2025 Edexcel IAL Pure Mathematics (YPM01) Practice Paper with Answers

First, rewrite the derivative by dividing each term in the numerator by \(\sqrt{x} = x^{1/2}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{5x^2}{x^{1/2}} - \frac{3}{x^{1/2}} = 5x^{3/2} - 3x^{-1/2}\)

Next, integrate each term with respect to \(x\) to find \(y\):
\(y = \int (5x^{3/2} - 3x^{-1/2}) \, \mathrm{d}x\)
\(y = \frac{5x^{5/2}}{5/2} - \frac{3x^{1/2}}{1/2} + C\)
\(y = 2x^{5/2} - 6x^{1/2} + C\)
where \(C\) is the constant of integration.

We are given that the curve passes through the point \((4, 50)\). Substitute \(x = 4\) and \(y = 50\) into the equation:
\(50 = 2(4)^{5/2} - 6(4)^{1/2} + C\)

Calculate the powers of 4:
\(4^{5/2} = (\sqrt{4})^5 = 2^5 = 32\)
\(4^{1/2} = \sqrt{4} = 2\)

Substitute these values back into the equation to find \(C\):
\(50 = 2(32) - 6(2) + C\)
\(50 = 64 - 12 + C\)
\(50 = 52 + C\)
\(C = -2\)

Therefore, the equation of the curve is:
\(y = 2x^{5/2} - 6x^{1/2} - 2\)

M1: Attempt to write \(\frac{5x^2 - 3}{\sqrt{x}}\) in the form \(Ax^p + Bx^q\), with at least one index correct.
A1: Correctly simplified expression: \(5x^{3/2} - 3x^{-1/2}\).
M1: Integrates their expression, raising at least one power of \(x\) by 1, and substitutes \(x = 4\) and \(y = 50\) to find a value for \(C\).
A1: Correct final equation \(y = 2x^{5/2} - 6x^{1/2} - 2\) or equivalent. (Must be in simplest form, with fractional indices simplified, e.g., \(\frac{5}{5/2}\) simplified to \(2\)).

(a) Rearranging the equation of \(l_1\) to the form \(y = mx + c\) gives \(3y = 2x + 6\), which simplifies to \(y = \frac{2}{3}x + 2\). Thus, the gradient of \(l_1\) is \(m_1 = \frac{2}{3}\). Since \(l_2\) is perpendicular to \(l_1\), its gradient is \(m_2 = -\frac{1}{m_1} = -\frac{3}{2}\). Using the equation of a straight line with gradient \(m_2 = -\frac{3}{2}\) and point \(B(1, 7)\) gives \(y - 7 = -\frac{3}{2}(x - 1)\). Multiplying by 2 yields \(2y - 14 = -3(x - 1)\), which expands to \(2y - 14 = -3x + 3\). Rearranging into the form \(ax + by + c = 0\) gives \(3x + 2y - 17 = 0\). (b) To find the coordinates of \(A\), solve the equations of \(l_1\) and \(l_2\) simultaneously. Multiplying \(2x - 3y = -6\) by 2 gives \(4x - 6y = -12\). Multiplying \(3x + 2y = 17\) by 3 gives \(9x + 6y = 51\). Adding these two equations gives \(13x = 39\), which simplifies to \(x = 3\). Substituting \(x = 3\) into \(3x + 2y = 17\) gives \(3(3) + 2y = 17\), which simplifies to \(9 + 2y = 17\), so \(2y = 8\) and \(y = 4\). Thus, the coordinates of \(A\) are \((3, 4)\). (c) The distance between \(A(3, 4)\) and \(B(1, 7)\) is given by \(AB = \sqrt{(3-1)^2 + (4-7)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}\).

(a) M1: Attempts to find the gradient of \(l_1\) by rearranging to \(y = mx + c\) or using \(m = -\frac{\text{coefficient of } x}{\text{coefficient of } y}\). M1: Uses the perpendicular gradient rule \(m_2 = -\frac{1}{m_1}\) and attempts to find the equation of the line using \(B(1, 7)\). A1: Correct equation in the required form: \(3x + 2y - 17 = 0\) (or any non-zero integer multiple). (b) M1: Attempts to solve the simultaneous equations by elimination or substitution. A1: Correctly finds either \(x = 3\) or \(y = 4\). A0.5: Correct coordinates \(A(3, 4)\) written together. (c) B1: For the exact length \(\sqrt{13}\).

(a) Rearranging the equation of \(l_2\) to \(y = -2x + 7\) shows that its gradient is \(-2\). Since \(l_1\) is perpendicular to \(l_2\), the gradient of \(l_1\) is \(-\frac{1}{-2} = \frac{1}{2}\). The gradient of \(l_1\) can also be expressed in terms of \(k\) using the points \(A\) and \(B\): \(m = \frac{(k-1) - 5}{4 - 2k} = \frac{k-6}{4-2k}\). Equating this to \(\frac{1}{2}\) gives \(\frac{k-6}{4-2k} = \frac{1}{2}\). Multiplying both sides by \(2(4-2k)\) gives \(2(k-6) = 4-2k\), which expands to \(2k - 12 = 4 - 2k\). Rearranging gives \(4k = 16\), hence \(k = 4\). (b) Substituting \(k = 4\) into the coordinates of \(B\) gives \(B(4, 3)\). The gradient of \(l_1\) is \(\frac{1}{2}\). The equation of \(l_1\) is \(y - 3 = \frac{1}{2}(x - 4)\), which simplifies to \(y = \frac{1}{2}x + 1\). (c) To find the intersection of \(l_1\) and \(l_2\), substitute \(y = \frac{1}{2}x + 1\) into \(2x + y - 7 = 0\). This gives \(2x + (\frac{1}{2}x + 1) - 7 = 0\), which simplifies to \(\frac{5}{2}x - 6 = 0\), so \(5x = 12\) and \(x = 2.4\). Substituting \(x = 2.4\) back into \(y = \frac{1}{2}x + 1\) gives \(y = 1.2 + 1 = 2.2\). The coordinates of the intersection point are \((2.4, 2.2)\) or \((\frac{12}{5}, \frac{11}{5})\).

(a) M1: Finds the gradient of \(l_2\) is \(-2\) and states the perpendicular gradient of \(l_1\) is \(\frac{1}{2}\). M1: Writes an expression for the gradient of \(l_1\) in terms of \(k\). A1: Equates their gradient expression to \(\frac{1}{2}\) and sets up a correct equation in \(k\). A0.5: Correctly solves to show \(k=4\) with no errors seen. (b) M1: Substitutes \(k=4\) to find coordinates for \(A\) or \(B\), and uses their perpendicular gradient to write a line equation. A1: Correct equation in the form \(y = mx + c\), which is \(y = \frac{1}{2}x + 1\). (c) B1: Correctly solves the simultaneous equations to find the intersection coordinates \((2.4, 2.2)\) or equivalent fractions.

\text{(a)} \\ The total surface area \(S\) of a closed box with square base \(x\) and height \(h\) is given by:\\ \[S = 2x^2 + 4xh\] \\ Given \(S = 5400\), we have:\\ \[2x^2 + 4xh = 5400\] \\ \[4xh = 5400 - 2x^2\] \\ \[h = \frac{5400 - 2x^2}{4x} = \frac{1350}{x} - \frac{1}{2}x\] \\ The volume \(V\) of the box is:\\ \[V = x^2 h\] \\ \[V = x^2 \left( \frac{1350}{x} - \frac{1}{2}x \right)\] \\ \[V = 1350x - \frac{1}{2}x^3 \quad \text{(as required)}\] \\\\ \text{(b)} \\ To find the maximum volume, differentiate \(V\) with respect to \(x\):\\ \[\frac{\mathrm{d}V}{\mathrm{d}x} = 1350 - \frac{3}{2}x^2\] \\ At the stationary point, set \(\frac{\mathrm{d}V}{\mathrm{d}x} = 0\):\\ \[1350 - \frac{3}{2}x^2 = 0\] \\ \[\frac{3}{2}x^2 = 1350\] \\ \[x^2 = 900\] \\ \[x = 30 \quad (\text{since } x > 0)\] \\\\ \text{(c)} \\ Find the second derivative to justify that it is a maximum:\\ \[\frac{\mathrm{d}^2V}{\mathrm{d}x^2} = -3x\] \\ When \(x = 30\):\\ \[\frac{\mathrm{d}^2V}{\mathrm{d}x^2} = -3(30) = -90 < 0\] \\ Since the second derivative is negative, the volume is a maximum at \(x = 30\).\\\\ The maximum volume is:\\ \[V = 1350(30) - \frac{1}{2}(30)^3 = 40500 - 13500 = 27000\text{ cm}^3\]

\text{(a)} \\ M1: Formulates a correct expression for the surface area of the box, \(2x^2 + 4xh = 5400\) (or equivalent).\\ M1: Rearranges the formula to express \(h\) in terms of \(x\), e.g., \(h = \frac{5400 - 2x^2}{4x}\).\\ M1: Substitutes their expression for \(h\) into the volume formula \(V = x^2 h\).\\ A1*: Correctly simplifies to the given answer \(V = 1350x - \frac{1}{2}x^3\) with no errors seen.\\ \\ \text{(b)} \\ M1: Correctly differentiates at least one term of \(V\) to find \(\frac{\mathrm{d}V}{\mathrm{d}x} = 1350 - \frac{3}{2}x^2\).\\ M1: Sets their \(\frac{\mathrm{d}V}{\mathrm{d}x} = 0\) and solves for \(x^2\) or \(x\).\\ A1: Obtains \(x = 30\) (accept \(\pm 30\) but must select positive value for final mark).\\ \\ \text{(c)} \\ M1: Differentiates again to find \(\frac{\mathrm{d}^2V}{\mathrm{d}x^2} = -3x\) and evaluates at \(x = 30\) to show it is negative, or uses another valid method to show it is a maximum.\\ A1: Correctly calculates the maximum volume of \(27000\text{ cm}^3\).

\text{(a)} \\ The height of the platform is the height at \(t = 0\):\\ \[H = 45\text{ m}\] \\\\ \text{(b)} \\ \[45 + 18t - 5t^2 = 45 - 5(t^2 - 3.6t)\] \\ Complete the square inside the bracket:\\ \[t^2 - 3.6t = (t - 1.8)^2 - 1.8^2 = (t - 1.8)^2 - 3.24\] \\ Substitute back:\\ \[H = 45 - 5[(t - 1.8)^2 - 3.24]\] \\ \[H = 45 - 5(t - 1.8)^2 + 16.2\] \\ \[H = 61.2 - 5(t - 1.8)^2\] \\ So, \(A = 61.2\), \(B = 5\), and \(C = 1.8\).\\\\ \text{(c)} \\ \text{(i) The maximum height corresponds to the maximum value of } H\text{, which is } 61.2\text{ m.}\\ \text{(ii) This maximum occurs when } t - 1.8 = 0 \implies t = 1.8\text{ seconds.}\\\
\text{(d)} \\ The rocket hits the ground when \(H = 0\):\\ \[61.2 - 5(t - 1.8)^2 = 0\] \\ \[5(t - 1.8)^2 = 61.2\] \\ \[(t - 1.8)^2 = 12.24\] \\ \[t - 1.8 = \pm \sqrt{12.24}\] \\ \[t = 1.8 \pm 3.49857...\] \\ Since \(t \ge 0\), we reject the negative root:\\ \[t = 1.8 + 3.49857... \approx 5.30\text{ seconds (to 3 s.f.)}\]

\text{(a)} \\ B1: 45 (or 45m). (Accept without working).\\ \\ \text{(b)} \\ M1: Attempts to factorise out \(-5\) from the quadratic terms, e.g., \(-5(t^2 - 3.6t)\) or equivalent.\\ M1: Correctly completes the square for \(t^2 - 3.6t\) to get \((t - 1.8)^2 - 3.24\) (or equivalent fraction \((t - 9/5)^2 - 81/25\)).\\ A1: Correctly expresses in the form \(61.2 - 5(t - 1.8)^2\) (or \(\frac{306}{5} - 5\left(t - \frac{9}{5}\right)^2\)).\\ \\ \text{(c)} \\ B1ft: Max height is \(61.2\text{ m}\) (or equivalent fraction) or follow through their value of \(A\).\\ B1ft: Time is \(1.8\text{ s}\) (or equivalent fraction) or follow through their value of \(C\).\\ \\ \text{(d)} \\ M1: Sets their completed square expression or the original expression to 0.\\ M1: Solves for \(t\) using a correct quadratic method (formula or square root method).\\ A1: \(t \approx 5.30\) only (must be 3 s.f.).

