May/June 2025 Paper 33 Worked Answers (A-Level Maths 9709 A2)

16 questions · 75 marks · 110 minutes

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1. Question 1b

   3 marksModulus Inequalities with a Parameter
   Step 1: $|3x - 2a| < x + 5a$ requires $x + 5a > 0$ (i.e. $x > -5a$); within this region, the inequality is equivalent to $-(x + 5a) < 3x - 2a < x + 5a$. Step 2: Right inequality: $3x - 2a < x + 5a \Rightarrow 2x < 7a \Rightarrow x < \dfrac{7a}{2}$. Step 3: Left inequality: $-(x + 5a) < 3x - 2a \Rightarrow -x - 5a < 3x - 2a \Rightarrow -3a < 4x \Rightarrow x > -\dfrac{3a}{4}$. Step 4: Combine: $-\dfrac{3a}{4} < x < \dfrac{7a}{2}$.
   Method:

   Convert the modulus inequality to a two-sided form, solve each linear inequality for $x$ in terms of $a$, and intersect.
   Examiner tips

   • $|f(x)| < g(x) \iff -g(x) < f(x) < g(x)$ when $g(x) > 0$
   • When parameters appear, keep $a$ symbolic until the end

2. Question 2

   4 marksLogarithm Equation
   Step 1: Use $2 \ln u = \ln u^2$ and combine: $\ln \dfrac{(2x+3)^2}{2x+5} = \ln(3x)$. Step 2: Equate arguments: $(2x+3)^2 = 3x(2x+5)$. Step 3: Expand: $4x^2 + 12x + 9 = 6x^2 + 15x$, so $2x^2 + 3x - 9 = 0$. Step 4: Factorise: $(2x - 3)(x + 3) = 0$, so $x = \dfrac{3}{2}$ or $x = -3$. Step 5: Check domain: $\ln(3x)$ needs $x > 0$, so reject $x = -3$. Final answer: $x = \dfrac{3}{2}$.
   Method:

   Apply log laws to form a single log on each side, equate the arguments, solve the resulting quadratic, and reject solutions outside the log domain.
   Examiner tips

   • Always check the domain of each logarithm and reject extraneous solutions
   • Use $a \ln b = \ln b^a$ before combining

3. Question 3

   4 marksDefinite Integral Using Double Angle Formula
   Step 1: Use the identity $\cos^2 5x = \dfrac{1 + \cos 10x}{2}$, so $3 \cos^2 5x = \dfrac{3}{2} + \dfrac{3}{2} \cos 10x$. Step 2: Antidifferentiate: $\displaystyle \int \left(\dfrac{3}{2} + \dfrac{3}{2} \cos 10x\right) dx = \dfrac{3}{2} x + \dfrac{3}{20} \sin 10x$. Step 3: Apply the upper limit $x = \pi/4$: $\dfrac{3}{2} \cdot \dfrac{\pi}{4} + \dfrac{3}{20} \sin \dfrac{10\pi}{4} = \dfrac{3\pi}{8} + \dfrac{3}{20} \sin \dfrac{5\pi}{2} = \dfrac{3\pi}{8} + \dfrac{3}{20}$ (since $\sin(5\pi/2) = 1$). Step 4: Apply the lower limit $x = \pi/5$: $\dfrac{3}{2} \cdot \dfrac{\pi}{5} + \dfrac{3}{20} \sin 2\pi = \dfrac{3\pi}{10} + 0$. Step 5: Subtract: $\left(\dfrac{3\pi}{8} - \dfrac{3\pi}{10}\right) + \dfrac{3}{20} = \dfrac{15\pi - 12\pi}{40} + \dfrac{3}{20} = \dfrac{3\pi}{40} + \dfrac{3}{20}$.
   Method:

   Apply the double-angle identity, integrate term-by-term, and substitute the limits separately before subtracting.
   Examiner tips

   • $\cos^2 \theta$ and $\sin^2 \theta$ usually call for the double-angle identity in integration
   • Work each limit separately to avoid confusing $\sin$ values

