May/June 2025 Paper 32 Worked Answers (A-Level Maths 9709 A2)

17 questions · 75 marks · 110 minutes

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1. Question 1

   5 marksExponential Equation Reducing to Quadratic
   Step 1: Multiply both sides by $(e^x - 3)$: $e^x + 2e^{-x} = 4e^x - 12$. Step 2: Multiply throughout by $e^x$ to remove the negative exponent: $e^{2x} + 2 = 4e^{2x} - 12 e^x$, hence $3e^{2x} - 12e^x - 2 = 0$. Step 3: Let $u = e^x$. Solve $3u^2 - 12u - 2 = 0$: $u = \dfrac{12 \pm \sqrt{144 + 24}}{6} = \dfrac{6 \pm \sqrt{42}}{3}$. Step 4: Reject the negative root (since $u = e^x > 0$). Take $u = \dfrac{6+\sqrt{42}}{3} \approx 4.1602$. Step 5: $x = \ln u \approx \ln(4.1602) \approx 1.426$ (3 d.p.).
   Method:

   Clear the denominator and multiply by $e^x$ to obtain a quadratic in $e^x$, solve it, reject the negative root, and take the natural log.
   Examiner tips

   • Always multiply by $e^x$ to convert $e^{-x}$ before forming a quadratic
   • Reject negative roots of the quadratic since $e^x > 0$

2. Question 2a

   4 marksBinomial Expansion of a Product
   Step 1: Use the binomial series $(1+u)^n = 1 + nu + \dfrac{n(n-1)}{2}u^2 + \dots$ with $n = -\dfrac{3}{2}$, $u = -2x$. Step 2: Linear term: $\left(-\dfrac{3}{2}\right)(-2x) = 3x$. Quadratic term: $\dfrac{(-3/2)(-5/2)}{2}(-2x)^2 = \dfrac{15/4}{2}\cdot 4x^2 = \dfrac{15}{2}x^2$. Step 3: Hence $(1-2x)^{-3/2} = 1 + 3x + \dfrac{15}{2}x^2 + \dots$. Step 4: Multiply by $(6-x)$, keeping terms up to $x^2$: $6\!\left(1 + 3x + \dfrac{15}{2}x^2\right) - x(1 + 3x) = 6 + 18x + 45x^2 - x - 3x^2$. Step 5: Simplify: $6 + 17x + 42x^2$.
   Method:

   Expand $(1-2x)^{-3/2}$ to three terms using the rational-index binomial series, then multiply by $(6-x)$ keeping only terms up to $x^2$.
   Examiner tips

   • Always expand $(1+u)^n$ first; substitute $u = -2x$ at the end
   • Track all $x^2$ contributions, including the cross term from $(-x) \cdot 3x$

3. Question 2b

   1 marksValidity of Binomial Expansion
   Step 1: The series $(1+u)^n$ for non-integer $n$ converges if and only if $|u| < 1$. Step 2: Here $u = -2x$, so the requirement is $|-2x| < 1$, i.e. $|x| < \dfrac{1}{2}$. Step 3: This is a strict inequality (the series does not converge at $|u| = 1$).
   Method:

   Apply the convergence condition $|u| < 1$ for the binomial series with $u = -2x$.
   Examiner tips

   • The validity condition is always $|u| < 1$ where $u$ is the bracket inside
   • It is a strict inequality

4. Question 4

   6 marksTrig Equation with Cot 2x Identity
   Step 1: Use the identity $\cot 2x = \dfrac{1-\tan^2 x}{2\tan x}$. The equation becomes $\dfrac{3}{\tan x} - \dfrac{4(1-\tan^2 x)}{2\tan x} = 3$, i.e. $\dfrac{3 - 2(1-\tan^2 x)}{\tan x} = 3$. Step 2: Simplify the numerator: $3 - 2 + 2\tan^2 x = 1 + 2\tan^2 x$, so $\dfrac{1 + 2\tan^2 x}{\tan x} = 3$. Step 3: Multiply by $\tan x$: $2\tan^2 x - 3\tan x + 1 = 0$. Step 4: Factorise: $(2\tan x - 1)(\tan x - 1) = 0$, so $\tan x = \dfrac{1}{2}$ or $\tan x = 1$. Step 5: In the interval $[0°, 180°]$: $\tan x = 1 \Rightarrow x = 45°$; $\tan x = \dfrac{1}{2} \Rightarrow x \approx 26.6°$. (The other branches lie outside the interval since $\tan$ is negative there.)
   Method:

