May/June 2025 Paper 12 Worked Answers (A-Level Maths 9709 AS)

24 questions · 75 marks · 110 minutes

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1. Question 2

   4 marksSimultaneous Equations
   Step 1: From the line: $2x = -y - 2$, so $x = \dfrac{-y-2}{2}$. Step 2: Substitute into the curve: $2 \cdot \dfrac{-y-2}{2} \cdot y + 3y^2 = 20$, giving $(-y-2)y + 3y^2 = 20$. Step 3: Expand: $-y^2 - 2y + 3y^2 = 20$, so $2y^2 - 2y - 20 = 0$, i.e. $y^2 - y - 10 = 0$. Step 4: By the sum of roots formula, the sum of the $y$-coordinates $= -(-1)/1 = 1$.
   Method:

   Rearrange the linear equation for $x$, substitute into the curve equation, solve the resulting quadratic in $y$.
   Examiner tips

   • Use the linear equation to eliminate one variable before substituting into the curve

2. Question 3

   4 marksBinomial Expansion
   Step 1: The general term is $\binom{5}{r}(qx^2)^{5-r}\left(\dfrac{3x}{q}\right)^r = \binom{5}{r}q^{5-r} \cdot \dfrac{3^r}{q^r} \cdot x^{2(5-r)+r}$. Step 2: For $x^7$: $10 - 2r + r = 7$, so $r = 3$. Step 3: The coefficient is $\binom{5}{3}q^{2} \cdot \dfrac{3^3}{q^3} = 10 \cdot \dfrac{27}{q} = \dfrac{270}{q}$. Step 4: Set $\dfrac{270}{q} = 810$, so $q = \dfrac{270}{810} = \dfrac{1}{3}$.
   Method:

   Write the general term, find which value of $r$ gives the required power of $x$, then equate the coefficient to the given value.
   Examiner tips

   • Track the power of $x$ carefully when the binomial has $x$ terms in both parts

3. Question 4a

   3 marksConnected Rates of Change
   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: $\dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{\mathrm{d}y}{\mathrm{d}x} \times \dfrac{\mathrm{d}x}{\mathrm{d}t} = (6b - 8) \times 4 = 24b - 32$.
   Method:

   Differentiate, evaluate at the given $x$, then multiply by $dx/dt$.
   Examiner tips

   • The chain rule connects rates: $\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}$

4. Question 4b

   2 marksStationary Points
   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 $c = 4$.
   Method:

   Differentiate, set equal to zero at $x=9$, solve for $c$.
   Examiner tips

   • Stationary points occur where $dy/dx = 0$

5. Question 5a

   2 marksTrigonometric Graphs
   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 $-5 + 2 = -3$.
   Method:

   Use the fact that $\cos$ ranges from $-1$ to $1$, multiply by the amplitude, then add the vertical shift.
   Examiner tips

   • The range of $a\cos(bx) + c$ is $[c-|a|, c+|a|]$

6. Question 5b

   2 marksTrigonometric Graphs
   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 $\dfrac{2\pi}{\pi} = 2$.
   Method:

   Calculate the period using $2\pi/b$, then divide the interval length by the period.
   Examiner tips

   • Period of $\cos(bx)$ is $\dfrac{2\pi}{b}$

7. Question 5c

   1 marksTrigonometric Equations
   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.
   Method:

   Sketch $y = 4\cos 2x + 3$ and $y = 2x - 1$ on the same axes and count the intersection points.
   Examiner tips

   • Rearranging to find intersections graphically is often the best approach for equations mixing trig and polynomial terms

8. Question 7a

   3 marksTrigonometric Identities
   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 $= 1 - \cos^2\theta - 3\cos^2\theta = 1 - 4\cos^2\theta$.
   Method:

   Replace $\tan\theta$ with $\sin\theta/\cos\theta$, multiply through by $\cos^2\theta$, simplify using the Pythagorean identity.
   Examiner tips

   • When proving identities involving $\tan$, converting to $\sin/\cos$ and clearing fractions is usually the best first step

9. Question 7b

   4 marksSolving Trigonometric Equations
   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.
   Method:

   Apply the identity, cross-multiply to form a quadratic in $\tan\theta$, solve, and find all solutions in the given range.
   Examiner tips

