A Level Mathematics 9709/22 May/June 2025 Past Paper with Answers

Question 1

Given that $\displaystyle\int_1^{9} \dfrac{10}{5x+3}\,\mathrm{d}x = \ln a$, what is the integer $a$?

1. $36$
2. $6$
3. $48$
4. $25$

Question 2

At what $x$-value does the graph of $y = |2x - 9|$ have its vertex?

1. $4.5$
2. $9$
3. $-4.5$
4. $2$

Question 3

Solve the inequality $|3x - 8| < 5x - 2$. What is the solution?

1. $x > \dfrac{5}{4}$
2. $x > 3$
3. $x > \dfrac{3}{4}$
4. $\dfrac{3}{4} < x < \dfrac{5}{4}$

Question 4 · Answer A

Step 1: Differentiate $\dfrac{8x}{2x+3}$ using the quotient rule: $\dfrac{8(2x+3) - 8x(2)}{(2x+3)^2} = \dfrac{24}{(2x+3)^2}$. Step 2: So $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{24}{(2x+3)^2} - 6$. Set to zero: $\dfrac{24}{(2x+3)^2} = 6$, so $(2x+3)^2 = 4$. Step 3: $2x + 3 = \pm 2$, giving $x = -\dfrac{1}{2}$ or $x = -\dfrac{5}{2}$. Step 4: The smaller $x$-value is $x = -\dfrac{5}{2}$. Substituting: $y = \dfrac{8(-5/2)}{2(-5/2)+3} - 6(-5/2) + 5 = \dfrac{-20}{-2} + 15 + 5 = 10 + 15 + 5 = 30$.

Question 5 · Answer A

Step 1: At intersection, $4\mathrm{e}^{-2x} = 1 + 0.5\sin 3x$. Step 2: Divide both sides by 4: $\mathrm{e}^{-2x} = 0.25 + 0.125\sin 3x$. Step 3: Take $\ln$ of both sides: $-2x = \ln(0.25 + 0.125\sin 3x)$. Step 4: Divide by $-2$: $x = -0.5\ln(0.25 + 0.125\sin 3x)$.

Question 6 · Answer A

Step 1: $x_0 = 0.5$. Step 2: $x_1 = -0.5\ln(0.25 + 0.125\sin 1.5) = -0.5\ln(0.25 + 0.125 \times 0.9975) = -0.5\ln(0.3747) = 0.4910\ldots$ Step 3: Continue iterating: $x_2 = 0.4912\ldots$, $x_3 = 0.4912\ldots$ Step 4: The iterations converge to $0.4912$ (4 s.f.).

Question 7 · Answer A

Step 1: The area is $\displaystyle\int_0^{0.4912}(4\mathrm{e}^{-2x} - 1 - 0.5\sin 3x)\,\mathrm{d}x$ since $4\mathrm{e}^{-2x}$ is above $1 + 0.5\sin 3x$ on this interval. Step 2: Integrate: $\left[-2\mathrm{e}^{-2x} - x + \dfrac{1}{6}\cos 3x\right]_0^{0.4912}$. Step 3: At $x = 0.4912$: $-2\mathrm{e}^{-0.9824} - 0.4912 + \dfrac{1}{6}\cos 1.4736 \approx -0.749 - 0.491 + 0.019 = -1.221$. Step 4: At $x = 0$: $-2(1) - 0 + \dfrac{1}{6}(1) = -2 + 0.1667 = -1.833$. Step 5: Area $= -1.221 - (-1.833) = 0.612 \approx 0.61$ (2 s.f.).

Question 8 · Answer A

Step 1: Since $(2x-1)$ is a factor, $p(1/2) = 0$: $\dfrac{a}{16} + \dfrac{b}{8} + \dfrac{13}{4} - \dfrac{35}{2} + 15 = 0$. Multiply by 16: $a + 2b + 52 - 280 + 240 = 0$, so $a + 2b = -12$. Step 2: Since $(x-3)$ is a factor, $p(3) = 0$: $81a + 27b + 117 - 105 + 15 = 0$, so $81a + 27b = -27$, i.e. $3a + b = -1$. Step 3: From $3a + b = -1$: $b = -1 - 3a$. Substituting into $a + 2b = -12$: $a + 2(-1-3a) = -12$, so $a - 2 - 6a = -12$, giving $-5a = -10$, hence $a = 2$.

Question 9 · Answer A

Step 1: First multiply the known factors: $(2x-1)(x-3) = 2x^2 - 7x + 3$. Step 2: Divide $p(x) = 2x^4 - 7x^3 + 13x^2 - 35x + 15$ by $2x^2 - 7x + 3$ using polynomial long division. Step 3: $\dfrac{2x^4 - 7x^3 + 13x^2 - 35x + 15}{2x^2 - 7x + 3} = x^2 + 5$. Step 4: So $p(x) = (2x-1)(x-3)(x^2 + 5)$.