\text{(a)} \\ The curve meets the \(x\)-axis where \(y = 0\):\\ \[12x^{1/2} - 3x = 0\] \\ \[3x^{1/2}(4 - x^{1/2}) = 0\] \\ This gives \(x^{1/2} = 0 \implies x = 0\) and \(x^{1/2} = 4 \implies x = 16\).\\ So the points of intersection are \((0,0)\) and \((16,0)\).\\ The curve is above the \(x\)-axis for \(0 < x < 16\). It starts at \((0,0)\), reaches a maximum, and returns to \((16,0)\) on the \(x\)-axis.\\ \\ \text{(b)} \\ The area of the region \(R\) is given by:\\ \[\text{Area} = \int_{0}^{16} (12x^{1/2} - 3x) \mathrm{d}x\] \\ \[= \left[ 12 \left( \frac{x^{3/2}}{3/2} \right) - \frac{3x^2}{2} \right]_{0}^{16}\] \\ \[= \left[ 8x^{3/2} - 1.5x^2 \right]_{0}^{16}\] \\ Substituting the upper limit \(x = 16\):\\ \[\text{Area} = 8(16)^{3/2} - 1.5(16)^2 - (0)\] \\ \[= 8(64) - 1.5(256)\] \\ \[= 512 - 384\] \\ \[= 128\] \\\\ \text{(c)} \\ The line \(x = k\) divides the area of \(128\) into two equal parts of area \(64\).\\ Therefore, \\ \[\int_{0}^{k} (12x^{1/2} - 3x) \mathrm{d}x = 64\] \\ \[\left[ 8x^{3/2} - 1.5x^2 \right]_{0}^{k} = 64\] \\ \[8k^{3/2} - 1.5k^2 = 64\] \\ \[8k^{3/2} - 1.5k^2 - 64 = 0 \quad \text{(as required)}\]

\text{(a)} \\ M1: Sets \(y = 0\) and attempts to solve for \(x\) to find the roots.\\ A1: Finds the roots \(x = 0\) and \(x = 16\) (accept \((0,0)\) and \((16,0)\)).\\ B1: Correct shape of curve in first quadrant starting at \((0,0)\) and ending at \((16,0)\) with a maximum in between.\\ \\ \text{(b)} \\ M1: Integrates at least one term of \(12x^{1/2} - 3x\) correctly, raising the power by 1.\\ A1: Correct integration: \(8x^{3/2} - 1.5x^2\) (or equivalent).\\ M1: Substitutes the limit 16 into their integrated expression (the 0 limit may be implied).\\ A1: Obtains the correct area of 128.\\ \\ \text{(c)} \\ M1: Sets up the equation \(\int_{0}^{k} (12x^{1/2} - 3x) \mathrm{d}x = 64\) (or equivalent, e.g., using limits from \(k\) to \(16\) equaling 64).\\ A1*: Correctly substitutes limits and simplifies to the given result \(8k^{3/2} - 1.5k^2 - 64 = 0\) with no errors seen.

(a) First, write the terms of the equation for \(C\) in index form:
\[y = \frac{4}{3}x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + 3\]

Differentiate with respect to \(x\):
\[\frac{dy}{dx} = \frac{4}{3} \cdot \frac{3}{2}x^{\frac{1}{2}} - 6 \cdot \frac{1}{2}x^{-\frac{1}{2}}\]
\[\frac{dy}{dx} = 2x^{\frac{1}{2}} - 3x^{-\frac{1}{2}}\]

(b) Find the \(y\)-coordinate of \(P\) by substituting \(x = 4\) into the equation of \(C\):
\[y_P = \frac{4}{3}(4)^{\frac{3}{2}} - 6(4)^{\frac{1}{2}} + 3 = \frac{4}{3}(8) - 6(2) + 3 = \frac{32}{3} - 12 + 3 = \frac{5}{3}\]
So \(P\) is \(\left(4, \frac{5}{3}\right)\).

Find the gradient of the tangent at \(P\) by substituting \(x = 4\) into \(\frac{dy}{dx}\):
\[m = 2(4)^{\frac{1}{2}} - 3(4)^{-\frac{1}{2}} = 2(2) - 3\left(\frac{1}{2}\right) = 4 - \frac{3}{2} = \frac{5}{2}\]

The equation of the tangent at \(P\) is:
\[y - \frac{5}{3} = \frac{5}{2}(x - 4)\]

The tangent meets the \(y\)-axis where \(x = 0\):
\[y - \frac{5}{3} = \frac{5}{2}(0 - 4)\]
\[y - \frac{5}{3} = -10\]
\[y = -10 + \frac{5}{3} = -\frac{25}{3}\]

Thus, the coordinates of \(Q\) are \(\left(0, -\frac{25}{3}\right)\).

(c) The given line is \(y - x + 7 = 0 \implies y = x - 7\), which has gradient \(1\).
Since the tangent at \(R\) is parallel to this line, the gradient of the tangent at \(R\) is also \(1\).

Set \(\frac{dy}{dx} = 1\):
\[2x^{\frac{1}{2}} - 3x^{-\frac{1}{2}} = 1\]

Let \(u = x^{\frac{1}{2}}\) (where \(u > 0\) since \(x > 0\)):
\[2u - \frac{3}{u} = 1\]
\[2u^2 - u - 3 = 0\]
\[(2u - 3)(u + 1) = 0\]

Since \(u > 0\), we select the positive root:
\[u = \frac{3}{2}\]

Since \(u = x^{\frac{1}{2}}\):
\[x^{\frac{1}{2}} = \frac{3}{2} \implies x = \frac{9}{4}\]

**Part (a): [3 Marks]**
- **M1**: For an attempt to differentiate. Decreases the power by 1 on at least one term (e.g., \(x^{\frac{3}{2}} \to A x^{\frac{1}{2}}\) or \(x^{\frac{1}{2}} \to B x^{-\frac{1}{2}}\)).
- **A1**: One correct term: either \(2x^{\frac{1}{2}}\) or \(-3x^{-\frac{1}{2}}\).
- **A1**: Fully correct simplified derivative: \(2x^{\frac{1}{2}} - 3x^{-\frac{1}{2}}\).

**Part (b): [3 Marks]**
- **M1**: Substitutes \(x = 4\) into the original equation to find the \(y\)-coordinate of \(P\) AND into their \(\frac{dy}{dx}\) to find the gradient.
- **M1**: Uses their \(y\)-coordinate and gradient to form the equation of the tangent, and substitutes \(x = 0\) to find the \(y\)-intercept.
- **A1**: Correct coordinates \(\left(0, -\frac{25}{3}\right)\) or equivalent exact coordinates.

**Part (c): [2 Marks]**
- **M1**: Sets their \(\frac{dy}{dx} = 1\) and attempts to solve the resulting equation for \(x\). Accept substitution \(u = \sqrt{x}\) leading to a three-term quadratic and solving for \(u\), or squaring both sides to form a quadratic in \(x\).
- **A1**: Correct exact \(x\)-coordinate of \(x = \frac{9}{4}\) (or \(2.25\)). Must reject extraneous solutions (such as \(x = 1\)) if introduced during the solving process.

### Part (a)
Using the Cosine Rule on triangle \(ABC\):
\[AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\]
Substitute the given values:
\[7^2 = x^2 + (x + 3)^2 - 2x(x + 3)\cos(60^\circ)\]
Since \(\cos(60^\circ) = 0.5\):
\[49 = x^2 + (x^2 + 6x + 9) - 2x(x + 3)(0.5)\]
\[49 = 2x^2 + 6x + 9 - x(x + 3)\]
\[49 = 2x^2 + 6x + 9 - x^2 - 3x\]
\[49 = x^2 + 3x + 9\]
Rearrange to form the quadratic equation:
\[x^2 + 3x - 40 = 0 \quad \text{(as required)}\]

### Part (b)
Solve the quadratic equation by factorisation:
\[(x + 8)(x - 5) = 0\]
This gives \(x = -8\) or \(x = 5\).
Since \(x\) represents a physical length, \(x > 0\).
Therefore, \(x = 5\).

### Part (c)
With \(x = 5\), the side lengths are:
\[AB = 5\text{ cm}\]
\[BC = 5 + 3 = 8\text{ cm}\]
Using the area of a triangle formula:
\[\text{Area} = \frac{1}{2}(AB)(BC)\sin(\angle ABC)\]
\[\text{Area} = \frac{1}{2} \times 5 \times 8 \times \sin(60^\circ)\]
Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[\text{Area} = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\text{ cm}^2\]
Thus, \(a = 10\).

### Part (d)
Using the Sine Rule on triangle \(ABC\):
\[\frac{\sin(\angle ACB)}{AB} = \frac{\sin(\angle ABC)}{AC}\]
\[\frac{\sin(\angle ACB)}{5} = \frac{\sin(60^\circ)}{7}\]
\[\sin(\angle ACB) = \frac{5 \sin(60^\circ)}{7} = \frac{5 \times \frac{\sqrt{3}}{2}}{7} = \frac{5\sqrt{3}}{14}\]
Calculating this value:
\[\sin(\angle ACB) \approx 0.618589\]
\[\angle ACB = \arcsin(0.618589) \approx 38.2132^\circ\]
To 1 decimal place, \(\angle ACB = 38.2^\circ\).

**Part (a)**
* **M1**: Attempts to apply the cosine rule with the given expressions. Must have \(7^2 = x^2 + (x + 3)^2 - 2x(x + 3)\cos(60^\circ)\).
* **A1**: Correct expansion of \((x + 3)^2\) to \(x^2 + 6x + 9\) and correct substitution of \(\cos(60^\circ) = 0.5\).
* **M1**: Simplifies the equation by expanding \(-2x(x + 3)(0.5)\) to \(-x^2 - 3x\) and gathering terms.
* **A1\***: Fully correct proof leading to \(x^2 + 3x - 40 = 0\) with no algebraic errors.

**Part (b)**
* **M1**: Attempts to solve the quadratic equation, either by factorisation to \((x + a)(x + b) = 0\) or by using the quadratic formula.
* **A1**: Correctly identifies \(x = 5\) and explicitly rejects \(x = -8\) because a length must be positive.

**Part (c)**
* **M1**: Uses the correct formula for the area of a triangle, \(\frac{1}{2}ab\sin C\), with \(AB = 5\) and \(BC = 8\).
* **A1**: Correct substitution of values: \(\frac{1}{2} \times 5 \times 8 \times \sin(60^\circ)\).
* **A1**: Obtains \(10\sqrt{3}\) (exact form required).

**Part (d)**
* **M1**: Correctly applies the sine rule (or cosine rule) to find angle \(ACB\). E.g., \(\frac{\sin(\angle ACB)}{5} = \frac{\sin(60^\circ)}{7}\).
* **A1**: Rearranges to get \(\sin(\angle ACB) = \frac{5\sqrt{3}}{14}\) or equivalent.
* **M1**: Applies inverse sine to find a value for the angle.
* **A1**: Obtains \(38.2^\circ\) (accept \(38.2\)).

### Part (a)
Since \(S\) lies on \(QR\), the total length of the side \(QR\) is:
\[QR = QS + SR = 7 + 9 = 16\text{ cm}\]
Using the Cosine Rule on the triangle \(PQR\) to find the angle \(\angle PQR\) (let \(\angle PQR = \theta\)):
\[PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos\theta\]
Substitute the known values:
\[15^2 = 12^2 + 16^2 - 2(12)(16)\cos\theta\]
\[225 = 144 + 256 - 384\cos\theta\]
\[225 = 400 - 384\cos\theta\]
\[384\cos\theta = 400 - 225\]
\[384\cos\theta = 175\]
\[\cos\theta = \frac{175}{384} \approx 0.455729\]
\[\theta = \arccos(0.455729) \approx 62.884^\circ\]
To 1 decimal place, the angle \(\angle PQR = 62.9^\circ\).

### Part (b)
Using the Cosine Rule on triangle \(PQS\) to find the length \(PS\):
\[PS^2 = PQ^2 + QS^2 - 2(PQ)(QS)\cos\theta\]
Substitute the exact value of \(\cos\theta = \frac{175}{384}\) (or its decimal equivalent):
\[PS^2 = 12^2 + 7^2 - 2(12)(7)\left(\frac{175}{384}\right)\]
\[PS^2 = 144 + 49 - 168 \times \frac{175}{384}\]
\[PS^2 = 193 - 76.5625 = 116.4375\]
\[PS = \sqrt{116.4375} \approx 10.7906\text{ cm}\]
To 2 decimal places, the length \(PS = 10.79\text{ cm}\).

