May/June 2025 Paper 22 Worked Answers (A-Level Maths 9709 AS)

14 questions · 50 marks · 75 minutes

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1. Question 1

   3 marksIntegration of Rational Functions
   Step 1: Integrate: $\displaystyle\int \dfrac{10}{5x+3}\,\mathrm{d}x = \dfrac{10}{5}\ln(5x+3) = 2\ln(5x+3)$. Step 2: Evaluate from 1 to 9: $2\ln(5(9)+3) - 2\ln(5(1)+3) = 2\ln 48 - 2\ln 8 = 2\ln\dfrac{48}{8} = 2\ln 6$. Step 3: Use the log law $2\ln 6 = \ln 6^2 = \ln 36$. So $a = 36$.
   Method:

   Integrate $\dfrac{8}{4x+1}$ to get $2\ln(4x+1)$, evaluate from 2 to 11, then simplify using log laws.
   Examiner tips

   • Remember that $\int \dfrac{k}{ax+b}\,\mathrm{d}x = \dfrac{k}{a}\ln|ax+b| + c$
   • Use $\ln p - \ln q = \ln\dfrac{p}{q}$ and $k\ln m = \ln m^k$ to simplify

2. Question 2a

   2 marksVertex of Modulus Graph
   Step 1: The vertex of $y = |2x - 9|$ occurs where the expression inside the modulus equals zero. Step 2: Set $2x - 9 = 0$, so $x = \dfrac{9}{2} = 4.5$. Step 3: At $x = 4.5$, $y = 0$, giving the vertex at $(4.5, 0)$.
   Method:

   Find where the expression inside the modulus equals zero to locate the vertex.
   Examiner tips

   • The vertex of $|ax + b|$ is at $x = -b/a$
   • The graph is V-shaped, opening upwards

3. Question 2b

   3 marksModulus Inequalities
   Step 1: Find the critical point by solving $-(3x-8) = 5x-2$: $-3x+8 = 5x-2$, so $10 = 8x$, giving $x = \dfrac{5}{4}$. Step 2: Check the other case: $3x-8 = 5x-2$, so $-6 = 2x$, giving $x = -3$. Reject since $5(-3)-2 = -17 < 0$. Step 3: For $|3x-8| < 5x-2$, the RHS must be positive and exceed the LHS. Testing $x = 2$: $|6-8| = 2$ and $10-2 = 8$, so $2 < 8$ (true). The solution is $x > \dfrac{5}{4}$.
   Method:

   Solve the two cases, reject the invalid critical point, and determine the solution region.
   Examiner tips

   • When solving $|f(x)| < g(x)$, always check that $g(x) > 0$ at critical points
   • Reject any solution where the RHS is negative

4. Question 3

   5 marksQuotient Rule and Stationary Points
   Step 1: Differentiate $\dfrac{8x}{2x+3}$ using the quotient rule: $\dfrac{8(2x+3) - 8x(2)}{(2x+3)^2} = \dfrac{24}{(2x+3)^2}$. Step 2: So $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{24}{(2x+3)^2} - 6$. Set to zero: $\dfrac{24}{(2x+3)^2} = 6$, so $(2x+3)^2 = 4$. Step 3: $2x + 3 = \pm 2$, giving $x = -\dfrac{1}{2}$ or $x = -\dfrac{5}{2}$. Step 4: The smaller $x$-value is $x = -\dfrac{5}{2}$. Substituting: $y = \dfrac{8(-5/2)}{2(-5/2)+3} - 6(-5/2) + 5 = \dfrac{-20}{-2} + 15 + 5 = 10 + 15 + 5 = 30$.
   Method:

   Apply quotient rule, set derivative to zero, solve for $x$, then find the corresponding $y$-values.
   Examiner tips

   • The quotient rule: $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{u}{v}\right) = \dfrac{u'v - uv'}{v^2}$
   • Remember to substitute back into the original equation to find $y$-values, not the derivative

5. Question 4a

   1 marksShowing an Iterative Equation
   Step 1: At intersection, $4\mathrm{e}^{-2x} = 1 + 0.5\sin 3x$. Step 2: Divide both sides by 4: $\mathrm{e}^{-2x} = 0.25 + 0.125\sin 3x$. Step 3: Take $\ln$ of both sides: $-2x = \ln(0.25 + 0.125\sin 3x)$. Step 4: Divide by $-2$: $x = -0.5\ln(0.25 + 0.125\sin 3x)$.
   Method:

   Set the two curve equations equal, divide by 4, take logarithms, and rearrange for $x$.
   Examiner tips

   • In 'show that' questions, every algebraic step must be shown clearly
   • Remember $\ln(\mathrm{e}^{-2x}) = -2x$

6. Question 4b

   3 marksIterative Methods
   Step 1: $x_0 = 0.5$. Step 2: $x_1 = -0.5\ln(0.25 + 0.125\sin 1.5) = -0.5\ln(0.25 + 0.125 \times 0.9975) = -0.5\ln(0.3747) = 0.4910\ldots$ Step 3: Continue iterating: $x_2 = 0.4912\ldots$, $x_3 = 0.4912\ldots$ Step 4: The iterations converge to $0.4912$ (4 s.f.).
   Method:

   Perform successive substitutions into the iterative formula until convergence to 4 s.f.
   Examiner tips

   • Keep full calculator accuracy between iterations
   • Continue until two consecutive values agree to the required number of significant figures

7. Question 4c

   4 marksArea Between Curves
   Step 1: The area is $\displaystyle\int_0^{0.4912}(4\mathrm{e}^{-2x} - 1 - 0.5\sin 3x)\,\mathrm{d}x$ since $4\mathrm{e}^{-2x}$ is above $1 + 0.5\sin 3x$ on this interval. Step 2: Integrate: $\left[-2\mathrm{e}^{-2x} - x + \dfrac{1}{6}\cos 3x\right]_0^{0.4912}$. Step 3: At $x = 0.4912$: $-2\mathrm{e}^{-0.9824} - 0.4912 + \dfrac{1}{6}\cos 1.4736 \approx -0.749 - 0.491 + 0.019 = -1.221$. Step 4: At $x = 0$: $-2(1) - 0 + \dfrac{1}{6}(1) = -2 + 0.1667 = -1.833$. Step 5: Area $= -1.221 - (-1.833) = 0.612 \approx 0.61$ (2 s.f.).
   Method:

   Integrate the difference (upper curve minus lower curve) from 0 to 0.4912 and evaluate numerically.
   Examiner tips

   • Check which curve is on top by comparing $y$-intercepts: $4 > 1$
   • Remember $\int \sin(ax)\,\mathrm{d}x = -\dfrac{1}{a}\cos(ax)$

8. Question 5a

   4 marksFactor and Remainder Theorem
   Step 1: Since $(2x-1)$ is a factor, $p(1/2) = 0$: $\dfrac{a}{16} + \dfrac{b}{8} + \dfrac{13}{4} - \dfrac{35}{2} + 15 = 0$. Multiply by 16: $a + 2b + 52 - 280 + 240 = 0$, so $a + 2b = -12$. Step 2: Since $(x-3)$ is a factor, $p(3) = 0$: $81a + 27b + 117 - 105 + 15 = 0$, so $81a + 27b = -27$, i.e. $3a + b = -1$. Step 3: From $3a + b = -1$: $b = -1 - 3a$. Substituting into $a + 2b = -12$: $a + 2(-1-3a) = -12$, so $a - 2 - 6a = -12$, giving $-5a = -10$, hence $a = 2$.
   Method:

   Use the factor theorem to form two simultaneous equations, then solve for $a$ and $b$.
   Examiner tips

   • If $(2x-1)$ is a factor, then $p(1/2) = 0$
   • Form two simultaneous equations and solve

9. Question 5b

   3 marksPolynomial Factorisation
   Step 1: First multiply the known factors: $(2x-1)(x-3) = 2x^2 - 7x + 3$. Step 2: Divide $p(x) = 2x^4 - 7x^3 + 13x^2 - 35x + 15$ by $2x^2 - 7x + 3$ using polynomial long division. Step 3: $\dfrac{2x^4 - 7x^3 + 13x^2 - 35x + 15}{2x^2 - 7x + 3} = x^2 + 5$. Step 4: So $p(x) = (2x-1)(x-3)(x^2 + 5)$.
   Method:

   Multiply the known factors, perform polynomial division, and express the complete factorisation.
   Examiner tips

   • Multiply the two known factors first to get a quadratic divisor
   • Check your answer by expanding all factors

10. Question 5c

    2 marksTrigonometric Equations from Polynomials
    Step 1: $p(\cot 2\theta) = 0$ when $\cot 2\theta = \dfrac{1}{2}$, $\cot 2\theta = 3$, or $\cot^2 2\theta + 5 = 0$ (no real solutions). Step 2: $\cot 2\theta = \dfrac{1}{2}$ means $\tan 2\theta = 2$, so $2\theta = \arctan 2 = 1.1071\ldots$, giving $\theta = 0.5536\ldots \approx 0.554$. Step 3: $\cot 2\theta = 3$ means $\tan 2\theta = \dfrac{1}{3}$, so $2\theta = \arctan\left(\dfrac{1}{3}\right) = 0.32175\ldots$, giving $\theta = 0.16088\ldots \approx 0.161$. Step 4: The least positive value is $\theta \approx 0.161$.
    Method:

    Set each factor to zero, reject complex roots, convert to $\tan$, and find the smallest positive $\theta$.
    Examiner tips

    • Remember $\cot = 1/\tan$
    • Check all possible roots and find the smallest positive $\theta$
    • Reject $x^2 + 5 = 0$ since it has no real solutions

