A Level Mathematics 9709/21 October/November 2025 Past Paper with Answers

Question 1

What is $\displaystyle\int 6\sin^2 x\,\mathrm{d}x$?

1. $3x - \dfrac{3}{2}\sin 2x + c$
2. $3x - 3\sin 2x + c$
3. $6x - 3\sin 2x + c$
4. $3x + \dfrac{3}{2}\sin 2x + c$

Question 2

What is the solution to $\mathrm{e}^{2x}(\mathrm{e}^{2x} - 8) = 48$?

1. $x = \dfrac{1}{2}\ln 12$
2. $x = \dfrac{1}{2}\ln 4$
3. $x = \ln 12$
4. $x = \dfrac{1}{2}\ln 12$ and $x = \dfrac{1}{2}\ln(-4)$

Question 3

What are the solutions to $|2x - 3| = |5x + 2|$?

1. $x = -\dfrac{5}{3}$ and $x = \dfrac{1}{7}$
2. $x = \dfrac{5}{3}$ and $x = -\dfrac{1}{7}$
3. $x = -\dfrac{5}{3}$ only
4. $x = \dfrac{1}{7}$ and $x = \dfrac{5}{3}$

Question 4 · Answer A

Step 1: Replace $x$ by $\sec\theta$: $\sec\theta = -\dfrac{5}{3}$ or $\sec\theta = \dfrac{1}{7}$. Step 2: $\sec\theta = \dfrac{1}{7}$ gives $\cos\theta = 7$, which is impossible since $|\cos\theta| \le 1$. Step 3: $\sec\theta = -\dfrac{5}{3}$ gives $\cos\theta = -\dfrac{3}{5}$. In the range $\pi < \theta < 2\pi$, $\cos\theta < 0$ only for $\pi < \theta < \dfrac{3\pi}{2}$. Step 4: $\theta = \pi + \cos^{-1}\!\left(\dfrac{3}{5}\right) = \pi + 0.9273\ldots = 4.069\ldots \approx 4.07$ (3 s.f.).

Question 5 · Answer A

Step 1: Write $\cot\theta = \dfrac{1}{\tan\theta}$ and use the compound angle formula: $\tan(\theta + 45^{\circ}) = \dfrac{\tan\theta + 1}{1 - \tan\theta}$. Step 2: The equation becomes $\dfrac{1}{\tan\theta} \times \dfrac{\tan\theta + 1}{1 - \tan\theta} = 7$, i.e. $\dfrac{1 + \tan\theta}{\tan\theta(1 - \tan\theta)} = 7$. Step 3: Cross-multiply: $1 + \tan\theta = 7\tan\theta - 7\tan^2\theta$, i.e. $7\tan^2\theta - 6\tan\theta + 1 = 0$. Step 4: Quadratic formula: $\tan\theta = \dfrac{6 \pm \sqrt{36 - 28}}{14} = \dfrac{6 \pm 2\sqrt{2}}{14} = \dfrac{3 \pm \sqrt{2}}{7}$. Step 5: $\tan\theta = \dfrac{3 - \sqrt{2}}{7} \Rightarrow \theta = 12.8^{\circ}$; $\tan\theta = \dfrac{3 + \sqrt{2}}{7} \Rightarrow \theta = 32.2^{\circ}$.

Question 6 · Answer A

Step 1: At $B$, $y = 0$: $8\mathrm{e}^{-x/2} - 1 = 0 \Rightarrow 8\mathrm{e}^{-x/2} = 1 \Rightarrow \mathrm{e}^{-x/2} = \dfrac{1}{8}$. Step 2: Take natural log: $-\dfrac{x}{2} = \ln\dfrac{1}{8} = -\ln 8 = -3\ln 2$. Step 3: Multiply by $-2$: $x = 6\ln 2$.

Question 7 · Answer A

Step 1: Line $AB$ has gradient $\dfrac{0 - 7}{6\ln 2 - 0} = -\dfrac{7}{6\ln 2}$, so $y = 7 - \dfrac{7x}{6\ln 2}$. Step 2: Area $= \displaystyle\int_0^{6\ln 2}\!\left[(8\mathrm{e}^{-x/2} - 1) - \left(7 - \dfrac{7x}{6\ln 2}\right)\right]\mathrm{d}x = \int_0^{6\ln 2}\!\left(8\mathrm{e}^{-x/2} - 8 + \dfrac{7x}{6\ln 2}\right)\mathrm{d}x$. Step 3: Integrate: $\left[-16\mathrm{e}^{-x/2} - 8x + \dfrac{7x^2}{12\ln 2}\right]_0^{6\ln 2}$. Step 4: At $x = 6\ln 2$: $-16\mathrm{e}^{-3\ln 2} - 48\ln 2 + \dfrac{7(6\ln 2)^2}{12\ln 2} = -2 - 48\ln 2 + 21\ln 2 = -2 - 27\ln 2$. At $x = 0$: $-16 - 0 + 0 = -16$. Step 5: Area $= (-2 - 27\ln 2) - (-16) = 14 - 27\ln 2$. Taking absolute value (since curve is above line): Area $= 27\ln 2 - 14$.

Question 8 · Answer A

Step 1: $\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = \sec^2\theta$ and $\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = \cos\theta - 6\sin^2\theta\cos\theta = \cos\theta(1 - 6\sin^2\theta)$. Step 2: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\cos\theta(1 - 6\sin^2\theta)}{\sec^2\theta} = \cos^3\theta(1 - 6\sin^2\theta)$. Step 3: Replace $\sin^2\theta = 1 - \cos^2\theta$: $\cos^3\theta(1 - 6 + 6\cos^2\theta) = \cos^3\theta(-5 + 6\cos^2\theta) = 6\cos^5\theta - 5\cos^3\theta$.

