9709/33 Oct/Nov 2025 Worked Answers (A-Level Maths 9709 A2)

14 questions · 75 marks · 110 minutes

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1. Question 1

   4 marksModulus Inequality (Two Solution Regions)
   Step 1: Square both sides: $(3x + 2)^2 < 9(2x - 1)^2$, valid because both are non-negative. Step 2: Expand: $9x^2 + 12x + 4 < 9(4x^2 - 4x + 1) = 36x^2 - 36x + 9$, giving $27x^2 - 48x + 5 > 0$. Step 3: Factorise: $(9x - 1)(3x - 5) > 0$. Critical values $x = \dfrac{1}{9}$ and $x = \dfrac{5}{3}$. Step 4: Sign chart: the upward-opening quadratic is positive outside its roots. Hence $x < \dfrac{1}{9}$ or $x > \dfrac{5}{3}$.
   Method:

   Square both sides, form the quadratic inequality, factorise, and select the correct sign regions.
   Examiner tips

   • $|f(x)| < |g(x)| \iff f(x)^2 < g(x)^2$
   • Sketch a sign chart for the factorised quadratic

2. Question 2

   3 marksPolynomial Long Division
   Step 1: Use the remainder theorem: remainder $= f(-1) = 3(-1)^4 - 2(-1)^2 = 3 - 2 = 1$. Step 2: Long division: $3x^4 / x = 3x^3$. Multiply: $3x^3(x+1) = 3x^4 + 3x^3$. Subtract: $-3x^3 - 2x^2$. Step 3: $-3x^3/x = -3x^2$. Multiply: $-3x^3 - 3x^2$. Subtract: $x^2$. Step 4: $x^2/x = x$. Multiply: $x^2 + x$. Subtract: $-x$. Step 5: $-x/x = -1$. Multiply: $-x - 1$. Subtract: $1$. So quotient $= 3x^3 - 3x^2 + x - 1$, remainder $= 1$.
   Method:

   Apply the remainder theorem to find the constant, then long-divide systematically.
   Examiner tips

   • Use $f(-1) =$ remainder as a quick check
   • Treat missing terms with coefficient $0$

3. Question 3

   4 marksExponential Equation as Log Quotient
   Step 1: Rearrange: $2^{3x - 4} \cdot 5^x = 3$, so $\dfrac{2^{3x}}{2^4} \cdot 5^x = 3$, giving $\dfrac{(2^3)^x \cdot 5^x}{16} = 3$, i.e. $(2^3 \cdot 5)^x = 3 \cdot 16 = 48$. Step 2: Simplify: $40^x = 48$. Step 3: Take natural logs: $x \ln 40 = \ln 48$, so $x = \dfrac{\ln 48}{\ln 40}$.
   Method:

   Combine all base-2 and base-5 terms into $40^x$, then take logs to express $x$ as a quotient.
   Examiner tips

   • Combine bases as $(2^3 \cdot 5)^x = 40^x$ to simplify
   • $\log_a b = \dfrac{\ln b}{\ln a}$

4. Question 5b

   4 marksSolving a Trig Equation Using Identity
   Step 1: Substitute the identity: $8 \sin^4 x - 6 \sin^2 x = 0$, factor $2 \sin^2 x (4 \sin^2 x - 3) = 0$. Step 2: Either $\sin x = 0$ or $\sin^2 x = \dfrac{3}{4}$, i.e. $\sin x = \pm \dfrac{\sqrt{3}}{2}$. Step 3: $\sin x = 0$ in $[-180^\circ, 180^\circ]$: $x = -180^\circ, 0^\circ, 180^\circ$. Step 4: $\sin x = \dfrac{\sqrt 3}{2}$: $x = 60^\circ, 120^\circ$. $\sin x = -\dfrac{\sqrt 3}{2}$: $x = -60^\circ, -120^\circ$. Step 5: All seven solutions: $\pm 180^\circ, \pm 120^\circ, \pm 60^\circ, 0^\circ$.
   Method:

   Use the identity to reduce, factor out $\sin^2 x$, solve both cases, and list all solutions in the interval.
   Examiner tips

