| Syllabus | Alternate Universe Cambridge 9709 |
| Subject | Mathematics (9709) A2 |
| Paper | 9709/32/M/J/25 · Paper 3, Variant 2 |
| Also listed as | 9709_s25_qp_32 · 9709_s25_ms_32 |
| Session | May/June 2025 |
| Questions | 17, multiple choice, with answers and explanations |
| Cost | Free |
9709/32/M/J/25 answer key
Preview the first questions
Question 1
Solve the equation $\dfrac{e^x + 2e^{-x}}{e^x - 3} = 4$, giving the answer correct to 3 decimal places.
- $x \approx 1.426$
- $x \approx 1.386$
- $x \approx 0.405$
- $x \approx 2.398$
Show the answer and explanation
Correct answer: A
Step 1: Multiply both sides by $(e^x - 3)$: $e^x + 2e^{-x} = 4e^x - 12$. Step 2: Multiply throughout by $e^x$ to remove the negative exponent: $e^{2x} + 2 = 4e^{2x} - 12 e^x$, hence $3e^{2x} - 12e^x - 2 = 0$. Step 3: Let $u = e^x$. Solve $3u^2 - 12u - 2 = 0$: $u = \dfrac{12 \pm \sqrt{144 + 24}}{6} = \dfrac{6 \pm \sqrt{42}}{3}$. Step 4: Reject the negative root (since $u = e^x > 0$). Take $u = \dfrac{6+\sqrt{42}}{3} \approx 4.1602$. Step 5: $x = \ln u \approx \ln(4.1602) \approx 1.426$ (3 d.p.).
Question 2
Find the expansion of $(6 - x)(1 - 2x)^{-3/2}$ in ascending powers of $x$, up to and including the term in $x^2$.
- $6 + 17x + 42x^2$
- $6 + 17x + 45x^2$
- $6 + 19x + 42x^2$
- $6 - 17x + 42x^2$
Show the answer and explanation
Correct answer: A
Step 1: Use the binomial series $(1+u)^n = 1 + nu + \dfrac{n(n-1)}{2}u^2 + \dots$ with $n = -\dfrac{3}{2}$, $u = -2x$. Step 2: Linear term: $\left(-\dfrac{3}{2}\right)(-2x) = 3x$. Quadratic term: $\dfrac{(-3/2)(-5/2)}{2}(-2x)^2 = \dfrac{15/4}{2}\cdot 4x^2 = \dfrac{15}{2}x^2$. Step 3: Hence $(1-2x)^{-3/2} = 1 + 3x + \dfrac{15}{2}x^2 + \dots$. Step 4: Multiply by $(6-x)$, keeping terms up to $x^2$: $6\!\left(1 + 3x + \dfrac{15}{2}x^2\right) - x(1 + 3x) = 6 + 18x + 45x^2 - x - 3x^2$. Step 5: Simplify: $6 + 17x + 42x^2$.
Question 3
For what set of values of $x$ is the binomial expansion of $(1 - 2x)^{-3/2}$ valid?
- $|x| < \dfrac{1}{2}$
- $|x| < 1$
- $|x| \le \dfrac{1}{2}$
- $|x| < 2$
Show the answer and explanation
Correct answer: A
Step 1: The series $(1+u)^n$ for non-integer $n$ converges if and only if $|u| < 1$. Step 2: Here $u = -2x$, so the requirement is $|-2x| < 1$, i.e. $|x| < \dfrac{1}{2}$. Step 3: This is a strict inequality (the series does not converge at $|u| = 1$).
Worked answers: questions 4 to 17
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Question 4 · Answer A
Step 1: Use the identity $\cot 2x = \dfrac{1-\tan^2 x}{2\tan x}$. The equation becomes $\dfrac{3}{\tan x} - \dfrac{4(1-\tan^2 x)}{2\tan x} = 3$, i.e. $\dfrac{3 - 2(1-\tan^2 x)}{\tan x} = 3$. Step 2: Simplify the numerator: $3 - 2 + 2\tan^2 x = 1 + 2\tan^2 x$, so $\dfrac{1 + 2\tan^2 x}{\tan x} = 3$. Step 3: Multiply by $\tan x$: $2\tan^2 x - 3\tan x + 1 = 0$. Step 4: Factorise: $(2\tan x - 1)(\tan x - 1) = 0$, so $\tan x = \dfrac{1}{2}$ or $\tan x = 1$. Step 5: In the interval $[0°, 180°]$: $\tan x = 1 \Rightarrow x = 45°$; $\tan x = \dfrac{1}{2} \Rightarrow x \approx 26.6°$. (The other branches lie outside the interval since $\tan$ is negative there.)
