A Level Mathematics 9709/12 May/June 2025 Past Paper with Answers

Question 1

Two curves $y = f(x)$ and $y = g(x)$ are related by a sequence of two transformations. The first transformation is a stretch with factor $3$ parallel to the $y$-axis. The second transformation is a translation of $\begin{pmatrix}0\\-5\end{pmatrix}$. If the point $(2, 4)$ lies on $y = f(x)$, what is the image of this point on $y = g(x)$?

1. $(2, 7)$
2. $(2, -1)$
3. $(2, 12)$
4. $(6, -1)$

Question 2

Find the points of intersection of the curve $2xy + 3y^2 = 20$ and the line $2x + y + 2 = 0$. What is the sum of the $y$-coordinates of the intersection points?

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

Question 3

The coefficient of $x^7$ in the expansion of $\left(qx^2 + \dfrac{3}{q}x\right)^5$ is $810$. What is the value of $q$?

1. $\dfrac{1}{3}$
2. $3$
3. $\dfrac{1}{9}$
4. $9$

Question 4 · Answer A

Step 1: Differentiate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{2}bx^{1/2} - 8$. Step 2: At $x = 16$: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{2}b(4) - 8 = 6b - 8$. Step 3: By the chain rule: $\dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{\mathrm{d}y}{\mathrm{d}x} \times \dfrac{\mathrm{d}x}{\mathrm{d}t} = (6b - 8) \times 4 = 24b - 32$.

Question 5 · Answer A

Step 1: Differentiate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{2}cx^{1/2} - 18$. Step 2: At a stationary point, $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$: $\dfrac{3}{2}c(9)^{1/2} - 18 = 0$. Step 3: $\dfrac{3}{2}c(3) = 18$, so $\dfrac{9c}{2} = 18$, giving $c = 4$.

Question 6 · Answer A

Step 1: The range of $\cos 3x$ is $[-1, 1]$. Step 2: The minimum of $5\cos 3x$ is $5(-1) = -5$. Step 3: The least value of $y = 5\cos 3x + 2$ is $-5 + 2 = -3$.

Question 7 · Answer A

Step 1: The period of $\cos 2x$ is $\dfrac{2\pi}{2} = \pi$. Step 2: In the interval $0 \le x \le 2\pi$, the number of complete cycles is $\dfrac{2\pi}{\pi} = 2$.

Question 8 · Answer A

Step 1: The equation $4\cos 2x + 3 = 2x - 1$ asks where the curve $y = 4\cos 2x + 3$ meets the line $y = 2x - 1$. Step 2: The curve oscillates between $-1$ and $7$ with period $\pi$, completing two cycles in $[0, 2\pi]$. Step 3: The line $y = 2x - 1$ passes through $(0, -1)$ and $(2\pi, 2\pi \cdot 2 - 1) \approx (6.28, 11.57)$, rising steadily. Step 4: By considering the graph, the line intersects the curve $3$ times.

Question 9 · Answer A

Step 1: At $P$: $5 - 2x = 3 \Rightarrow x = 1$. Check: $\dfrac{6}{\sqrt{4}} = 3$. The line meets the $x$-axis where $5 - 2x = 0$, i.e. $x = \dfrac{5}{2}$. Step 2: Area under the curve from $0$ to $1$: $\int_0^1 6(3x+1)^{-1/2}\,dx = 6 \cdot \dfrac{2}{3}(3x+1)^{1/2}\Big|_0^1 = 4(\sqrt{4} - \sqrt{1}) = 4$. Step 3: Area under the line from $1$ to $\dfrac{5}{2}$: $\int_1^{5/2}(5-2x)\,dx = \left[5x - x^2\right]_1^{5/2} = \left(\dfrac{25}{2} - \dfrac{25}{4}\right) - (5 - 1) = \dfrac{25}{4} - 4 = \dfrac{9}{4}$. Step 4: Total area $= 4 + \dfrac{9}{4} = \dfrac{25}{4}$.

Question 10 · Answer A

Step 1: Write $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, so the expression becomes $\dfrac{\frac{\sin\theta}{\cos\theta} + 7}{\frac{\sin^2\theta}{\cos^2\theta} - 3}$. Step 2: Multiply numerator and denominator by $\cos^2\theta$: numerator $= \sin\theta\cos\theta + 7\cos^2\theta$, denominator $= \sin^2\theta - 3\cos^2\theta$. Step 3: Use $\sin^2\theta = 1 - \cos^2\theta$: denominator $= 1 - \cos^2\theta - 3\cos^2\theta = 1 - 4\cos^2\theta$.

Question 11 · Answer A

Step 1: Use $\sin^2\theta = 1 - \cos^2\theta$: $2(1 - \cos^2\theta) + 3\cos\theta - 3 = 0$. Step 2: Simplify: $-2\cos^2\theta + 3\cos\theta - 1 = 0$, i.e. $2\cos^2\theta - 3\cos\theta + 1 = 0$. Step 3: Factorise: $(2\cos\theta - 1)(\cos\theta - 1) = 0$. Step 4: $\cos\theta = \frac{1}{2}$ gives $\theta = 60°, 300°$. $\cos\theta = 1$ gives $\theta = 0°$ (but $360°$ boundary). Total: $3$ solutions.

Question 12 · Answer A

Step 1: The centre is $(5, -2)$ and radius $= \sqrt{25 + 4 - 9} = \sqrt{20}$. Step 2: Substitute $y = 0$: $x^2 - 10x + 9 = 0$, so $(x-1)(x-9) = 0$. Step 3: The points of intersection are $(1, 0)$ and $(9, 0)$. Step 4: Distance $= 9 - 1 = 8$.

