| Syllabus | Alternate Universe Cambridge 9709 |
| Subject | Mathematics (9709) A2 |
| Paper | 9709/32/O/N/25 · Paper 3, Variant 2 |
| Also listed as | 9709_w25_qp_32 · 9709_w25_ms_32 |
| Session | October/November 2025 |
| Questions | 15, multiple choice, with answers and explanations |
| Cost | Free |
9709/32/O/N/25 answer key
Preview the first questions
Question 1
For $a > 0$, the set of values of $x$ satisfying $|x + 3a| \ge a - 2x$ is:
- $x \ge -\dfrac{2a}{3}$
- $x \le -\dfrac{2a}{3}$
- $-\dfrac{2a}{3} \le x \le 4a$
- $x \ge 4a$ only
Show the answer and explanation
Correct answer: A
Step 1: Squaring requires care since $a - 2x$ can be negative. For $x \ge \dfrac{a}{2}$, $a - 2x \le 0$ and the inequality $|x + 3a| \ge a - 2x$ is automatic. Step 2: For $x < \dfrac{a}{2}$, square: $(x+3a)^2 \ge (a-2x)^2$, i.e. $(x+3a)^2 - (a-2x)^2 \ge 0$, which factors as $(4a - x)(3x + 2a) \ge 0$. Step 3: Critical values $x = 4a$ and $x = -\dfrac{2a}{3}$. Sign chart: positive on $-\dfrac{2a}{3} \le x \le 4a$. Step 4: Combine the two cases: from the squared method, $-\dfrac{2a}{3} \le x \le 4a$ together with $x \ge \dfrac{a}{2}$ (automatic). Since the trivial region $x > \dfrac{a}{2}$ already extends past $4a$, the full solution is $x \ge -\dfrac{2a}{3}$.
Question 2
Solve $3 \times 2^{x+1} = 4 \times 3^{2x-3}$. Give your answer correct to 3 s.f.
- $x \approx 2.46$
- $x \approx -2.46$
- $x \approx 1.50$
- $x \approx 0.405$
Show the answer and explanation
Correct answer: A
Step 1: Take natural logs of both sides: $\ln 3 + (x+1) \ln 2 = \ln 4 + (2x-3) \ln 3$. Step 2: Expand: $\ln 3 + x \ln 2 + \ln 2 = \ln 4 + 2x \ln 3 - 3 \ln 3$. Step 3: Collect $x$ terms on one side: $x(\ln 2 - 2 \ln 3) = \ln 4 - 3 \ln 3 - \ln 3 - \ln 2 = \ln 4 - 4 \ln 3 - \ln 2 = \ln \dfrac{4}{81 \cdot 2} = \ln \dfrac{2}{81}$. Step 4: $\ln 2 - 2 \ln 3 = \ln(2/9)$. Hence $x = \dfrac{\ln(2/81)}{\ln(2/9)} \approx \dfrac{-3.7012}{-1.5041} \approx 2.46$.
Question 3
The line $\operatorname{Re}(z) = 2$ and the circle $|z - (3 + i)| = 2$ intersect at two points. The lower one is:
- $2 + (1 - \sqrt{3})\, i$
- $2 + (1 + \sqrt{3})\, i$
- $3 - i$
- $2 - \sqrt{3} \, i$
Show the answer and explanation
Correct answer: A
Step 1: The circle $|z - (3 + i)| = 2$ corresponds to $(x - 3)^2 + (y - 1)^2 = 4$. Step 2: Substitute $x = 2$: $(2 - 3)^2 + (y - 1)^2 = 4$, so $1 + (y - 1)^2 = 4$, giving $(y - 1)^2 = 3$. Step 3: $y - 1 = \pm \sqrt{3}$, so $y = 1 + \sqrt{3}$ (upper) or $y = 1 - \sqrt{3}$ (lower). Step 4: The lower intersection point is $z = 2 + (1 - \sqrt{3}) i$.
Worked answers: questions 4 to 15
Why each answer is right. The full questions are waiting in every Paperverse world, with your mark at the end.