Question 5 · Answer A
Step 1: Let $(x + iy)^2 = -1 - 4\sqrt{5}\,i$. Expanding: $x^2 - y^2 + 2ixy = -1 - 4\sqrt{5}\,i$. Step 2: Equate real and imaginary parts: $x^2 - y^2 = -1$ and $2xy = -4\sqrt{5}$, hence $xy = -2\sqrt{5}$, so $y = -\dfrac{2\sqrt{5}}{x}$. Step 3: Substitute: $x^2 - \dfrac{20}{x^2} = -1$. Multiply by $x^2$: $x^4 + x^2 - 20 = 0$. Step 4: Factorise: $(x^2 - 4)(x^2 + 5) = 0$. Since $x$ is real, $x^2 = 4$, so $x = \pm 2$. Step 5: Then $y = -\dfrac{2\sqrt{5}}{x} = \mp \sqrt{5}$. Hence the two square roots are $2 - \sqrt{5}\,i$ and $-2 + \sqrt{5}\,i$, i.e. $\pm(2 - \sqrt{5}\,i)$. Verify: $(2 - \sqrt{5}\,i)^2 = 4 - 4\sqrt{5}\,i + 5 i^2 = -1 - 4\sqrt{5}\,i$. ✓
Question 6 · Answer A
Step 1: Define $f(x) = |x - 2| - 2 \sin\!\left(\tfrac{1}{2}x\right)$. A root of $|x-2| = 2 \sin(x/2)$ corresponds to $f(x) = 0$. Step 2: At $x = 1$: $|1-2| = 1$ and $2 \sin(0.5) \approx 2(0.4794) \approx 0.9589$. So $f(1) \approx 1 - 0.9589 \approx 0.0411 > 0$. Step 3: At $x = 1.5$: $|1.5-2| = 0.5$ and $2 \sin(0.75) \approx 2(0.6816) \approx 1.3633$. So $f(1.5) \approx 0.5 - 1.3633 \approx -0.8633 < 0$. Step 4: Since $f$ is continuous on $[1, 1.5]$ and changes sign, by the Intermediate Value Theorem there is a root in $(1, 1.5)$.
Question 7 · Answer A
Step 1: With calculator in radians, compute $x_2 = 2 - 2 \sin(1.03/2) = 2 - 2 \sin(0.515) \approx 2 - 2(0.4926) \approx 1.0149$. Step 2: $x_3 = 2 - 2 \sin(1.0149/2) \approx 2 - 2(0.4860) \approx 1.0281$. Step 3: $x_4 \approx 1.0166$, $x_5 \approx 1.0266$, $x_6 \approx 1.0179$, $x_7 \approx 1.0255$, ... Step 4: The iterates oscillate around $1.02$, with sign change in $(1.015, 1.025)$. Hence the root to 2 d.p. is $1.02$.
Question 8 · Answer A
Step 1: Match $R \cos(\theta - \alpha) = R \cos\alpha \cos\theta + R \sin\alpha \sin\theta$ with $7 \sin\theta + 24 \cos\theta$. So $R \cos\alpha = 24$ and $R \sin\alpha = 7$. Step 2: $R = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25$. Step 3: $\tan\alpha = \dfrac{R \sin\alpha}{R \cos\alpha} = \dfrac{7}{24}$. Step 4: $\alpha = \tan^{-1}(7/24) \approx 0.2838$ radians (4 d.p.).
Question 9 · Answer A
Step 1: Substitute $\theta = x/3$: $7 \sin(x/3) + 24 \cos(x/3) = 25 \cos(x/3 - 0.2838)$. The equation becomes $25 \cos(x/3 - 0.2838) = 24.5$, so $\cos(x/3 - 0.2838) = 0.98$. Step 2: Take inverse cosine: $x/3 - 0.2838 = \pm \cos^{-1}(0.98) \approx \pm 0.2003$ (principal values). Step 3: Solve: $x/3 = 0.2838 + 0.2003 = 0.4841$ or $x/3 = 0.2838 - 0.2003 = 0.0835$. Step 4: Multiply by 3: $x \approx 1.45$ or $x \approx 0.250$. Both lie in $0 < x < \pi$.
