A Level Mathematics 9709/22 October/November 2025 Past Paper with Answers

Question 1

Solve the equation $\ln(4x + 7) - \ln(x - 1) = 3$, giving your answer in exact form. Which of the following is the correct solution?

1. $x = \dfrac{\mathrm{e}^3 + 7}{4 - \mathrm{e}^3}$
2. $x = \dfrac{\mathrm{e}^3 + 7}{\mathrm{e}^3 - 4}$
3. $x = \dfrac{7 - \mathrm{e}^3}{\mathrm{e}^3 - 4}$
4. $x = \dfrac{\mathrm{e}^3 + 7}{4 + \mathrm{e}^3}$

Question 2

How many solutions does the equation $2\tan^2\theta + 3\sec\theta = 18$ have for $-180^\circ < \theta < 180^\circ$?

1. $2$
2. $3$
3. $4$
4. $5$

Question 3

Solve the inequality $|3x - 4| \leqslant |2x + 5|$. What is the solution set?

1. $-\dfrac{1}{5} \leqslant x \leqslant 9$
2. $x \leqslant 9$
3. $x \geqslant -\dfrac{1}{5}$
4. $-9 \leqslant x \leqslant \dfrac{1}{5}$

Question 4 · Answer A

Step 1: Let $x = 7^{0.01N}$. The inequality becomes $|3x - 4| \leqslant |2x + 5|$, which has solution $-\dfrac{1}{5} \leqslant x \leqslant 9$. Step 2: Since $7^{0.01N} > 0$ always, the binding constraint is $7^{0.01N} \leqslant 9$. Step 3: Take $\log_7$: $0.01N \leqslant \log_7 9 = \dfrac{\ln 9}{\ln 7} = 1.1292\ldots$ Step 4: $N \leqslant 112.92\ldots$, so the largest integer is $N = 112$.

Question 5 · Answer A

Step 1: Divide $x^4$ by $x^2$ to get $x^2$. Multiply: $x^2(x^2+3) = x^4 + 3x^2$. Step 2: Subtract: $(x^4 - 10x^3 + 20x^2 - 30x + 40) - (x^4 + 3x^2) = -10x^3 + 17x^2 - 30x + 40$. Step 3: Divide $-10x^3$ by $x^2$ to get $-10x$. Multiply: $-10x(x^2+3) = -10x^3 - 30x$. Step 4: Subtract: $(-10x^3 + 17x^2 - 30x + 40) - (-10x^3 - 30x) = 17x^2 + 40$. Step 5: Divide $17x^2$ by $x^2$ to get $17$. Multiply: $17(x^2+3) = 17x^2 + 51$. Step 6: Subtract: $(17x^2 + 40) - (17x^2 + 51) = -11$. Quotient: $x^2 - 10x + 17$, Remainder: $-11$.

Question 6 · Answer A

Step 1: $p(x) + 11 = 0$ means $(x^2 + 3)(x^2 - 10x + 17) = 0$. Step 2: $x^2 + 3 = 0$ gives $x^2 = -3$, which has no real roots. Step 3: Solve $x^2 - 10x + 17 = 0$ using the quadratic formula: $x = \dfrac{10 \pm \sqrt{100 - 68}}{2} = \dfrac{10 \pm \sqrt{32}}{2} = \dfrac{10 \pm 4\sqrt{2}}{2} = 5 \pm 2\sqrt{2}$.

Question 7 · Answer A

Step 1: Differentiate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -8\sin 2x + 8\cos x$. Step 2: Use $\sin 2x = 2\sin x\cos x$: $-16\sin x\cos x + 8\cos x = 0$. Step 3: Factor: $8\cos x(1 - 2\sin x) = 0$. Step 4: $\cos x = 0 \Rightarrow x = \pi/2$ (check: this is a minimum). $\sin x = 1/2 \Rightarrow x = \pi/6$ or $5\pi/6$ (these are maxima). Step 5: At $x = \pi/6$: $y = 4\cos(\pi/3) + 8\sin(\pi/6) = 4(1/2) + 8(1/2) = 2 + 4 = 6$. Step 6: At $x = 5\pi/6$: $y = 4\cos(5\pi/3) + 8\sin(5\pi/6) = 4(1/2) + 8(1/2) = 2 + 4 = 6$. Both maximum points have $y$-coordinate $6$.