Question 4 · Answer A
Step 1: $\arg z$ is maximised at the point of the region closest in angle to the imaginary axis from the right. This corresponds to the tangent line $y = mx$ from the origin to the circle (chosen on the upper side). Step 2: Distance from centre $(3, 1)$ to line $mx - y = 0$ equals $\dfrac{|3m - 1|}{\sqrt{m^2 + 1}}$. Set $= 2$: $(3m - 1)^2 = 4(m^2 + 1) \Rightarrow 9m^2 - 6m + 1 = 4m^2 + 4 \Rightarrow 5m^2 - 6m - 3 = 0$. Step 3: $m = \dfrac{6 + \sqrt{36 + 60}}{10} = \dfrac{6 + \sqrt{96}}{10} = \dfrac{3 + 2\sqrt 6}{5} \approx 1.580$. Step 4: $\arg z = \tan^{-1}(1.580) \approx 1.006 \approx 1.01$ rad. The tangent point at $x = (9 - \sqrt 6)/5 \approx 1.31$ does satisfy $x \le 2$, so this is achievable.
Question 5 · Answer A
Step 1: Use integration by parts with $u = \tan^{-1} x$, $dv = x \, dx$. So $du = \dfrac{1}{1+x^2} dx$ and $v = \dfrac{x^2}{2}$. Step 2: $\int x \tan^{-1} x \, dx = \dfrac{x^2}{2} \tan^{-1} x - \dfrac{1}{2} \int \dfrac{x^2}{1+x^2} dx$. Step 3: Use $\dfrac{x^2}{1+x^2} = 1 - \dfrac{1}{1+x^2}$, so $\int \dfrac{x^2}{1+x^2} dx = x - \tan^{-1} x$. Step 4: Antiderivative: $\dfrac{x^2}{2} \tan^{-1} x - \dfrac{1}{2}(x - \tan^{-1} x) = \dfrac{x^2}{2} \tan^{-1} x - \dfrac{x}{2} + \dfrac{1}{2} \tan^{-1} x$. Step 5: At $x = 1$: $\dfrac{1}{2} \cdot \dfrac{\pi}{4} - \dfrac{1}{2} + \dfrac{1}{2} \cdot \dfrac{\pi}{4} = \dfrac{\pi}{4} - \dfrac{1}{2}$. At $x = 0$: $0 - 0 + 0 = 0$. Integral $= \dfrac{\pi}{4} - \dfrac{1}{2}$.
Question 6 · Answer A
Step 1: $(x-3)^2$ a factor means both $f(3) = 0$ and $f'(3) = 0$. Step 2: $f(3) = 2(27) - 4(9) + 3p + q = 54 - 36 + 3p + q = 18 + 3p + q = 0$, so $3p + q = -18$. Step 3: $f'(x) = 6x^2 - 8x + p$. $f'(3) = 54 - 24 + p = 30 + p = 0$, so $p = -30$. Step 4: Substitute back: $3(-30) + q = -18 \Rightarrow q = -18 + 90 = 72$.
Question 7 · Answer A
Step 1: Compute $x_1 = \dfrac{1}{2} \tan^{-1}\left(\dfrac{1}{2 \sin 1 - 1}\right) = \dfrac{1}{2} \tan^{-1}\left(\dfrac{1}{0.6829}\right) \approx \dfrac{1}{2}(0.9716) \approx 0.4858$. Step 2: $x_2 = \dfrac{1}{2} \tan^{-1}\left(\dfrac{1}{2 \sin 0.9716 - 1}\right) \approx 0.4967$. Step 3: $x_3 \approx 0.4883$, $x_4 \approx 0.4947$, $x_5 \approx 0.4895$. Step 4: Sequence oscillates and converges to $0.49$ (2 d.p.).
Question 8 · Answer A
Step 1: Normal parallel to the $y$-axis means the normal has undefined slope, hence the tangent is horizontal: $\dfrac{dy}{dx} = 0$. Step 2: Set the numerator $x^2 + 2xy = 0$: $x(x + 2y) = 0$. So $x = 0$ or $x = -2y$. Step 3: Case $x = 0$: $2y^3 = 16 \Rightarrow y^3 = 8 \Rightarrow y = 2$. Point $(0, 2)$. Step 4: Case $x = -2y$: $2y^3 - 3(4y^2)y - (-2y)^3 = 2y^3 - 12y^3 + 8y^3 = -2y^3 = 16$, so $y^3 = -8 \Rightarrow y = -2$, giving $x = -2(-2) = 4$. Point $(4, -2)$.