Question 10 · Answer A
Step 1: Separate variables: $\dfrac{dx}{4x+3} = \dfrac{\cos 2\theta}{\sin 2\theta} d\theta = \cot 2\theta \, d\theta$. Step 2: Integrate: $\dfrac{1}{4}\ln|4x+3| = \dfrac{1}{2}\ln|\sin 2\theta| + c$. Step 3: Apply initial condition $x = 0$, $\theta = \pi/12$ (so $2\theta = \pi/6$ and $\sin 2\theta = 1/2$): $\dfrac{1}{4}\ln 3 = \dfrac{1}{2}\ln(1/2) + c$, hence $c = \dfrac{1}{4}\ln 3 + \dfrac{1}{2}\ln 2$. Step 4: Multiply through by 4: $\ln(4x+3) = 2 \ln|\sin 2\theta| + \ln 3 + 2 \ln 2 = \ln(12 \sin^2 2\theta)$. Step 5: Exponentiate: $4x + 3 = 12 \sin^2 2\theta$, so $x = \dfrac{12 \sin^2 2\theta - 3}{4}$.
Question 11 · Answer A
Step 1: A vector line equation uses one point on the line and a direction vector. Step 2: Use $A$ as the point: position vector $(1, -4, 2)$. Step 3: The direction is $\vec{AB} = \vec{OB} - \vec{OA} = (-2-1, 1-(-4), 3-2) = (-3, 5, 1)$. Step 4: So $\mathbf{r} = (1, -4, 2) + \lambda(-3, 5, 1)$. (Equivalent forms with different point or scaled direction are also valid lines.)
Question 12 · Answer A
Step 1: Compute $\vec{AB} = \vec{OB} - \vec{OA} = (-3, 5, 1)$ and $\vec{AC} = \vec{OC} - \vec{OA} = (1, 7, 3)$. Step 2: Scalar product: $\vec{AB} \cdot \vec{AC} = (-3)(1) + (5)(7) + (1)(3) = -3 + 35 + 3 = 35$. Step 3: Magnitudes: $|\vec{AB}| = \sqrt{9+25+1} = \sqrt{35}$ and $|\vec{AC}| = \sqrt{1+49+9} = \sqrt{59}$. Step 4: Hence $\cos BAC = \dfrac{35}{\sqrt{35}\sqrt{59}} = \dfrac{35}{\sqrt{2065}} = \dfrac{\sqrt{35}}{\sqrt{59}} = \sqrt{\dfrac{35}{59}}$.
Question 13 · Answer A
Step 1: Area of a triangle using two sides and included angle: $\text{Area} = \tfrac{1}{2}|\vec{AB}||\vec{AC}|\sin BAC$. Step 2: From $\cos^2 BAC = \dfrac{35}{59}$: $\sin^2 BAC = 1 - \dfrac{35}{59} = \dfrac{24}{59}$, so $\sin BAC = \sqrt{\dfrac{24}{59}}$ (positive in a triangle). Step 3: Substitute: $\text{Area} = \tfrac{1}{2} \sqrt{35} \cdot \sqrt{59} \cdot \sqrt{\dfrac{24}{59}} = \tfrac{1}{2} \sqrt{35 \cdot 24}$ (the $59$ cancels). Step 4: Simplify: $\sqrt{35 \cdot 24} = \sqrt{840} = \sqrt{4 \cdot 210} = 2\sqrt{210}$. Hence Area $= \tfrac{1}{2} \cdot 2\sqrt{210} = \sqrt{210}$.
