A Level Mathematics 9709/13 October/November 2025 Past Paper with Answers

Question 1

What is the coefficient of $x^2$ in the expansion of $\left(2 - \dfrac{x}{2}\right)^6$?

1. $60$
2. $-60$
3. $96$
4. $15$

Question 2

Given that $\left(2 - \dfrac{x}{2}\right)^6 = 64 - 96x + 60x^2 - 20x^3 + \ldots$, find the coefficient of $x^3$ in the expansion of $(3 - x + 2x^3)\left(2 - \dfrac{x}{2}\right)^6$.

1. $8$
2. $-8$
3. $68$
4. $-60$

Question 3

What is the solution of $\tan^{-1}(3x - 1) = -\dfrac{\pi}{4}$?

1. $x = 0$
2. $x = \dfrac{2}{3}$
3. $x = -\dfrac{2}{3}$
4. $x = \dfrac{1}{3}$

Question 4 · Answer A

Step 1: Replace $\cos^2\theta$ with $1 - \sin^2\theta$: $3(1 - \sin^2\theta) = 2\sin\theta + 2$. Step 2: Rearrange: $3\sin^2\theta + 2\sin\theta - 1 = 0$. Step 3: Factorise: $(3\sin\theta - 1)(\sin\theta + 1) = 0$, so $\sin\theta = \dfrac{1}{3}$ or $\sin\theta = -1$. Step 4: For $\sin\theta = \dfrac{1}{3}$ in $[0, 2\pi]$: two solutions ($\theta \approx 0.3398$ and $\theta \approx \pi - 0.3398 \approx 2.802$). Step 5: For $\sin\theta = -1$: one solution ($\theta = \dfrac{3\pi}{2}$). Step 6: Total: $3$ solutions.

Question 5 · Answer A

Step 1: Compute $8^{2/3} = (8^{1/3})^2 = 2^2 = 4$. Step 2: Compute $(8 - 2)^2 = 6^2 = 36$. Step 3: $f(8) = \dfrac{1}{2} \times 4 \times 36 = 72$.

Question 6 · Answer A

Step 1: Gradient $= \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{12.719 - 12}{5.1 - 5} = \dfrac{0.719}{0.1} = 7.19$. Step 2: Correct to 1 decimal place: $7.2$.

Question 7 · Answer A

Step 1: As the second point gets closer to $A$ (from $D$ to $C$ to $B$), the chord gradient approaches $30$: $30.389 \to 30.153 \to 30.030$. Step 2: In the limit, the gradient of the tangent at $A$ is $f'(8) = 30$.

Question 8 · Answer A

Step 1: For an AP, the common difference is constant: $k - 24 = (k - 7) - k = -7$. Step 2: So $k - 24 = -7$, giving $k = 17$ and $d = -7$. Step 3: The 20th term: $a + 19d = 24 + 19(-7) = 24 - 133 = -109$.

Question 9 · Answer A

Step 1: For a GP, $\dfrac{k}{12} = \dfrac{k-3}{k}$, so $k^2 = 12(k - 3) = 12k - 36$. Step 2: $k^2 - 12k + 36 = 0$, so $(k - 6)^2 = 0$, giving $k = 6$. Step 3: Common ratio $r = \dfrac{6}{12} = \dfrac{1}{2}$. Since $|r| < 1$, the sum to infinity exists. Step 4: $S_\infty = \dfrac{a}{1 - r} = \dfrac{12}{1 - 1/2} = \dfrac{12}{1/2} = 24$.

Question 10 · Answer A

Step 1: Apply the cosine rule: $(6\sqrt{3})^2 = 6^2 + 6^2 - 2(6)(6)\cos\theta$. Step 2: $108 = 72 - 72\cos\theta$, so $72\cos\theta = -36$, $\cos\theta = -\dfrac{1}{2}$. Step 3: $\theta = \dfrac{2\pi}{3}$.

