October/November 2025 Paper 23 Worked Answers (A-Level Maths 9709 AS)

14 questions · 50 marks · 75 minutes

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1. Question 1

   3 marksIntegration Using Trigonometric Identities
   Step 1: Use the double angle identity $\sin^2 x = \dfrac{1 - \cos 2x}{2}$, so $6\sin^2 x = 3(1 - \cos 2x) = 3 - 3\cos 2x$. Step 2: Integrate term by term: $\displaystyle\int (3 - 3\cos 2x)\,\mathrm{d}x = 3x - 3 \times \dfrac{\sin 2x}{2} + c = 3x - \dfrac{3}{2}\sin 2x + c$.
   Method:

   Replace $\sin^2 x$ using the double angle identity, then integrate each term.
   Examiner tips

   • When integrating $\sin^2 x$ or $\cos^2 x$, always use the double angle identity first
   • Remember that $\int \cos kx\,\mathrm{d}x = \frac{1}{k}\sin kx + c$

2. Question 2

   3 marksSolving Exponential Equations
   Step 1: Expand: $\mathrm{e}^{4x} - 8\mathrm{e}^{2x} = 48$, i.e. $\mathrm{e}^{4x} - 8\mathrm{e}^{2x} - 48 = 0$. Step 2: Substitute $u = \mathrm{e}^{2x}$: $u^2 - 8u - 48 = 0$. Step 3: Factorise: $(u - 12)(u + 4) = 0$, so $u = 12$ or $u = -4$. Step 4: Since $\mathrm{e}^{2x} > 0$ for all real $x$, reject $u = -4$. Therefore $\mathrm{e}^{2x} = 12$, giving $x = \dfrac{1}{2}\ln 12$.
   Method:

   Substitute $u = \mathrm{e}^{2x}$, solve the resulting quadratic, reject the negative root, then take logarithms.
   Examiner tips

   • Recognise that $\mathrm{e}^{4x} = (\mathrm{e}^{2x})^2$ to form a quadratic
   • Always reject negative values when solving for $\mathrm{e}^{f(x)}$

3. Question 3a

   3 marksModulus Equations
   Step 1: When $|a| = |b|$, either $a = b$ or $a = -b$. Step 2: Case 1: $2x - 3 = 5x + 2 \Rightarrow -3x = 5 \Rightarrow x = -\dfrac{5}{3}$. Step 3: Case 2: $2x - 3 = -(5x + 2) \Rightarrow 2x - 3 = -5x - 2 \Rightarrow 7x = 1 \Rightarrow x = \dfrac{1}{7}$.
   Method:

   Consider the two cases $2x - 3 = 5x + 2$ and $2x - 3 = -(5x + 2)$, and solve each.
   Examiner tips

   • When solving $|f(x)| = |g(x)|$, always consider both cases
   • Alternatively, squaring both sides also works: $(2x-3)^2 = (5x+2)^2$

4. Question 3b

   3 marksModulus Trigonometric Equations
   Step 1: Replace $x$ by $\sec\theta$: $\sec\theta = -\dfrac{5}{3}$ or $\sec\theta = \dfrac{1}{7}$. Step 2: $\sec\theta = \dfrac{1}{7}$ gives $\cos\theta = 7$, which is impossible since $|\cos\theta| \le 1$. Step 3: $\sec\theta = -\dfrac{5}{3}$ gives $\cos\theta = -\dfrac{3}{5}$. In the range $\pi < \theta < 2\pi$, $\cos\theta < 0$ only for $\pi < \theta < \dfrac{3\pi}{2}$. Step 4: $\theta = \pi + \cos^{-1}\!\left(\dfrac{3}{5}\right) = \pi + 0.9273\ldots = 4.069\ldots \approx 4.07$ (3 s.f.).
   Method:

   Substitute $\sec\theta$ for $x$, convert to $\cos\theta$, reject the impossible value, and find $\theta$ in the given range.
   Examiner tips

   • Always check whether $\cos\theta$ values are in the valid range $[-1, 1]$
   • Be careful with the quadrant when the range is specified