Question 13 · Answer A

Step 1: $CA = \sqrt{(5-1)^2 + (-2-0)^2} = \sqrt{16+4} = \sqrt{20}$. Similarly $CB = \sqrt{(5-9)^2 + (-2)^2} = \sqrt{20}$. Step 2: The midpoint $M$ of $AB$ is $(5, 0)$. $CM = 2$ (vertical distance). $AM = 4$ (half the chord). Step 3: $\tan(\angle ACM) = \dfrac{AM}{CM} = \dfrac{4}{2} = 2$, so $\angle ACM = \tan^{-1}(2) = 1.1071$ rad. Step 4: $\angle ACB = 2\angle ACM = 2(1.1071) = 2.21$ rad (3 s.f.).

Question 14 · Answer A

Step 1: Smaller sector area $= \dfrac{1}{2}r^2\theta = \dfrac{1}{2}(20)(2.21) = 22.1$. Step 2: Triangle $ACB$ area: base $AB = 8$, height from $C$ to $AB$ is $2$, so area $= \dfrac{1}{2}(8)(2) = 8$. Step 3: Smaller segment $= 22.1 - 8 = 14.1$. Step 4: Circle area $= \pi(20) = 62.8$. Step 5: Larger segment $= 62.8 - 14.1 = 48.7$.

Question 15 · Answer A

Step 1: Integrate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \int -18x^{-3}\,dx = \dfrac{-18x^{-2}}{-2} + c = \dfrac{9}{x^2} + c$. Step 2: At the stationary point $(-3, 10)$, $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$: $\dfrac{9}{9} + c = 0$, so $1 + c = 0$, giving $c = -1$. Step 3: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{9}{x^2} - 1$.

Question 16 · Answer A

Step 1: Set $\dfrac{dy}{dx} = 0$: $\dfrac{9}{x^2} - 1 = 0$, so $x^2 = 9$, giving $x = 3$ or $x = -3$. Step 2: The other stationary point is at $x = 3$. Step 3: $\dfrac{d^2y}{dx^2} = -\dfrac{18}{x^3}$. At $x = 3$: $\dfrac{d^2y}{dx^2} = -\dfrac{18}{27} = -\dfrac{2}{3} < 0$. Step 4: Since $\dfrac{d^2y}{dx^2} < 0$, the stationary point at $x = 3$ is a maximum.

Question 17 · Answer A

Step 1: Integrate: $y = \int \left(\dfrac{9}{x^2} - 1\right)dx = -\dfrac{9}{x} - x + d$. Step 2: Substitute $(-3, 10)$: $10 = -\dfrac{9}{-3} - (-3) + d = 3 + 3 + d = 6 + d$. Step 3: $d = 10 - 6 = 4$.

Question 18 · Answer A

Step 1: The gradient of the tangent is $-\dfrac{7}{16}$. Step 2: The gradient of the normal is the negative reciprocal: $-\dfrac{1}{-7/16} = \dfrac{16}{7}$.

Question 19 · Answer A

Step 1: For an AP, the common difference is constant: $k^2 - 3k = 6k - k^2$. Step 2: $2k^2 - 9k = 0$, so $k(2k - 9) = 0$. Step 3: Since $k > 0$: $k = \dfrac{9}{2}$.

Question 20 · Answer A

Step 1: $a = \dfrac{27}{2}$, $d = \dfrac{9}{2}$. Step 2: $S_{20} = \dfrac{20}{2}\left(2 \times \dfrac{27}{2} + 19 \times \dfrac{9}{2}\right) = 10\left(27 + \dfrac{171}{2}\right) = 10 \times \dfrac{225}{2} = \dfrac{2250}{2} = 1125$.

Question 21 · Answer A

Step 1: $ar^3 = 24$ and $ar^5 = 8$. Dividing: $r^2 = 8/24 = 1/3$, so $r = 1/\sqrt{3}$ (positive). Step 2: $a = \dfrac{24}{r^3} = \dfrac{24}{(1/\sqrt{3})^3} = 24 \times 3\sqrt{3} = 72\sqrt{3}$. Step 3: $S_\infty = \dfrac{72\sqrt{3}}{1 - 1/\sqrt{3}} = \dfrac{72\sqrt{3} \times \sqrt{3}}{\sqrt{3} - 1} = \dfrac{216}{\sqrt{3} - 1}$.

Question 22 · Answer A

Step 1: $x^2 + 6x + 5 = (x + 3)^2 - 9 + 5 = (x + 3)^2 - 4$. Step 2: So $a = 3$ and $b = -4$.

Question 23 · Answer A

Step 1: Using the completed square form: $y = (x+3)^2 - 4$. Step 2: Rearrange: $(x+3)^2 = y + 4$, so $x + 3 = \pm\sqrt{y + 4}$. Step 3: Since $x \le -3$, we need $x + 3 \le 0$, so take the negative root: $x = -\sqrt{y+4} - 3$. Step 4: Swap $x$ and $y$: $f^{-1}(x) = -\sqrt{x + 4} - 3$.

Question 24 · Answer A

Step 1: $gf(x) = g(f(x)) = -(f(x)) - 5 = -(x^2 + 6x + 5) - 5 = -x^2 - 6x - 10$. Step 2: Using completed square: $gf(x) = -((x+3)^2 - 4) - 5 = -(x+3)^2 + 4 - 5 = -(x+3)^2 - 1$. Step 3: To find $(gf)^{-1}$: let $y = -(x+3)^2 - 1$. Then $(x+3)^2 = -y - 1$, so $x + 3 = -\sqrt{-y-1}$ (negative root since $x \le -3$). Step 4: $x = -\sqrt{-y-1} - 3$. Swap: $(gf)^{-1}(x) = -\sqrt{-x-1} - 3$.