Question 9 · Answer A
Step 1: Substitute the identity: $\cos^3 x \sin 4x = 4 \sin x \cos^3 x (2 \cos^3 x - \cos x) = 8 \sin x \cos^6 x - 4 \sin x \cos^4 x$. Step 2: Antiderivatives: $\int 8 \sin x \cos^6 x \, dx = -\dfrac{8}{7} \cos^7 x$ and $\int 4 \sin x \cos^4 x \, dx = -\dfrac{4}{5} \cos^5 x$. So total antiderivative: $-\dfrac{8}{7} \cos^7 x + \dfrac{4}{5} \cos^5 x$. Step 3: Evaluate at $x = \pi/4$: $\cos(\pi/4) = \dfrac{1}{\sqrt 2}$, $\cos^7 = \dfrac{1}{8\sqrt 2}$, $\cos^5 = \dfrac{1}{4\sqrt 2}$. Value: $-\dfrac{8}{7} \cdot \dfrac{1}{8\sqrt 2} + \dfrac{4}{5} \cdot \dfrac{1}{4\sqrt 2} = \dfrac{-1}{7\sqrt 2} + \dfrac{1}{5\sqrt 2} = \dfrac{-5 + 7}{35\sqrt 2} = \dfrac{2}{35\sqrt 2} = \dfrac{\sqrt 2}{35}$. Step 4: At $x = 0$: $-\dfrac{8}{7} + \dfrac{4}{5} = \dfrac{-40 + 28}{35} = -\dfrac{12}{35}$. Step 5: Subtract: $\dfrac{\sqrt 2}{35} - \left(-\dfrac{12}{35}\right) = \dfrac{\sqrt 2 + 12}{35}$.
Question 10 · Answer A
Step 1: Numerator and denominator have equal degree. $(x+2a)(x+3a) = x^2 + 5ax + 6a^2$. So $x^2 + 4ax + 6a^2 = (x^2 + 5ax + 6a^2) - ax$, giving $f(x) = 1 - \dfrac{ax}{(x+2a)(x+3a)}$. Step 2: Decompose $\dfrac{-ax}{(x+2a)(x+3a)} = \dfrac{B}{x+2a} + \dfrac{C}{x+3a}$, so $-ax = B(x+3a) + C(x+2a)$. Step 3: Substitute $x = -2a$: $-a(-2a) = 2a^2 = B(-2a + 3a) = aB \Rightarrow B = 2a$. Step 4: Substitute $x = -3a$: $-a(-3a) = 3a^2 = C(-3a + 2a) = -aC \Rightarrow C = -3a$. Step 5: Hence $f(x) = 1 + \dfrac{2a}{x + 2a} - \dfrac{3a}{x + 3a}$.
Question 11 · Answer A
Step 1: Antiderivative: $\int \left(1 + \dfrac{2a}{x+2a} - \dfrac{3a}{x+3a}\right) dx = x + 2a \ln|x+2a| - 3a \ln|x+3a|$. Step 2: At $x = a$: $a + 2a \ln(3a) - 3a \ln(4a)$. Step 3: At $x = -a$: $-a + 2a \ln(a) - 3a \ln(2a)$. Step 4: Difference: $(a - (-a)) + 2a (\ln 3a - \ln a) - 3a (\ln 4a - \ln 2a) = 2a + 2a \ln 3 - 3a \ln 2$. Step 5: Simplify: $2a + a \ln(3^2) - a \ln(2^3) = 2a + a (\ln 9 - \ln 8) = a\left(2 + \ln \dfrac{9}{8}\right)$.