   Substitute the double-angle identity for $\cot 2x$, simplify into a quadratic in $\tan x$, factorise, and read off the values in $[0°, 180°]$.
   Examiner tips

   • Convert all trig terms to $\tan x$ using a single identity
   • Both roots of the quadratic must be tested in the given interval

5. Question 5

   5 marksSquare Roots of a Complex Number
   Step 1: Let $(x + iy)^2 = -1 - 4\sqrt{5}\,i$. Expanding: $x^2 - y^2 + 2ixy = -1 - 4\sqrt{5}\,i$. Step 2: Equate real and imaginary parts: $x^2 - y^2 = -1$ and $2xy = -4\sqrt{5}$, hence $xy = -2\sqrt{5}$, so $y = -\dfrac{2\sqrt{5}}{x}$. Step 3: Substitute: $x^2 - \dfrac{20}{x^2} = -1$. Multiply by $x^2$: $x^4 + x^2 - 20 = 0$. Step 4: Factorise: $(x^2 - 4)(x^2 + 5) = 0$. Since $x$ is real, $x^2 = 4$, so $x = \pm 2$. Step 5: Then $y = -\dfrac{2\sqrt{5}}{x} = \mp \sqrt{5}$. Hence the two square roots are $2 - \sqrt{5}\,i$ and $-2 + \sqrt{5}\,i$, i.e. $\pm(2 - \sqrt{5}\,i)$. Verify: $(2 - \sqrt{5}\,i)^2 = 4 - 4\sqrt{5}\,i + 5 i^2 = -1 - 4\sqrt{5}\,i$. ✓
   Method:

   Set $(x+iy)^2$ equal to the given complex number, equate real and imaginary parts to form a quartic in $x$, solve for real $x$, then back-substitute to find $y$.
   Examiner tips

   • Always verify by squaring the candidate root
   • The two square roots of any non-zero complex number are negatives of each other

6. Question 6b

   2 marksLocating a Root by Sign Change
   Step 1: Define $f(x) = |x - 2| - 2 \sin\!\left(\tfrac{1}{2}x\right)$. A root of $|x-2| = 2 \sin(x/2)$ corresponds to $f(x) = 0$. Step 2: At $x = 1$: $|1-2| = 1$ and $2 \sin(0.5) \approx 2(0.4794) \approx 0.9589$. So $f(1) \approx 1 - 0.9589 \approx 0.0411 > 0$. Step 3: At $x = 1.5$: $|1.5-2| = 0.5$ and $2 \sin(0.75) \approx 2(0.6816) \approx 1.3633$. So $f(1.5) \approx 0.5 - 1.3633 \approx -0.8633 < 0$. Step 4: Since $f$ is continuous on $[1, 1.5]$ and changes sign, by the Intermediate Value Theorem there is a root in $(1, 1.5)$.
   Method:

   Form $f(x)$ as the difference of the two sides, evaluate at $x = 1$ and $x = 1.5$, observe the sign change, and conclude that a root lies in the interval.
   Examiner tips

   • Set the calculator to radians
   • State the conclusion in words: a sign change indicates a root in the interval

7. Question 6c

   3 marksIterative Method to 2 Decimal Places
   Step 1: With calculator in radians, compute $x_2 = 2 - 2 \sin(1.03/2) = 2 - 2 \sin(0.515) \approx 2 - 2(0.4926) \approx 1.0149$. Step 2: $x_3 = 2 - 2 \sin(1.0149/2) \approx 2 - 2(0.4860) \approx 1.0281$. Step 3: $x_4 \approx 1.0166$, $x_5 \approx 1.0266$, $x_6 \approx 1.0179$, $x_7 \approx 1.0255$, ... Step 4: The iterates oscillate around $1.02$, with sign change in $(1.015, 1.025)$. Hence the root to 2 d.p. is $1.02$.
   Method:

   Iterate using a calculator in radians, recording each iterate to 4 d.p., until the sequence has clearly converged to 2 d.p.
   Examiner tips

   • Set calculator to radians
   • Show intermediate iterates to 4 d.p. to justify the 2 d.p. answer

8. Question 7a

   3 marksR-Formula (Harmonic Form)
   Step 1: Match $R \cos(\theta - \alpha) = R \cos\alpha \cos\theta + R \sin\alpha \sin\theta$ with $7 \sin\theta + 24 \cos\theta$. So $R \cos\alpha = 24$ and $R \sin\alpha = 7$. Step 2: $R = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25$. Step 3: $\tan\alpha = \dfrac{R \sin\alpha}{R \cos\alpha} = \dfrac{7}{24}$. Step 4: $\alpha = \tan^{-1}(7/24) \approx 0.2838$ radians (4 d.p.).
   Method:

   Expand $R\cos(\theta-\alpha)$, match coefficients with the original expression, take Pythagorean sum for $R$, and take inverse tangent for $\alpha$.
   Examiner tips

   • $R^2$ is always the sum of squares of the two coefficients
   • Match terms: for $R\cos(\theta - \alpha)$, $\tan\alpha = (\text{coefficient of }\sin\theta)/(\text{coefficient of }\cos\theta)$

9. Question 7b

   4 marksSolving Equation Using Harmonic Form
   Step 1: Substitute $\theta = x/3$: $7 \sin(x/3) + 24 \cos(x/3) = 25 \cos(x/3 - 0.2838)$. The equation becomes $25 \cos(x/3 - 0.2838) = 24.5$, so $\cos(x/3 - 0.2838) = 0.98$. Step 2: Take inverse cosine: $x/3 - 0.2838 = \pm \cos^{-1}(0.98) \approx \pm 0.2003$ (principal values). Step 3: Solve: $x/3 = 0.2838 + 0.2003 = 0.4841$ or $x/3 = 0.2838 - 0.2003 = 0.0835$. Step 4: Multiply by 3: $x \approx 1.45$ or $x \approx 0.250$. Both lie in $0 < x < \pi$.
   Method:

   Rewrite using the harmonic form, isolate the cosine, take both branches of the inverse cosine, then multiply by 3 and select solutions in the given interval.
   Examiner tips

   • Use both $+\cos^{-1}$ and $-\cos^{-1}$ to capture all solutions
   • Multiply through by 3 at the end since the inner argument is $x/3$

10. Question 8

    7 marksSeparable Differential Equation
    Step 1: Separate variables: $\dfrac{dx}{4x+3} = \dfrac{\cos 2\theta}{\sin 2\theta} d\theta = \cot 2\theta \, d\theta$. Step 2: Integrate: $\dfrac{1}{4}\ln|4x+3| = \dfrac{1}{2}\ln|\sin 2\theta| + c$. Step 3: Apply initial condition $x = 0$, $\theta = \pi/12$ (so $2\theta = \pi/6$ and $\sin 2\theta = 1/2$): $\dfrac{1}{4}\ln 3 = \dfrac{1}{2}\ln(1/2) + c$, hence $c = \dfrac{1}{4}\ln 3 + \dfrac{1}{2}\ln 2$. Step 4: Multiply through by 4: $\ln(4x+3) = 2 \ln|\sin 2\theta| + \ln 3 + 2 \ln 2 = \ln(12 \sin^2 2\theta)$. Step 5: Exponentiate: $4x + 3 = 12 \sin^2 2\theta$, so $x = \dfrac{12 \sin^2 2\theta - 3}{4}$.
    Method:

    Separate the variables, integrate both sides (using $\int \cot 2\theta = \dfrac{1}{2}\ln|\sin 2\theta|$), apply the initial condition to find the constant, then exponentiate and solve for $x$.
    Examiner tips

    • Multiply your log equation by the LCM (here, 4) before exponentiating to avoid fractional exponents
    • Combine all log constants before exponentiating to avoid algebra slips

11. Question 9a

    2 marksVector Equation of a Line
    Step 1: A vector line equation uses one point on the line and a direction vector. Step 2: Use $A$ as the point: position vector $(1, -4, 2)$. Step 3: The direction is $\vec{AB} = \vec{OB} - \vec{OA} = (-2-1, 1-(-4), 3-2) = (-3, 5, 1)$. Step 4: So $\mathbf{r} = (1, -4, 2) + \lambda(-3, 5, 1)$. (Equivalent forms with different point or scaled direction are also valid lines.)
    Method:

    Use point $A$ and the direction vector $\vec{AB} = \vec{OB} - \vec{OA}$ to write the line in the form $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$.
    Examiner tips

    • Always write $\mathbf{r} = ...$, not $\lambda = ...$
    • Direction is the difference of position vectors

