A Level Mathematics 9709/33 October/November 2025 Past Paper with Answers

Question 1

Solve $|3x + 2| < 3 |2x - 1|$.

1. $x < \dfrac{1}{9}$ or $x > \dfrac{5}{3}$
2. $\dfrac{1}{9} < x < \dfrac{5}{3}$
3. $x < \dfrac{1}{3}$ or $x > 5$
4. $-\dfrac{1}{9} < x < \dfrac{5}{3}$

Question 2

Divide $3x^4 - 2x^2$ by $x + 1$. The quotient and remainder are:

1. Quotient $3x^3 - 3x^2 + x - 1$, remainder $1$
2. Quotient $3x^3 - 3x^2 + x - 1$, remainder $-1$
3. Quotient $3x^3 + 3x^2 - x + 1$, remainder $1$
4. Quotient $3x^3 - 2x$, remainder $0$

Question 3

Solve $2^{3x - 4} = \dfrac{3}{5^x}$ in the form $\dfrac{\ln m}{\ln n}$ with integer $m, n$:

1. $x = \dfrac{\ln 48}{\ln 40}$
2. $x = \dfrac{\ln 3}{\ln 40}$
3. $x = \dfrac{\ln 16}{\ln 40}$
4. $x = \dfrac{\ln 48}{\ln 5}$

Question 4 · Answer A

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: $\pm 180^\circ, \pm 120^\circ, \pm 60^\circ, 0^\circ$.

Question 5 · Answer A

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: $-\dfrac{\pi^2}{144} + \dfrac{\pi \sqrt 3}{24} + \dfrac{1}{8} - \dfrac{1}{4} = -\dfrac{\pi^2}{144} + \dfrac{\pi \sqrt 3}{24} - \dfrac{1}{8}$.

Question 6 · Answer A

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 $z = -\dfrac{16}{5} - \dfrac{12}{5} i$.

Question 7 · Answer A

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 $0.35$.

Question 8 · Answer A

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 $l_2: \mathbf{r} = (-1, 3, -6) + \mu(0, 1, 2)$.

Question 9 · Answer A

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.

Question 10 · Answer A

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 $l_1$: $(3 - 4, 1 - 2, -6 - 8) = (-1, -1, -14)$.

Question 11 · Answer A

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 $A = 1$.

Question 12 · Answer A

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 $y = \dfrac{k - 1}{3(k + 1)} = \dfrac{e^{\tan 3x - 1} - 1}{3(e^{\tan 3x - 1} + 1)}$.

Question 13 · Answer A

Step 1: Differentiate by product rule, with chain rule on the square root: $\dfrac{dy}{dx} = (2 \sec^2 x \tan x) \sqrt{3 + 2 \tan x} + \sec^2 x \cdot \dfrac{2 \sec^2 x}{2\sqrt{3 + 2 \tan x}} = 2 \sec^2 x \tan x \sqrt{3 + 2 \tan x} + \dfrac{\sec^4 x}{\sqrt{3 + 2 \tan x}}$. Step 2: Set $= 0$ and multiply by $\sqrt{3 + 2 \tan x}$: $2 \sec^2 x \tan x (3 + 2 \tan x) + \sec^4 x = 0$. Step 3: Divide by $\sec^2 x$ and use $\sec^2 x = 1 + \tan^2 x$: $2 \tan x (3 + 2 \tan x) + (1 + \tan^2 x) = 0$, i.e. $6 \tan x + 4 \tan^2 x + 1 + \tan^2 x = 0$, so $5 \tan^2 x + 6 \tan x + 1 = 0$. Step 4: Factorise: $(5 \tan x + 1)(\tan x + 1) = 0$. So $\tan x = -\dfrac{1}{5}$ or $\tan x = -1$. Step 5: In the OPEN interval $(-\pi/4, \pi/4)$, $\tan x = -1$ corresponds to the boundary $x = -\pi/4$. The interior critical value is $\tan x = -\dfrac{1}{5}$, giving $x = -\arctan(1/5) \approx -0.197$.

Question 14 · Answer A

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 $= \dfrac{1}{2} \cdot \dfrac{2}{3} u^{3/2} \Big|_1^5 = \dfrac{1}{3} (5^{3/2} - 1^{3/2}) = \dfrac{5\sqrt 5 - 1}{3}$.