   • Never divide by $\sin^2 x$ — you lose the $\sin x = 0$ solutions
   • Always list ALL solutions in the given range

5. Question 6

   6 marksDefinite Integral by Parts (Twice)
   Step 1: Apply integration by parts with $u = x^2$, $dv = \sin 2x \, dx$: $du = 2x\, dx$, $v = -\dfrac{1}{2} \cos 2x$. So $\int x^2 \sin 2x \, dx = -\dfrac{x^2}{2} \cos 2x + \int x \cos 2x \, dx$. Step 2: Apply parts again to $\int x \cos 2x\, dx$ with $u = x$, $dv = \cos 2x\, dx$: $du = dx$, $v = \dfrac{1}{2} \sin 2x$. $\int x \cos 2x\, dx = \dfrac{x}{2} \sin 2x - \dfrac{1}{2} \int \sin 2x \, dx = \dfrac{x}{2} \sin 2x + \dfrac{1}{4} \cos 2x$. Step 3: Antiderivative: $-\dfrac{x^2}{2} \cos 2x + \dfrac{x}{2} \sin 2x + \dfrac{1}{4} \cos 2x$. Step 4: At $x = \pi/6$: $\cos(\pi/3) = 1/2$, $\sin(\pi/3) = \sqrt 3 /2$. Value: $-\dfrac{\pi^2/36}{2} \cdot \dfrac{1}{2} + \dfrac{\pi/6}{2} \cdot \dfrac{\sqrt 3}{2} + \dfrac{1}{4} \cdot \dfrac{1}{2} = -\dfrac{\pi^2}{144} + \dfrac{\pi \sqrt 3}{24} + \dfrac{1}{8}$. Step 5: At $x = 0$: $0 + 0 + \dfrac{1}{4}$. Subtract: $-\dfrac{\pi^2}{144} + \dfrac{\pi \sqrt 3}{24} + \dfrac{1}{8} - \dfrac{1}{4} = -\dfrac{\pi^2}{144} + \dfrac{\pi \sqrt 3}{24} - \dfrac{1}{8}$.
   Method:

   Apply integration by parts twice to get the antiderivative, then evaluate at the limits using exact values $\cos \pi/3 = 1/2$ and $\sin \pi/3 = \sqrt 3/2$.
   Examiner tips

   • Track signs carefully when integrating $\sin 2x$ (gives $-\dfrac{1}{2}\cos 2x$)
   • After applying integration by parts twice, evaluate ALL three boundary terms

6. Question 7

   6 marksComplex Equation in z and z*
   Step 1: Multiply through by $(2 - i)$: $5z - (2-i)(x^2 + y^2) + (20 + 8i)(2 - i) = 0$. Step 2: $(20+8i)(2-i) = 40 - 20i + 16i - 8i^2 = 40 + 8 - 4i = 48 - 4i$. Step 3: Substitute $z = x+iy$: $5(x+iy) - (2-i)(x^2+y^2) + 48 - 4i = 0$. Expand $(2-i)(x^2+y^2) = 2(x^2+y^2) - i(x^2+y^2)$. Step 4: Real part: $5x - 2(x^2+y^2) + 48 = 0$. Imaginary part: $5y + (x^2+y^2) - 4 = 0$. Step 5: From imaginary: $x^2 + y^2 = 4 - 5y$. Substitute into real: $5x - 2(4 - 5y) + 48 = 0 \Rightarrow 5x + 10y + 40 = 0 \Rightarrow x + 2y + 8 = 0$, so $x = -2y - 8$. Step 6: Substitute back: $(-2y-8)^2 + y^2 = 4 - 5y \Rightarrow 5y^2 + 37y + 60 = 0 \Rightarrow (y + 5)(5y + 12) = 0$, giving $y = -5$ or $y = -\dfrac{12}{5}$. So $z = 2 - 5i$ or $z = -\dfrac{16}{5} - \dfrac{12}{5} i$.
   Method:

   Clear the fraction, expand, equate real and imaginary parts to zero, eliminate $|z|^2$ to obtain a linear equation, and solve.
   Examiner tips

   • $z z^*$ is always real (equals $|z|^2$)
   • Eliminating $x^2 + y^2$ between the two equations gives a linear constraint

