May/June 2025 Paper 13 Worked Answers (A-Level Maths 9709 AS)

20 questions · 75 marks · 110 minutes

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1. Question 1

   4 marksTangents and Normals
   Step 1: Differentiate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4 - \dfrac{64}{x^3}$. Step 2: At $x = -2$: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4 - \dfrac{64}{-8} = 4 + 8 = 12$. Step 3: Tangent equation: $y - (-16) = 12(x - (-2))$, so $y + 16 = 12(x + 2)$. Step 4: $y = 12x + 24 - 16 = 12x + 8$. The $y$-intercept is $8$.
   Method:

   Differentiate using the power rule, evaluate the gradient at the point, then use the point-slope formula to find the tangent equation.
   Examiner tips

   • Rewrite fractions as negative powers before differentiating
   • Be very careful with signs when substituting negative $x$-values into cubic denominators

2. Question 2

   4 marksConvergence and Sum to Infinity
   Step 1: The common ratio is $r = \dfrac{6\sin^3\alpha}{3\sin^2\alpha} = 2\sin\alpha$. Step 2: $S_\infty = \dfrac{3\sin^2\alpha}{1 - 2\sin\alpha} = \dfrac{1}{4}$. Step 3: Cross-multiply: $12\sin^2\alpha = 1 - 2\sin\alpha$, so $12\sin^2\alpha + 2\sin\alpha - 1 = 0$. Step 4: Factorise: $(6\sin\alpha - 1)(2\sin\alpha + 1) = 0$, giving $\sin\alpha = \dfrac{1}{6}$ or $\sin\alpha = -\dfrac{1}{2}$. Step 5: Since $0 < \alpha < \dfrac{\pi}{6}$, we need $\sin\alpha > 0$, so $\sin\alpha = \dfrac{1}{6}$.
   Method:

   Find common ratio, apply sum to infinity formula, solve the resulting quadratic in $\sin\alpha$, and reject the invalid root.
   Examiner tips

   • Always check that the common ratio satisfies $|r| < 1$ for the sum to infinity to exist
   • Reject solutions outside the given domain

3. Question 3

   4 marksDefinite Integration
   Step 1: Integrate: $\displaystyle\int \dfrac{b}{(2x-1)^2}\,\mathrm{d}x = -\dfrac{b}{2}(2x-1)^{-1} = -\dfrac{b}{2(2x-1)}$. Step 2: Evaluate $\left[-\dfrac{b}{2(2x-1)} + 4x\right]_1^3 = \left(-\dfrac{b}{10} + 12\right) - \left(-\dfrac{b}{2} + 4\right)$. Step 3: Simplify: $-\dfrac{b}{10} + \dfrac{b}{2} + 8 = \dfrac{-b + 5b}{10} + 8 = \dfrac{4b}{10} + 8 = \dfrac{2b}{5} + 8$. Step 4: Set equal to $14$: $\dfrac{2b}{5} = 6$, so $b = 15$.
   Method:

   Integrate using the reverse chain rule, evaluate the definite integral, and solve the resulting equation for the constant.
   Examiner tips

   • When integrating $(ax+b)^n$, remember to divide by the coefficient of $x$ inside the bracket
   • Be careful with signs when evaluating definite integrals

4. Question 4a

   3 marksBinomial Expansion
   Step 1: The general term is $\binom{5}{r}(2)^{5-r}\left(-\dfrac{3x}{2}\right)^r$. Step 2: For $x^2$, set $r = 2$: $\binom{5}{2}(2)^3\left(-\dfrac{3}{2}\right)^2 = 10 \times 8 \times \dfrac{9}{4} = 10 \times 18 = 180$. Step 3: The coefficient of $x^2$ is $180$ (positive, since $(-3/2)^2$ is positive).
   Method:

   Apply the binomial theorem systematically for $r = 0, 1, 2$ to find the first three terms.
   Examiner tips

   • Identify $a$ and $b$ carefully, including any negative signs and fractions
   • Check each term by verifying the powers add up correctly