Question 11 · Answer A

Step 1: Major sector angle $= 2\pi - \dfrac{2\pi}{3} = \dfrac{4\pi}{3}$. Step 2: Major sector area $= \dfrac{1}{2}r^2\theta = \dfrac{1}{2}(25)\left(\dfrac{4\pi}{3}\right) = \dfrac{50\pi}{3}$. Step 3: Triangle $OAB$ area $= \dfrac{1}{2}r^2\sin\left(\dfrac{2\pi}{3}\right) = \dfrac{1}{2}(25)\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{25\sqrt{3}}{4}$. Step 4: Major segment $=$ major sector $+$ triangle $= \dfrac{50\pi}{3} + \dfrac{25\sqrt{3}}{4}$.

Question 12 · Answer A

Step 1: Start with $y = x^3$. Apply stretch factor 3 parallel to $y$-axis: $y = 3x^3$. Step 2: Then translate by $\begin{pmatrix} 2 \\ 6 \end{pmatrix}$: replace $x$ by $(x-2)$ and add 6, giving $y = 3(x-2)^3 + 6$. Step 3: This matches the target, confirming: stretch factor 3 in $y$, then translate $\begin{pmatrix} 2 \\ 6 \end{pmatrix}$.

Question 13 · Answer A

Step 1: Write $f(x) = 5(x - 4)^3 + 7$ in the form $a(x + b)^3 + c$. Step 2: Compare: $x - 4 = x + (-4)$, so $b = -4$. Step 3: Also $a = 5$ and $c = 7$.

Question 14 · Answer A

Step 1: Let $y = \dfrac{2}{4x - 3} + \dfrac{1}{2}$. Rearrange: $y - \dfrac{1}{2} = \dfrac{2}{4x - 3}$. Step 2: $4x - 3 = \dfrac{2}{y - 1/2} = \dfrac{2}{(2y - 1)/2} = \dfrac{4}{2y - 1}$. Step 3: $4x = \dfrac{4}{2y - 1} + 3$, so $x = \dfrac{1}{2y - 1} + \dfrac{3}{4}$. Step 4: Swap $x$ and $y$: $g^{-1}(x) = \dfrac{1}{2x - 1} + \dfrac{3}{4}$.

Question 15 · Answer A

Step 1: Centre $= \left(\dfrac{5 + (-3)}{2}, \dfrac{-1 + 5}{2}\right) = (1, 2)$. Step 2: Radius $= $ distance from centre to either endpoint $= \sqrt{(5 - 1)^2 + (-1 - 2)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.

Question 16 · Answer A

Step 1: The normal to the tangent has gradient $2$ (negative reciprocal of $-1/2$). Step 2: Line through centre $(3, 1)$ with gradient $2$: $y - 1 = 2(x - 3)$, so $y = 2x - 5$. Step 3: Substitute into the circle equation: $(x - 3)^2 + (2x - 5 - 1)^2 = 20$. Step 4: $(x - 3)^2 + 4(x - 3)^2 = 20$, so $5(x - 3)^2 = 20$, $(x - 3)^2 = 4$. Step 5: $x = 3 \pm 2$, giving $x = 1$ or $x = 5$.

Question 17 · Answer A

Step 1: Volume $= \pi\int_1^8\left[\left(\dfrac{9}{2}\right)^2 - \left(\dfrac{x}{2} + \dfrac{4}{x}\right)^2\right]\mathrm{d}x$. Step 2: Expand $\left(\dfrac{x}{2} + \dfrac{4}{x}\right)^2 = \dfrac{x^2}{4} + 4 + \dfrac{16}{x^2}$. Step 3: Integrand $= \dfrac{81}{4} - \dfrac{x^2}{4} - 4 - \dfrac{16}{x^2} = \dfrac{65}{4} - \dfrac{x^2}{4} - \dfrac{16}{x^2}$. Step 4: Integrate: $\int_1^8\left(\dfrac{65}{4} - \dfrac{x^2}{4} - \dfrac{16}{x^2}\right)\mathrm{d}x = \left[\dfrac{65x}{4} - \dfrac{x^3}{12} + \dfrac{16}{x}\right]_1^8$. Step 5: At $x = 8$: $\dfrac{520}{4} - \dfrac{512}{12} + 2 = 130 - \dfrac{128}{3} + 2 = 132 - \dfrac{128}{3} = \dfrac{396 - 128}{3} = \dfrac{268}{3}$. Step 6: At $x = 1$: $\dfrac{65}{4} - \dfrac{1}{12} + 16 = \dfrac{195}{12} - \dfrac{1}{12} + \dfrac{192}{12} = \dfrac{386}{12} = \dfrac{193}{6}$. Step 7: Difference $= \dfrac{268}{3} - \dfrac{193}{6} = \dfrac{536 - 193}{6} = \dfrac{343}{6}$. Step 8: Volume $= \dfrac{343\pi}{6}$.

