A Level Mathematics 9709/31 May/June 2025 Past Paper with Answers

Question 1

Find the set of values of $x$ for which $|2x - 1| < |x + 3|$.

1. $-\dfrac{2}{3} < x < 4$
2. $x < -\dfrac{2}{3}$ or $x > 4$
3. $-4 < x < \dfrac{2}{3}$
4. $x < 4$

Question 2

Given that $3 \ln a - \dfrac{1}{2} \ln(b + 4) = 2$, express $b$ in terms of $a$.

1. $b = \dfrac{a^6}{e^4} - 4$
2. $b = \dfrac{a^3}{e^2} - 4$
3. $b = \dfrac{a^6}{e^4} + 4$
4. $b = \dfrac{a^3}{e^4} - 4$

Question 3

Let $z = x + iy$ where $x, y$ are real. Given that $\dfrac{z + 5i}{z - 5}$ is real and $|z| = \sqrt{17}$, one solution for $z$ is:

1. $z = 4 - i$
2. $z = 4 + i$
3. $z = -1 + 4i$
4. $z = 5 + i\sqrt{2}$

Question 4 · Answer A

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 $y = \dfrac{6}{e} x - 6 + 3 = \dfrac{6}{e} x - 3$.

Question 5 · Answer A

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 $p + q = 1$.

Question 6 · Answer A

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 $-\pi < \theta \le \pi$.

Question 7 · Answer A

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 $2$.

Question 8 · Answer A

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.

Question 9 · Answer A

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.).

Question 10 · Answer A

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 $\mathbf{r} = \mathbf{a} + \lambda \overrightarrow{AB} = (\mathbf{i} + 2\mathbf{j} + \mathbf{k}) + \lambda(3\mathbf{i} - \mathbf{j} + 4\mathbf{k})$.

Question 11 · Answer A

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.

Question 12 · Answer A

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.

Question 13 · Answer A

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: $(x + 5)(x^2 - 3) + x = x^3 - 3x + 5x^2 - 15 + x = x^3 + 5x^2 - 2x - 15$. ✓

Question 14 · Answer A

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: $y^2 = \dfrac{x^2}{6} + \dfrac{5x}{3} + \dfrac{1}{6} \ln|x^2 - 3|$.

Question 15 · Answer A

Step 1: Apply the product rule with $u = \cos x$ and $v = (\sin 2x)^{1/2}$. $\dfrac{du}{dx} = -\sin x$ and $\dfrac{dv}{dx} = \dfrac{2 \cos 2x}{2(\sin 2x)^{1/2}} = \dfrac{\cos 2x}{\sqrt{\sin 2x}}$. Step 2: $\dfrac{dy}{dx} = -\sin x \sqrt{\sin 2x} + \dfrac{\cos x \cos 2x}{\sqrt{\sin 2x}}$. Step 3: Set $\dfrac{dy}{dx} = 0$ and multiply by $\sqrt{\sin 2x}$ (positive on the open interval): $-\sin x \sin 2x + \cos x \cos 2x = 0$. Step 4: Recognise the LHS as $\cos(x + 2x) = \cos 3x$. So $\cos 3x = 0$. Step 5: For $0 \le x \le \dfrac{\pi}{2}$, we have $0 \le 3x \le \dfrac{3\pi}{2}$. Solutions of $\cos 3x = 0$: $3x = \dfrac{\pi}{2}$ (and $\dfrac{3\pi}{2}$ which gives the boundary). The interior critical point is $x = \dfrac{\pi}{6}$, giving the maximum.

Question 16 · Answer A

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: $V = 2\pi \left[\dfrac{u^4}{4}\right]_0^1 = 2\pi \cdot \dfrac{1}{4} = \dfrac{\pi}{2}$.