11. Question 6a

    3 marksLogarithmic Equations with No Solution
    Step 1: Substitute $y = \mathrm{e}^{-1}$, so $\ln y = -1$: $(x^2 - 3)(-1) + 6x = 14$. Step 2: Simplify: $-x^2 + 3 + 6x = 14$, so $x^2 - 6x + 11 = 0$. Step 3: Discriminant $= (-6)^2 - 4(1)(11) = 36 - 44 = -8 < 0$. Step 4: Since the discriminant is negative, there are no real solutions for $x$, so no point on the curve has $y = \mathrm{e}^{-1}$. For comparison, $y = \mathrm{e}^2$ gives $\ln y = 2$: $2(x^2-3) + 6x = 14 \Rightarrow 2x^2 + 6x - 20 = 0 \Rightarrow x^2 + 3x - 10 = 0$, discriminant $= 49 > 0$ (real solutions exist). $y = 1$ gives $\ln y = 0$: $6x = 14$, so $x = 7/3$ (real solution). $y = \mathrm{e}$ gives $\ln y = 1$: $x^2 - 3 + 6x = 14 \Rightarrow x^2 + 6x - 17 = 0$, discriminant $= 104 > 0$ (real solutions).
    Method:

    Substitute $y = \mathrm{e}^{-1}$, simplify to a quadratic in $x$, and show the discriminant is negative.
    Examiner tips

    • Remember $\ln(\mathrm{e}^k) = k$
    • A negative discriminant means no real solutions for the quadratic

12. Question 6b

    6 marksImplicit Differentiation and Tangent Lines
    Step 1: Differentiate $(x^2-3)\ln y + 6x = 14$ implicitly: $2x\ln y + (x^2-3)\cdot\dfrac{1}{y}\cdot\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6 = 0$. Step 2: Substitute $x = 2$, $y = \mathrm{e}^2$: $2(2)(2) + (4-3)\cdot\dfrac{1}{\mathrm{e}^2}\cdot\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6 = 0$. Step 3: $8 + \dfrac{1}{\mathrm{e}^2}\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6 = 0$, so $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -14\mathrm{e}^2$.
    Method:

    Differentiate implicitly, substitute the point, solve for the gradient, then write the tangent equation.
    Examiner tips

    • Use the product rule when differentiating $(x^2-3)\ln y$
    • Remember that $\dfrac{\mathrm{d}}{\mathrm{d}x}(\ln y) = \dfrac{1}{y}\dfrac{\mathrm{d}y}{\mathrm{d}x}$

13. Question 7a

    6 marksR-Formula (Harmonic Form)
    Step 1: Expand $\sin(\theta + 30^\circ) = \sin\theta\cos 30^\circ + \cos\theta\sin 30^\circ = \dfrac{\sqrt{3}}{2}\sin\theta + \dfrac{1}{2}\cos\theta$. Step 2: $4\cos\theta\sin(\theta + 30^\circ) = 2\sqrt{3}\cos\theta\sin\theta + 2\cos^2\theta$. Step 3: Use double angle identities: $2\sqrt{3}\cos\theta\sin\theta = \sqrt{3}\sin 2\theta$ and $2\cos^2\theta = \cos 2\theta + 1$. Step 4: So $4\cos\theta\sin(\theta+30^\circ) = \sqrt{3}\sin 2\theta + \cos 2\theta + 1$. Step 5: Express $\sqrt{3}\sin 2\theta + \cos 2\theta$ as $R\cos(2\theta - \alpha)$: $R = \sqrt{(\sqrt{3})^2 + 1^2} = 2$, and $\tan\alpha = \dfrac{\sqrt{3}}{1}$, so $\alpha = 60^\circ$. Step 6: Result: $2\cos(2\theta - 60^\circ) + 1$, so $R = 2$, $k = 1$, and $R + k = 3$.
    Method:

    Expand the compound angle, convert to double angles, then apply the R-formula to express in the required form.
    Examiner tips

    • Expand the compound angle first, then use products-to-double-angles
    • For $a\sin\phi + b\cos\phi = R\cos(\phi - \alpha)$, use $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = a/b$

14. Question 7b

    5 marksSolving Trigonometric Equations Using R-Formula
    Step 1: Write $12\cos 2\varphi\sin(2\varphi+30^\circ) = 3 \times 4\cos 2\varphi\sin(2\varphi+30^\circ)$. Step 2: Using the identity with $\theta = 2\varphi$: $3[2\cos(4\varphi-60^\circ)+1] = 5$. Step 3: Simplify: $6\cos(4\varphi-60^\circ) + 3 = 5$, so $\cos(4\varphi-60^\circ) = \dfrac{1}{3}$. Step 4: For $0^\circ < \varphi < 90^\circ$, we have $-60^\circ < 4\varphi - 60^\circ < 300^\circ$. Step 5: $4\varphi - 60^\circ = 70.53^\circ \Rightarrow \varphi = 32.6^\circ$. Step 6: $4\varphi - 60^\circ = 360^\circ - 70.53^\circ = 289.47^\circ \Rightarrow \varphi = 87.4^\circ$. Step 7: Both solutions are in the range, so there are $2$ solutions.
    Method:

    Factor, apply the R-formula result with $\theta = 2\varphi$, solve the cosine equation, and find all solutions in the range.
    Examiner tips

    • Always check the range of the substituted variable ($4\varphi - 60^\circ$) to find all valid solutions
    • Use the identity from part (a) by replacing $\theta$ with $2\varphi$