5. Question 4b

   3 marksBinomial Expansion
   Step 1: We need $2 - \dfrac{3x}{2} = 1.985$. Step 2: $\dfrac{3x}{2} = 2 - 1.985 = 0.015$. Step 3: $x = \dfrac{0.015 \times 2}{3} = \dfrac{0.03}{3} = 0.01$.
   Method:

   Set the bracket expression equal to 1.985, solve for x, then substitute into the expansion.
   Examiner tips

   • Always verify your substitution by checking that the bracket expression gives the target value

6. Question 5

   6 marksSolving Trigonometric Equations
   Step 1: Replace $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$: $\dfrac{5\sin^2\theta}{\cos\theta} = 2 + 7\cos\theta$. Step 2: Multiply by $\cos\theta$: $5\sin^2\theta = 2\cos\theta + 7\cos^2\theta$. Step 3: Use $\sin^2\theta = 1 - \cos^2\theta$: $5 - 5\cos^2\theta = 2\cos\theta + 7\cos^2\theta$. Step 4: Rearrange: $12\cos^2\theta + 2\cos\theta - 5 = 0$. Step 5: Discriminant $= 4 + 240 = 244 > 0$, giving two real values of $\cos\theta$: $\cos\theta = \dfrac{-2 \pm \sqrt{244}}{24}$. Step 6: $\cos\theta \approx 0.568$ or $\cos\theta \approx -0.734$. Both satisfy $|\cos\theta| \le 1$. Step 7: Each value gives two solutions in $-180^\circ < \theta < 180^\circ$ (since $\cos\theta = k$ gives $\theta = \pm\cos^{-1}(k)$). Step 8: Total solutions $= 4$.
   Method:

   Convert to a quadratic in $\cos\theta$ using identities, solve, and find all solutions in the given range.
   Examiner tips

   • Always express everything in terms of one trig function when solving
   • Remember that $\cos\theta = k$ gives $\theta = \pm \cos^{-1}(k)$ in $(-180^\circ, 180^\circ)$

7. Question 6

   6 marksArithmetic Progressions
   Step 1: $N$th term: $a + 2(N-1) = 55$, so $a = 57 - 2N$. Step 2: Sum of first $3N$ terms: $\dfrac{3N}{2}(2a + 2(3N-1)) = 5760$, so $3N(a + 3N - 1) = 5760$. Step 3: Substitute $a = 57 - 2N$: $3N(57 - 2N + 3N - 1) = 3N(56 + N) = 5760$. Step 4: $N^2 + 56N = 1920$, so $N^2 + 56N - 1920 = 0$. Step 5: $(N + 80)(N - 24) = 0$, giving $N = 24$ (rejecting $N = -80$). Step 6: $a = 57 - 2(24) = 57 - 48 = 9$.
   Method:

   Form two simultaneous equations from the nth term and sum formulas, eliminate a, solve the quadratic for N, then find a.
   Examiner tips

   • With two unknowns, you need two equations — use the $n$th term and sum formulas
   • Always reject negative values for $N$

8. Question 7a

   3 marksStationary Points
   Step 1: $y$ is decreasing when $\dfrac{\mathrm{d}y}{\mathrm{d}x} < 0$, i.e., $2x^2 + 7x - 15 < 0$. Step 2: Factorise: $(2x - 3)(x + 5) < 0$, giving critical values $x = \dfrac{3}{2}$ and $x = -5$. Step 3: Since the coefficient of $x^2$ is positive, the parabola opens upward, so the expression is negative between the roots. Step 4: Therefore $-5 < x < \dfrac{3}{2}$.
   Method:

   Set the derivative less than zero, factorise the quadratic, and identify the interval where it is negative.
   Examiner tips

   • Decreasing means $dy/dx < 0$, not $\le 0$
   • For a positive quadratic, it is negative between its roots

9. Question 7b

   4 marksFinding the equation of a curve from its derivative
   Step 1: Factorise the derivative: $3x^2 + 10x - 8 = (3x - 2)(x + 4) = 0$, so $x = \frac{2}{3}$ or $x = -4$. Step 2: The maximum is at $x = -4$ (check second derivative or signs). Step 3: At $x = -4$: $y = (-4)^3 + 5(-4)^2 - 8(-4) + c = -64 + 80 + 32 + c = 48 + c$. Step 4: $48 + c = 34$, so $c = -14$.
   Method:

   Identify the x-coordinate of the maximum, integrate the derivative, then substitute the known point to find the constant.
   Examiner tips

   • The maximum occurs where the derivative changes from positive to negative
   • Don't forget the constant of integration when integrating

10. Question 9a

    4 marksPerpendicular lines
    Step 1: Gradient of $PR = \dfrac{m - 4}{1 - (-11)} = \dfrac{m - 4}{12}$. Step 2: Gradient of $QR = \dfrac{m - (-1)}{1 - 4} = \dfrac{m + 1}{-3}$. Step 3: For perpendicular lines, the product of gradients $= -1$: $\dfrac{(m-4)(m+1)}{-36} = -1$. Step 4: $(m-4)(m+1) = 36$. $m^2 - 3m - 4 = 36$. $m^2 - 3m - 40 = 0$. Step 5: $(m - 8)(m + 5) = 0$, so $m = 8$ or $m = -5$. Step 6: Since $m > 0$, the positive value is $m = 8$.
    Method:

    Express gradients of PR and QR in terms of the unknown, apply the perpendicular condition, and solve the resulting quadratic.
    Examiner tips

    • Perpendicular lines have gradients whose product is $-1$
    • Be careful with signs when computing gradients with negative coordinates

11. Question 9b

    5 marksEquation of a tangent line
    Step 1: The circle has equation $x^2 + y^2 + 2gx + 2fy + c_0 = 0$. Substituting $P(-13, 5)$: $169 + 25 - 26g + 10f + c_0 = 0$, i.e., $-26g + 10f + c_0 = -194$. Step 2: Substituting $Q(5, 1)$: $25 + 1 + 10g + 2f + c_0 = 0$, i.e., $10g + 2f + c_0 = -26$. Step 3: Substituting $R(2, 10)$: $4 + 100 + 4g + 20f + c_0 = 0$, i.e., $4g + 20f + c_0 = -104$. Step 4: Solving the system: centre is $(-4, 3)$, giving the circle $x^2 + y^2 + 8x - 6y - 60 = 0$. Step 5: Gradient of radius from $(-4, 3)$ to $R(2, 10)$: $\dfrac{10-3}{2+4} = \dfrac{7}{6}$. Step 6: Tangent is perpendicular to the radius, so tangent gradient $= -\dfrac{6}{7}$. Step 7: Tangent: $y - 10 = -\dfrac{6}{7}(x - 2)$, giving $6x + 7y - 82 = 0$. Step 8: $a + b + c = 6 + 7 + (-82) = -69$.
    Method:

    Find the circle through the three points, determine the centre, compute the radius gradient to R, take its negative reciprocal for the tangent, and form the equation.
    Examiner tips

    • The tangent to a circle is always perpendicular to the radius at the point of tangency
    • When finding the circle equation, use the general form and substitute three points

12. Question 10a

    4 marksStationary Points
    Step 1: Differentiate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{48}{(3x-7)^2} + 3$. Step 2: Set $= 0$: $\dfrac{48}{(3x-7)^2} = 3$, so $(3x-7)^2 = 16$. Step 3: $3x - 7 = \pm 4$, giving $x = \dfrac{11}{3}$ or $x = 1$. Step 4: Both values are valid (neither equals $7/3$), so there are $2$ stationary points.
    Method:

    Differentiate using the chain rule, set equal to zero, solve the resulting equation to find both stationary points.
    Examiner tips

    • Remember to consider both positive and negative square roots
    • Check that solutions don't make the original function undefined

13. Question 10b

    3 marksStationary Points
    Step 1: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -18(2x-5)^{-2} + 2$. Step 2: $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 72(2x-5)^{-3} = \dfrac{72}{(2x-5)^3}$. Step 3: At $x = 1$: $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{72}{(2-5)^3} = \dfrac{72}{-27} = -\dfrac{8}{3} < 0$. Step 4: Since $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0$, the stationary point at $x = 1$ is a maximum.
    Method:

    Differentiate again to find the second derivative, evaluate at each stationary point, and use the sign to determine maximum or minimum.
    Examiner tips

    • The second derivative test: $f''(x) > 0$ means minimum, $f''(x) < 0$ means maximum
    • Be careful with cubes of negative numbers

14. Question 10c_i

    1 marksGraph Transformations
    Step 1: The minimum of $C$ at $(4, 6)$ becomes the maximum of $C_1$ after reflection in the $x$-axis (reflection flips max/min). Step 2: Apply translation to $(4, 6)$: $(4-3, 6+7) = (1, 13)$. Step 3: Apply reflection in the $x$-axis: $(1, -13)$. Step 4: The maximum point of $C_1$ is $(1, -13)$.
    Method:

    Identify which point of C becomes the maximum of C1 after reflection, apply the translation then reflection.
    Examiner tips

    • Reflection in the $x$-axis swaps maxima and minima
    • Apply transformations in the given order

15. Question 10c_ii

    3 marksGraph Transformations
    Step 1: Translation by $\begin{pmatrix} -3 \\ 7 \end{pmatrix}$: replace $x$ with $x + 3$ and add $7$ to the function. Step 2: After translation: $y = \dfrac{9}{2(x+3)-5} + 2(x+3) - 5 + 7 = \dfrac{9}{2x+1} + 2x + 6 - 5 + 7 = \dfrac{9}{2x+1} + 2x + 8$. Step 3: Reflect in $x$-axis: $y = -\left(\dfrac{9}{2x+1} + 2x + 8\right) = -\dfrac{9}{2x+1} - 2x - 8$. Step 4: So $p = -9$, $q = -2$, $r = -8$. $p + q + r = -9 - 2 - 8 = -19$.
    Method:

    Apply the translation algebraically by replacing x with x+3 and adding 7, simplify, then negate for the reflection.
    Examiner tips

    • Translation by $\begin{pmatrix} h \\ k \end{pmatrix}$ means replace $x$ by $x - h$ and $y$ by $y - k$
    • Reflection in the $x$-axis means replace $y$ by $-y$

16. Question 11a

    4 marksDiscriminant
    Step 1: Complete the square: $f(x) = (x + 2a)^2 - 4a^2 + a$. Step 2: The minimum value is $-4a^2 + a = -33$. Step 3: $4a^2 - a - 33 = 0$. Step 4: Factorise: $(4a + 11)(a - 3) = 0$, so $a = 3$ or $a = -\dfrac{11}{4}$. Step 5: The positive value is $a = 3$.
    Method:

    Complete the square to find the minimum of f(x) in terms of a, set equal to -33, and solve the resulting quadratic in a.
    Examiner tips

    • Completing the square reveals the minimum value of a quadratic with positive leading coefficient
    • The minimum of $(x+p)^2 + q$ is $q$

17. Question 11b

    6 marksComposite Functions
    Step 1: From $g^{-1}(x) = \sqrt[3]{2x - 4}$, find $g(x)$. Let $y = g^{-1}(x) = (2x-4)^{1/3}$. Then $y^3 = 2x - 4$, so $x = \dfrac{y^3 + 4}{2}$. Thus $g(x) = \dfrac{x^3 + 4}{2}$. Step 2: $g(0) = \dfrac{0 + 4}{2} = 2$. Step 3: $g(g(0)) = g(2) = \dfrac{8 + 4}{2} = 6$. Step 4: $f(g(g(0))) = f(6) = 36 + 24a + a = 36 + 25a$. Step 5: Set equal to $96$: $36 + 25a = 96$, so $25a = 60$, $a = \dfrac{12}{5}$.
    Method:

    Invert the given inverse function to find g, evaluate the composite step by step, then solve for a.
    Examiner tips

    • To find $g$ from $g^{-1}$, swap $x$ and $y$ in the inverse function
    • $fgg(0)$ means $f(g(g(0)))$ — evaluate from the inside out