### Part (c)
First, we find the total area of triangle \(PQR\):
\[\text{Area of } PQR = \frac{1}{2}(PQ)(QR)\sin\theta\]
Using \(\theta \approx 62.884^\circ\):
\[\sin\theta \approx 0.89012\]
\[\text{Area of } PQR = \frac{1}{2} \times 12 \times 16 \times 0.89012 \approx 85.451\text{ cm}^2\]
Since \(S\) divides the base \(QR\) into segments \(QS = 7\) and \(SR = 9\), the area of triangle \(PSR\) is proportional to the base length \(SR\):
\[\text{Area of } PSR = \frac{SR}{QR} \times \text{Area of } PQR\]
\[\text{Area of } PSR = \frac{9}{16} \times 85.451 \approx 48.066\text{ cm}^2\]
To 1 decimal place, the area of triangle \(PSR = 48.1\text{ cm}^2\).

*(Alternatively, use the Sine Rule to find \(\angle PRQ = 45.4^\circ\) and calculate \(\text{Area} = \frac{1}{2} \times PR \times SR \times \sin(\angle PRQ) = \frac{1}{2} \times 15 \times 9 \times \sin(45.4^\circ) \approx 48.1\text{ cm}^2\).*

### Part (d)
The shortest distance from \(P\) to the line \(QR\) is the perpendicular height, \(h\), of the triangle \(PQR\) relative to base \(QR\).
Using the right-angled triangle formed by the perpendicular from \(P\) to \(QR\):
\[h = PQ \sin\theta\]
\[h = 12 \times \sin(62.884^\circ) = 12 \times 0.89012 \approx 10.6814\text{ cm}\]
To 2 decimal places, the shortest distance is \(10.68\text{ cm}\).

*(Alternatively, using the area formula: \(\frac{1}{2} \times QR \times h = \text{Area of } PQR \implies \frac{1}{2} \times 16 \times h = 85.451 \implies h \approx 10.68\text{ cm}\).*

**Part (a)**
* **M1**: Obtains \(QR = 16\) and sets up the correct Cosine Rule equation to find \(\angle PQR\): \(15^2 = 12^2 + 16^2 - 2(12)(16)\cos\theta\).
* **A1**: Correct numerical expansion: \(225 = 400 - 384\cos\theta\).
* **M1**: Correct rearrangement to find \(\cos\theta = \frac{175}{384}\) (or equivalent decimal \(\approx 0.456\)).
* **A1**: Obtains \(62.9^\circ\) (accept \(62.9\)).

**Part (b)**
* **M1**: Applies the Cosine Rule in triangle \(PQS\) using \(PQ = 12\), \(QS = 7\) and their angle from part (a).
* **A1**: Correctly substitutes values: \(PS^2 = 12^2 + 7^2 - 2(12)(7)\cos(62.9^\circ)\).
* **M1**: Performs calculation to find \(PS^2 \approx 116.4\) and takes the square root.
* **A1**: Obtains \(10.79\) (accept \(10.79\text{ cm}\)).

**Part (c)**
* **M1**: Uses the correct formula for the area of triangle \(PQR\): \(\frac{1}{2} \times 12 \times 16 \times \sin(62.9^\circ)\) (or uses another valid method to find area of \(PSR\)).
* **M1**: Multiplies the total area of \(PQR\) by \(\frac{9}{16}\) to find the area of \(PSR\), or uses a direct formula for the area of \(PSR\).
* **A1**: Obtains \(48.1\) (accept \(48.1\text{ cm}^2\)).

**Part (d)**
* **M1**: Sets up a correct trigonometric or area relation to find the perpendicular height, e.g., \(h = 12\sin(62.9^\circ)\) or \(0.5 \times 16 \times h = 85.45\).
* **A1**: Obtains \(10.68\) (accept \(10.68\text{ cm}\)).

Using the formula for the sum of the first \( n \) terms of a geometric series: \( S_2 = a + ar = a(1 + r) = 15 \) [Equation 1]. Using the formula for the sum to infinity: \( S_{\infty} = \frac{a}{1 - r} = 27 \implies a = 27(1 - r) \) [Equation 2]. Substituting Equation 2 into Equation 1 gives: \( 27(1 - r)(1 + r) = 15 \implies 27(1 - r^2) = 15 \implies 1 - r^2 = \frac{15}{27} = \frac{5}{9} \). Solving for \( r^2 \) yields \( r^2 = 1 - \frac{5}{9} = \frac{4}{9} \), which gives \( r = \pm \frac{2}{3} \). If \( r = \frac{2}{3} \), then \( a = 27\left(1 - \frac{2}{3}\right) = 9 \). If \( r = -\frac{2}{3} \), then \( a = 27\left(1 - \left(-\frac{2}{3}\right)\right) = 45 \). Therefore, the possible solutions are \( r = \frac{2}{3}, a = 9 \) and \( r = -\frac{2}{3}, a = 45 \).

M1: Attempts to write two equations in terms of \( a \) and \( r \) using the sum of the first two terms and sum to infinity formulas. M1: Eliminates \( a \) to obtain a quadratic equation in terms of \( r \) (e.g., \( 27(1-r^2) = 15 \)). A1: Solves to find both correct values of \( r = \pm \frac{2}{3} \). M1: Substitutes at least one value of \( r \) back to find a corresponding value of \( a \). A1: Correctly identifies both pairs of solutions: \( r = \frac{2}{3}, a = 9 \) and \( r = -\frac{2}{3}, a = 45 \).

The \( n \)-th term of an arithmetic series is given by \( u_n = a + (n - 1)d \). Since the 3rd term is 19, we have \( a + 2d = 19 \) [Equation 1]. The sum of the first \( n \) terms is given by \( S_n = \frac{n}{2}(2a + (n - 1)d) \). Since the sum of the first 10 terms is 290, we have \( \frac{10}{2}(2a + 9d) = 290 \implies 5(2a + 9d) = 290 \implies 2a + 9d = 58 \) [Equation 2]. We solve these equations simultaneously. Multiplying Equation 1 by 2 gives \( 2a + 4d = 38 \) [Equation 3]. Subtracting Equation 3 from Equation 2 gives \( 5d = 20 \implies d = 4 \). Substituting \( d = 4 \) back into Equation 1 gives \( a + 2(4) = 19 \implies a = 11 \). The sum of the first 20 terms is \( S_{20} = \frac{20}{2}(2a + 19d) = 10(2(11) + 19(4)) = 10(22 + 76) = 10(98) = 980 \).

M1: Formulates two linear equations in terms of \( a \) and \( d \) using \( u_3 \) and \( S_{10} \). A1: Obtains two correct equations equivalent to \( a + 2d = 19 \) and \( 2a + 9d = 58 \). M1: Solves the simultaneous equations to find the values of \( a \) and \( d \). A1: Correctly finds \( a = 11 \) and \( d = 4 \). A1: Correctly calculates \( S_{20} = 980 \).

(a) Using the binomial theorem:
\((2 + kx)^6 = 2^6 + \binom{6}{1}(2^5)(kx)^1 + \binom{6}{2}(2^4)(kx)^2 + \dots\)
\(= 64 + 6 \times 32 \times kx + 15 \times 16 \times k^2 x^2 + \dots\)
\(= 64 + 192kx + 240k^2x^2\)

(b) Consider the expansion of \((3 - x)(64 + 192kx + 240k^2x^2 + \dots)\)

The terms containing \(x^2\) are:
\(3 \times 240k^2x^2 - x \times 192kx = 720k^2x^2 - 192kx^2 = (720k^2 - 192k)x^2\)

Since the coefficient of \(x^2\) is zero:
\(720k^2 - 192k = 0\)

Since \(k\) is non-zero, we can divide by \(k\):
\(720k = 192\)
\(k = \frac{192}{720} = \frac{4}{15}\)

(a)
M1: An attempt at the binomial expansion with correct structure for at least two terms (e.g., \(2^6 + \dots\) or \(\binom{6}{1}2^5(kx)\)).
A1: Correct first two terms: \(64 + 192kx\).
A1: Correct third term: \(240k^2x^2\) (must be fully simplified).

(b)
M1: Correct algebraic expression for the coefficient of \(x^2\) in terms of \(k\), showing the combination of two terms: \(3(240k^2) - 1(192k)\).
A1ft: Correct equation \(720k^2 - 192k = 0\) (or equivalent based on their coefficients from part a).
M1: An attempt to solve the quadratic equation to find a non-zero value for \(k\).
A1: \(k = \frac{4}{15}\) (or exact equivalent).

(a) Since \((x - 2)\) is a factor, by the factor theorem:
\(f(2) = 0\)
\(2(2)^3 + a(2)^2 + b(2) - 6 = 0\)
\(16 + 4a + 2b - 6 = 0\)
\(4a + 2b = -10 \implies 2a + b = -5\) (Equation 1)

Since the remainder when divided by \((x + 2)\) is \(-4\), by the remainder theorem:
\(f(-2) = -4\)
\(2(-2)^3 + a(-2)^2 + b(-2) - 6 = -4\)
\(-16 + 4a - 2b - 6 = -4\)
\(4a - 2b = 18 \implies 2a - b = 9\) (Equation 2)

Adding Equation 1 and Equation 2:
\(4a = 4 \implies a = 1\)

Substituting \(a = 1\) into Equation 1:
\(2(1) + b = -5 \implies b = -7\)

(b) Substituting \(a = 1\) and \(b = -7\) into \(f(x)\):
\(f(x) = 2x^3 + x^2 - 7x - 6\)

Since \((x - 2)\) is a factor, we can express \(f(x)\) as:
\(2x^3 + x^2 - 7x - 6 = (x - 2)(2x^2 + px + q)\)

By comparing coefficients or using algebraic division:
\(2x^2 + px + q = 2x^2 + 5x + 3\)

Factorising the quadratic term:
\(2x^2 + 5x + 3 = (2x + 3)(x + 1)\)

Therefore, the fully factorised form of \(f(x)\) is:
\(f(x) = (x - 2)(2x + 3)(x + 1)\)

(a)
M1: Attempts to use the factor theorem by setting \(f(2) = 0\).
A1: Obtains a correct simplified linear equation in \(a\) and \(b\), e.g., \(2a + b = -5\) or \(4a + 2b = -10\).
M1: Attempts to use the remainder theorem by setting \(f(-2) = -4\).
A1: Obtains a correct simplified linear equation in \(a\) and \(b\), e.g., \(2a - b = 9\) or \(4a - 2b = 18\).
A1: Solves simultaneously to find both \(a = 1\) and \(b = -7\).

(b)
M1: Attempts algebraic division or inspection to find a quadratic factor \(2x^2 + px + q\) (where \(p\) and \(q\) are constants) of \(f(x)\).
A1: Fully factorised expression \((x - 2)(2x + 3)(x + 1)\) (any order of factors, must be correct).

(a)
The volume \(V\) of the bin is given by:
\[V = \text{length} \times \text{width} \times \text{height} = 3x \times x \times h = 3x^2h\]
We are given that \(V = 432\text{ cm}^3\), so:
\[3x^2h = 432 \implies h = \frac{144}{x^2}\]

The total surface area \(A\) of the bin is made up of:
- The rectangular base: \(3x \times x = 3x^2\)
- Two outer side walls: \(2 \times (3x \times h) = 6xh\)
- Two outer end walls: \(2 \times (x \times h) = 2xh\)
- One internal partition: \(1 \times (x \times h) = xh\)

Summing these areas gives:
\[A = 3x^2 + 6xh + 2xh + xh = 3x^2 + 9xh\]

Substitute \(h = \frac{144}{x^2}\) into the expression for \(A\):
\[A = 3x^2 + 9x\left(\frac{144}{x^2}\right) = 3x^2 + \frac{1296}{x}\]
(as required)

(b)
To find the value of \(x\) for which \(A\) is a minimum, we differentiate \(A\) with respect to \(x\):
\[\frac{\mathrm{d}A}{\mathrm{d}x} = 6x - \frac{1296}{x^2}\]
Set \[\frac{\mathrm{d}A}{\mathrm{d}x} = 0\] to find the stationary point:
\[6x - \frac{1296}{x^2} = 0\]
\[6x^3 = 1296\]
\[x^3 = 216\]
\[x = 6\]

(c)
Substitute \(x = 6\) into the expression for \(A\):
\[A_{\text{min}} = 3(6)^2 + \frac{1296}{6} = 108 + 216 = 324\text{ cm}^2\]

To show that this is a minimum, find the second derivative \(\frac{\mathrm{d}^2A}{\mathrm{d}x^2}\):
\[\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 6 + \frac{2592}{x^3}\]
Substitute \(x = 6\):
\[\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 6 + \frac{2592}{216} = 6 + 12 = 18\]
Since \(\frac{\mathrm{d}^2A}{\mathrm{d}x^2} > 0\), the value of \(A\) is a minimum.

