pa^3}{9} + \dfrac{7a^3}{3} + qa^3 = \left(\dfrac{1+p}{9} + \dfrac{7}{3} + q\right)a^3 = -a^3$. Step 3: Multiply through by $9$: $1 + p + 21 + 9q = -9$, giving $p + 9q = -31$. Step 4: Solve the system: from $4p + q = 16$, $q = 16 - 4p$. Substitute: $p + 9(16 - 4p) = -31 \Rightarrow -35p = -175 \Rightarrow p = 5$, then $q = 16 - 20 = -4$. So $p + q = 1$.
   Method:

   Apply the remainder theorem at the two roots of the divisors to obtain a pair of linear equations in $p$ and $q$, then solve simultaneously.
   Examiner tips

   • Substitute carefully — keep $a$ symbolic and divide by $a^3$ at the end
   • Check both equations after solving

6. Question 6a

   2 marksMultiplication of Complex Numbers in Polar Form
   Step 1: Multiplication in polar form multiplies moduli and adds arguments: $|wz| = |w| \cdot |z| = 2 \cdot 4 = 8$. Step 2: $\arg(wz) = \arg w + \arg z = \dfrac{\pi}{6} + \dfrac{\pi}{3} = \dfrac{\pi + 2\pi}{6} = \dfrac{\pi}{2}$. Step 3: Therefore $wz = 8 e^{i \pi/2}$, which is already in the principal range $-\pi < \theta \le \pi$.
   Method:

   Multiply moduli and add arguments to express the product in polar form.
   Examiner tips

   • Always check the result is in the principal range $-\pi < \theta \le \pi$
   • Common identity: $e^{i\alpha} \cdot e^{i\beta} = e^{i(\alpha+\beta)}$

7. Question 6c

   2 marksGeometric Effect of Complex Multiplication
   Step 1: Multiplication by a complex number $r e^{i\theta}$ in the Argand diagram has two geometric effects: enlargement (scale factor $r$) and rotation by angle $\theta$ (anticlockwise if positive). Step 2: For $\omega = 2 e^{i \pi/2}$, the modulus is $r = 2$ and the argument is $\theta = \dfrac{\pi}{2}$. Step 3: Hence multiplying by $\omega$ rotates the point about the origin through $\dfrac{\pi}{2}$ anticlockwise and enlarges its distance from the origin by a factor of $2$.
   Method:

   Read off the modulus as the enlargement factor and the argument as the (anticlockwise) rotation angle.
   Examiner tips

   • Multiplication by $i$ alone is rotation by $\pi/2$ anticlockwise
   • The modulus of the multiplier is the scale factor of the enlargement

8. Question 7a

   4 marksR-Formula (Harmonic Form)
   Step 1: Expand $5 \sin\left(x + \dfrac{\pi}{6}\right) = 5 \sin x \cos\dfrac{\pi}{6} + 5 \cos x \sin\dfrac{\pi}{6} = \dfrac{5\sqrt{3}}{2} \sin x + \dfrac{5}{2} \cos x$. Step 2: Subtracting $4 \cos x$: $\dfrac{5\sqrt{3}}{2} \sin x + \left(\dfrac{5}{2} - 4\right) \cos x = \dfrac{5\sqrt{3}}{2} \sin x - \dfrac{3}{2} \cos x$. Step 3: Match $R \sin(x - \alpha) = R \cos\alpha \sin x - R \sin\alpha \cos x$. So $R \cos\alpha = \dfrac{5\sqrt{3}}{2}$ and $R \sin\alpha = \dfrac{3}{2}$. Step 4: $R^2 = \dfrac{75}{4} + \dfrac{9}{4} = \dfrac{84}{4} = 21$, so $R = \sqrt{21}$. Step 5: $\tan\alpha = \dfrac{3/2}{5\sqrt{3}/2} = \dfrac{3}{5\sqrt{3}} = \dfrac{\sqrt{3}}{5} \approx 0.3464$, so $\alpha = \tan^{-1}(0.3464) \approx 0.333$ radians.
   Method:

   Expand the compound angle, collect $\sin x$ and $\cos x$ terms, equate to $R \cos\alpha$ and $R \sin\alpha$, then compute $R$ via Pythagoras and $\alpha$ via $\tan^{-1}$.
   Examiner tips

   • Always expand the compound angle first before matching coefficients
   • Use $\tan\alpha = \dfrac{\text{coefficient of }\cos x}{\text{coefficient of }\sin x}$ for the form $R \sin(x - \alpha)$