4. Question 4b

   3 marksQuadratic Equation with Complex Roots
   Step 1: Since the coefficients are real, complex roots come in conjugate pairs. The other root is $\bar{z} = 3 e^{-i\pi/4}$. Step 2: For a monic quadratic with roots $z, \bar{z}$: $b = -(z + \bar{z})$ and $c = z \bar{z}$. Step 3: $z + \bar{z} = 3 e^{i\pi/4} + 3 e^{-i\pi/4} = 6 \cos(\pi/4) = 6 \cdot \dfrac{\sqrt{2}}{2} = 3\sqrt{2}$. So $b = -3\sqrt{2}$. Step 4: $z \bar{z} = 3 \cdot 3 \cdot e^{i\pi/4 - i\pi/4} = 9$. So $c = 9$.
   Method:

   Identify the conjugate as the other root, then use Vieta's formulas: $b = -(z + \bar z)$ and $c = z \bar z$.
   Examiner tips

   • $z + \bar{z} = 2 \operatorname{Re}(z)$ and $z \bar{z} = |z|^2$
   • Real coefficients $\Rightarrow$ conjugate pairs

5. Question 5b

   2 marksImplicit Differentiation Gradients
   Step 1: At $x = 0$, the curve equation becomes $0 \cdot y + y^2 \cdot 1 = 4$, so $y^2 = 4$, giving $y = 2$ or $y = -2$. Step 2: At $x = 0$, the gradient simplifies: $\dfrac{dy}{dx} = \dfrac{y^2 - y}{2y} = \dfrac{y(y-1)}{2y} = \dfrac{y - 1}{2}$. Step 3: At $y = 2$: $\dfrac{2 - 1}{2} = \dfrac{1}{2}$. Step 4: At $y = -2$: $\dfrac{-2 - 1}{2} = -\dfrac{3}{2}$.
   Method:

   Determine the $y$-values at $x = 0$ from the curve, then evaluate the given gradient formula at each point.
   Examiner tips

   • Always find ALL points on the curve at the given $x$-value
   • Simplify the gradient formula before substituting

6. Question 6

   6 marksComplex Numbers from Real and Modulus Conditions
   Step 1: Write $z = x + iy$. Then $\dfrac{z + 4}{z + 4i} = \dfrac{(x+4) + iy}{x + i(y+4)}$. Multiply numerator and denominator by $x - i(y+4)$. Step 2: Numerator becomes $(x+4)x + y(y+4) + i\big[xy - (y+4)(x+4)\big]$. Step 3: For the quotient to be real, the imaginary part vanishes: $xy - (y+4)(x+4) = 0 \Rightarrow xy - xy - 4x - 4y - 16 = 0 \Rightarrow x + y = -4$. Step 4: Use $|z|^2 = x^2 + y^2 = 10$. With $y = -x - 4$: $x^2 + (x+4)^2 = 10 \Rightarrow 2x^2 + 8x + 6 = 0 \Rightarrow x^2 + 4x + 3 = 0 \Rightarrow (x+1)(x+3) = 0$, so $x = -1$ or $x = -3$. Step 5: Corresponding $y$: at $x = -1$, $y = -3$; at $x = -3$, $y = -1$. So $z = -1 - 3i$ or $z = -3 - i$.
   Method:

   Set $z = x + iy$, rationalise the quotient, set the imaginary part of the numerator to zero, and combine with the modulus condition to solve.
   Examiner tips

   • Realness of a quotient $\Rightarrow$ imaginary part of numerator (after rationalising) is zero
   • Always combine with the modulus condition to get a unique pair of solutions