(a)
- **M1**: Sets up a correct equation for the volume: \(3x^2h = 432\) or equivalent.
- **A1**: Correctly expresses \(h\) in terms of \(x\): \(h = \frac{144}{x^2}\) or equivalent.
- **M1**: Writes down a correct expression for the total surface area \(A\) in terms of \(x\) and \(h\): \(A = 3x^2 + 9xh\).
- **A1* (cso)**: Substitutes \(h\) and shows clear algebraic steps to reach the given formula \(A = 3x^2 + \frac{1296}{x}\) with no errors.

(b)
- **M1**: Differentiates \(A = 3x^2 + 1296x^{-1}\) to obtain \(\frac{\mathrm{d}A}{\mathrm{d}x} = kx \pm Cx^{-2}\) where \(k\) and \(C\) are non-zero constants.
- **A1**: Correct derivative: \(\frac{\mathrm{d}A}{\mathrm{d}x} = 6x - \frac{1296}{x^2}\).
- **M1**: Sets their derivative equal to \(0\) and solves to find a value for \(x^3\).
- **A1**: Obtains \(x = 6\).

(c)
- **M1**: Obtains a correct expression for the second derivative \(\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 6 + \frac{2592}{x^3}\).
- **A1**: Finds the minimum surface area is \(324\text{ cm}^2\).
- **A1**: Substitutes \(x = 6\) (or explains using \(x > 0\)) to show \(\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 18 > 0\), concluding it is a minimum.

(a)
The perimeter of the window consists of the bottom edge (length \(2x\)), the two vertical sides (each of length \(y\)), and the semi-circular arc (length \(\pi x\)).
\[P = 2x + 2y + \pi x\]
We are given that \(P = 12\), so:
\[2x + 2y + \pi x = 12\]
\[2y = 12 - (2 + \pi)x \implies y = 6 - \left(1 + \frac{\pi}{2}\right)x\]

The area of the window \(A\) is the sum of the area of the rectangle and the area of the semi-circle:
\[A = 2xy + \frac{1}{2}\pi x^2\]
Substitute \(y = 6 - \left(1 + \frac{\pi}{2}\right)x\) into the area equation:
\[A = 2x\left[6 - \left(1 + \frac{\pi}{2}\right)x\right] + \frac{1}{2}\pi x^2\]
\[A = 12x - (2 + \pi)x^2 + \frac{1}{2}\pi x^2\]
\[A = 12x - 2x^2 - \pi x^2 + \frac{1}{2}\pi x^2\]
\[A = 12x - 2x^2 - \frac{1}{2}\pi x^2\]
\[A = 12x - \left(2 + \frac{\pi}{2}\right)x^2\]
(as required)

(b)
To find the maximum, differentiate \(A\) with respect to \(x\):
\[\frac{\mathrm{d}A}{\mathrm{d}x} = 12 - 2\left(2 + \frac{\pi}{2}\right)x = 12 - (4 + \pi)x\]
Set \(\frac{\mathrm{d}A}{\mathrm{d}x} = 0\):
\[12 - (4 + \pi)x = 0\]
\[(4 + \pi)x = 12\]
\[x = \frac{12}{4 + \pi}\]

(c)
Substitute \(x = \frac{12}{4 + \pi}\) into the expression for \(A\):
\[A = 12\left(\frac{12}{4 + \pi}\right) - \left(2 + \frac{\pi}{2}\right)\left(\frac{12}{4 + \pi}\right)^2\]
Since \(2 + \frac{\pi}{2} = \frac{4 + \pi}{2}\):
\[A = \frac{144}{4 + \pi} - \left(\frac{4 + \pi}{2}\right)\frac{144}{(4 + \pi)^2}\]
\[A = \frac{144}{4 + \pi} - \frac{72}{4 + \pi} = \frac{72}{4 + \pi}\]
Using a calculator:
\[A \approx \frac{72}{7.14159...} \approx 10.1\text{ m}^2\text{ (to 3 s.f.)}\]

To justify that this is a maximum:
\[\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = -(4 + \pi)\]
Since \(4 + \pi > 0\), \(\frac{\mathrm{d}^2A}{\mathrm{d}x^2} < 0\) for all \(x\).
Thus, the value of \(A\) is a maximum.

(a)
- **M1**: Writes down an expression for the perimeter: \(2x + 2y + \pi x = 12\).
- **A1**: Correctly expresses \(y\) in terms of \(x\), e.g., \(y = 6 - x - \frac{\pi}{2}x\).
- **M1**: Writes down a correct expression for the total area: \(A = 2xy + \frac{1}{2}\pi x^2\).
- **A1* (cso)**: Substitutes \(y\) and shows clear algebraic steps to reach the given formula \(A = 12x - \left(2 + \frac{\pi}{2}\right)x^2\) with no errors.

(b)
- **M1**: Differentiates \(A\) with respect to \(x\) to get \(\frac{\mathrm{d}A}{\mathrm{d}x} = 12 - kx\), where \(k\) is a constant in terms of \(\pi\).
- **A1**: Correct derivative: \(\frac{\mathrm{d}A}{\mathrm{d}x} = 12 - (4 + \pi)x\).
- **M1**: Sets their derivative equal to \(0\) and solves for \(x\).
- **A1**: Exact value: \(x = \frac{12}{4 + \pi}\).

(c)
- **M1**: Substitutes their value of \(x\) from (b) into the area equation to find \(A\).
- **A1**: Obtains \(10.1\) (accept \(10.08\) or better, or exact \(\frac{72}{4+\pi}\) followed by \(10.1\)).
- **A1**: Finds \(\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = -(4 + \pi) < 0\) and concludes it is a maximum.

Let the two consecutive odd integers be represented as \(2n - 1\) and \(2n + 1\), where \(n\) is an integer. The sum of the squares of these two integers is given by: \((2n - 1)^2 + (2n + 1)^2\). Expanding both terms, we get: \((2n - 1)^2 = 4n^2 - 4n + 1\) and \((2n + 1)^2 = 4n^2 + 4n + 1\). Adding these expressions together yields: \((4n^2 - 4n + 1) + (4n^2 + 4n + 1) = 8n^2 + 2\). We can rewrite this result as \(8(n^2) + 2\). Since \(n^2\) is an integer, \(8(n^2)\) must be a multiple of 8. Therefore, the sum is always 2 more than a multiple of 8. (Alternatively, choosing consecutive odd integers as \(2k + 1\) and \(2k + 3\) yields \((2k + 1)^2 + (2k + 3)^2 = 8k^2 + 16k + 10 = 8(k^2 + 2k + 1) + 2\), which leads to the same conclusion).

M1: Set up correct algebraic expressions for two consecutive odd integers, such as \(2n-1\) and \(2n+1\). M1: Attempt to expand both squared expressions. A1: Correctly simplify the sum to \(8n^2 + 2\) (or \(8k^2 + 16k + 10\) if using \(2k+1\) and \(2k+3\)). M1: Factorise or structure the expression to clearly isolate a multiple of 8 plus 2, e.g., \(8(n^2) + 2\) or \(8(k^2+2k+1) + 2\). A1.5: Complete proof with a concluding statement explaining that since the term in the bracket is an integer, the expression is 2 more than a multiple of 8.

First, apply the power law of logarithms to the first term: \(2\log_5(x - 2) = \log_5(x - 2)^2\). Next, express the constant 1 as a logarithm to base 5: \(1 = \log_5 5\). Substitute these into the equation: \\log_5(x - 2)^2 - \log_5(x + 10) = \log_5 5 - \log_5 2\\. Apply the division law of logarithms to both sides: \\log_5 \left(\frac{(x-2)^2}{x+10}\right) = \log_5 \left(\frac{5}{2}\right)\\. Since the bases are identical, equate the arguments: \\\frac{(x - 2)^2}{x + 10} = 2.5\\\. Multiply out the denominator: \\2(x - 2)^2 = 5(x + 10)\\. Expand and simplify into a quadratic equation: \\2(x^2 - 4x + 4) = 5x + 50 \implies 2x^2 - 8x + 8 = 5x + 50 \implies 2x^2 - 13x - 42 = 0\\. Solve the quadratic equation by factorisation: \(2x - 21)(x + 2) = 0\\, which gives potential solutions \\x = 10.5\\ or \\x = -2\\. We must check the constraints: \\x - 2 > 0 \implies x > 2\\ for the term \\\log_5(x-2)\\\ to be defined. Therefore, \\x = -2\\ is rejected. The only valid solution is \\x = 10.5\\.

M1: Use the power law of logarithms correctly on the first term. M1: Express the constant 1 as \(\log_5 5\) or combine constants into a single log term. M1: Apply the subtraction law of logarithms to combine terms on one or both sides. A1: Formulate a correct three-term quadratic equation, e.g., \(2x^2 - 13x - 42 = 0\). M1: Solve the quadratic equation to find two roots. A1.5: Give the final answer as \(x = 10.5\) (or \(\frac{21}{2}\)) only, with a clear rejection of the extraneous root \(x = -2\).

Start with the logarithmic equation: \(\log_2 y - \log_2 x = 3\). Apply the quotient rule of logarithms: \(\log_2 \left(\frac{y}{x}\right) = 3\). Convert this to exponential form: \(\frac{y}{x} = 2^3 \implies y = 8x\) (Equation 1). Now look at the second equation: \(2^y = 8^{x+1}\). Express 8 with base 2: \(2^y = (2^3)^{x+1} \implies 2^y = 2^{3x+3}\). Equating the indices gives: \(y = 3x + 3\) (Equation 2). Substitute Equation 1 into Equation 2: \(8x = 3x + 3 \implies 5x = 3 \implies x = 0.6\). Substitute \(x = 0.6\) back into Equation 1 to find \(y\): \(y = 8(0.6) = 4.8\). Both \(x = 0.6\) and \(y = 4.8\) are positive, so they are valid within the domains of the original logarithms.

M1: Apply the subtraction law of logarithms to obtain \(\log_2(y/x) = 3\). A1: Correctly rewrite this to get \(y = 8x\) (or equivalent). M1: Express both sides of the exponential equation with base 2, leading to \(2^y = 2^{3(x+1)}\). A1: Equate indices to obtain \(y = 3x + 3\) (or equivalent). M1: Solve the resulting system of linear equations for one of the variables. A1.5: Correctly find both solutions: \(x = 0.6\) (or \(\frac{3}{5}\)) and \(y = 4.8\) (or \(\frac{24}{5}\)).

(a)
We start with the given equation:
\[6\cos^2 \theta - \cos \theta \tan \theta = 5\]
Using the trigonometric identity \(\tan \theta \equiv \frac{\sin \theta}{\cos \theta}\):
\[6\cos^2 \theta - \cos \theta \left( \frac{\sin \theta}{\cos \theta} \right) = 5\]
\[6\cos^2 \theta - \sin \theta = 5\quad (\text{provided } \cos \theta \neq 0)\]
Now substitute the identity \[\cos^2 \theta \equiv 1 - \sin^2 \theta\] into the equation:
\[6(1 - \sin^2 \theta) - \sin \theta = 5\]
\[6 - 6\sin^2 \theta - \sin \theta = 5\]
Rearranging to make the coefficient of \(\sin^2 \theta\) positive:
\[6\sin^2 \theta + \sin \theta - 1 = 0\quad \text{(as required)}\]

(b)
Let \(\theta = 2x - 30^\circ\). Since \(0 \le x < 360^\circ\), we have:
\[0 \le 2x < 720^\circ \implies -30^\circ \le 2x - 30^\circ < 690^\circ\]
Thus, we solve
\[6\sin^2 \theta + \sin \theta - 1 = 0\quad \text{for } -30^\circ \le \theta < 690^\circ\]
Factoring the quadratic equation:
\[(3\sin \theta - 1)(2\sin \theta + 1) = 0\]
This gives two cases:

**Case 1:** \(\sin \theta = \frac{1}{3}\)
Since \(\sin \theta > 0\), the principal value is:
\[\theta = \arcsin\left(\frac{1}{3}\right) \approx 19.47^\circ\]
The solutions for \(\theta\) in the interval \(-30^\circ \le \theta < 690^\circ\) are:
- \(\theta_1 \approx 19.47^\circ\)
- \(\theta_2 \approx 180^\circ - 19.47^\circ = 160.53^\circ\)
- \(\theta_3 \approx 19.47^\circ + 360^\circ = 379.47^\circ\)
- \(\theta_4 \approx 160.53^\circ + 360^\circ = 520.53^\circ\)

**Case 2:** \(\sin \theta = -\frac{1}{2}\)
Since \(\sin \theta < 0\), the solutions for \(\theta\) in the interval \(-30^\circ \le \theta < 690^\circ\) are:
- \(\theta_5 = -30^\circ\)
- \(\theta_6 = 180^\circ - (-30^\circ) = 210^\circ\)
- \(\theta_7 = -30^\circ + 360^\circ = 330^\circ\)
- \(\theta_8 = 210^\circ + 360^\circ = 570^\circ\)
*(Note: \(330^\circ + 360^\circ = 690^\circ\) is outside the interval because \(\theta < 690^\circ\))*

Now, calculate the values of \(x\) using \(x = \frac{\theta + 30^\circ}{2}\):
- From \(\theta = -30^\circ \implies x = 0^\circ\)
- From \(\theta = 19.47^\circ \implies x \approx 24.7^\circ\)
- From \(\theta = 160.53^\circ \implies x \approx 95.3^\circ\)
- From \(\theta = 210^\circ \implies x = 120^\circ\)
- From \(\theta = 330^\circ \implies x = 180^\circ\)
- From \(\theta = 379.47^\circ \implies x \approx 204.7^\circ\)
- From \(\theta = 520.53^\circ \implies x \approx 275.3^\circ\)
- From \(\theta = 570^\circ \implies x = 300^\circ\)

**Part (a): 3 marks**
- **M1**: Substitutes \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) into the given equation to obtain an equation containing only sine and cosine terms.
- **M1**: Uses the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to obtain an equation in terms of \(\sin \theta\) only.
- **A1\***: Fully correct algebraic working leading to the given quadratic equation \(6\sin^2 \theta + \sin \theta - 1 = 0\) with no errors or omissions. Must state "\(= 0\)".

