May/June 2025 Paper 11 Worked Answers (A-Level Maths 9709 AS)

22 questions · 75 marks · 110 minutes

Question papers and mark schemes are copyright Cambridge International. We do not reproduce them: the worked answers here are written by The Practice Book. Have the paper open alongside. Get the official paper from Cambridge

1. Question 1

   4 marksTrigonometric Equations
   Step 1: Multiply both sides by $\sin\theta$: $8\sin^2\theta = 2\sin\theta + 3$, giving $8\sin^2\theta - 2\sin\theta - 3 = 0$. Step 2: Factorise: $(2\sin\theta + 1)(4\sin\theta - 3) = 0$, so $\sin\theta = -\frac{1}{2}$ or $\sin\theta = \frac{3}{4}$. Step 3: For $\sin\theta = -\frac{1}{2}$: $\theta = -30^\circ$ and $\theta = -150^\circ$ (two solutions in the range). Step 4: For $\sin\theta = \frac{3}{4}$: $\theta = \sin^{-1}(0.75) \approx 48.6^\circ$ and $\theta = 180^\circ - 48.6^\circ \approx 131.4^\circ$ (two solutions). Step 5: Total number of solutions = $4$.
   Method:

   Multiply through by $\sin\theta$, form and solve a quadratic, then find all solutions in the range.
   Examiner tips

   • Always multiply through to clear fractions involving trig functions
   • Remember that $\sin\theta = k$ gives two solutions in a $360^\circ$ range when $|k| < 1$

2. Question 2a

   2 marksTangents and Normals
   Step 1: Substitute $x = 4$ into $\dfrac{\mathrm{d}y}{\mathrm{d}x}$: $3(2(4)-3)^3 - 8(4)^{1/2} = 3(5)^3 - 8(2) = 375 - 16 = 359$. Step 2: The gradient of the normal is the negative reciprocal of the tangent gradient: $-\dfrac{1}{359}$.
   Method:

   Substitute $x=4$ into the derivative to get the tangent gradient, then take the negative reciprocal.
   Examiner tips

   • The normal is perpendicular to the tangent, so $m_{\text{normal}} = -1/m_{\text{tangent}}$

3. Question 2b

   4 marksIntegration - Finding Curve Equation
   Step 1: Integrate $4(2x-5)^3$: $\dfrac{4}{2 \times 4}(2x-5)^4 = \dfrac{1}{2}(2x-5)^4$. Step 2: Integrate $-9x^{1/2}$: $-9 \times \dfrac{x^{3/2}}{3/2} = -6x^{3/2}$. Step 3: So $y = \dfrac{1}{2}(2x-5)^4 - 6x^{3/2} + c$. Step 4: Substitute $(4, -\frac{11}{2})$: $-\dfrac{11}{2} = \dfrac{1}{2}(3)^4 - 6(8) + c = \dfrac{81}{2} - 48 + c = -\dfrac{15}{2} + c$. Step 5: $c = -\dfrac{11}{2} + \dfrac{15}{2} = 2$.
   Method:

   Integrate each term, then substitute the known point to find c.
   Examiner tips

   • When integrating $(ax+b)^n$, remember to divide by the coefficient of $x$ inside the bracket

4. Question 3a

   5 marksGeometric Progressions
   Step 1: Let the first term be $a$ and common ratio $r$. Then $ar^2 = 12$ so $a = 12/r^2$. Step 2: Sum of first 3 terms: $a + ar + ar^2 = 28$, so $12/r^2 + 12/r + 12 = 28$. Step 3: Multiply by $r^2$: $12 + 12r + 12r^2 = 28r^2$, giving $16r^2 - 12r - 12 = 0$, i.e. $4r^2 - 3r - 3 = 0$. Step 4: Using the quadratic formula: $r = \dfrac{3 \pm \sqrt{9 + 48}}{8} = \dfrac{3 \pm \sqrt{57}}{8}$. Since $r < 0$: $r = \dfrac{3 - \sqrt{57}}{8} \approx -0.5686$. Step 5: $a = 12/(-0.5686)^2 \approx 37.11$. Step 6: Tenth term $= 37.11 \times (-0.5686)^9 \approx -0.901$ (3 s.f.).
   Method:

   Form simultaneous equations from the third term and sum, solve for $r$ (negative), find $a$, compute $ar^9$.
   Examiner tips

   • Be careful with signs when $r$ is negative — odd powers preserve the negative sign

5. Question 3b

   2 marksSum to Infinity of GP
   Step 1: The sum to infinity formula is $S_\infty = \dfrac{a}{1 - r}$, valid when $|r| < 1$. Step 2: Here $a = 48$, $r = -\dfrac{2}{3}$, so $|r| = \dfrac{2}{3} < 1$. Step 3: $S_\infty = \dfrac{48}{1 - (-2/3)} = \dfrac{48}{5/3} = \dfrac{48 \times 3}{5} = \dfrac{144}{5}$.
   Method:

   Substitute into the sum to infinity formula, simplifying the double negative carefully.
   Examiner tips

   • When $r$ is negative, $1 - r$ becomes $1 + |r|$

6. Question 5a_i

   2 marksBinomial Expansion
   Step 1: Use the binomial expansion $(a + b)^n = \sum_{r=0}^{n} \binom{n}{r}a^{n-r}b^r$. Step 2: First term: $\binom{4}{0} \cdot 3^{4} \cdot (-qx)^{0} = 81$. Step 3: Second term: $\binom{4}{1} \cdot 3^{3} \cdot (-qx)^{1} = 4(27)(-qx) = -108qx$. Step 4: Third term: $\binom{4}{2} \cdot 3^{2} \cdot (-qx)^{2} = 6(9)(q^{2}x^{2}) = 54q^{2}x^{2}$. Step 5: First 3 terms: $81 - 108qx + 54q^{2}x^{2}$.
   Method:

   Apply binomial theorem for $r = 0, 1, 2$.
   Examiner tips

   • Be careful with signs when the second term in the bracket is negative

7. Question 5a_ii

   2 marksBinomial Expansion
   Step 1: First term: $\binom{5}{0}(1)^5 = 1$. Step 2: Second term: $\binom{5}{1}(1)^4\left(-\dfrac{x}{3}\right) = 5\left(-\dfrac{x}{3}\right) = -\dfrac{5x}{3}$. Step 3: Third term: $\binom{5}{2}(1)^3\left(-\dfrac{x}{3}\right)^2 = 10 \times \dfrac{x^2}{9} = \dfrac{10x^2}{9}$. Step 4: First 3 terms: $1 - \dfrac{5x}{3} + \dfrac{10x^2}{9}$.
   Method:

   Apply binomial theorem for first three terms, carefully handling the fraction.
   Examiner tips

   • When the bracket contains a fraction, remember to raise the entire fraction to the required power

8. Question 5b

   3 marksBinomial Expansion - Coefficient
   Step 1: The $x^2$ coefficient in the product comes from three pairs: (constant from first) $\times$ ($x^2$ term from second) + ($x$ term from first) $\times$ ($x$ term from second) + ($x^2$ term from first) $\times$ (constant from second). Step 2: $= 32 \times \dfrac{3}{2} + (-80p)(-2) + 80p^2 \times 1 = 48 + 160p + 80p^2$. Step 3: Set equal to 93: $80p^2 + 160p + 48 = 93$, so $80p^2 + 160p - 45 = 0$. Step 4: Divide by 5: $16p^2 + 32p - 9 = 0$, giving $(4p - 1)(4p + 9) = 0$. Step 5: $p = \dfrac{1}{4}$ or $p = -\dfrac{9}{4}$.
   Method:

   Collect all contributions to the $x^2$ term in the product, set equal to 93, solve the resulting quadratic.
   Examiner tips

   • When finding the coefficient of $x^2$ in a product of expansions, consider all pairs of terms that multiply to give $x^2$

9. Question 6a

   4 marksSimultaneous Equations
   Step 1: Substitute $y = 11x + 3$ into $2x^2 - 2xy + 2 = 0$: $2x^2 - 2x(11x + 3) + 2 = 0$. Step 2: Expand: $2x^2 - 22x^2 - 6x + 2 = 0$, giving $-20x^2 - 6x + 2 = 0$. Step 3: Multiply by $-1$: $20x^2 + 6x - 2 = 0$, divide by 2: $10x^2 + 3x - 1 = 0$. Step 4: Factorise: $(5x - 1)(2x + 1) = 0$, so $x = \dfrac{1}{5}$ or $x = -\dfrac{1}{2}$.
   Method:

   Substitute the line equation into the curve, simplify, and solve the quadratic.
   Examiner tips

   • Always simplify the quadratic before attempting to factorise

10. Question 6b

    5 marksDiscriminant
    Step 1: Substituting $y = 4x + 3$ into $2x^2 - kxy + 2 = 0$ gives $(2 - 4k)x^2 - 3kx + 2 = 0$. Step 2: For no intersection, the discriminant must be negative: $(-3k)^2 - 4(2 - 4k)(2) < 0$. Step 3: Expand: $9k^2 - 8(2 - 4k) < 0$, so $9k^2 - 16 + 32k < 0$, giving $9k^2 + 32k - 16 < 0$. Step 4: Factorise: $(k + 4)(9k - 4) < 0$. Step 5: The quadratic is negative between its roots: $-4 < k < \dfrac{4}{9}$.
    Method:

    Form discriminant condition, expand and simplify to quadratic in $k$, solve the inequality.
    Examiner tips

    • For a quadratic inequality, find the roots first, then determine the sign in each region

11. Question 7a

    4 marksConnected Rates of Change
    Step 1: Differentiate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x - \dfrac{8}{x^3}$. Step 2: At $x = 2$: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 12 - \dfrac{8}{8} = 12 - 1 = 11$. Step 3: Use the chain rule: $\dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{\mathrm{d}y}{\mathrm{d}x} \times \dfrac{\mathrm{d}x}{\mathrm{d}t}$. Step 4: $-7 = 11 \times \dfrac{\mathrm{d}x}{\mathrm{d}t}$, so $\dfrac{\mathrm{d}x}{\mathrm{d}t} = -\dfrac{7}{11}$.
    Method:

    Differentiate, evaluate at $x=2$, apply the chain rule with the given rate.
    Examiner tips

    • 'Decreasing' means the rate is negative

12. Question 7b

    5 marksStationary Points
    Step 1: Differentiate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 8x - \dfrac{18}{x^3}$. Step 2: Set $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$: $8x = \dfrac{18}{x^3}$, so $8x^4 = 18$, giving $x^4 = \dfrac{9}{4}$. Step 3: $x^2 = \dfrac{3}{2}$ (taking the positive root of $x^4$), so $x = \pm\sqrt{\dfrac{3}{2}}$. Step 4: Substitute into $y$: $y = 4\left(\dfrac{3}{2}\right) + 9\left(\dfrac{2}{3}\right) - 8 = 6 + 6 - 8 = 4$. Step 5: Both stationary points have $y = 4$. The second derivative $8 + 54/x^4 > 0$ for all $x$, so both are minima.
    Method:

    Find $x$ by setting derivative to zero, substitute back to find $y$, use second derivative to classify.
    Examiner tips

    • Use $x^2$ rather than $x$ when substituting back — it avoids dealing with surds

13. Question 8

    8 marksCircle Geometry and Tangents
    Step 1: Rewrite the circle in standard form: $(x-3)^2 + (y+5)^2 = 9 + 25 + 27 = 61$. Centre $(3, -5)$, radius $\sqrt{61}$. Step 2: Substitute $x = -2$: $4 + y^2 + 12 + 10y - 27 = 0$, so $y^2 + 10y - 11 = 0$, giving $(y - 1)(y + 11) = 0$. So $P(-2, 1)$ and $Q(-2, -11)$. Step 3: Gradient of $CP = \dfrac{-5-1}{3-(-2)} = -\dfrac{6}{5}$. Tangent at $P$ is perpendicular: gradient $= \dfrac{5}{6}$. Step 4: Tangent at $P$: $y - 1 = \dfrac{5}{6}(x + 2)$. Step 5: Gradient of $CQ = \dfrac{-5-(-11)}{3-(-2)} = \dfrac{6}{5}$. Tangent at $Q$: $y + 11 = -\dfrac{5}{6}(x + 2)$. Step 6: Find intersection of tangents. From tangent at $P$: $y = \dfrac{5}{6}(x+2) + 1$. From tangent at $Q$: $y = -\dfrac{5}{6}(x+2) - 11$. Set equal: $\dfrac{5}{6}(x+2) + 1 = -\dfrac{5}{6}(x+2) - 11$, so $\dfrac{10}{6}(x+2) = -12$, giving $x + 2 = -\dfrac{36}{5}$, so $x = -\dfrac{46}{5}$. Step 7: $y = \dfrac{5}{6}(-\dfrac{36}{5}) + 1 = -6 + 1 = -5$. Tangents meet at $(-\dfrac{46}{5}, -5)$. Step 8: Base $PQ = |1 - (-11)| = 12$. Height = distance from $(-\dfrac{46}{5}, -5)$ to line $x = -2$: $|-\dfrac{46}{5} + 2| = \dfrac{36}{5}$. Step 9: Area $= \dfrac{1}{2} \times 12 \times \dfrac{36}{5} = \dfrac{216}{5}$.
    Method:

    Find circle centre/radius, locate P and Q, find tangent equations using perpendicularity, find tangent intersection, compute area.
    Examiner tips

    • A tangent to a circle is perpendicular to the radius at the point of contact

14. Question 10b

    2 marksInverse Functions
    Step 1: The domain of $g^{-1}$ equals the range of $g$. Step 2: Since $g(x) = 3\sqrt{x+2} - 5$ with $x \ge -2$: when $x = -2$, $g(-2) = 3(0) - 5 = -5$. As $x \to \infty$, $g(x) \to \infty$. Step 3: So the range of $g$ is $g(x) \ge -5$, meaning the domain of $g^{-1}$ is $x \ge -5$.
    Method:

    Find the range of $g$ by evaluating at the boundary of its domain.
    Examiner tips

    • Domain of inverse = range of original, and vice versa

15. Question 10c

    2 marksInverse Functions
    Step 1: Let $y = 3\sqrt{x+2} - 5$. Add 5: $y + 5 = 3\sqrt{x+2}$. Step 2: Divide by 3: $\dfrac{y+5}{3} = \sqrt{x+2}$. Step 3: Square both sides: $\left(\dfrac{y+5}{3}\right)^2 = x + 2$. Step 4: Subtract 2: $x = \left(\dfrac{y+5}{3}\right)^2 - 2$. Step 5: Swap $x$ and $y$: $g^{-1}(x) = \left(\dfrac{x+5}{3}\right)^2 - 2$.
    Method:

    Rearrange step by step: add 5, divide by 3, square, subtract 2, then swap variables.
    Examiner tips

    • Work through each algebraic step carefully, especially when squaring

16. Question 10d

    1 marksRange of Inverse Function
    Step 1: The range of $g^{-1}$ equals the domain of $g$. Step 2: The domain of $g$ is $x \ge -2$. Step 3: Therefore the range of $g^{-1}$ is $g^{-1}(x) \ge -2$.
    Method:

    The range of the inverse is the domain of the original function.
    Examiner tips

    • Range of inverse = domain of original

17. Question 10e

    1 marksComposite Functions
    Step 1: First find $h(4) = 4 - 2 = 2$. Step 2: Then find $g^{-1}(2) = \left(\dfrac{2+5}{3}\right)^2 - 2 = \left(\dfrac{7}{3}\right)^2 - 2 = \dfrac{49}{9} - 2 = \dfrac{49 - 18}{9} = \dfrac{31}{9}$.
    Method:

    Evaluate $h(4) = 2$, then substitute into $g^{-1}$.
    Examiner tips

    • In composite functions $g^{-1}h(4)$, apply $h$ first, then $g^{-1}$

18. Question 10f

    1 marksComposite Functions - Domain/Range
    Step 1: For $hg^{-1}(x)$ to be defined, the output of $g^{-1}$ must lie within the domain of $h$. Step 2: The range of $g^{-1}$ is $y \ge -2$, which includes values in $[-2, 0)$. Step 3: The domain of $h$ is $x \ge 0$, so values from $g^{-1}$ in $[-2, 0)$ are not valid inputs for $h$. Step 4: Therefore $hg^{-1}$ cannot be formed.
    Method:

    The range of $g^{-1}$ ($\ge -2$) is not contained in the domain of $h$ ($\ge 0$).
    Examiner tips

    • For composite functions, always check: range of inner function must be subset of domain of outer function