Question 14 · Answer A
Step 1: Write $x^2 = q(1 + 4x^2) + r$ where $q$ is constant (the quotient) and $r$ is constant (the remainder, of degree less than the divisor's leading $x^2$ term). Step 2: Compare $x^2$ coefficients: $1 = 4q$, so $q = \dfrac{1}{4}$. Step 3: Compare constants: $0 = q + r = \dfrac{1}{4} + r$, so $r = -\dfrac{1}{4}$. Step 4: Verify: $\dfrac{1}{4}(1 + 4x^2) - \dfrac{1}{4} = \dfrac{1}{4} + x^2 - \dfrac{1}{4} = x^2$. ✓
Question 15 · Answer A
Step 1: Apply integration by parts with $u = \tan^{-1}(2x)$ ($du = \dfrac{2}{1+4x^2}dx$) and $dv = x\,dx$ ($v = \dfrac{x^2}{2}$): $\int x \tan^{-1}(2x)\,dx = \dfrac{x^2}{2}\tan^{-1}(2x) - \int \dfrac{x^2}{1+4x^2}dx$. Step 2: Use $\dfrac{x^2}{1+4x^2} = \dfrac{1}{4} - \dfrac{1/4}{1+4x^2}$ (from polynomial division). Integrate: $\int \dfrac{x^2}{1+4x^2}dx = \dfrac{x}{4} - \dfrac{1}{8}\tan^{-1}(2x)$. Step 3: Combine: antiderivative $F(x) = \dfrac{x^2}{2}\tan^{-1}(2x) - \dfrac{x}{4} + \dfrac{1}{8}\tan^{-1}(2x)$. Step 4: Evaluate $F(0.5)$: $\dfrac{0.25}{2}\tan^{-1}(1) - \dfrac{0.5}{4} + \dfrac{1}{8}\tan^{-1}(1) = \dfrac{1}{8} \cdot \dfrac{\pi}{4} - \dfrac{1}{8} + \dfrac{1}{8} \cdot \dfrac{\pi}{4} = \dfrac{\pi}{32} - \dfrac{1}{8} + \dfrac{\pi}{32} = \dfrac{\pi}{16} - \dfrac{1}{8}$. Step 5: $F(0) = 0$. So the definite integral $= \dfrac{\pi}{16} - \dfrac{1}{8}$.
Question 16 · Answer A
Step 1: Use the product rule on $y = 5 \sin 2x \cdot \cos^2 x$ with $\dfrac{d}{dx}(\cos^2 x) = -2 \sin x \cos x = -\sin 2x$: $\dfrac{dy}{dx} = 5(2\cos 2x) \cos^2 x + 5 \sin 2x \cdot (-\sin 2x) \cdot ...$. Better: rewrite $y = 5(2\sin x \cos x)\cos^2 x = 10 \sin x \cos^3 x$. Step 2: Differentiate: $\dfrac{dy}{dx} = 10 \cos x \cdot \cos^3 x + 10 \sin x \cdot 3 \cos^2 x (-\sin x) = 10 \cos^4 x - 30 \sin^2 x \cos^2 x$. Step 3: Set $\dfrac{dy}{dx} = 0$: $10 \cos^2 x (\cos^2 x - 3 \sin^2 x) = 0$. With $0 < x < \pi/2$, $\cos x \ne 0$, so $\cos^2 x = 3 \sin^2 x$. Step 4: Hence $\dfrac{\sin^2 x}{\cos^2 x} = \dfrac{1}{3}$, i.e. $\tan^2 x = \dfrac{1}{3}$, so $\tan x = \dfrac{1}{\sqrt{3}}$ (positive in $(0, \pi/2)$). Step 5: Therefore $x = \dfrac{\pi}{6}$. (Confirm $M$ by checking it is a maximum, not a minimum.)
Question 17 · Answer A
Step 1: Rewrite using double-angle: $5 \sin 2x \cos^2 x = 5 \cdot 2 \sin x \cos x \cdot \cos^2 x = 10 \sin x \cos^3 x$. Step 2: Substitute $u = \cos x$, so $du = -\sin x \, dx$. The integral becomes $\displaystyle\int 10 \sin x \cos^3 x \, dx = -10 \int u^3 \, du$. Step 3: Update limits: $x = 0 \Rightarrow u = 1$ and $x = \pi/4 \Rightarrow u = \cos(\pi/4) = \dfrac{1}{\sqrt{2}}$. Step 4: Evaluate: $-10 \left[\dfrac{u^4}{4}\right]_{1}^{1/\sqrt{2}} = -10 \left(\dfrac{(1/\sqrt{2})^4}{4} - \dfrac{1}{4}\right) = -10 \left(\dfrac{1/4}{4} - \dfrac{1}{4}\right) = -10 \left(\dfrac{1}{16} - \dfrac{1}{4}\right)$. Step 5: Simplify: $-10 \cdot \left(-\dfrac{3}{16}\right) = \dfrac{30}{16} = \dfrac{15}{8}$.