Cambridge AS & A Level. 2025. Mathematics (9709). May/June - Paper 1 (12) · Pure Mathematics 1

The diagram shows the graphs with equations $y = \mathrm{f}(x)$ and $y = \mathrm{g}(x)$.

Describe fully a sequence of two transformations which transforms the graph of $y = \mathrm{f}(x)$ to the graph of $y = \mathrm{g}(x)$. Make clear the order in which the transformations should be applied.

Find the coordinates of the points of intersection of the curve and the line with equations

$$2xy + 5y^2 = 24 \quad \text{and} \quad 2x + y + 4 = 0.$$

The coefficient of $x^7$ in the expansion of $\left(px^2 + \dfrac{4}{p}x\right)^5$ is 1280.

Find the value of the constant $p$.

A point $P$ is moving along the curve with equation $y = ax^{\frac{3}{2}} - 12x$ in such a way that the $x$-coordinate of $P$ is increasing at a constant rate of 5 units per second.

**(a)** Find the rate at which the $y$-coordinate of $P$ is changing when $x = 9$. Give your answer in terms of the constant $a$.

A point $P$ is moving along the curve with equation $y = ax^{\frac{3}{2}} - 12x$ in such a way that the $x$-coordinate of $P$ is increasing at a constant rate of 5 units per second.

**(b)** Given that the curve has a minimum point when $x = \frac{1}{4}$, find the value of $a$.

The equation of a curve is $y = 4\cos 2x + 3$ for $0 \le x \le 2\pi$.

**(a)** State the greatest and least possible values of $y$.

The equation of a curve is $y = 4\cos 2x + 3$ for $0 \le x \le 2\pi$.

**(b)** Sketch the curve.

The equation of a curve is $y = 4\cos 2x + 3$ for $0 \le x \le 2\pi$.

**(c)** Hence determine the number of solutions of the equation $4\cos 2x + 3 = 2x - 1$ for $0 \le x \le 2\pi$.

(In part (b) you sketched $y = 4\cos 2x + 3$: two cycles with maxima 7 at $x = 0, \pi, 2\pi$ and minima −1 at $x = \frac{\pi}{2}, \frac{3\pi}{2}$.)

The diagram shows the curve with equation $y = \dfrac{9}{(5x+4)^{\frac{1}{2}}}$ and the line $y = 6 - 3x$. The line and the curve intersect at the point $P$ which has $y$-coordinate 3.

Find the area of the shaded region.

**(a)** Prove the identity $\dfrac{\tan\theta + 7}{\tan^2\theta - 3} \equiv \dfrac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta}$.

**(b)** Hence solve the equation $\dfrac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta} = \dfrac{5}{\tan\theta}$ for $0° \le \theta \le 180°$.

(In part (a) you proved that $\dfrac{\tan\theta + 7}{\tan^2\theta - 3} \equiv \dfrac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta}$.)

The diagram shows the circle with equation $x^2 + y^2 - 14x + 8y + 36 = 0$ and the line $y = -2$. The line intersects the circle at the points $A$ and $B$. The centre of the circle is $C$.

**(a)** Find the coordinates of $A$, $B$ and $C$.

The diagram shows the circle with equation $x^2 + y^2 - 14x + 8y + 36 = 0$ and the line $y = -2$. The line intersects the circle at the points $A$ and $B$. The centre of the circle is $C$.

**(b)** Find the angle $ACB$ in radians. Give your answer correct to 3 significant figures. (In part (a) you found $A(2, -2)$, $B(12, -2)$ and $C(7, -4)$.)

(answer in radians)

The diagram shows the circle with equation $x^2 + y^2 - 14x + 8y + 36 = 0$ and the line $y = -2$. The line intersects the circle at the points $A$ and $B$. The centre of the circle is $C$.

**(c)** The chord $AB$ divides the circle into two segments.

Find the area of the larger segment. (In part (a) you found $A(2, -2)$, $B(12, -2)$ and $C(7, -4)$.) (In part (b) you found angle $ACB = 2.38$ radians, to 3 significant figures.)

The equation of a curve is such that $\dfrac{d^2y}{dx^2} = -\dfrac{24}{x^3}$. It is given that the curve has a stationary point at $(-2, 19)$.

**(a)** Find an expression for $\dfrac{dy}{dx}$.

The equation of a curve is such that $\dfrac{d^2y}{dx^2} = -\dfrac{24}{x^3}$. It is given that the curve has a stationary point at $(-2, 19)$.

**(b)** Find the $x$-coordinate of the other stationary point of the curve, and determine the nature of this stationary point.

(In part (a) you found $\dfrac{dy}{dx} = \dfrac{12}{x^2} - 3$.)

The equation of a curve is such that $\dfrac{d^2y}{dx^2} = -\dfrac{24}{x^3}$. It is given that the curve has a stationary point at $(-2, 19)$.

**(c)** Find the equation of the curve.

(In part (a) you found $\dfrac{dy}{dx} = \dfrac{12}{x^2} - 3$.)

The equation of a curve is such that $\dfrac{d^2y}{dx^2} = -\dfrac{24}{x^3}$. It is given that the curve has a stationary point at $(-2, 19)$.

**(d)** Find the equation of the normal to the curve at the point where $\dfrac{dy}{dx} = -\dfrac{9}{4}$ and $x$ is positive. Express your answer in the form $px + qy + r = 0$, where $p$, $q$ and $r$ are integers.

(In parts (a) and (c) you found $\dfrac{dy}{dx} = \dfrac{12}{x^2} - 3$ and $y = -\dfrac{12}{x} - 3x + 7$.)

**(a)** The first, second and third terms of an arithmetic progression are $4k$, $k^2$ and $8k$ respectively, where $k$ is a non-zero constant.

**(i)** Find the value of $k$.

**(a)** The first, second and third terms of an arithmetic progression are $4k$, $k^2$ and $8k$ respectively, where $k$ is a non-zero constant.

**(ii)** Find the sum of the first 20 terms of the progression.

(In part (i) you found $k = 6$.)