7. Question 7a

   3 marksPartial Fractions
   Step 1: Set up $\dfrac{3a - 5x}{(3a + 2x)(2a - x)} = \dfrac{A}{3a + 2x} + \dfrac{B}{2a - x}$. Step 2: Multiply through: $3a - 5x = A(2a - x) + B(3a + 2x)$. Step 3: Substitute $x = 2a$: $3a - 10a = B(3a + 4a) \Rightarrow -7a = 7aB \Rightarrow B = -1$. Step 4: Substitute $x = -\dfrac{3a}{2}$: $3a + \dfrac{15a}{2} = A\left(2a + \dfrac{3a}{2}\right) \Rightarrow \dfrac{21a}{2} = \dfrac{7a}{2} A \Rightarrow A = 3$. Step 5: Therefore $f(x) = \dfrac{3}{3a + 2x} - \dfrac{1}{2a - x}$.
   Method:

   Set up partial fraction form, multiply through, substitute the roots of each linear factor to find $A$ and $B$.
   Examiner tips

   • Cover-up rule is fastest for distinct linear factors
   • Verify by reconstructing the numerator

8. Question 7b

   4 marksBinomial Expansion via Partial Fractions
   Step 1: $\dfrac{3}{3a + 2x} = \dfrac{3}{3a} \cdot \dfrac{1}{1 + \frac{2x}{3a}} = \dfrac{1}{a} \left(1 - \dfrac{2x}{3a} + \dfrac{4x^2}{9a^2} + \cdots\right) = \dfrac{1}{a} - \dfrac{2x}{3a^2} + \dfrac{4x^2}{9a^3}$. Step 2: $-\dfrac{1}{2a - x} = -\dfrac{1}{2a} \cdot \dfrac{1}{1 - \frac{x}{2a}} = -\dfrac{1}{2a}\left(1 + \dfrac{x}{2a} + \dfrac{x^2}{4a^2} + \cdots\right) = -\dfrac{1}{2a} - \dfrac{x}{4a^2} - \dfrac{x^2}{8a^3}$. Step 3: Constant: $\dfrac{1}{a} - \dfrac{1}{2a} = \dfrac{1}{2a}$. Step 4: $x$-coefficient: $-\dfrac{2}{3a^2} - \dfrac{1}{4a^2} = -\dfrac{8 + 3}{12 a^2} = -\dfrac{11}{12 a^2}$. Step 5: $x^2$-coefficient: $\dfrac{4}{9 a^3} - \dfrac{1}{8 a^3} = \dfrac{32 - 9}{72 a^3} = \dfrac{23}{72 a^3}$.
   Method:

   Factor each denominator, expand each term up to $x^2$, then combine the constants, $x$-, and $x^2$-coefficients.
   Examiner tips

   • Always factor the constant out of each linear factor before applying $(1+u)^{-1}$
   • Check $x$ and $x^2$ coefficients separately

9. Question 7c

   1 marksValidity of a Binomial Expansion
   Step 1: $\dfrac{1}{1 + 2x/(3a)}$ is valid for $|2x/(3a)| < 1$, i.e. $|x| < \dfrac{3a}{2}$. Step 2: $\dfrac{1}{1 - x/(2a)}$ is valid for $|x/(2a)| < 1$, i.e. $|x| < 2a$. Step 3: Both expansions must converge simultaneously, so we need the tighter condition: $|x| < \dfrac{3a}{2}$ (since $\dfrac{3a}{2} < 2a$).
   Method:

   Identify the convergence region of each binomial expansion separately, then intersect them.
   Examiner tips

   • When two binomial expansions are added, take the smaller (tighter) convergence region
   • Always factor the constant out first to find the convergence parameter

10. Question 8b

    4 marksTrigonometric Equation Using Identity
    Step 1: Substitute the identity: $4 \cot 2x \operatorname{cosec} 2x = 5 \sec 2x$, i.e. $\dfrac{4 \cos 2x}{\sin^2 2x} = \dfrac{5}{\cos 2x}$. Step 2: Cross-multiply: $4 \cos^2 2x = 5 \sin^2 2x$, so $\tan^2 2x = \dfrac{4}{5}$. Step 3: $\tan 2x = \pm \dfrac{2}{\sqrt{5}} \approx \pm 0.8944$. For $0 < x < 90^\circ$, $0 < 2x < 180^\circ$. Step 4: $2x = \arctan(2/\sqrt 5) \approx 41.81^\circ$, so $x \approx 20.9^\circ$. Also $2x = 180^\circ - 41.81^\circ = 138.19^\circ$, giving $x \approx 69.1^\circ$.
    Method:

    Use the given identity to reduce both sides to expressions in $\sin 2x$ and $\cos 2x$, simplify to $\tan^2 2x =$ constant, and find both solutions in the range.
    Examiner tips

    • When the equation reduces to $\tan^2 \phi = k$, expect TWO solutions in any half-revolution interval
    • Convert all to a single trig function before solving

11. Question 9a

    2 marksVector Equation of a Line
    Step 1: Direction vector $\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (2-1, -1-3, 3-(-2)) = (1, -4, 5)$. Step 2: A vector equation is $\mathbf{r} = \overrightarrow{OB} + \lambda \overrightarrow{BC} = (\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}) + \lambda(\mathbf{i} - 4\mathbf{j} + 5\mathbf{k})$.
    Method:

    Compute $\overrightarrow{BC}$ as the difference of position vectors, then write $\mathbf{r} = \overrightarrow{OB} + \lambda \overrightarrow{BC}$.
    Examiner tips

    • Either $B$ or $C$ can be used as the position vector
    • $\overrightarrow{BC}$ and $-\overrightarrow{BC}$ are both valid direction vectors

12. Question 9b

    4 marksFoot of Perpendicular from a Point to a Line
    Step 1: A general point on $l$ is $P = (1 + \lambda, 3 - 4\lambda, -2 + 5\lambda)$. So $\overrightarrow{AP} = (\lambda, 1 - 4\lambda, -2 + 5\lambda)$. Step 2: Perpendicularity: $\overrightarrow{AP} \cdot (1, -4, 5) = 0$. Step 3: $\lambda \cdot 1 + (1 - 4\lambda)(-4) + (-2 + 5\lambda)(5) = 0 \Rightarrow \lambda - 4 + 16\lambda - 10 + 25\lambda = 0 \Rightarrow 42\lambda = 14 \Rightarrow \lambda = \dfrac{1}{3}$. Step 4: $\overrightarrow{OP} = \left(1 + \dfrac{1}{3}, 3 - \dfrac{4}{3}, -2 + \dfrac{5}{3}\right) = \left(\dfrac{4}{3}, \dfrac{5}{3}, -\dfrac{1}{3}\right)$.
    Method:

    Parameterise the foot of perpendicular on the line, set the scalar product of $\overrightarrow{AP}$ with the direction vector to zero, solve for $\lambda$, and substitute back.
    Examiner tips

    • Distinguish $\overrightarrow{AP}$ from $\overrightarrow{OP}$
    • Don't forget to substitute $\lambda$ back to find $\overrightarrow{OP}$

13. Question 9c

    2 marksReflection of a Point in a Line
    Step 1: $P$ is the midpoint of $A$ and its reflection $D$, so $\overrightarrow{OP} = \dfrac{\overrightarrow{OA} + \overrightarrow{OD}}{2}$, giving $\overrightarrow{OD} = 2\overrightarrow{OP} - \overrightarrow{OA}$. Step 2: Compute $2 \overrightarrow{OP} = \left(\dfrac{8}{3}, \dfrac{10}{3}, -\dfrac{2}{3}\right)$. Step 3: Subtract $\overrightarrow{OA} = (1, 2, 0)$: $\overrightarrow{OD} = \left(\dfrac{8}{3} - 1, \dfrac{10}{3} - 2, -\dfrac{2}{3}\right) = \left(\dfrac{5}{3}, \dfrac{4}{3}, -\dfrac{2}{3}\right)$.
    Method:

    Use the midpoint relation to express the reflection as $\overrightarrow{OD} = 2 \overrightarrow{OP} - \overrightarrow{OA}$.
    Examiner tips