12. Question 9b

    4 marksCosine of Angle Using Scalar Product
    Step 1: Compute $\vec{AB} = \vec{OB} - \vec{OA} = (-3, 5, 1)$ and $\vec{AC} = \vec{OC} - \vec{OA} = (1, 7, 3)$. Step 2: Scalar product: $\vec{AB} \cdot \vec{AC} = (-3)(1) + (5)(7) + (1)(3) = -3 + 35 + 3 = 35$. Step 3: Magnitudes: $|\vec{AB}| = \sqrt{9+25+1} = \sqrt{35}$ and $|\vec{AC}| = \sqrt{1+49+9} = \sqrt{59}$. Step 4: Hence $\cos BAC = \dfrac{35}{\sqrt{35}\sqrt{59}} = \dfrac{35}{\sqrt{2065}} = \dfrac{\sqrt{35}}{\sqrt{59}} = \sqrt{\dfrac{35}{59}}$.
    Method:

    Calculate the scalar product of $\vec{AB}$ and $\vec{AC}$, divide by the product of their magnitudes, and simplify the surd.
    Examiner tips

    • $\cos\theta = \dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}$ — denominator is the product of moduli, not their squares
    • Simplify $\dfrac{35}{\sqrt{35}\sqrt{59}} = \sqrt{\dfrac{35}{59}}$ for an exact form

13. Question 9c

    3 marksExact Area of a Triangle from Vectors
    Step 1: Area of a triangle using two sides and included angle: $\text{Area} = \tfrac{1}{2}|\vec{AB}||\vec{AC}|\sin BAC$. Step 2: From $\cos^2 BAC = \dfrac{35}{59}$: $\sin^2 BAC = 1 - \dfrac{35}{59} = \dfrac{24}{59}$, so $\sin BAC = \sqrt{\dfrac{24}{59}}$ (positive in a triangle). Step 3: Substitute: $\text{Area} = \tfrac{1}{2} \sqrt{35} \cdot \sqrt{59} \cdot \sqrt{\dfrac{24}{59}} = \tfrac{1}{2} \sqrt{35 \cdot 24}$ (the $59$ cancels). Step 4: Simplify: $\sqrt{35 \cdot 24} = \sqrt{840} = \sqrt{4 \cdot 210} = 2\sqrt{210}$. Hence Area $= \tfrac{1}{2} \cdot 2\sqrt{210} = \sqrt{210}$.
    Method:

    Apply Area $= \tfrac{1}{2}|\vec{AB}||\vec{AC}|\sin BAC$, find $\sin$ from $\cos$ via the Pythagorean identity, and simplify the resulting surd.
    Examiner tips

    • Convert $\cos$ to $\sin$ exactly using $\sin^2 + \cos^2 = 1$, then simplify the surd
    • Watch out for fortuitous cancellations like the $\sqrt{59}$ in this problem

14. Question 10a

    2 marksPolynomial Division
    Step 1: Write $x^2 = q(1 + 4x^2) + r$ where $q$ is constant (the quotient) and $r$ is constant (the remainder, of degree less than the divisor's leading $x^2$ term). Step 2: Compare $x^2$ coefficients: $1 = 4q$, so $q = \dfrac{1}{4}$. Step 3: Compare constants: $0 = q + r = \dfrac{1}{4} + r$, so $r = -\dfrac{1}{4}$. Step 4: Verify: $\dfrac{1}{4}(1 + 4x^2) - \dfrac{1}{4} = \dfrac{1}{4} + x^2 - \dfrac{1}{4} = x^2$. ✓
    Method:

    Write the division as an identity, compare coefficients of like powers of $x$ to find both the quotient and remainder.
    Examiner tips

    • When the dividend has the same degree as the divisor, the quotient is a non-zero constant
    • Always verify by multiplying back

15. Question 10b

    6 marksDefinite Integral by Parts with Inverse Tan
    Step 1: Apply integration by parts with $u = \tan^{-1}(2x)$ ($du = \dfrac{2}{1+4x^2}dx$) and $dv = x\,dx$ ($v = \dfrac{x^2}{2}$): $\int x \tan^{-1}(2x)\,dx = \dfrac{x^2}{2}\tan^{-1}(2x) - \int \dfrac{x^2}{1+4x^2}dx$. Step 2: Use $\dfrac{x^2}{1+4x^2} = \dfrac{1}{4} - \dfrac{1/4}{1+4x^2}$ (from polynomial division). Integrate: $\int \dfrac{x^2}{1+4x^2}dx = \dfrac{x}{4} - \dfrac{1}{8}\tan^{-1}(2x)$. Step 3: Combine: antiderivative $F(x) = \dfrac{x^2}{2}\tan^{-1}(2x) - \dfrac{x}{4} + \dfrac{1}{8}\tan^{-1}(2x)$. Step 4: Evaluate $F(0.5)$: $\dfrac{0.25}{2}\tan^{-1}(1) - \dfrac{0.5}{4} + \dfrac{1}{8}\tan^{-1}(1) = \dfrac{1}{8} \cdot \dfrac{\pi}{4} - \dfrac{1}{8} + \dfrac{1}{8} \cdot \dfrac{\pi}{4} = \dfrac{\pi}{32} - \dfrac{1}{8} + \dfrac{\pi}{32} = \dfrac{\pi}{16} - \dfrac{1}{8}$. Step 5: $F(0) = 0$. So the definite integral $= \dfrac{\pi}{16} - \dfrac{1}{8}$.
    Method:

    Use integration by parts with $u = \tan^{-1}(2x)$, then integrate $\dfrac{x^2}{1+4x^2}$ using the quotient/remainder from (a), assemble the antiderivative, and substitute the limits.
    Examiner tips

    • Choosing $u = \tan^{-1}(2x)$ is essential — letting $dv = \tan^{-1}(2x) dx$ is much harder
    • Use polynomial division to reduce $\dfrac{x^2}{1+4x^2}$ to a polynomial plus a fraction with a simple antiderivative

16. Question 11a

    6 marksMaximum of a Trigonometric Curve
    Step 1: Use the product rule on $y = 5 \sin 2x \cdot \cos^2 x$ with $\dfrac{d}{dx}(\cos^2 x) = -2 \sin x \cos x = -\sin 2x$: $\dfrac{dy}{dx} = 5(2\cos 2x) \cos^2 x + 5 \sin 2x \cdot (-\sin 2x) \cdot ...$. Better: rewrite $y = 5(2\sin x \cos x)\cos^2 x = 10 \sin x \cos^3 x$. Step 2: Differentiate: $\dfrac{dy}{dx} = 10 \cos x \cdot \cos^3 x + 10 \sin x \cdot 3 \cos^2 x (-\sin x) = 10 \cos^4 x - 30 \sin^2 x \cos^2 x$. Step 3: Set $\dfrac{dy}{dx} = 0$: $10 \cos^2 x (\cos^2 x - 3 \sin^2 x) = 0$. With $0 < x < \pi/2$, $\cos x \ne 0$, so $\cos^2 x = 3 \sin^2 x$. Step 4: Hence $\dfrac{\sin^2 x}{\cos^2 x} = \dfrac{1}{3}$, i.e. $\tan^2 x = \dfrac{1}{3}$, so $\tan x = \dfrac{1}{\sqrt{3}}$ (positive in $(0, \pi/2)$). Step 5: Therefore $x = \dfrac{\pi}{6}$. (Confirm $M$ by checking it is a maximum, not a minimum.)
    Method:

    Rewrite $y$ using the double-angle identity, differentiate, set the derivative equal to zero, and reduce to a single trig equation.
    Examiner tips

    • Rewriting $5 \sin 2x \cos^2 x = 10 \sin x \cos^3 x$ often simplifies the differentiation
    • Always verify the stationary point is a maximum (not minimum or boundary)

17. Question 11b

    5 marksIntegration by Substitution (Trig)
    Step 1: Rewrite using double-angle: $5 \sin 2x \cos^2 x = 5 \cdot 2 \sin x \cos x \cdot \cos^2 x = 10 \sin x \cos^3 x$. Step 2: Substitute $u = \cos x$, so $du = -\sin x \, dx$. The integral becomes $\displaystyle\int 10 \sin x \cos^3 x \, dx = -10 \int u^3 \, du$. Step 3: Update limits: $x = 0 \Rightarrow u = 1$ and $x = \pi/4 \Rightarrow u = \cos(\pi/4) = \dfrac{1}{\sqrt{2}}$. Step 4: Evaluate: $-10 \left[\dfrac{u^4}{4}\right]_{1}^{1/\sqrt{2}} = -10 \left(\dfrac{(1/\sqrt{2})^4}{4} - \dfrac{1}{4}\right) = -10 \left(\dfrac{1/4}{4} - \dfrac{1}{4}\right) = -10 \left(\dfrac{1}{16} - \dfrac{1}{4}\right)$. Step 5: Simplify: $-10 \cdot \left(-\dfrac{3}{16}\right) = \dfrac{30}{16} = \dfrac{15}{8}$.
    Method:

    Rewrite the integrand using the double-angle identity, substitute $u = \cos x$, change the limits, integrate the polynomial in $u$, and evaluate.
    Examiner tips

    • Always change limits when substituting, rather than reverting to $x$ at the end
    • $\cos^4(\pi/4) = (1/\sqrt{2})^4 = 1/4$ — keep this exact