9. Question 7b

   4 marksSolving Trigonometric Equations Using R-Formula
   Step 1: Replace $x$ by $2\theta$ in the identity to get $\sqrt{21} \sin(2\theta - 0.333) = \sqrt{7}$, so $\sin(2\theta - 0.333) = \dfrac{\sqrt{7}}{\sqrt{21}} = \dfrac{1}{\sqrt{3}}$. Step 2: Let $\phi = 2\theta - 0.333$. The reference angle is $\sin^{-1}(1/\sqrt{3}) \approx 0.6155$. Step 3: For $0 \le \theta \le \pi$, we need $\phi \in [-0.333, 2\pi - 0.333]$. Solutions are $\phi = 0.6155$ and $\phi = \pi - 0.6155 = 2.526$. Step 4: $\phi = 0.6155 \Rightarrow 2\theta = 0.948 \Rightarrow \theta = 0.47$ (2 d.p.). $\phi = 2.526 \Rightarrow 2\theta = 2.859 \Rightarrow \theta = 1.43$ (2 d.p.).
   Method:

   Apply the harmonic form to reduce the equation to a single sine, find all solutions for $\sin\phi = c$ in the appropriate $\phi$ range, then translate back to $\theta$.
   Examiner tips

   • Don't forget the supplementary angle $\pi - \phi$ when solving $\sin\phi = c$ in $[0, \pi]$
   • Always check the candidate $\theta$ values lie in the given range

10. Question 8a

    2 marksVector Equation of a Line Through Two Points
    Step 1: A line through points with position vectors $\mathbf{a}$ and $\mathbf{b}$ has direction vector $\mathbf{b} - \mathbf{a}$. Step 2: Compute $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (4-1)\mathbf{i} + (1-2)\mathbf{j} + (5-1)\mathbf{k} = 3\mathbf{i} - \mathbf{j} + 4\mathbf{k}$. Step 3: A vector equation is $\mathbf{r} = \mathbf{a} + \lambda \overrightarrow{AB} = (\mathbf{i} + 2\mathbf{j} + \mathbf{k}) + \lambda(3\mathbf{i} - \mathbf{j} + 4\mathbf{k})$.
    Method:

    Subtract position vectors to find the direction, then write the equation in $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$ form.
    Examiner tips

    • The position vector in the equation can be either $A$ or $B$ — both are valid points on the line
    • Direction vector is unique up to a scalar multiple

11. Question 8c

    3 marksAcute Angle Between Two Lines
    Step 1: Compute the scalar product: $\mathbf{d}_1 \cdot \mathbf{d}_2 = (3)(1) + (1)(2) + (2)(3) = 3 + 2 + 6 = 11$. Step 2: Compute the moduli: $|\mathbf{d}_1| = \sqrt{9 + 1 + 4} = \sqrt{14}$ and $|\mathbf{d}_2| = \sqrt{1 + 4 + 9} = \sqrt{14}$. Step 3: Use $\cos\theta = \dfrac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|} = \dfrac{11}{14} \approx 0.7857$. Step 4: $\theta = \cos^{-1}(0.7857) \approx 38.2^\circ$. Since this is already acute, no further adjustment is needed.
    Method:

    Compute the scalar product and the moduli, divide and take the inverse cosine; if the angle is obtuse, subtract from $180^\circ$ to get the acute angle.
    Examiner tips

    • If the dot product is negative, take its absolute value to get the acute angle
    • $|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$

12. Question 9c

    3 marksIterative Methods to Find a Root
    Step 1: Compute $a_1 = \exp\left(\dfrac{1}{6}\left(\dfrac{5}{4} + 3\right)\right) = \exp\left(\dfrac{1}{6} \cdot \dfrac{17}{4}\right) = \exp(0.7083) \approx 2.0306$. Step 2: $a_2 = \exp\left(\dfrac{1}{6}\left(\dfrac{5}{2.0306^2} + 3\right)\right) \approx \exp(0.7021) \approx 2.0180$. Step 3: $a_3 \approx 2.0231$, $a_4 \approx 2.0210$, $a_5 \approx 2.0218$. The sequence stabilises around $2.021$, which to 2 d.p. is $2.02$. Step 4: Verify with a sign-change argument: $f(2.015)$ and $f(2.025)$ have opposite signs in $f(x) = x - \exp((1/6)(5/x^2 + 3))$, confirming the root is in $(2.015, 2.025)$, i.e. $a = 2.02$ to 2 d.p.
    Method:

    Apply the iteration starting from $a_0 = 2$, retaining at least 4 d.p. each step until convergence to 3 d.p., then round to 2 d.p.
    Examiner tips

    • Keep at least 4 d.p. through the iteration to maintain accuracy
    • Stop when two successive values agree to 3 d.p.