7. Question 8d

   3 marksIterative Method for a Logarithmic Root
   Step 1: Compute $p_1 = \dfrac{1}{5} \exp\left(\dfrac{1}{2}\right) = \dfrac{e^{0.5}}{5} \approx \dfrac{1.6487}{5} \approx 0.3297$. Step 2: $p_2 = \dfrac{1}{5} \exp\left(\dfrac{1}{5 \cdot 0.3297}\right) = \dfrac{1}{5} \exp(0.6066) \approx 0.3668$. Step 3: $p_3 \approx 0.3450$, $p_4 \approx 0.3571$, $p_5 \approx 0.3502$, $p_6 \approx 0.3541$. Step 4: The sequence oscillates and stabilises around $0.354$, which to 2 d.p. is $0.35$.
   Method:

   Apply the iteration repeatedly, retaining at least 4 d.p., until two successive values agree to 3 d.p.
   Examiner tips

   • Oscillating sequences require more iterations to confirm convergence
   • Keep at least 4 d.p. to check convergence to 2 d.p.

8. Question 9a

   3 marksVector Equations of Two Lines
   Step 1: $l_1$ has position $(3, 1, -6)$ and direction $(2, 1, 4)$, so $\mathbf{r} = (3, 1, -6) + \lambda(2, 1, 4)$. Step 2: For $l_2$, the direction vector has no $x$-component: $(0, b, c)$. It must be perpendicular to $(3, -2, 1)$, so $(0)(3) + b(-2) + c(1) = -2b + c = 0$, giving $c = 2b$. Step 3: Take $b = 1$, $c = 2$: direction $(0, 1, 2)$. So $l_2: \mathbf{r} = (-1, 3, -6) + \mu(0, 1, 2)$.
   Method:

   Write $l_1$ directly, then for $l_2$ set the $x$-component to zero and use the scalar-product condition to find the remaining components.
   Examiner tips

   • $\mathbf{a} \cdot \mathbf{b} = 0 \iff \mathbf{a} \perp \mathbf{b}$
   • Combine constraints sequentially

9. Question 9b

   3 marksAcute Angle Between Two Lines
   Step 1: Scalar product: $(2)(0) + (1)(1) + (4)(2) = 0 + 1 + 8 = 9$. Step 2: Moduli: $|\mathbf{d}_1| = \sqrt{4 + 1 + 16} = \sqrt{21}$, $|\mathbf{d}_2| = \sqrt{0 + 1 + 4} = \sqrt{5}$. Step 3: $\cos\theta = \dfrac{9}{\sqrt{21} \cdot \sqrt{5}} = \dfrac{9}{\sqrt{105}} \approx \dfrac{9}{10.247} \approx 0.8783$. Step 4: $\theta = \cos^{-1}(0.8783) \approx 28.6^\circ$ (or $0.498$ rad). This is already acute.
   Method:

   Compute scalar product, moduli, divide, take inverse cosine.
   Examiner tips

   • Take the absolute value of the cosine if it is negative
   • $\sqrt{21 \cdot 5} = \sqrt{105}$

10. Question 9c

    3 marksIntersection Point of Two Lines
    Step 1: Equate $x$-components: $3 + 2\lambda = -1$, giving $\lambda = -2$. Step 2: Equate $z$-components: $-6 + 4\lambda = -6 + 2\mu$, so $4(-2) = 2\mu$, giving $\mu = -4$. Step 3: Verify $y$: $1 + \lambda = 1 + (-2) = -1$, and $3 + \mu = 3 + (-4) = -1$. ✓ Step 4: Substitute $\lambda = -2$ into $l_1$: $(3 - 4, 1 - 2, -6 - 8) = (-1, -1, -14)$.
    Method:

    Equate two pairs of components to find $\lambda$ and $\mu$, verify the third, and substitute back to get the intersection.
    Examiner tips

    • Always verify the third equation — if it fails, the lines are skew
    • Substitute the parameter back to recover the position vector