Question 8 · Answer A

Step 1: The line $AB$ is $y = 6$ (horizontal line between the two maxima). Step 2: Area under line = $6 \times \left(\dfrac{5\pi}{6} - \dfrac{\pi}{6}\right) = 6 \times \dfrac{2\pi}{3} = 4\pi$. Step 3: Area under curve: $\displaystyle\int_{\pi/6}^{5\pi/6} (4\cos 2x + 8\sin x)\,\mathrm{d}x = \left[2\sin 2x - 8\cos x\right]_{\pi/6}^{5\pi/6}$. Step 4: At $x = 5\pi/6$: $2\sin(5\pi/3) - 8\cos(5\pi/6) = 2(-\sqrt{3}/2) - 8(-\sqrt{3}/2) = -\sqrt{3} + 4\sqrt{3} = 3\sqrt{3}$. Step 5: At $x = \pi/6$: $2\sin(\pi/3) - 8\cos(\pi/6) = 2(\sqrt{3}/2) - 8(\sqrt{3}/2) = \sqrt{3} - 4\sqrt{3} = -3\sqrt{3}$. Step 6: Integral $= 3\sqrt{3} - (-3\sqrt{3}) = 6\sqrt{3}$. Step 7: Shaded area $= 4\pi - 6\sqrt{3}$.

Question 9 · Answer A

Step 1: $\displaystyle\int \tfrac{1}{2}\mathrm{e}^{2x}\,\mathrm{d}x = \tfrac{1}{2} \times \tfrac{1}{2}\mathrm{e}^{2x} = \tfrac{1}{4}\mathrm{e}^{2x}$. Step 2: $\displaystyle\int \tfrac{1}{4}\mathrm{e}^{-x}\,\mathrm{d}x = \tfrac{1}{4} \times (-1)\mathrm{e}^{-x} = -\tfrac{1}{4}\mathrm{e}^{-x}$. Step 3: Antiderivative: $\tfrac{1}{4}\mathrm{e}^{2x} - \tfrac{1}{4}\mathrm{e}^{-x}$.

Question 10 · Answer A

Step 1: $f(1.0) = \tfrac{1}{2}\ln(10 + \tfrac{1}{2}\mathrm{e}^{-1} + \tfrac{1}{2}\mathrm{e}^{-4}) - 1 \approx 1.161 - 1 = 0.161 > 0$. Step 2: $f(1.2) = \tfrac{1}{2}\ln(10 + \tfrac{1}{2}\mathrm{e}^{-1.2} + \tfrac{1}{2}\mathrm{e}^{-4.8}) - 1.2 \approx 1.157 - 1.2 = -0.043 < 0$. Step 3: The sign change from positive to negative confirms a root exists between $1.0$ and $1.2$.

Question 11 · Answer A

Step 1: $a_0 = 1$. Step 2: $a_1 = \tfrac{1}{2}\ln(10 + \tfrac{1}{2}\mathrm{e}^{-1} + \tfrac{1}{2}\mathrm{e}^{-4}) = 1.1613\ldots$ Step 3: $a_2 = \tfrac{1}{2}\ln(10 + \tfrac{1}{2}\mathrm{e}^{-1.1613} + \tfrac{1}{2}\mathrm{e}^{-4.6452}) = 1.1578\ldots$ Step 4: $a_3 = 1.1586\ldots$, $a_4 = 1.1584\ldots$ Step 5: Converges to $a = 1.159$ (4 s.f.).

Question 12 · Answer A

Step 1: Differentiate implicitly: $10xy + 5x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 8\mathrm{e}^{2y}\dfrac{\mathrm{d}y}{\mathrm{d}x} - 7 = 0$. Step 2: Solve for $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{7 - 10xy}{5x^2 + 8\mathrm{e}^{2y}}$. Step 3: Find $x$ when $y = 0$: $0 + 4 - 7x + 10 = 0$, so $x = 2$. Step 4: Substitute $(2, 0)$: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{7 - 0}{20 + 8} = \dfrac{7}{28} = \dfrac{1}{4}$.

Note: "$\dfrac{7}{28}$" gives $\dfrac{7}{28}$, which equals $\dfrac{1}{4}$ but is not fully simplified. The correct simplified answer is $\dfrac{1}{4}$.

Question 13 · Answer A

Step 1: A tangent parallel to the $y$-axis would require $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ to be undefined, which happens when the denominator is zero. Step 2: The denominator is $5x^2 + 8\mathrm{e}^{2y}$. Since $5x^2 \geqslant 0$ for all real $x$ and $8\mathrm{e}^{2y} > 0$ for all real $y$, the sum is always at least $8$. Step 3: Therefore $5x^2 + 8\mathrm{e}^{2y} \geqslant 8 > 0$, the denominator is never zero, $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ is always defined, and no tangent can be vertical.