   • When solving $\tan\theta = k$ in $[0^\circ, 180^\circ]$, there is exactly one solution for each value of $k$

10. Question 8a

    3 marksEquation of a Circle
    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 $= 9 - 1 = 8$.
    Method:

    Substitute $y$ from the line into the circle equation, solve the quadratic in $x$, find the intersection points.
    Examiner tips

    • Complete the square to find the centre and radius, then substitute the line equation

11. Question 8b

    2 marksEquation of a Circle
    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.).
    Method:

    Use the perpendicular from the centre to the chord to find the half-angle, then double it.
    Examiner tips

    • Use the isosceles triangle formed by two radii and the chord — the perpendicular from the centre bisects both the chord and the angle

12. Question 8c

    4 marksProblems with Segments and Composite Shapes
    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 $= 62.8 - 14.1 = 48.7$.
    Method:

    Calculate the smaller sector area, subtract the triangle area to get the smaller segment, then subtract from the total circle area.
    Examiner tips

    • The larger segment = total circle area - smaller segment
    • Smaller segment = sector area - triangle area

13. Question 9a

    3 marksIntegration and Differentiation
    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: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{9}{x^2} - 1$.
    Method:

    Integrate the second derivative, use $dy/dx = 0$ at the stationary point to find $c$.
    Examiner tips

    • At a stationary point, $dy/dx = 0$ — use this to find the constant of integration

14. Question 9b

    2 marksStationary Points
    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.
    Method:

    Solve $dy/dx = 0$, identify the other root, evaluate the second derivative there to classify.
    Examiner tips

    • Remember that $d^2y/dx^2 < 0$ means maximum, $d^2y/dx^2 > 0$ means minimum

15. Question 9c

    3 marksFinding the Equation of a Curve
    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: $d = 10 - 6 = 4$.
    Method:

    Integrate $dy/dx$, substitute the known point, solve for the constant.
    Examiner tips

    • Be very careful with signs when substituting negative values of $x$

16. Question 9d

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

    The normal gradient is the negative reciprocal of the given tangent gradient.
    Examiner tips

    • The gradient of the normal is $-1/m$ where $m$ is the tangent gradient

17. Question 10a_i

    2 marksArithmetic Progressions
    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 $k > 0$: $k = \dfrac{9}{2}$.
    Method:

    Equate the two differences to form an equation in $k$, solve.
    Examiner tips

    • In an AP, $b - a = c - b$ for consecutive terms $a, b, c$

18. Question 10a_ii

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

    Substitute $a$, $d$, and $n$ into the AP sum formula and evaluate.
    Examiner tips

    • Always double-check arithmetic when working with fractions in the AP formula

19. Question 10b

    5 marksGeometric Progressions - Sum to Infinity
    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: $S_\infty = \dfrac{72\sqrt{3}}{1 - 1/\sqrt{3}} = \dfrac{72\sqrt{3} \times \sqrt{3}}{\sqrt{3} - 1} = \dfrac{216}{\sqrt{3} - 1}$.
    Method:

    Divide the two given terms to find $r^2$, take the positive root, find $a$, then compute $S_\infty$ and simplify.
    Examiner tips

    • When $r$ involves a surd, multiply numerator and denominator by $\sqrt{b}$ to simplify the $S_\infty$ expression

20. Question 11a

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

    Complete the square by halving the $x$-coefficient, squaring, and adjusting.
    Examiner tips

    • $(x + a)^2 = x^2 + 2ax + a^2$, so $a$ is half the coefficient of $x$

21. Question 11b_i

    3 marksInverse Functions
    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 $y$: $f^{-1}(x) = -\sqrt{x + 4} - 3$.
    Method:

    Use the completed square form, rearrange for $x$, choose the negative root based on the domain, swap variables.
    Examiner tips

    • The domain of $f$ determines which root to take when finding the inverse

22. Question 11b_ii

    4 marksComposite and Inverse Functions
    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: $(gf)^{-1}(x) = -\sqrt{-x-1} - 3$.
    Method:

    Compute $gf(x)$, complete the square, rearrange for the inverse using the negative root.
    Examiner tips

    • Find $gf(x)$ first, then treat it as a single function to invert