Question 10 · Answer A

Step 1: $p(\cot 2\theta) = 0$ when $\cot 2\theta = \dfrac{1}{2}$, $\cot 2\theta = 3$, or $\cot^2 2\theta + 5 = 0$ (no real solutions). Step 2: $\cot 2\theta = \dfrac{1}{2}$ means $\tan 2\theta = 2$, so $2\theta = \arctan 2 = 1.1071\ldots$, giving $\theta = 0.5536\ldots \approx 0.554$. Step 3: $\cot 2\theta = 3$ means $\tan 2\theta = \dfrac{1}{3}$, so $2\theta = \arctan\left(\dfrac{1}{3}\right) = 0.32175\ldots$, giving $\theta = 0.16088\ldots \approx 0.161$. Step 4: The least positive value is $\theta \approx 0.161$.

Question 11 · Answer A

Step 1: Substitute $y = \mathrm{e}^{-1}$, so $\ln y = -1$: $(x^2 - 3)(-1) + 6x = 14$. Step 2: Simplify: $-x^2 + 3 + 6x = 14$, so $x^2 - 6x + 11 = 0$. Step 3: Discriminant $= (-6)^2 - 4(1)(11) = 36 - 44 = -8 < 0$. Step 4: Since the discriminant is negative, there are no real solutions for $x$, so no point on the curve has $y = \mathrm{e}^{-1}$.

For comparison, $y = \mathrm{e}^2$ gives $\ln y = 2$: $2(x^2-3) + 6x = 14 \Rightarrow 2x^2 + 6x - 20 = 0 \Rightarrow x^2 + 3x - 10 = 0$, discriminant $= 49 > 0$ (real solutions exist). $y = 1$ gives $\ln y = 0$: $6x = 14$, so $x = 7/3$ (real solution). $y = \mathrm{e}$ gives $\ln y = 1$: $x^2 - 3 + 6x = 14 \Rightarrow x^2 + 6x - 17 = 0$, discriminant $= 104 > 0$ (real solutions).

Question 12 · Answer A

Step 1: Differentiate $(x^2-3)\ln y + 6x = 14$ implicitly: $2x\ln y + (x^2-3)\cdot\dfrac{1}{y}\cdot\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6 = 0$. Step 2: Substitute $x = 2$, $y = \mathrm{e}^2$: $2(2)(2) + (4-3)\cdot\dfrac{1}{\mathrm{e}^2}\cdot\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6 = 0$. Step 3: $8 + \dfrac{1}{\mathrm{e}^2}\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6 = 0$, so $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -14\mathrm{e}^2$.

Question 13 · Answer A

Step 1: Expand $\sin(\theta + 30^\circ) = \sin\theta\cos 30^\circ + \cos\theta\sin 30^\circ = \dfrac{\sqrt{3}}{2}\sin\theta + \dfrac{1}{2}\cos\theta$. Step 2: $4\cos\theta\sin(\theta + 30^\circ) = 2\sqrt{3}\cos\theta\sin\theta + 2\cos^2\theta$. Step 3: Use double angle identities: $2\sqrt{3}\cos\theta\sin\theta = \sqrt{3}\sin 2\theta$ and $2\cos^2\theta = \cos 2\theta + 1$. Step 4: So $4\cos\theta\sin(\theta+30^\circ) = \sqrt{3}\sin 2\theta + \cos 2\theta + 1$. Step 5: Express $\sqrt{3}\sin 2\theta + \cos 2\theta$ as $R\cos(2\theta - \alpha)$: $R = \sqrt{(\sqrt{3})^2 + 1^2} = 2$, and $\tan\alpha = \dfrac{\sqrt{3}}{1}$, so $\alpha = 60^\circ$. Step 6: Result: $2\cos(2\theta - 60^\circ) + 1$, so $R = 2$, $k = 1$, and $R + k = 3$.

Question 14 · Answer A

Step 1: Write $12\cos 2\varphi\sin(2\varphi+30^\circ) = 3 \times 4\cos 2\varphi\sin(2\varphi+30^\circ)$. Step 2: Using the identity with $\theta = 2\varphi$: $3[2\cos(4\varphi-60^\circ)+1] = 5$. Step 3: Simplify: $6\cos(4\varphi-60^\circ) + 3 = 5$, so $\cos(4\varphi-60^\circ) = \dfrac{1}{3}$. Step 4: For $0^\circ < \varphi < 90^\circ$, we have $-60^\circ < 4\varphi - 60^\circ < 300^\circ$. Step 5: $4\varphi - 60^\circ = 70.53^\circ \Rightarrow \varphi = 32.6^\circ$. Step 6: $4\varphi - 60^\circ = 360^\circ - 70.53^\circ = 289.47^\circ \Rightarrow \varphi = 87.4^\circ$. Step 7: Both solutions are in the range, so there are $2$ solutions.