**Part (b): 5 marks**
- **M1**: Attempt to solve the quadratic equation to find two values for \(\sin(2x - 30^\circ)\). (e.g., through factorisation, quadratic formula, or calculator).
- **A1**: Correct values of \(\sin(2x - 30^\circ) = \frac{1}{3}\) and \(\sin(2x - 30^\circ) = -\frac{1}{2}\).
- **M1**: Attempt to find at least one value of \(2x - 30^\circ\) leading to a value for \(x\).
- **A1**: Any four of the correct eight solutions: \(x = 0^\circ, 24.7^\circ, 95.3^\circ, 120^\circ, 180^\circ, 204.7^\circ, 275.3^\circ, 300^\circ\) (accept values rounded to 1 d.p. where appropriate, or integer equivalent for exact values).
- **A1**: All eight correct solutions: \(x = 0^\circ, 24.7^\circ, 95.3^\circ, 120^\circ, 180^\circ, 204.7^\circ, 275.3^\circ, 300^\circ\) and no extra solutions within the range \(0 \le x < 360^\circ\).

(a) Let \(f(x) = 3\ln(x) + 2x - 5\). Evaluating the function at the endpoints: \(f(1.7) = 3\ln(1.7) + 2(1.7) - 5 \approx -0.0081\) and \(f(1.8) = 3\ln(1.8) + 2(1.8) - 5 \approx 0.3633\). Since there is a change of sign and \(f(x)\) is continuous over the interval \([1.7, 1.8]\), there is a root \(\alpha\) in this interval. (b) Using the iterative formula with \(x_1 = 1.7\): \(x_2 = \frac{5 - 3\ln(1.7)}{2} \approx 1.704058 \approx 1.7041\) (to 4 d.p.) and \(x_3 = \frac{5 - 3\ln(1.704058)}{2} \approx 1.700483 \approx 1.7005\) (to 4 d.p.). (c) To show that \(\alpha = 1.702\) to 3 d.p., we evaluate \(f(x)\) at the bounds \(1.7015\) and \(1.7025\): \(f(1.7015) = 3\ln(1.7015) + 2(1.7015) - 5 \approx -0.0025\) and \(f(1.7025) = 3\ln(1.7025) + 2(1.7025) - 5 \approx 0.0013\). Since there is a change of sign and \(f(x)\) is continuous on \([1.7015, 1.7025]\), the root \(\alpha\) must lie in the interval \((1.7015, 1.7025)\). Thus, \(\alpha = 1.702\) correct to 3 decimal places.

(a) M1: For attempting to evaluate \(f(1.7)\) and \(f(1.8)\). This requires substituting both values into \(f(x)\) and evaluating at least one to 2 decimal places. A0.5: Evaluates both correctly (\(f(1.7) \approx -0.01\) and \(f(1.8) \approx 0.36\)), states that there is a change of sign in a continuous function, and concludes there is a root in the interval. (b) M1: For a correct attempt to find \(x_2\) using the iterative formula. Must show the substitution or \(1.704...\). A0.5: Both \(x_2 = 1.7041\) and \(x_3 = 1.7005\) correct to 4 decimal places. (c) M1: Choosing the appropriate bounds \(1.7015\) and \(1.7025\) and attempting to evaluate \(f(x)\) at these bounds. A0.5: Correct values \(f(1.7015) \approx -0.0025\) and \(f(1.7025) \approx 0.0013\), concluding with a change of sign argument and reference to continuity.

(a) Set \(D = 3\), giving \(4 + 2\cos(0.5t) - 0.15t = 3 \implies 2\cos(0.5t) - 0.15t + 1 = 0\). Let \(f(t) = 2\cos(0.5t) - 0.15t + 1\). Evaluating the function: \(f(3.6) = 2\cos(1.8) - 0.15(3.6) + 1 \approx 0.0056\) and \(f(3.7) = 2\cos(1.85) - 0.15(3.7) + 1 \approx -0.1062\). Since there is a change of sign and \(f(t)\) is continuous over \([3.6, 3.7]\), there is a root \(\alpha\) in this interval. (b) Rearranging: \(2\cos(0.5t) = 0.15t - 1 \implies \cos(0.5t) = 0.075t - 0.5 \implies 0.5t = \arccos(0.075t - 0.5) \implies t = 2\arccos(0.075t - 0.5)\). (c) Using the iterative formula with \(t_1 = 3.6\): \(t_2 = 2\arccos(0.075(3.6) - 0.5) = 2\arccos(-0.23) \approx 3.60488 \approx 3.605\) (to 3 d.p.) and \(t_3 = 2\arccos(0.075(3.60488) - 0.5) \approx 3.60410 \approx 3.604\) (to 3 d.p.).

(a) M1: For attempting to evaluate \(f(t)\) at \(t = 3.6\) and \(t = 3.7\) (requires radian mode). A0.5: Correct values \(f(3.6) \approx 0.01\) and \(f(3.7) \approx -0.11\), and a correct conclusion with reference to sign change and continuity. (b) M1: Attempting to isolate \(\cos(0.5t)\). A0.5: Showing all steps clearly to arrive at the given iterative formula. (c) M1: Substituting \(t_1 = 3.6\) into the iterative formula to find \(t_2\). A0.5: Both \(t_2 = 3.605\) and \(t_3 = 3.604\) correct to 3 decimal places.

(a) Let \(u = x^2 - 3\) and \(v = \mathrm{e}^{-2x}\).
Differentiating with respect to \(x\):
\(\frac{\mathrm{d}u}{\mathrm{d}x} = 2x\)
\(\frac{\mathrm{d}v}{\mathrm{d}x} = -2\mathrm{e}^{-2x}\)

Using the product rule:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = (x^2 - 3)(-2\mathrm{e}^{-2x}) + (2x)(\mathrm{e}^{-2x})\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{-2x}(-2x^2 + 6 + 2x)\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^{-2x}(x^2 - x - 3)\)

(b) For stationary points, set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
Since \(\mathrm{e}^{-2x} \neq 0\), we must have:
\(x^2 - x - 3 = 0\)

Using the quadratic formula:
\(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-3)}}{2(1)} = \frac{1 \pm \sqrt{13}}{2}\)

To find the corresponding \(y\)-coordinates, substitute \(x\) back into the original curve equation:
From the quadratic, we know \(x^2 - 3 = x\). This simplifies the substitution:
\(y = (x^2 - 3)\mathrm{e}^{-2x} = x\mathrm{e}^{-2x}\)

For \(x = \frac{1+\sqrt{13}}{2}\):
\(y = \left(\frac{1+\sqrt{13}}{2}\right)\mathrm{e}^{-2\left(\frac{1+\sqrt{13}}{2}\right)} = \frac{1+\sqrt{13}}{2}\mathrm{e}^{-(1+\sqrt{13})}\)

For \(x = \frac{1-\sqrt{13}}{2}\):
\(y = \left(\frac{1-\sqrt{13}}{2}\right)\mathrm{e}^{-2\left(\frac{1-\sqrt{13}}{2}\right)} = \frac{1-\sqrt{13}}{2}\mathrm{e}^{-(1-\sqrt{13})}\)

So the exact coordinates of the stationary points are:
\(\left(\frac{1+\sqrt{13}}{2}, \frac{1+\sqrt{13}}{2}\mathrm{e}^{-(1+\sqrt{13})}\right)\) and \(\left(\frac{1-\sqrt{13}}{2}, \frac{1-\sqrt{13}}{2}\mathrm{e}^{-(1-\sqrt{13})}\right)\)

(a)
M1: Applies the product rule to differentiate \((x^2-3)\mathrm{e}^{-2x}\). Condone minor sign errors, but must have the structure \(A x \mathrm{e}^{-2x} + B(x^2-3)\mathrm{e}^{-2x}\) with non-zero constants \(A\) and \(B\).
A1: Correct unsimplified expression, e.g., \(2x\mathrm{e}^{-2x} - 2(x^2-3)\mathrm{e}^{-2x}\).
A1: Achieves the fully simplified form: \(-2\mathrm{e}^{-2x}(x^2 - x - 3)\) or equivalent (e.g., \(2\mathrm{e}^{-2x}(3 + x - x^2)\)).

(b)
M1: Sets their \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solves the resulting quadratic equation in \(x\).
A1: Correct \(x\)-coordinates: \(x = \frac{1 \pm \sqrt{13}}{2}\).
M1: Sustitutes at least one of their \(x\)-values back into the equation of curve \(C\) to find a corresponding \(y\)-coordinate.
A1: One correct \(y\)-coordinate in exact simplified form.
A1: Both coordinate pairs fully correct and written in exact form.

(a) Differentiate term by term using the chain rule:
For the first term, let \(u = \tan(2x)\), so \(y_1 = 2u^2\):
\(\frac{\mathrm{d}}{\mathrm{d}x}(2\tan^2(2x)) = 2 \cdot 2\tan(2x) \cdot \frac{\mathrm{d}}{\mathrm{d}x}(\tan(2x))\)
\(= 4\tan(2x) \cdot 2\sec^2(2x) = 8\tan(2x)\sec^2(2x)\)

For the second term, using the standard derivative of \(\sec(kx)\):
\(\frac{\mathrm{d}}{\mathrm{d}x}(-8\sec(2x)) = -8 \cdot 2\sec(2x)\tan(2x) = -16\sec(2x)\tan(2x)\)

Combine the terms:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 8\tan(2x)\sec^2(2x) - 16\sec(2x)\tan(2x)\)

Factorise out the common factor \(8\tan(2x)\sec(2x)\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 8\tan(2x)\sec(2x)[\sec(2x) - 2]\) (as required).

(b) To find the stationary point, set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(8\tan(2x)\sec(2x)[\sec(2x) - 2] = 0\)

Since \(0 < x < \frac{\pi}{4}\), we have \(0 < 2x < \frac{\pi}{2}\).
In this interval:
\(\tan(2x) > 0\) (so \(\tan(2x) \neq 0\))
\(\sec(2x) \ge 1\) (so \(\sec(2x) \neq 0\))

Thus, we must have:
\(\sec(2x) - 2 = 0 \implies \sec(2x) = 2\)
\(\cos(2x) = \frac{1}{2}\)

Since \(0 < 2x < \frac{\pi}{2}\):
\(2x = \frac{\pi}{3} \implies x = \frac{\pi}{6}\)

Now, substitute \(x = \frac{\pi}{6}\) back into the equation of \(C\) to find \(y\):
\(y = 2\tan^2\left(2 \cdot \frac{\pi}{6}\right) - 8\sec\left(2 \cdot \frac{\pi}{6}\right)\)
\(y = 2\tan^2\left(\frac{\pi}{3}\right) - 8\sec\left(\frac{\pi}{3}\right)\)

We know that \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\) and \(\sec\left(\frac{\pi}{3}\right) = 2\):
\(y = 2(\sqrt{3})^2 - 8(2) = 2(3) - 16 = 6 - 16 = -10\)

Therefore, the exact coordinates of the stationary point are \(\left(\frac{\pi}{6}, -10\right)\).