11. Question 10a

    2 marksPartial Fractions of $\dfrac{2}{1 - 9y^2}$
    Step 1: Factor: $1 - 9y^2 = (1 - 3y)(1 + 3y)$. Step 2: Set $\dfrac{2}{(1-3y)(1+3y)} = \dfrac{A}{1+3y} + \dfrac{B}{1-3y}$. Step 3: Multiply through: $2 = A(1 - 3y) + B(1 + 3y)$. Step 4: At $y = 1/3$: $2 = 0 + B(1 + 1) = 2B$, so $B = 1$. At $y = -1/3$: $2 = A(1 + 1) = 2A$, so $A = 1$.
    Method:

    Factor by difference of squares and apply cover-up to find the two constants.
    Examiner tips

    • $1 - k^2 y^2 = (1 - ky)(1 + ky)$ — difference of squares
    • Cover-up rule is fastest

12. Question 10b

    6 marksSolving a Separable DE Using Partial Fractions
    Step 1: Separate variables: $\dfrac{2 \, dy}{1 - 9y^2} = \sec^2 3x \, dx$. Step 2: Use partial fractions: $\dfrac{2}{1 - 9y^2} = \dfrac{1}{1 + 3y} + \dfrac{1}{1 - 3y}$. Integrate: $\dfrac{1}{3}\ln|1 + 3y| - \dfrac{1}{3}\ln|1 - 3y| = \dfrac{1}{3} \ln\left|\dfrac{1 + 3y}{1 - 3y}\right|$. Step 3: Integrate RHS: $\int \sec^2 3x \, dx = \dfrac{1}{3} \tan 3x$. Step 4: Apply IC $y = 0, x = \pi/12$: $\dfrac{1}{3} \ln 1 = \dfrac{1}{3} \tan(\pi/4) + C = \dfrac{1}{3} + C$, so $C = -\dfrac{1}{3}$. Step 5: $\ln\left|\dfrac{1+3y}{1-3y}\right| = \tan 3x - 1$. Exponentiate: $\dfrac{1+3y}{1-3y} = e^{\tan 3x - 1}$. Step 6: Let $k = e^{\tan 3x - 1}$: $1 + 3y = k(1 - 3y)$, so $3y(1 + k) = k - 1$, giving $y = \dfrac{k - 1}{3(k + 1)} = \dfrac{e^{\tan 3x - 1} - 1}{3(e^{\tan 3x - 1} + 1)}$.
    Method:

    Separate, decompose using partial fractions, integrate, apply the initial condition, exponentiate, and solve algebraically for $y$.
    Examiner tips

    • $\sec^2 3x = \dfrac{1}{\cos^2 3x}$ — recognise this in the RHS
    • $\int \sec^2 (kx) dx = \dfrac{1}{k} \tan(kx)$

13. Question 11b

    6 marksArea by Substitution
    Step 1: Substitute $u = 3 + 2 \tan x$. Then $\dfrac{du}{dx} = 2 \sec^2 x$, so $\sec^2 x \, dx = \dfrac{du}{2}$. Step 2: Limits: at $x = -\pi/4$, $\tan(-\pi/4) = -1$, so $u = 3 - 2 = 1$. At $x = \pi/4$, $\tan(\pi/4) = 1$, so $u = 3 + 2 = 5$. Step 3: Area $= \displaystyle \int_{-\pi/4}^{\pi/4} \sec^2 x \sqrt{3 + 2 \tan x} \, dx = \int_1^5 \sqrt u \cdot \dfrac{du}{2} = \dfrac{1}{2} \int_1^5 u^{1/2} \, du$. Step 4: $\int u^{1/2} \, du = \dfrac{2}{3} u^{3/2}$, so Area $= \dfrac{1}{2} \cdot \dfrac{2}{3} u^{3/2} \Big|_1^5 = \dfrac{1}{3} (5^{3/2} - 1^{3/2}) = \dfrac{5\sqrt 5 - 1}{3}$.
    Method:

    Use the substitution to convert into $\int \sqrt u \, du/2$, change limits, and evaluate.
    Examiner tips

    • Always change limits when substituting (don't substitute back to $x$ unnecessarily)
    • $\int u^{1/2} du = \dfrac{2}{3} u^{3/2}$