5. Question 4

   5 marksCompound Angle Equations
   Step 1: Write $\cot\theta = \dfrac{1}{\tan\theta}$ and use the compound angle formula: $\tan(\theta + 45^{\circ}) = \dfrac{\tan\theta + 1}{1 - \tan\theta}$. Step 2: The equation becomes $\dfrac{1}{\tan\theta} \times \dfrac{\tan\theta + 1}{1 - \tan\theta} = 7$, i.e. $\dfrac{1 + \tan\theta}{\tan\theta(1 - \tan\theta)} = 7$. Step 3: Cross-multiply: $1 + \tan\theta = 7\tan\theta - 7\tan^2\theta$, i.e. $7\tan^2\theta - 6\tan\theta + 1 = 0$. Step 4: Quadratic formula: $\tan\theta = \dfrac{6 \pm \sqrt{36 - 28}}{14} = \dfrac{6 \pm 2\sqrt{2}}{14} = \dfrac{3 \pm \sqrt{2}}{7}$. Step 5: $\tan\theta = \dfrac{3 - \sqrt{2}}{7} \Rightarrow \theta = 12.8^{\circ}$; $\tan\theta = \dfrac{3 + \sqrt{2}}{7} \Rightarrow \theta = 32.2^{\circ}$.
   Method:

   Convert to $\tan\theta$ using identities, form a quadratic, solve using the quadratic formula, and find both angles.
   Examiner tips

   • The addition formula for $\tan$ with $45^{\circ}$ simplifies nicely since $\tan 45^{\circ} = 1$
   • After forming the quadratic, check the discriminant before using the formula

6. Question 5a

   2 marksExponential Curve Intersections
   Step 1: At $B$, $y = 0$: $8\mathrm{e}^{-x/2} - 1 = 0 \Rightarrow 8\mathrm{e}^{-x/2} = 1 \Rightarrow \mathrm{e}^{-x/2} = \dfrac{1}{8}$. Step 2: Take natural log: $-\dfrac{x}{2} = \ln\dfrac{1}{8} = -\ln 8 = -3\ln 2$. Step 3: Multiply by $-2$: $x = 6\ln 2$.
   Method:

   Set $y = 0$, isolate the exponential, take logarithms, and simplify using log laws.
   Examiner tips

   • For 'show that' questions, every step must be clearly justified
   • Use $\ln\frac{1}{a} = -\ln a$ and $\ln a^n = n\ln a$

7. Question 5b

   5 marksArea Between Curve and Line
   Step 1: Line $AB$ has gradient $\dfrac{0 - 7}{6\ln 2 - 0} = -\dfrac{7}{6\ln 2}$, so $y = 7 - \dfrac{7x}{6\ln 2}$. Step 2: Area $= \displaystyle\int_0^{6\ln 2}\!\left[(8\mathrm{e}^{-x/2} - 1) - \left(7 - \dfrac{7x}{6\ln 2}\right)\right]\mathrm{d}x = \int_0^{6\ln 2}\!\left(8\mathrm{e}^{-x/2} - 8 + \dfrac{7x}{6\ln 2}\right)\mathrm{d}x$. Step 3: Integrate: $\left[-16\mathrm{e}^{-x/2} - 8x + \dfrac{7x^2}{12\ln 2}\right]_0^{6\ln 2}$. Step 4: At $x = 6\ln 2$: $-16\mathrm{e}^{-3\ln 2} - 48\ln 2 + \dfrac{7(6\ln 2)^2}{12\ln 2} = -2 - 48\ln 2 + 21\ln 2 = -2 - 27\ln 2$. At $x = 0$: $-16 - 0 + 0 = -16$. Step 5: Area $= (-2 - 27\ln 2) - (-16) = 14 - 27\ln 2$. Taking absolute value (since curve is above line): Area $= 27\ln 2 - 14$.
   Method:

   Find line $AB$, set up the integral of (curve $-$ line), integrate, and evaluate using exact values.
   Examiner tips

   • Be careful with the signs when evaluating definite integrals
   • Remember $\mathrm{e}^{-3\ln 2} = 2^{-3} = \frac{1}{8}$