(a)
M1: Applies chain rule to differentiate \(\tan^2(2x)\) to obtain \(k\tan(2x)\sec^2(2x)\) where \(k\) is a constant.
A1: Correct derivative of the first term: \(8\tan(2x)\sec^2(2x)\).
M1: Correctly differentiates \(-8\sec(2x)\) to obtain \(-16\sec(2x)\tan(2x)\).
A1*: Correctly factors out the terms to show the given target expression. Must include at least one intermediate step of combination before the final result.

(b)
M1: Deduces that for the stationary point, \(\sec(2x) = 2\) (or \(\cos(2x) = \frac{1}{2}\)) in the given domain.
A1: Finds \(x = \frac{\pi}{6}\) (must be exact in radians; reject degrees).
M1: Substitutes their exact \(x = \frac{\pi}{6}\) back into the equation of \(C\).
A1: Obtains \(y = -10\) and presents the answer as coordinates \(\left(\frac{\pi}{6}, -10\right)\).

(a) Using the quotient rule with \(u = \ln(3x)\) and \(v = 2x^2 + 1\):
\(\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{3x} \cdot 3 = \frac{1}{x}\)
\(\frac{\mathrm{d}v}{\mathrm{d}x} = 4x\)

Applying the quotient rule formula \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\frac{\mathrm{d}u}{\mathrm{d}x} - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(2x^2 + 1)\left(\frac{1}{x}\right) - 4x\ln(3x)}{(2x^2 + 1)^2}\)

(b) To find the equation of the tangent, we first need the coordinates of the point of contact and the gradient at that point.

When \(x = \frac{1}{3}\):
\(y = \frac{\ln\left(3 \cdot \frac{1}{3}\right)}{2\left(\frac{1}{3}\right)^2 + 1} = \frac{\ln(1)}{\frac{2}{9} + 1} = \frac{0}{\frac{11}{9}} = 0\)
So the point of contact is \(\left(\frac{1}{3}, 0\right)\).

Now substitute \(x = \frac{1}{3}\) into \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find the gradient \(m\):
\(Numerator = \left(2\left(\frac{1}{9}\right) + 1\right)\left(3\right) - 4\left(\frac{1}{3}\right)\ln(1)\)
\(Numerator = \left(\frac{11}{9}\right)(3) - 0 = \frac{11}{3}\)

\(Denominator = \left(2\left(\frac{1}{9}\right) + 1\right)^2 = \left(\frac{11}{9}\right)^2 = \frac{121}{81}\)

\(m = \frac{11/3}{121/81} = \frac{11}{3} \cdot \frac{81}{121} = \frac{1}{1} \cdot \frac{27}{11} = \frac{27}{11}\)

Using the straight line equation \(y - y_1 = m(x - x_1)\):
\(y - 0 = \frac{27}{11}\left(x - \frac{1}{3}\right)\)
\(11y = 27\left(x - \frac{1}{3}\right)\)
\(11y = 27x - 9\)

Rearranging into the form \(ax + by + c = 0\):
\(27x - 11y - 9 = 0\)

(a)
M1: Applies the quotient rule. Look for a structure of \(\frac{(2x^2+1)(P) -
\ln(3x)(Q)}{(2x^2+1)^2}\) where \(P = \frac{A}{x}\) and \(Q = Bx\).
A1: Correct differentiation of \(\ln(3x)\) to get \(\frac{1}{x}\).
A1: Correct derivative expression: \(\frac{(2x^2 + 1)\left(\frac{1}{x}\right) - 4x\ln(3x)}{(2x^2 + 1)^2}\) (or equivalent).

(b)
B1: Identifies the \(y\)-coordinate at \(x = \frac{1}{3}\) is \(0\).
M1: Substitutes \(x = \frac{1}{3}\) into their derivative to evaluate the gradient of the tangent.
A1: Obtains gradient \(m = \frac{27}{11}\) (or equivalent exact fraction).
M1: Uses the point-slope formula with their \(\left(\frac{1}{3}, 0\right)\) and their gradient \(m\) to form an equation.
A1: Correct equation in the required form: \(27x - 11y - 9 = 0\) (or any non-zero integer multiple, such as \(-27x + 11y + 9 = 0\)).

**(a)** Using polynomial long division or equating coefficients: \[2x^3 - 5x^2 + 7 \equiv (ax^2 + bx + c)(x - 2) + d\] Expanding the right-hand side: \[(ax^2 + bx + c)(x - 2) + d = ax^3 + (b - 2a)x^2 + (c - 2b)x + (d - 2c)\] Comparing coefficients: \(x^3\) term: \(a = 2\); \(x^2\) term: \(b - 2a = -5 \implies b = -1\); \(x\) term: \(c - 2b = 0 \implies c = -2\); Constant term: \(d - 2c = 7 \implies d = 3\). Thus, \[\text{f}(x) = 2x^2 - x - 2 + \frac{3}{x-2}\] **(b)** Using the expression from part (a): \[\int_{3}^{5} \text{f}(x) \, \text{d}x = \int_{3}^{5} \left( 2x^2 - x - 2 + \frac{3}{x-2} \right) \text{d}x\] Integrate term by term: \[= \left[ \frac{2}{3}x^3 - \frac{1}{2}x^2 - 2x + 3\ln|x-2| \right]_{3}^{5}\] Substitute upper limit \(x = 5\): \[\left(\frac{250}{3} - \frac{25}{2} - 10 + 3\ln 3\right) = \frac{365}{6} + 3\ln 3\] Substitute lower limit \(x = 3\): \[\left(\frac{54}{3} - \frac{9}{2} - 6 + 3\ln 1\right) = 18 - \frac{9}{2} - 6 = \frac{15}{2} = \frac{45}{6}\] Subtract the lower limit from the upper limit: \[\left(\frac{365}{6} + 3\ln 3\right) - \frac{45}{6} = \frac{320}{6} + 3\ln 3 = \frac{160}{3} + 3\ln 3\] Thus, \(A = \frac{160}{3}\) and \(B = 3\).

**Part (a)** - **M1**: Attempts to use algebraic long division or equates coefficients to find the values of at least two constants. - **A1**: Correctly identifies any two constants from \(a=2, b=-1, c=-2, d=3\). - **A1**: Fully correct expression \(2x^2 - x - 2 + \frac{3}{x-2}\) (or all four values correctly stated). **Part (b)** - **M1**: Integrates the polynomial and logarithmic parts of their part (a) expression, achieving an expression of the form \(px^3 + qx^2 + rx + s\ln|x-2|\) where \(p, q, r, s \ne 0\). - **A1ft**: Fully correct integration of their expression, e.g., \(\frac{2}{3}x^3 - \frac{1}{2}x^2 - 2x + 3\ln|x-2|\). - **M1**: Substitutes the limits \(5\) and \(3\) into their integrated expression. - **M1**: Subtracts the lower limit term from the upper limit term and uses the log law \(\ln 1 = 0\) to simplify. - **A1**: Correct rational part of the answer, \(A = \frac{160}{3}\). - **A1**: Correct logarithmic coefficient, \(B = 3\), leading to \(\frac{160}{3} + 3\ln 3\).

**(a)** Let \(u = \text{e}^x\). The equation becomes: \[u^2 - 5u + 4 = 0 \implies (u-1)(u-4) = 0\] This gives \(u = 1\) or \(u = 4\). Since \(u = \text{e}^x\): For \(u = 1\), \(\text{e}^x = 1 \implies x = 0\); For \(u = 4\), \(\text{e}^x = 4 \implies x = \ln 4\). So the solutions are \(x = 0\) and \(x = \ln 4\). **(b)** The finite region \(R\) is bounded by the curve \(C\) and the \(x\)-axis. The intersections with the \(x\)-axis are at \(x = 0\) and \(x = \ln 4\). For \(0 < x < \ln 4\), \(y < 0\). The curve lies below the \(x\)-axis. Thus, the area of \(R\) is given by: \[\text{Area} = -\int_{0}^{\ln 4} (\text{e}^{2x} - 5\text{e}^x + 4) \, \text{d}x = \int_{0}^{\ln 4} (5\text{e}^x - \text{e}^{2x} - 4) \, \text{d}x\] Find the antiderivative: \[\int (5\text{e}^x - \text{e}^{2x} - 4) \, \text{d}x = 5\text{e}^x - \frac{1}{2}\text{e}^{2x} - 4x\] Evaluate at the upper limit \(x = \ln 4\): \[5\text{e}^{\ln 4} - \frac{1}{2}\text{e}^{2\ln 4} - 4\ln 4 = 5(4) - \frac{1}{2}(16) - 4(2\ln 2) = 12 - 8\ln 2\] Evaluate at the lower limit \(x = 0\): \[5\text{e}^{0} - \frac{1}{2}\text{e}^{0} - 4(0) = 5 - \frac{1}{2} = \frac{9}{2}\] Subtract the lower limit value from the upper limit value: \[\text{Area} = (12 - 8\ln 2) - \frac{9}{2} = \frac{15}{2} - 8\ln 2\] Thus, \(P = \frac{15}{2}\) and \(Q = 8\).

**Part (a)** - **M1**: Recognises the equation as a quadratic in \(\text{e}^x\) and attempts to factorise or solve \(u^2 - 5u + 4 = 0\). - **A1**: Obtains \(\text{e}^x = 1\) and \(\text{e}^x = 4\). - **A1**: Deduces \(x = 0\) and \(x = \ln 4\) (accept \(x = 2\ln 2\)). **Part (b)** - **M1**: Formulates a definite integral with limits \(0\) and \(\ln 4\) (or their roots from part a) to find the area. - **M1**: Attempts to integrate \(\text{e}^{2x} - 5\text{e}^x + 4\) to obtain an expression of the form \(a\text{e}^{2x} + b\text{e}^x + cx\). - **A1**: Correct integration: \(\frac{1}{2}\text{e}^{2x} - 5\text{e}^x + 4x\) (or its negative, \(5\text{e}^x - \frac{1}{2}\text{e}^{2x} - 4x\)). - **M1**: Substitutes their limits \(0\) and \(\ln 4\) into their integrated expression. - **M1**: Correctly evaluates exponential terms, e.g., \(\text{e}^{\ln 4} = 4\) and \(\text{e}^{2\ln 4} = 16\), and uses the log identity \(\ln 4 = 2\ln 2\). - **A1**: Obtains the exact area of \(\frac{15}{2} - 8\ln 2\) (or states \(P = \frac{15}{2}\) and \(Q = 8\)).

(a) The graph of \( y = 4 - |3x - 6| \) is an inverted V-shape.

The local maximum (vertex) occurs when the expression inside the modulus is zero:
\( 3x - 6 = 0 \Rightarrow x = 2 \)
At \( x = 2 \), \( y = 4 - 0 = 4 \), so the local maximum point is at \( (2, 4) \).

To find the y-intercept, set \( x = 0 \):
\( y = 4 - |-6| = 4 - 6 = -2 \), so the y-intercept is at \( (0, -2) \).

To find the x-intercepts, set \( y = 0 \):
\( 4 - |3x - 6| = 0 \Rightarrow |3x - 6| = 4 \)
This gives two cases:
\( 3x - 6 = 4 \Rightarrow 3x = 10 \Rightarrow x = \frac{10}{3} \)
\( 3x - 6 = -4 \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3} \)
So the x-intercepts are at \( \left(\frac{2}{3}, 0\right) \) and \( \left(\frac{10}{3}, 0\right) \).

(b) To solve \( 4 - |3x - 6| > 2x - 1 \), we first find the critical values by solving the equation:
\( 4 - |3x - 6| = 2x - 1 \Rightarrow |3x - 6| = 5 - 2x \)

Case 1:
\( 3x - 6 = 5 - 2x \Rightarrow 5x = 11 \Rightarrow x = \frac{11}{5} \)

Case 2:
\( -(3x - 6) = 5 - 2x \Rightarrow -3x + 6 = 5 - 2x \Rightarrow x = 1 \)

By analyzing the graph, the line \( y = 2x - 1 \) intersects the graph of \( y = \mathrm{g}(x) \) at \( x = 1 \) and \( x = \frac{11}{5} \). The graph of \( y = \mathrm{g}(x) \) is strictly above the line between these two critical values.

Therefore, the solution to the inequality is:
\( 1 < x < \frac{11}{5} \)

(a)
B1: Inverted V-shape located in the correct quadrants (vertex in the first quadrant, y-intercept on the negative y-axis, and two positive x-intercepts).
B1: Vertex correctly labeled as \( (2, 4) \).
B1: y-intercept correctly labeled as \( (0, -2) \) or \( -2 \) marked on the y-axis.
B1: x-intercepts correctly labeled as \( \left(\frac{2}{3}, 0\right) \) and \( \left(\frac{10}{3}, 0\right) \) or equivalent on the x-axis.