Question 9 · Answer A

Step 1: Find where the curve crosses the $x$-axis: $y = 0 \Rightarrow \sin\theta(1 - 2\sin^2\theta) = 0$. Since $0 < \theta < \dfrac{\pi}{2}$, $\sin\theta \ne 0$, so $\sin^2\theta = \dfrac{1}{2}$, giving $\theta = \dfrac{\pi}{4}$. Step 2: At $\theta = \dfrac{\pi}{4}$: $x = \tan\dfrac{\pi}{4} = 1$ and $y = 0$. The point is $(1, 0)$. Step 3: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6\left(\dfrac{1}{\sqrt{2}}\right)^{\!5} - 5\left(\dfrac{1}{\sqrt{2}}\right)^{\!3} = \dfrac{6}{4\sqrt{2}} - \dfrac{5}{2\sqrt{2}} = \dfrac{6 - 10}{4\sqrt{2}} = -\dfrac{4}{4\sqrt{2}} = -\dfrac{1}{\sqrt{2}}$. Step 4: Normal gradient $= -\dfrac{1}{(-1/\sqrt{2})} = \sqrt{2}$. Step 5: Normal equation: $y - 0 = \sqrt{2}(x - 1)$, i.e. $y = \sqrt{2}\,x - \sqrt{2}$.

Question 10 · Answer A

Step 1: Since $(x + 2)$ is a factor, $p(-2) = 0$. Step 2: $p(-2) = 2(-2)^4 + k(-2)^3 + k(-2)^2 + 17(-2) + 18 = 32 - 8k + 4k - 34 + 18 = 0$. Step 3: $-4k + 16 = 0 \Rightarrow k = 4$.

Question 11 · Answer A

Step 1: Divide $p(x) = 2x^4 + 4x^3 + 4x^2 + 17x + 18$ by $(x + 2)$ using polynomial long division or synthetic division. Step 2: $2x^4 \div x = 2x^3$. Multiply: $2x^3(x + 2) = 2x^4 + 4x^3$. Subtract: $4x^2 + 17x + 18$. Step 3: $4x^2 \div x = 0x^2$... Actually: remainder after first step is $0x^3 + 4x^2 + 17x + 18$. Next: $0x^2(x+2) = 0$. Then $4x^2 \div x = 4x$... Correction: the quotient has no $x^2$ term. Continuing: $4x(x+2) = 4x^2 + 8x$. Subtract: $9x + 18$. Then $9(x + 2) = 9x + 18$. Remainder $= 0$. Step 4: Quotient is $2x^3 + 0x^2 + 4x + 9 = 2x^3 + 4x + 9$. Step 5: For the non-integer root: $2x^3 + 4x + 9 = 0 \Rightarrow x^3 = -2x - 4.5 \Rightarrow x = \sqrt[3]{-2x - 4.5}$.

Question 12 · Answer A

Step 1: Let $f(x) = 2x^3 + 4x + 9$. Step 2: $f(-1.4) = 2(-1.4)^3 + 4(-1.4) + 9 = 2(-2.744) - 5.6 + 9 = -5.488 - 5.6 + 9 = -2.088$. Step 3: $f(-1.0) = 2(-1)^3 + 4(-1) + 9 = -2 - 4 + 9 = 3$. Step 4: Since $f(-1.4) < 0$ and $f(-1.0) > 0$, there is a sign change, confirming a root between $-1.4$ and $-1.0$.

Question 13 · Answer A

Step 1: $x_0 = -1$. Step 2: $x_1 = \sqrt[3]{-2(-1) - 4.5} = \sqrt[3]{2 - 4.5} = \sqrt[3]{-2.5} = -1.3572\ldots$ Step 3: $x_2 = \sqrt[3]{-2(-1.3572) - 4.5} = \sqrt[3]{2.7145 - 4.5} = \sqrt[3]{-1.7855} = -1.2134\ldots$ Step 4: $x_3 = \sqrt[3]{-2(-1.2134) - 4.5} = \sqrt[3]{2.4268 - 4.5} = \sqrt[3]{-2.0732} = -1.2740\ldots$ Step 5: Continue iterating until values agree to 3 s.f.: $\beta = -1.26$.

Question 14 · Answer A

Step 1: Let $u = 4\mathrm{e}^{1-2x}$ and $v = \sqrt{3x-1} = (3x-1)^{1/2}$. Step 2: $u' = -8\mathrm{e}^{1-2x}$ and $v' = \dfrac{3}{2\sqrt{3x-1}}$. Step 3: Product rule: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -8\mathrm{e}^{1-2x}\sqrt{3x-1} + 4\mathrm{e}^{1-2x} \times \dfrac{3}{2\sqrt{3x-1}}$. Step 4: Set $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$. Factor out $\dfrac{\mathrm{e}^{1-2x}}{\sqrt{3x-1}}$: $-8(3x-1) + 6 = 0$. Step 5: $-24x + 8 + 6 = 0 \Rightarrow -24x + 14 = 0 \Rightarrow x = \dfrac{7}{12}$. Step 6: $y = 4\mathrm{e}^{1 - 7/6}\sqrt{7/4 - 1} = 4\mathrm{e}^{-1/6}\sqrt{3/4} = 4 \times \dfrac{\sqrt{3}}{2} \times \mathrm{e}^{-1/6} = 2\sqrt{3}\,\mathrm{e}^{-1/6}$.