Question 12 · Answer A
Step 1: Separate variables: $\dfrac{250}{500 - h^2} dh = dt$. Step 2: Integrate using $\int \dfrac{1}{a^2 - h^2} dh = \dfrac{1}{2a} \ln\left|\dfrac{a + h}{a - h}\right|$ with $a = \sqrt{500} = 10\sqrt 5$. So $\int \dfrac{250}{500 - h^2} dh = \dfrac{250}{20\sqrt 5} \ln\left|\dfrac{10\sqrt 5 + h}{10\sqrt 5 - h}\right| = \dfrac{5\sqrt 5}{2} \ln|\cdots|$. Step 3: Apply IC $h = 0, t = 0$: constant $= 0$. Step 4: At $h = 20$: $\dfrac{10\sqrt 5 + 20}{10\sqrt 5 - 20} = \dfrac{\sqrt 5 + 2}{\sqrt 5 - 2}$. Rationalise: multiply by $\dfrac{\sqrt 5 + 2}{\sqrt 5 + 2}$ to get $\dfrac{(\sqrt 5 + 2)^2}{5 - 4} = (\sqrt 5 + 2)^2 = 9 + 4\sqrt 5 \approx 17.944$. Step 5: $t = \dfrac{5\sqrt 5}{2} \ln(9 + 4\sqrt 5) \approx \dfrac{5(2.236)}{2} \cdot 2.887 \approx 5.59 \cdot 2.887 \approx 16.1$ s.
Question 13 · Answer A
Step 1: $M = \dfrac{1}{2}(A + B) = \dfrac{1}{2}((4 + 2), (-2 + 8), (0 + 4)) = (3, 3, 2)$. Step 2: $\overrightarrow{MB} = B - M = (2 - 3, 8 - 3, 4 - 2) = (-1, 5, 2)$. Step 3: $\overrightarrow{MC} = C - M = (-2 - 3, 0 - 3, 6 - 2) = (-5, -3, 4)$.
Question 14 · Answer A
Step 1: Scalar product: $\overrightarrow{MB} \cdot \overrightarrow{MC} = (-1)(-5) + (5)(-3) + (2)(4) = 5 - 15 + 8 = -2$. Step 2: Moduli: $|\overrightarrow{MB}| = \sqrt{1 + 25 + 4} = \sqrt{30}$. $|\overrightarrow{MC}| = \sqrt{25 + 9 + 16} = \sqrt{50} = 5\sqrt 2$. Step 3: $\cos(\angle CMB) = \dfrac{-2}{\sqrt{30} \cdot 5\sqrt 2} = \dfrac{-2}{5 \sqrt{60}} = \dfrac{-2}{5 \cdot 2\sqrt{15}} = \dfrac{-1}{5\sqrt{15}}$. Step 4: Rationalise: $\dfrac{-1}{5\sqrt{15}} = \dfrac{-\sqrt{15}}{75}$.
Question 15 · Answer A
Step 1: $\sin^2(\angle CMB) = 1 - \cos^2(\angle CMB) = 1 - \dfrac{15}{5625} = 1 - \dfrac{1}{375} = \dfrac{374}{375}$. So $\sin(\angle CMB) = \sqrt{\dfrac{374}{375}}$. Step 2: Area of triangle $MBC$ = $\dfrac{1}{2} |\overrightarrow{MB}| |\overrightarrow{MC}| \sin(\angle CMB) = \dfrac{1}{2} \cdot \sqrt{30} \cdot 5\sqrt 2 \cdot \sqrt{\dfrac{374}{375}} = \dfrac{5}{2} \sqrt{60 \cdot \dfrac{374}{375}} = \dfrac{5}{2} \sqrt{\dfrac{22440}{375}}$. Step 3: $\dfrac{60}{375} = \dfrac{4}{25}$, so $60 \cdot \dfrac{374}{375} = \dfrac{4 \cdot 374}{25} = \dfrac{1496}{25}$. Hence $\sqrt{1496/25} = \dfrac{2\sqrt{374}}{5}$. Area of $MBC = \dfrac{5}{2} \cdot \dfrac{2\sqrt{374}}{5} = \sqrt{374}$. Step 4: Since $M$ is the midpoint of $AB$, triangle $AMC$ has the same area as $MBC$. Hence area of $ABC$ = $2 \cdot$ area of $MBC$ = $2\sqrt{374}$.