(b)
M1: Attempts to find critical values by forming at least one correct equation without modulus, e.g., \( 3x - 6 = 5 - 2x \) or \( -(3x - 6) = 5 - 2x \).
A1: Both correct critical values: \( x = 1 \) and \( x = \frac{11}{5} \) (or \( 2.2 \)).
M1: Selects the inside region between their two critical values.
A1: Fully correct inequality: \( 1 < x < \frac{11}{5} \) (or equivalent notation).

(a) First evaluate \( \mathrm{g}(1) \):
\( \mathrm{g}(1) = |1 - 2| = |-1| = 1 \)
Now evaluate \( f(\mathrm{g}(1)) = f(1) \):
\( f(1) = \mathrm{e}^{2(1)} - 6 = \mathrm{e}^2 - 6 \)

(b) Since the domain of \( f \) is \( x \in \mathbb{R} \), we have \( \mathrm{e}^{2x} > 0 \).
Thus, \( \mathrm{e}^{2x} - 6 > -6 \).
Therefore, the range of \( f \) is \( f(x) > -6 \) (or \( y > -6 \)).

(c) The equation \( \mathrm{gf}(x) = 5 \) becomes:
\( |f(x) - 2| = 5 \Rightarrow |(\mathrm{e}^{2x} - 6) - 2| = 5 \Rightarrow |\mathrm{e}^{2x} - 8| = 5 \)

This leads to two possible cases:

Case 1:
\( \mathrm{e}^{2x} - 8 = 5 \Rightarrow \mathrm{e}^{2x} = 13 \)
Taking the natural logarithm of both sides:
\( 2x = \ln(13) \Rightarrow x = \frac{1}{2}\ln(13) \)

Case 2:
\( \mathrm{e}^{2x} - 8 = -5 \Rightarrow \mathrm{e}^{2x} = 3 \)
Taking the natural logarithm of both sides:
\( 2x = \ln(3) \Rightarrow x = \frac{1}{2}\ln(3) \)

Both solutions are valid. The exact values are \( x = \frac{1}{2}\ln(13) \) and \( x = \frac{1}{2}\ln(3) \) (or equivalent forms such as \( \ln(\sqrt{13}) \) and \( \ln(\sqrt{3}) \)).

(a)
M1: Evaluates \( \mathrm{g}(1) = 1 \) and attempts to substitute this value into \( f(x) \).
A1: Correct exact value \( \mathrm{e}^2 - 6 \).

(b)
B1: Correct range written in acceptable notation: \( f(x) > -6 \) or \( y > -6 \) or \( (-6, \infty) \). Must be strictly greater than.

(c)
M1: Correctly forms the composite equation \( |(\mathrm{e}^{2x} - 6) - 2| = 5 \).
M1: Correctly splits the modulus equation into two linear-style exponential equations: \( \mathrm{e}^{2x} - 8 = 5 \) and \( \mathrm{e}^{2x} - 8 = -5 \).
A1: Obtains \( \mathrm{e}^{2x} = 13 \) and \( \mathrm{e}^{2x} = 3 \).
dM1: Uses natural logarithms correctly on an equation of the form \( \mathrm{e}^{2x} = k \) (where \( k > 0 \)) to find \( x \). Dependent on the first M mark.
A1: Both correct exact solutions: \( x = \frac{1}{2}\ln(13) \) and \( x = \frac{1}{2}\ln(3) \) or equivalent single logarithms.

(a) Starting with the Left Hand Side (LHS):
\$$\text{LHS} = \frac{1 - \cos 2\theta + \sin 2\theta}{1 + \cos 2\theta + \sin 2\theta}\$$

Using the double-angle identities:
\(1 - \cos 2\theta = 2\sin^2\theta\)
\(1 + \cos 2\theta = 2\cos^2\theta\)
\(\sin 2\theta = 2\sin\theta\cos\theta\)

Substitute these into the expression:
\$$\text{LHS} = \frac{2\sin^2\theta + 2\sin\theta\cos\theta}{2\cos^2\theta + 2\sin\theta\cos\theta}\$$

Factorise the numerator and the denominator:
\$$\text{LHS} = \frac{2\sin\theta(\sin\theta + \cos\theta)}{2\cos\theta(\cos\theta + \sin\theta)}\$$

Cancelling the common factor of \(2(\sin\theta + \cos\theta)\) gives:
\$$\text{LHS} = \frac{\sin\theta}{\cos\theta} = \tan\theta\$$
This completes the proof.

(b) By substituting \(\theta = 2x\) into the identity from part (a), the equation simplifies to:
\$$\tan 2x = 3\cot 2x - 2\$$

Since \(\cot 2x = \frac{1}{\tan 2x}\):
\$$\tan 2x = \frac{3}{\tan 2x} - 2\$$

Multiplying through by \(\tan 2x\) and rearranging:
\$$\tan^2 2x + 2\tan 2x - 3 = 0\$$

Factorising this quadratic equation:
\$$(\tan 2x + 3)(\tan 2x - 1) = 0\$$

This gives two sets of solutions for the interval \(0 \le 2x < 2
\pi\):

1) \(\tan 2x = 1\)
\$$2x = \frac{\pi}{4}, \frac{5\pi}{4}\$$
\$$x = \frac{\pi}{8}, \frac{5\pi}{8}\$$
In decimals, this is \(x \approx 0.393, 1.96\).

2) \(\tan 2x = -3\)
\$$2x = \pi - \arctan(3) \approx 1.89254\$$
\$$2x = 2\pi - \arctan(3) \approx 5.03414\$$
\$$x \approx 0.94627, 2.51707\$$

Rounding to 3 significant figures, the solutions are:
\(x = \frac{\pi}{8}\) (or 0.393), 0.946, \(\frac{5\pi}{8}\) (or 1.96), 2.52.

**Part (a)**
* **M1**: Attempts to use double-angle formulae to rewrite either \(1 - \cos 2\theta\) as \(2\sin^2\theta\) or \(1 + \cos 2\theta\) as \(2\cos^2\theta\).
* **A1**: Obtains correct expressions for both numerator and denominator in factorised form: \(2\sin\theta(\sin\theta + \cos\theta)\) and \(2\cos\theta(\cos\theta + \sin\theta)\).
* **M1**: Shows factorisation and cancellation of the common brackets.
* **A1\***: Fully correct proof leading to \(\tan\theta\) with no errors seen.

**Part (b)**
* **M1**: Correctly identifies that the LHS of the equation can be replaced by \(\tan 2x\).
* **M1**: Employs \(\cot 2x = \frac{1}{\tan 2x}\) to form a quadratic equation in \(\tan 2x\): \(\tan^2 2x + 2\tan 2x - 3 = 0\).
* **A1**: Solves the quadratic to get \(\tan 2x = 1\) and \(\tan 2x = -3\).
* **M1**: Solves either equation to find at least two correct values of \(x\) within the interval.
* **A1**: Correctly finds all four solutions: \(x = \frac{\pi}{8}\) (or 0.393), 0.946, \(\frac{5\pi}{8}\) (or 1.96), 2.52 (awrt 3 s.f.). Deduct 1 mark for any extra solutions within the interval.

First, we state the volume of revolution formula about the \(x\)-axis: \(V = \pi \int y^2 \text{d}x\). Since the curve is given parametrically, we use \(V = \pi \int y^2 \frac{\text{d}x}{\text{d}t} \text{d}t\). Differentiating \(x\) with respect to \(t\) gives: \(\frac{\text{d}x}{\text{d}t} = \frac{1}{t+1}\). Squaring \(y\) gives: \(y^2 = \frac{t^2}{t+1}\). Thus, the integrand is: \(y^2 \frac{\text{d}x}{\text{d}t} = \frac{t^2}{(t+1)^2}\). Next, we find the limits for \(t\). When \(x = 0\), we have \(\ln(t+1) = 0 \implies t+1 = 1 \implies t = 0\). When \(x = \ln 3\), we have \(\ln(t+1) = \ln 3 \implies t+1 = 3 \implies t = 2\). Therefore, the volume is: \(V = \pi \int_{0}^{2} \frac{t^2}{(t+1)^2} \text{d}t\). To evaluate this integral, we use the substitution \(u = t + 1\). Then \(\text{d}u = \text{d}t\). The limits of integration become: when \(t = 0\), \(u = 1\); when \(t = 2\), \(u = 3\). Substituting these into the integral: \(V = \pi \int_{1}^{3} \frac{(u-1)^2}{u^2} \text{d}u = \pi \int_{1}^{3} \frac{u^2 - 2u + 1}{u^2} \text{d}u = \pi \int_{1}^{3} \left(1 - \frac{2}{u} + u^{-2}\right) \text{d}u\). Integrating each term with respect to \(u\) yields: \(V = \pi \left[ u - 2\ln|u| - \frac{1}{u} \right]_{1}^{3}\). Evaluating this at the limits: \(V = \pi \left[ \left(3 - 2\ln 3 - \frac{1}{3}\right) - \left(1 - 2\ln 1 - 1\right) \right] = \pi \left[ \left(\frac{8}{3} - 2\ln 3\right) - 0 \right] = \pi \left(\frac{8}{3} - \ln 9\right)\). This is in the required form \(\pi(a - \ln b)\) where \(a = \frac{8}{3}\) and \(b = 9\).

M1: Attempts to use the parametric volume of revolution formula \(V = \pi \int y^2 \frac{\text{d}x}{\text{d}t} \text{d}t\) with their expressions. A1: Correctly expresses the integral as \(\pi \int \frac{t^2}{(t+1)^2} \text{d}t\) (limits and \(\pi\) not required for this mark). B1: Correctly finds the limits for \(t\) to be \(t = 0\) and \(t = 2\). M1: Employs a valid algebraic manipulation or substitution method (such as \(u = t + 1\)) to write the integrand in a form that can be integrated. A1: Correctly integrates to obtain \(\left[ u - 2\ln|u| - \frac{1}{u} \right]\) or equivalent in terms of \(t\). A1.5: Substitutes the correct limits to obtain the final exact volume of \(\pi\left(\frac{8}{3} - \ln 9\right)\) or specifies \(a = \frac{8}{3}\) and \(b = 9\).

First, we find \(\frac{\text{d}x}{\text{d}\theta}\) and \(\frac{\text{d}y}{\text{d}\theta}\) using the quotient rule. For \(x = \frac{3\theta}{1 + \theta}\): \(\frac{\text{d}x}{\text{d}\theta} = \frac{3(1+\theta) - 3\theta(1)}{(1+\theta)^2} = \frac{3}{(1+\theta)^2}\). For \(y = \frac{\theta^2}{1 + \theta}\): \(\frac{\text{d}y}{\text{d}\theta} = \frac{2\theta(1+\theta) - \theta^2(1)}{(1+\theta)^2} = \frac{2\theta + \theta^2}{(1+\theta)^2}\). Next, we find the gradient of the tangent \(\frac{\text{d}y}{\text{d}x}\) using the chain rule: \(\frac{\text{d}y}{\text{d}x} = \frac{\text{d}y/\text{d}\theta}{\text{d}x/\text{d}\theta} = \frac{2\theta + \theta^2}{3}\). At the point where \(\theta = 2\), the gradient of the tangent is: \(m_T = \frac{2(2) + 2^2}{3} = \frac{8}{3}\). Therefore, the gradient of the normal, \(m_N\), is: \(m_N = -\frac{1}{m_T} = -\frac{3}{8}\). We also need the coordinates of the point where \(\theta = 2\): \(x = \frac{3(2)}{1+2} = 2\) and \(y = \frac{2^2}{1+2} = \frac{4}{3}\). Using the equation of a straight line, the equation of the normal is: \(y - \frac{4}{3} = -\frac{3}{8}(x - 2)\). Multiplying through by 24 to clear the fractions: \(24y - 32 = -9(x - 2) \implies 24y - 32 = -9x + 18 \implies 9x + 24y - 50 = 0\). Thus, the equation of the normal in the required form is \(9x + 24y - 50 = 0\).