    • $P$ is the midpoint of $A$ and the reflected point $D$
    • $\overrightarrow{OD} = 2\overrightarrow{OP} - \overrightarrow{OA}$ is the standard reflection formula

14. Question 10a

    8 marksSeparable Differential Equation by Integration by Parts
    Step 1: Use the double-angle identity $\sin 4y = 2 \sin 2y \cos 2y$, so $\dfrac{\sin 4y}{\sin 2y} = 2 \cos 2y$. Step 2: Separate variables: $2 \cos 2y \, dy = x \sin 3x \, dx$. Step 3: Integrate LHS: $\int 2 \cos 2y \, dy = \sin 2y$. Step 4: Integrate RHS by parts with $u = x, dv = \sin 3x \, dx$: $u = x, du = dx, v = -\dfrac{1}{3}\cos 3x$. So $\int x \sin 3x \, dx = -\dfrac{x}{3} \cos 3x + \dfrac{1}{3} \int \cos 3x \, dx = -\dfrac{x}{3} \cos 3x + \dfrac{1}{9} \sin 3x + C$. Step 5: Apply IC $y = \pi/12, x = \pi/2$: $\sin(\pi/6) = \dfrac{1}{2}$. RHS at $x=\pi/2$: $-\dfrac{\pi/2}{3} \cos(3\pi/2) + \dfrac{1}{9} \sin(3\pi/2) + C = 0 - \dfrac{1}{9} + C$. So $\dfrac{1}{2} = -\dfrac{1}{9} + C \Rightarrow C = \dfrac{11}{18}$.
    Method:

    Reduce $\sin 4y / \sin 2y$ to $2 \cos 2y$ using the double-angle identity, separate variables, integrate the RHS by parts, and apply the initial condition.
    Examiner tips

    • Use $\sin 4y = 2 \sin 2y \cos 2y$ to make the LHS easily separable
    • Always check signs carefully when integrating by parts

15. Question 10b

    2 marksParticular Solution Values from a Differential Equation
    Step 1: At $x = 0$: $\sin 2y = 0 + 0 + \dfrac{11}{18} \approx 0.6111$. Step 2: For $0 < y < \dfrac{\pi}{2}$, we have $0 < 2y < \pi$. Solutions of $\sin 2y = 0.6111$ in this range: $2y = \arcsin(0.6111) \approx 0.6587$ and $2y = \pi - 0.6587 \approx 2.483$. Step 3: $y \approx \dfrac{0.6587}{2} = 0.329$ (3 s.f.) and $y \approx \dfrac{2.483}{2} = 1.241 \approx 1.24$ (3 s.f.).
    Method:

    Substitute $x = 0$ to reduce to $\sin 2y = c$, find both solutions for $2y \in (0, \pi)$, then halve.
    Examiner tips

    • $\sin 2y = c$ in $(0, \pi)$ has exactly two solutions when $0 < c < 1$
    • Halve $2y$ to get $y$

16. Question 11d

    3 marksIterative Method to 4 Decimal Places
    Step 1: Starting from $x_0 = 0.9$: $x_1 = \tfrac{1}{2}(\pi - \tan^{-1}(3.6)) \approx \tfrac{1}{2}(3.14159 - 1.29920) \approx 0.920872$. Step 2: $x_2 = \tfrac{1}{2}(\pi - \tan^{-1}(3.683...)) \approx 0.917944$. Step 3: $x_3 \approx 0.918347$, $x_4 \approx 0.918292$, $x_5 \approx 0.918300$. Step 4: The sequence stabilises around $0.91830$, which to 4 d.p. is $0.9183$.
    Method:

    Apply the iteration in radians from $x_0 = 0.9$, retaining at least 6 d.p., until two successive values agree to 5 d.p.
    Examiner tips

    • Always work in radians for $\tan^{-1}$ unless told otherwise
    • Keep at least 6 d.p. through the iteration