Question 18 · Answer A

Step 1: Factor out the coefficient of $x^2$: $3(x^2 + 4x) + 7$. Step 2: Complete the square inside the bracket: $3(x^2 + 4x + 4 - 4) + 7 = 3((x + 2)^2 - 4) + 7$. Step 3: Expand: $3(x + 2)^2 - 12 + 7 = 3(x + 2)^2 - 5$. Step 4: So $a = 3$, $b = 2$, $c = -5$.

Question 19 · Answer A

Step 1: Since $(x + 2)^2 \ge 0$ for all real $x$, the minimum value of $3(x+2)^2$ is $0$. Step 2: Therefore the minimum value of $f(x) = 3(x+2)^2 - 5$ is $0 - 5 = -5$, achieved when $x = -2$. Step 3: Range: $f(x) \ge -5$.

Question 20 · Answer A

Step 1: By Vieta's formulas for $px^2 + 4x - 8 = 0$: sum of roots $= -4/p$ and product of roots $= -8/p$. Step 2: Sum: $4m + (-6m) = -2m = -4/p$, so $m = 2/p$. Step 3: Product: $(4m)(-6m) = -24m^2 = -8/p$, so $24m^2 = 8/p$, giving $3m^2 = 1/p$. Step 4: Substitute $m = 2/p$: $3 \times 4/p^2 = 1/p$, so $12/p^2 = 1/p$, giving $p = 12$. Step 5: Then $m = 2/12 = 1/6$.

Question 21 · Answer A

Step 1: Differentiate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-24}{(3x - 8)^2} + \dfrac{6}{(x - 1)^2}$. Step 2: At $x = 3$: $\dfrac{-24}{(9 - 8)^2} + \dfrac{6}{(3 - 1)^2} = \dfrac{-24}{1} + \dfrac{6}{4} = -24 + \dfrac{3}{2} = -\dfrac{45}{2}$.

Question 22 · Answer A

Step 1: Set $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$: $\dfrac{-24}{(3x-8)^2} + \dfrac{6}{(x-1)^2} = 0$. Step 2: Rearrange: $\dfrac{6}{(x-1)^2} = \dfrac{24}{(3x-8)^2}$, so $\dfrac{(3x-8)^2}{(x-1)^2} = 4$. Step 3: Take square roots: $\dfrac{3x - 8}{x - 1} = \pm 2$. Step 4: Case $+2$: $3x - 8 = 2x - 2$, so $x = 6$. Step 5: Case $-2$: $3x - 8 = -2x + 2$, so $5x = 10$, $x = 2$. Step 6: $p + q = 2 + 6 = 8$.

Question 23 · Answer A

Step 1: Find $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$. Differentiate $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -24(3x-8)^{-2} + 6(x-1)^{-2}$. Step 2: $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{144}{(3x-8)^3} - \dfrac{12}{(x-1)^3}$. Step 3: At $x = 2$: $\dfrac{144}{(6-8)^3} - \dfrac{12}{(2-1)^3} = \dfrac{144}{-8} - \dfrac{12}{1} = -18 - 12 = -30$. Step 4: Since $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -30 < 0$, the stationary point at $x = 2$ is a maximum.