13. Question 10a

    3 marksPolynomial Long Division
    Step 1: Divide $x^3$ by $x^2$ to get $x$. Multiply $x \cdot (x^2 - 3) = x^3 - 3x$ and subtract: $(x^3 + 5x^2 - 2x - 15) - (x^3 - 3x) = 5x^2 + x - 15$. Step 2: Divide $5x^2$ by $x^2$ to get $5$. Multiply $5 \cdot (x^2 - 3) = 5x^2 - 15$ and subtract: $(5x^2 + x - 15) - (5x^2 - 15) = x$. Step 3: Since $\deg(x) < \deg(x^2 - 3)$, the division stops. Quotient $= x + 5$, remainder $= x$. Step 4: Verification: $(x + 5)(x^2 - 3) + x = x^3 - 3x + 5x^2 - 15 + x = x^3 + 5x^2 - 2x - 15$. ✓
    Method:

    Apply standard polynomial long division, dividing leading terms, multiplying, subtracting, until the remainder has lower degree than the divisor.
    Examiner tips

    • Always verify by reconstructing: dividend = (quotient)(divisor) + remainder
    • Watch the sign of each subtraction step carefully

14. Question 10b

    5 marksSeparable Differential Equations
    Step 1: Separate variables: $6y \, dy = \dfrac{x^3 + 5x^2 - 2x - 15}{x^2 - 3} dx = \left(x + 5 + \dfrac{x}{x^2 - 3}\right) dx$ using the given decomposition. Step 2: Integrate both sides: $\int 6y \, dy = 3y^2$ and $\int (x + 5) dx = \dfrac{x^2}{2} + 5x$, $\int \dfrac{x}{x^2 - 3} dx = \dfrac{1}{2} \ln|x^2 - 3|$ (recognising $f'/f$ form). Step 3: Combine: $3y^2 = \dfrac{x^2}{2} + 5x + \dfrac{1}{2} \ln|x^2 - 3| + C$. Step 4: Apply initial condition $y = 2, x = 2$: $12 = 2 + 10 + \dfrac{1}{2}\ln 1 + C = 12 + C$, so $C = 0$. Step 5: Divide by 3: $y^2 = \dfrac{x^2}{6} + \dfrac{5x}{3} + \dfrac{1}{6} \ln|x^2 - 3|$.
    Method:

    Use the given decomposition to write $\dfrac{dy}{dx}$ as a sum of integrable terms, separate variables, integrate using known forms, and apply the initial condition.
    Examiner tips

    • Recognise $\int \dfrac{f'(x)}{f(x)} dx = \ln|f(x)| + C$ patterns to avoid lengthier substitution
    • Always check the initial condition gives a sensible value of $C$

15. Question 11b

    5 marksVolume of Revolution by Substitution
    Step 1: Volume of revolution: $V = \pi \int_0^{\pi/2} y^2 \, dx = \pi \int_0^{\pi/2} \cos^2 x \sin 2x \, dx$. Step 2: Use $\sin 2x = 2 \sin x \cos x$: integrand becomes $2 \cos^3 x \sin x$, so $V = 2\pi \int_0^{\pi/2} \cos^3 x \sin x \, dx$. Step 3: Substitute $u = \cos x$, $du = -\sin x \, dx$. When $x = 0$, $u = 1$; when $x = \pi/2$, $u = 0$. $V = 2\pi \int_1^0 u^3 (-du) = 2\pi \int_0^1 u^3 \, du$. Step 4: Evaluate: $V = 2\pi \left[\dfrac{u^4}{4}\right]_0^1 = 2\pi \cdot \dfrac{1}{4} = \dfrac{\pi}{2}$.
    Method:

    Set up $V = \pi \int y^2 \, dx$, use $\sin 2x = 2 \sin x \cos x$, substitute $u = \cos x$, and integrate.
    Examiner tips

    • $\sin 2x$ inside an integral with $\cos x$ powers usually invites the substitution $u = \cos x$
    • Reverse limits if the substitution flips them, or change the sign of $du$