8. Question 6a

   4 marksParametric Differentiation
   Step 1: $\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = \sec^2\theta$ and $\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = \cos\theta - 6\sin^2\theta\cos\theta = \cos\theta(1 - 6\sin^2\theta)$. Step 2: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\cos\theta(1 - 6\sin^2\theta)}{\sec^2\theta} = \cos^3\theta(1 - 6\sin^2\theta)$. Step 3: Replace $\sin^2\theta = 1 - \cos^2\theta$: $\cos^3\theta(1 - 6 + 6\cos^2\theta) = \cos^3\theta(-5 + 6\cos^2\theta) = 6\cos^5\theta - 5\cos^3\theta$.
   Method:

   Differentiate parametrically, divide, and simplify using $\sin^2\theta = 1 - \cos^2\theta$.
   Examiner tips

   • For parametric differentiation, $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta}$
   • The chain rule gives $\frac{\mathrm{d}}{\mathrm{d}\theta}(\sin^3\theta) = 3\sin^2\theta\cos\theta$

9. Question 6b

   5 marksNormal to a Parametric Curve
   Step 1: Find where the curve crosses the $x$-axis: $y = 0 \Rightarrow \sin\theta(1 - 2\sin^2\theta) = 0$. Since $0 < \theta < \dfrac{\pi}{2}$, $\sin\theta \ne 0$, so $\sin^2\theta = \dfrac{1}{2}$, giving $\theta = \dfrac{\pi}{4}$. Step 2: At $\theta = \dfrac{\pi}{4}$: $x = \tan\dfrac{\pi}{4} = 1$ and $y = 0$. The point is $(1, 0)$. Step 3: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6\left(\dfrac{1}{\sqrt{2}}\right)^{\!5} - 5\left(\dfrac{1}{\sqrt{2}}\right)^{\!3} = \dfrac{6}{4\sqrt{2}} - \dfrac{5}{2\sqrt{2}} = \dfrac{6 - 10}{4\sqrt{2}} = -\dfrac{4}{4\sqrt{2}} = -\dfrac{1}{\sqrt{2}}$. Step 4: Normal gradient $= -\dfrac{1}{(-1/\sqrt{2})} = \sqrt{2}$. Step 5: Normal equation: $y - 0 = \sqrt{2}(x - 1)$, i.e. $y = \sqrt{2}\,x - \sqrt{2}$.
   Method:

   Find $\theta$ for the $x$-intercept, evaluate the gradient, take the negative reciprocal for the normal, and write the line equation.
   Examiner tips

   • The normal is perpendicular to the tangent, so its gradient is $-1/m$ where $m$ is the tangent gradient
   • Check your value of $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ carefully when raising to powers

10. Question 7a

    2 marksFactor Theorem
    Step 1: Since $(x + 2)$ is a factor, $p(-2) = 0$. Step 2: $p(-2) = 2(-2)^4 + k(-2)^3 + k(-2)^2 + 17(-2) + 18 = 32 - 8k + 4k - 34 + 18 = 0$. Step 3: $-4k + 16 = 0 \Rightarrow k = 4$.
    Method:

    Apply the factor theorem by substituting $x = -2$ and solving for $k$.
    Examiner tips

    • Be careful with signs when substituting negative values into polynomials
    • Check: $p(-2) = 32 - 32 + 16 - 34 + 18 = 0$ confirms $k = 4$

11. Question 7b

    4 marksPolynomial Division and Root Rearrangement
    Step 1: Divide $p(x) = 2x^4 + 4x^3 + 4x^2 + 17x + 18$ by $(x + 2)$ using polynomial long division or synthetic division. Step 2: $2x^4 \div x = 2x^3$. Multiply: $2x^3(x + 2) = 2x^4 + 4x^3$. Subtract: $4x^2 + 17x + 18$. Step 3: $4x^2 \div x = 0x^2$... Actually: remainder after first step is $0x^3 + 4x^2 + 17x + 18$. Next: $0x^2(x+2) = 0$. Then $4x^2 \div x = 4x$... Correction: the quotient has no $x^2$ term. Continuing: $4x(x+2) = 4x^2 + 8x$. Subtract: $9x + 18$. Then $9(x + 2) = 9x + 18$. Remainder $= 0$. Step 4: Quotient is $2x^3 + 0x^2 + 4x + 9 = 2x^3 + 4x + 9$. Step 5: For the non-integer root: $2x^3 + 4x + 9 = 0 \Rightarrow x^3 = -2x - 4.5 \Rightarrow x = \sqrt[3]{-2x - 4.5}$.
    Method:

    Divide $p(x)$ by $(x+2)$ to find the cubic factor, then rearrange the cubic equation into the given iterative form.
    Examiner tips

    • In polynomial division, if a power of $x$ is missing in the quotient, its coefficient is zero
    • To show the iterative formula, isolate $x^3$ and take cube roots

12. Question 7c

    2 marksSign Change Method
    Step 1: Let $f(x) = 2x^3 + 4x + 9$. Step 2: $f(-1.4) = 2(-1.4)^3 + 4(-1.4) + 9 = 2(-2.744) - 5.6 + 9 = -5.488 - 5.6 + 9 = -2.088$. Step 3: $f(-1.0) = 2(-1)^3 + 4(-1) + 9 = -2 - 4 + 9 = 3$. Step 4: Since $f(-1.4) < 0$ and $f(-1.0) > 0$, there is a sign change, confirming a root between $-1.4$ and $-1.0$.
    Method:

    Evaluate $f(x)$ at $x = -1.4$ and $x = -1.0$, observe the sign change, and conclude.
    Examiner tips

    • Always state the values and conclude with a sign change statement
    • Be careful with negative numbers raised to odd powers

13. Question 7d

    3 marksIterative Methods
    Step 1: $x_0 = -1$. Step 2: $x_1 = \sqrt[3]{-2(-1) - 4.5} = \sqrt[3]{2 - 4.5} = \sqrt[3]{-2.5} = -1.3572\ldots$ Step 3: $x_2 = \sqrt[3]{-2(-1.3572) - 4.5} = \sqrt[3]{2.7145 - 4.5} = \sqrt[3]{-1.7855} = -1.2134\ldots$ Step 4: $x_3 = \sqrt[3]{-2(-1.2134) - 4.5} = \sqrt[3]{2.4268 - 4.5} = \sqrt[3]{-2.0732} = -1.2740\ldots$ Step 5: Continue iterating until values agree to 3 s.f.: $\beta = -1.26$.
    Method:

    Apply the iterative formula repeatedly from $x_0 = -1$ until convergence to 3 significant figures.
    Examiner tips

    • Keep at least 4 decimal places during iterations to avoid premature rounding errors
    • Continue iterating until two successive values agree to the required accuracy

14. Question 8

    6 marksProduct Rule with Exponential and Square Root
    Step 1: Let $u = 4\mathrm{e}^{1-2x}$ and $v = \sqrt{3x-1} = (3x-1)^{1/2}$. Step 2: $u' = -8\mathrm{e}^{1-2x}$ and $v' = \dfrac{3}{2\sqrt{3x-1}}$. Step 3: Product rule: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -8\mathrm{e}^{1-2x}\sqrt{3x-1} + 4\mathrm{e}^{1-2x} \times \dfrac{3}{2\sqrt{3x-1}}$. Step 4: Set $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$. Factor out $\dfrac{\mathrm{e}^{1-2x}}{\sqrt{3x-1}}$: $-8(3x-1) + 6 = 0$. Step 5: $-24x + 8 + 6 = 0 \Rightarrow -24x + 14 = 0 \Rightarrow x = \dfrac{7}{12}$. Step 6: $y = 4\mathrm{e}^{1 - 7/6}\sqrt{7/4 - 1} = 4\mathrm{e}^{-1/6}\sqrt{3/4} = 4 \times \dfrac{\sqrt{3}}{2} \times \mathrm{e}^{-1/6} = 2\sqrt{3}\,\mathrm{e}^{-1/6}$.
    Method:

    Differentiate using the product rule, set equal to zero, factor and solve for $x$, then substitute back to find $y$.
    Examiner tips

    • When the product rule gives a sum of two terms, factor out common factors before solving
    • Be careful with the exponent: $1 - 2 \times \frac{7}{12} = -\frac{1}{6}$