October/November 2025 Paper 22 Worked Answers (A-Level Maths 9709 AS)

13 questions · 50 marks · 75 minutes

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1. Question 1

   4 marksSolving Logarithmic Equations
   Step 1: Combine logarithms: $\ln\!\left(\dfrac{4x+7}{x-1}\right) = 3$. Step 2: Exponentiate: $\dfrac{4x+7}{x-1} = \mathrm{e}^3$. Step 3: Cross-multiply: $4x + 7 = \mathrm{e}^3(x - 1) = \mathrm{e}^3 x - \mathrm{e}^3$. Step 4: Rearrange: $4x - \mathrm{e}^3 x = -\mathrm{e}^3 - 7$, so $x(4 - \mathrm{e}^3) = -(\mathrm{e}^3 + 7)$. Step 5: Since $4 - \mathrm{e}^3 < 0$, divide and flip: $x = \dfrac{\mathrm{e}^3 + 7}{\mathrm{e}^3 - 4}$.
   Method:

   Combine logs, exponentiate, cross-multiply, and solve for $x$ in exact form.
   Examiner tips

   • Always check the domain: the answer must satisfy $x - 2 > 0$ and $3x + 5 > 0$
   • Leave the answer in exact form involving $\mathrm{e}^4$

2. Question 2

   5 marksTrigonometric Equations Using Identities
   Step 1: Use the identity $\tan^2\theta = \sec^2\theta - 1$ to get $2(\sec^2\theta - 1) + 3\sec\theta = 18$. Step 2: Simplify: $2\sec^2\theta + 3\sec\theta - 20 = 0$. Step 3: Factorise: $(2\sec\theta - 5)(\sec\theta + 4) = 0$. Step 4: $\sec\theta = \frac{5}{2} \Rightarrow \cos\theta = \frac{2}{5} \Rightarrow \theta = \pm 66.4^\circ$ (2 solutions). Step 5: $\sec\theta = -4 \Rightarrow \cos\theta = -\frac{1}{4} \Rightarrow \theta = \pm 104.5^\circ$ (2 solutions). Total: 4 solutions.
   Method:

   Substitute the identity, solve the quadratic in $\sec\theta$, then convert to $\cos\theta$ to find all angles.
   Examiner tips

   • Always check that solutions lie within the given range
   • Remember that $\cos\theta$ can be negative, giving solutions in the second and third quadrants

3. Question 3a

   4 marksModulus Inequality with Two Modulus Expressions
   Step 1: Square both sides (valid since both sides are non-negative): $(3x-4)^2 \leqslant (2x+5)^2$. Step 2: Alternatively, find critical values. Set $3x - 4 = 2x + 5$: $x = 9$. Step 3: Set $3x - 4 = -(2x + 5)$: $5x = -1$, so $x = -\dfrac{1}{5}$. Step 4: Test a value between $-\dfrac{1}{5}$ and $9$, e.g. $x = 0$: $|{-4}| = 4 \leqslant |5| = 5$. True. Step 5: Solution: $-\dfrac{1}{5} \leqslant x \leqslant 9$.
   Method:

   Find the two critical values where $|3x-4| = |2x+5|$ and determine the solution interval.
   Examiner tips

   • When solving $|f(x)| \leqslant |g(x)|$, find where the expressions are equal and test the regions
   • Remember to include the equality (non-strict inequality)

4. Question 3b

   3 marksLogarithmic Substitution in Modulus Inequality
   Step 1: Let $x = 7^{0.01N}$. The inequality becomes $|3x - 4| \leqslant |2x + 5|$, which has solution $-\dfrac{1}{5} \leqslant x \leqslant 9$. Step 2: Since $7^{0.01N} > 0$ always, the binding constraint is $7^{0.01N} \leqslant 9$. Step 3: Take $\log_7$: $0.01N \leqslant \log_7 9 = \dfrac{\ln 9}{\ln 7} = 1.1292\ldots$ Step 4: $N \leqslant 112.92\ldots$, so the largest integer is $N = 112$.
   Method:

   Substitute $x = 7^{0.01N}$, apply the upper bound from part (a), solve with logarithms, and take the floor.
   Examiner tips

   • Since $7^{0.01N}$ is always positive, the lower bound $x \geqslant -1/5$ is automatically satisfied
   • Use change of base formula: $\log_7 9 = \ln 9 / \ln 7$

5. Question 4a

   3 marksPolynomial Division
   Step 1: Divide $x^4$ by $x^2$ to get $x^2$. Multiply: $x^2(x^2+3) = x^4 + 3x^2$. Step 2: Subtract: $(x^4 - 10x^3 + 20x^2 - 30x + 40) - (x^4 + 3x^2) = -10x^3 + 17x^2 - 30x + 40$. Step 3: Divide $-10x^3$ by $x^2$ to get $-10x$. Multiply: $-10x(x^2+3) = -10x^3 - 30x$. Step 4: Subtract: $(-10x^3 + 17x^2 - 30x + 40) - (-10x^3 - 30x) = 17x^2 + 40$. Step 5: Divide $17x^2$ by $x^2$ to get $17$. Multiply: $17(x^2+3) = 17x^2 + 51$. Step 6: Subtract: $(17x^2 + 40) - (17x^2 + 51) = -11$. Quotient: $x^2 - 10x + 17$, Remainder: $-11$.
   Method:

   Perform polynomial long division of $p(x)$ by $x^2 + 3$.
   Examiner tips

   • Check your work by expanding $(x^2+3)(x^2-10x+17) + (-11)$ to verify you get $p(x)$
   • Be careful with signs during subtraction steps

6. Question 4b

   3 marksSolving Polynomial Equations Using Factorisation
   Step 1: $p(x) + 11 = 0$ means $(x^2 + 3)(x^2 - 10x + 17) = 0$. Step 2: $x^2 + 3 = 0$ gives $x^2 = -3$, which has no real roots. Step 3: Solve $x^2 - 10x + 17 = 0$ using the quadratic formula: $x = \dfrac{10 \pm \sqrt{100 - 68}}{2} = \dfrac{10 \pm \sqrt{32}}{2} = \dfrac{10 \pm 4\sqrt{2}}{2} = 5 \pm 2\sqrt{2}$.
   Method:

   Use the given factorisation, identify which factor can equal zero for real $x$, and solve using the quadratic formula.
   Examiner tips

   • $x^2 + 3 > 0$ for all real $x$, so it contributes no real roots
   • Simplify surds fully: $\sqrt{32} = 4\sqrt{2}$

7. Question 5a

   5 marksStationary Points of Trigonometric Functions
   Step 1: Differentiate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -8\sin 2x + 8\cos x$. Step 2: Use $\sin 2x = 2\sin x\cos x$: $-16\sin x\cos x + 8\cos x = 0$. Step 3: Factor: $8\cos x(1 - 2\sin x) = 0$. Step 4: $\cos x = 0 \Rightarrow x = \pi/2$ (check: this is a minimum). $\sin x = 1/2 \Rightarrow x = \pi/6$ or $5\pi/6$ (these are maxima). Step 5: At $x = \pi/6$: $y = 4\cos(\pi/3) + 8\sin(\pi/6) = 4(1/2) + 8(1/2) = 2 + 4 = 6$. Step 6: At $x = 5\pi/6$: $y = 4\cos(5\pi/3) + 8\sin(5\pi/6) = 4(1/2) + 8(1/2) = 2 + 4 = 6$. Both maximum points have $y$-coordinate $6$.
   Method:

   Differentiate, use double angle identity, factorise, find critical points, and evaluate $y$ at the maxima.
   Examiner tips

   • Confirm maximum/minimum nature by checking the second derivative or examining nearby values
   • Use exact values: $\cos(\pi/3) = 1/2$, $\sin(\pi/6) = 1/2$

8. Question 5b

   6 marksArea Between Curve and Line
   Step 1: The line $AB$ is $y = 6$ (horizontal line between the two maxima). Step 2: Area under line = $6 \times \left(\dfrac{5\pi}{6} - \dfrac{\pi}{6}\right) = 6 \times \dfrac{2\pi}{3} = 4\pi$. Step 3: Area under curve: $\displaystyle\int_{\pi/6}^{5\pi/6} (4\cos 2x + 8\sin x)\,\mathrm{d}x = \left[2\sin 2x - 8\cos x\right]_{\pi/6}^{5\pi/6}$. Step 4: At $x = 5\pi/6$: $2\sin(5\pi/3) - 8\cos(5\pi/6) = 2(-\sqrt{3}/2) - 8(-\sqrt{3}/2) = -\sqrt{3} + 4\sqrt{3} = 3\sqrt{3}$. Step 5: At $x = \pi/6$: $2\sin(\pi/3) - 8\cos(\pi/6) = 2(\sqrt{3}/2) - 8(\sqrt{3}/2) = \sqrt{3} - 4\sqrt{3} = -3\sqrt{3}$. Step 6: Integral $= 3\sqrt{3} - (-3\sqrt{3}) = 6\sqrt{3}$. Step 7: Shaded area $= 4\pi - 6\sqrt{3}$.
   Method:

   Find the area under the horizontal line $AB$ and subtract the definite integral of the curve between the two $x$-values.
   Examiner tips

   • The region is between the line (above) and the curve (below), so subtract curve area from line area
   • Use exact trigonometric values throughout

9. Question 6a

   4 marksDefinite Integration Leading to an Equation
   Step 1: $\displaystyle\int \tfrac{1}{2}\mathrm{e}^{2x}\,\mathrm{d}x = \tfrac{1}{2} \times \tfrac{1}{2}\mathrm{e}^{2x} = \tfrac{1}{4}\mathrm{e}^{2x}$. Step 2: $\displaystyle\int \tfrac{1}{4}\mathrm{e}^{-x}\,\mathrm{d}x = \tfrac{1}{4} \times (-1)\mathrm{e}^{-x} = -\tfrac{1}{4}\mathrm{e}^{-x}$. Step 3: Antiderivative: $\tfrac{1}{4}\mathrm{e}^{2x} - \tfrac{1}{4}\mathrm{e}^{-x}$.
   Method:

   Integrate, substitute limits, simplify, and rearrange to the given form of $a$.
   Examiner tips

   • In a 'show that' question, every step must be clearly justified
   • Remember $\int \mathrm{e}^{kx}\,\mathrm{d}x = \frac{1}{k}\mathrm{e}^{kx} + C$

10. Question 6b

    2 marksSign Change Method for Root Location
    Step 1: $f(1.0) = \tfrac{1}{2}\ln(10 + \tfrac{1}{2}\mathrm{e}^{-1} + \tfrac{1}{2}\mathrm{e}^{-4}) - 1 \approx 1.161 - 1 = 0.161 > 0$. Step 2: $f(1.2) = \tfrac{1}{2}\ln(10 + \tfrac{1}{2}\mathrm{e}^{-1.2} + \tfrac{1}{2}\mathrm{e}^{-4.8}) - 1.2 \approx 1.157 - 1.2 = -0.043 < 0$. Step 3: The sign change from positive to negative confirms a root exists between $1.0$ and $1.2$.
    Method:

    Evaluate $f(a)$ at $a = 1.0$ and $a = 1.2$, show a sign change occurs, and conclude the root lies between them.
    Examiner tips

    • Always state the sign change explicitly and what it implies
    • Show enough working that the sign of each evaluation is clear

11. Question 6c

    3 marksIterative Method for Solving Equations
    Step 1: $a_0 = 1$. Step 2: $a_1 = \tfrac{1}{2}\ln(10 + \tfrac{1}{2}\mathrm{e}^{-1} + \tfrac{1}{2}\mathrm{e}^{-4}) = 1.1613\ldots$ Step 3: $a_2 = \tfrac{1}{2}\ln(10 + \tfrac{1}{2}\mathrm{e}^{-1.1613} + \tfrac{1}{2}\mathrm{e}^{-4.6452}) = 1.1578\ldots$ Step 4: $a_3 = 1.1586\ldots$, $a_4 = 1.1584\ldots$ Step 5: Converges to $a = 1.159$ (4 s.f.).
    Method:

    Apply the iterative formula repeatedly starting from $a_0 = 1$ until convergence to 4 s.f.
    Examiner tips

    • Continue iterating until the 4th significant figure stabilises
    • Use the full calculator value at each step, not a rounded value

12. Question 7a

    6 marksImplicit Differentiation
    Step 1: Differentiate implicitly: $10xy + 5x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 8\mathrm{e}^{2y}\dfrac{\mathrm{d}y}{\mathrm{d}x} - 7 = 0$. Step 2: Solve for $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{7 - 10xy}{5x^2 + 8\mathrm{e}^{2y}}$. Step 3: Find $x$ when $y = 0$: $0 + 4 - 7x + 10 = 0$, so $x = 2$. Step 4: Substitute $(2, 0)$: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{7 - 0}{20 + 8} = \dfrac{7}{28} = \dfrac{1}{4}$. Note: "$\dfrac{7}{28}$" gives $\dfrac{7}{28}$, which equals $\dfrac{1}{4}$ but is not fully simplified. The correct simplified answer is $\dfrac{1}{4}$.
    Method:

    Differentiate implicitly, rearrange for $\mathrm{d}y/\mathrm{d}x$, find the $x$-coordinate when $y = 0$, and evaluate.
    Examiner tips

    • When differentiating $5x^2 y$, use the product rule: $\frac{\mathrm{d}}{\mathrm{d}x}(5x^2 y) = 10xy + 5x^2 \frac{\mathrm{d}y}{\mathrm{d}x}$
    • Remember $\frac{\mathrm{d}}{\mathrm{d}x}(\mathrm{e}^{2y}) = 2\mathrm{e}^{2y}\frac{\mathrm{d}y}{\mathrm{d}x}$ by chain rule

13. Question 7b

    2 marksProving No Vertical Tangent Exists
    Step 1: A tangent parallel to the $y$-axis would require $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ to be undefined, which happens when the denominator is zero. Step 2: The denominator is $5x^2 + 8\mathrm{e}^{2y}$. Since $5x^2 \geqslant 0$ for all real $x$ and $8\mathrm{e}^{2y} > 0$ for all real $y$, the sum is always at least $8$. Step 3: Therefore $5x^2 + 8\mathrm{e}^{2y} \geqslant 8 > 0$, the denominator is never zero, $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ is always defined, and no tangent can be vertical.
    Method:

    Show that the denominator of $\mathrm{d}y/\mathrm{d}x$ is always positive, so the gradient is always defined and never infinite.
    Examiner tips

    • $\mathrm{e}^{2y} > 0$ for all real $y$ is a key property of the exponential function
    • $x^2 \geqslant 0$ for all real $x$