M1: Applies the quotient rule (or product rule) correctly to find \(\frac{\text{d}x}{\text{d}\theta}\) or \(\frac{\text{d}y}{\text{d}\theta}\). A1: Correct expressions for both \(\frac{\text{d}x}{\text{d}\theta} = \frac{3}{(1+\theta)^2}\) and \(\frac{\text{d}y}{\text{d}\theta} = \frac{\theta^2 + 2\theta}{(1+\theta)^2}\) (can be unsimplified). M1: Uses \(\frac{\text{d}y}{\text{d}x} = \frac{\text{d}y/\text{d}\theta}{\text{d}x/\text{d}\theta}\) and substitutes \(\theta = 2\) to find the gradient of the tangent. A1: Obtains the correct tangent gradient of \(\frac{8}{3}\) and deduces the normal gradient is \(-\frac{3}{8}\). B1: Correctly calculates the coordinates of the point when \(\theta = 2\) to be \((2, \frac{4}{3})\). M1: Employs \(y - y_1 = m_N(x - x_1)\) with their coordinates and normal gradient. A0.5: Obtains the correct equation of the normal in the form \(9x + 24y - 50 = 0\) (or any integer multiple thereof).

Part (a):
Rewrite the function by factoring out 4:
\(f(x) = (4 - x)^{1/2} = 4^{1/2} \left(1 - \frac{x}{4}\right)^{1/2} = 2 \left(1 - \frac{x}{4}\right)^{1/2}\)

Apply the binomial expansion formula \((1 + y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots\) with \(n = \frac{1}{2}\) and \(y = -\frac{x}{4}\):
\(\left(1 - \frac{x}{4}\right)^{1/2} = 1 + \left(\frac{1}{2}\right)\left(-\frac{x}{4}\right) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2}\left(-\frac{x}{4}\right)^2 + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6}\left(-\frac{x}{4}\right)^3 + \dots\)
\(= 1 - \frac{x}{8} - \frac{x^2}{128} - \frac{x^3}{1024} + \dots\)

Multiply the expansion by 2:
\(f(x) \approx 2\left(1 - \frac{x}{8} - \frac{x^2}{128} - \frac{x^3}{1024}\right) = 2 - \frac{x}{4} - \frac{x^2}{64} - \frac{x^3}{512}\)

Part (b):
Substitute \(x = 0.04\) into the original expression for \(f(x)\):
\(f(0.04) = \sqrt{4 - 0.04} = \sqrt{3.96} = \sqrt{\frac{396}{100}} = \frac{\sqrt{36 \times 11}}{10} = \frac{6\sqrt{11}}{10} = \frac{3\sqrt{11}}{5}\)

Therefore, \(\sqrt{11} = \frac{5}{3} f(0.04)\).

Using the binomial expansion to find \(f(0.04)\):
\(f(0.04) \approx 2 - \frac{0.04}{4} - \frac{(0.04)^2}{64} - \frac{(0.04)^3}{512} = 2 - 0.01 - 0.000025 - 0.000000125 = 1.989974875\)

Now calculate the approximation for \(\sqrt{11}\):
\(\sqrt{11} \approx \frac{5}{3}(1.989974875) = 3.3166247916...\)
To 6 decimal places, this is \(3.316625\).

Part (a):
M1: For taking out a factor of \(4^{1/2}\) (or 2) to write the expression in the form \(2(1 - x/4)^{1/2}\).
M1: For an attempt to expand \((1 - x/4)^{1/2}\) using the binomial expansion formula with at least two terms correct.
A1: For a correct unsimplified expansion of \((1 - x/4)^{1/2}\).
A1: For simplifying up to the term in \(x^2\) to obtain \(2 - \frac{x}{4} - \frac{x^2}{64}\).
A1: For obtaining the correct simplified \(x^3\) term: \(-\frac{x^3}{512}\).

Part (b):
M1: For substituting \(x = 0.04\) into their expansion and showing/using the relationship \(\sqrt{11} = \frac{5}{3} f(0.04)\).
A1: For obtaining \(3.316625\) (must be correct to 6 decimal places; direct calculator verification of \(\sqrt{11}\) earns 0 marks, work must be shown).

Let \(u = \ln(x)\). Then \(\frac{du}{dx} = \frac{1}{x} \implies dx = x \, du\).

Change the limits of integration:
When \(x = 1\), \(u = \ln(1) = 0\).
When \(x = e\), \(u = \ln(e) = 1\).

Substitute these into the integral:
\( \int_{1}^{e} \frac{\ln(x)}{x(1 + \ln(x))^2} \, dx = \int_{0}^{1} \frac{u}{x(1 + u)^2} (x \, du) = \int_{0}^{1} \frac{u}{(1 + u)^2} \, du \)

To integrate \(\frac{u}{(1 + u)^2}\), rewrite the numerator as \(u = (1 + u) - 1\):
\( \int_{0}^{1} \frac{(1 + u) - 1}{(1 + u)^2} \, du = \int_{0}^{1} \left( \frac{1}{1 + u} - \frac{1}{(1 + u)^2} \right) \, du \)

Now integrate term by term:
\( \left[ \ln|1 + u| + \frac{1}{1 + u} \right]_{0}^{1} \)

Evaluate at the limits:
At \(u = 1\): \(\ln(2) + \frac{1}{2}\)
At \(u = 0\): \(\ln(1) + 1 = 1\)

Subtracting the lower limit from the upper limit:
\( \left( \ln(2) + \frac{1}{2} \right) - (0 + 1) = \ln(2) - \frac{1}{2} \)
Thus, \(a = 2\) and \(b = \frac{1}{2}\).

M1: For attempting the substitution \(u = \ln(x)\) to find \(du = \frac{1}{x} dx\) or equivalent.
A1: For obtaining \(\int \frac{u}{(1+u)^2} \, du\) (limits not required for this mark).
B1: For changing limits of integration correctly from \(x = 1 \to u = 0\) and \(x = e \to u = 1\).
M1: For attempting to integrate \(\frac{u}{(1+u)^2}\) by rewriting as \(\frac{1}{1+u} - (1+u)^{-2}\) or using a second substitution like \(w = 1+u\).
A1: For correct integrated expression \(\ln|1+u| + \frac{1}{1+u}\).
M1: For substituting limits \(0\) and \(1\) and subtracting.
A1: For the exact final answer \(\ln(2) - \frac{1}{2}\) or equivalent exact form (e.g. \(a=2, b=0.5\)).

(a) Separating the variables:
\(\int \frac{1}{N(400 - N)} \mathrm{d}N = \int \frac{1}{800} \mathrm{d}t\)

Using partial fractions:
\(\frac{1}{N(400 - N)} = \frac{A}{N} + \frac{B}{400 - N}\)
\(1 = A(400 - N) + BN\)
Setting \(N = 0\) gives \(A = \frac{1}{400}\).
Setting \(N = 400\) gives \(B = \frac{1}{400}\).

Substituting this back into the integral:
\(\frac{1}{400} \int \left( \frac{1}{N} + \frac{1}{400 - N} \right) \mathrm{d}N = \int \frac{1}{800} \mathrm{d}t\)
Multiply by 400:
\(\int \left( \frac{1}{N} + \frac{1}{400 - N} \right) \mathrm{d}N = \int 0.5 \mathrm{d}t\)
\(\ln|N| - \ln|400 - N| = 0.5t + C\)
\(\ln\left( \frac{N}{400 - N} \right) = 0.5t + C\)

Using the initial condition \(t = 0, N = 100\):
\(\ln\left( \frac{100}{300} \right) = C \implies C = \ln\left(\frac{1}{3}\right) = -\ln 3\)

So,
\(\ln\left( \frac{N}{400 - N} \right) = 0.5t - \ln 3\)

Exponentiating both sides:
\(\frac{N}{400 - N} = \mathrm{e}^{0.5t - \ln 3} = \frac{1}{3}\mathrm{e}^{0.5t}\)
\(3N = (400 - N)\mathrm{e}^{0.5t}\)
Multiply by \(\mathrm{e}^{-0.5t}\):
\(3N\mathrm{e}^{-0.5t} = 400 - N\)
\(N(3\mathrm{e}^{-0.5t} + 1) = 400\)
\(N = \frac{400}{1 + 3\mathrm{e}^{-0.5t}}\)

Thus, \(\lambda = 1\) and \(\mu = 3\).

(b) When \(N = 300\):
\(300 = \frac{400}{1 + 3\mathrm{e}^{-0.5t}}\)
\(1 + 3\mathrm{e}^{-0.5t} = \frac{4}{3}\)
\(3\mathrm{e}^{-0.5t} = \frac{1}{3}\)
\(\mathrm{e}^{-0.5t} = \frac{1}{9}\)
\(-0.5t = \ln\left(\frac{1}{9}\right) = -\ln 9\)
\(0.5t = \ln 9\)
\(t = 2\ln 9 = \ln 81 \approx 4.39\text{ years}\)

(a)
- M1: Attempts to separate variables and write \(\int \frac{1}{N(400 - N)} \mathrm{d}N = \int \frac{1}{800} \mathrm{d}t\).
- M1: Uses partial fractions to split the integrand into \(\frac{A}{N} + \frac{B}{400-N}\) and finds \(A = B = \frac{1}{400}\).
- A1: Integrates correctly to obtain \(\ln|N| - \ln|400 - N| = 0.5t + C'\) (or equivalent with \(\frac{1}{400}\) and \(\frac{1}{800}\)).
- M1: Uses the boundary condition \(t = 0, N = 100\) to evaluate their constant of integration.
- A1: Obtains the correct constant \(C = -\ln 3\) (or equivalent for other forms).
- M1: Removes logarithms correctly by exponentiating and attempts to make \(N\) the subject.
- A1: Correctly completes the proof to show \(N = \frac{400}{1 + 3\mathrm{e}^{-0.5t}}\) with \(\lambda = 1\) and \(\mu = 3\).

(b)
- M1: Sets \(N = 300\) and attempts to solve for \(\mathrm{e}^{-0.5t}\) or \(t\).
- A1: Obtains \(\mathrm{e}^{-0.5t} = \frac{1}{9}\) or equivalent.
- A1: Evaluates \(t = 2\ln 9 = \ln 81\) to give \(t \approx 4.39\) (accept 4.39 only).

(a) Given \(V = \frac{1}{9}\pi h^3\), differentiating with respect to \(h\) gives:
\(\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{1}{3}\pi h^2\)

The rate of change of volume \(\frac{\mathrm{d}V}{\mathrm{d}t}\) is:
\(\frac{\mathrm{d}V}{\mathrm{d}t} = \text{Rate in} - \text{Rate out} = 3\pi - \frac{1}{3}\pi h^2 = \frac{\pi(9 - h^2)}{3}\)

Using the chain rule:
\(\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}h}{\mathrm{d}V} \times \frac{\mathrm{d}V}{\mathrm{d}t} = \frac{3}{\pi h^2} \times \frac{\pi(9 - h^2)}{3} = \frac{9 - h^2}{h^2}\)

(b) Separating the variables:
\(\int \frac{h^2}{9 - h^2} \mathrm{d}h = \int 1 \mathrm{d}t\)

Rewrite the integrand:
\(\frac{h^2}{9 - h^2} = \frac{h^2 - 9 + 9}{9 - h^2} = -1 + \frac{9}{9 - h^2}\)

Using partial fractions on \(\frac{9}{9 - h^2}\):
\(\frac{9}{(3 - h)(3 + h)} = \frac{D}{3 - h} + \frac{E}{3 + h}\)
\(9 = D(3 + h) + E(3 - h)\)
Setting \(h = 3\) gives \(D = 1.5\).
Setting \(h = -3\) gives \(E = 1.5\).

Now, integrating both sides:
\(\int \left( -1 + \frac{1.5}{3 - h} + \frac{1.5}{3 + h} \right) \mathrm{d}h = t + C\)
\(-h - 1.5\ln|3 - h| + 1.5\ln|3 + h| = t + C\)
\(-h + 1.5\ln\left( \frac{3 + h}{3 - h} \right) = t + C\)

At \(t = 0, h = 1\):
\(-1 + 1.5\ln\left(\frac{4}{2}\right) = C \implies C = -1 + 1.5\ln 2\)

So,
\(t = 1 - h + 1.5\ln\left( \frac{3 + h}{3 - h} \right) - 1.5\ln 2\)

At \(h = 2\):
\(t = 1 - 2 + 1.5\ln\left(\frac{5}{1}\right) - 1.5\ln 2\)
\(t = -1 + 1.5\ln 5 - 1.5\ln 2\)
\(t = -1 + 1.5\ln(2.5)\)

(a)
- M1: Differentiates \(V = \frac{1}{9}\pi h^3\) to obtain \(\frac{\mathrm{d}V}{\mathrm{d}h} = k h^2\).
- B1: Writes down \(\frac{\mathrm{d}V}{\mathrm{d}t} = 3\pi - \frac{1}{3}\pi h^2\).
- M1: Employs the chain rule \(\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}h}{\mathrm{d}V} \times \frac{\mathrm{d}V}{\mathrm{d}t}\).
- A1*: Fully correct proof showing all intermediate steps leading to \(\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{9 - h^2}{h^2}\).

(b)
- M1: Separates the variables to obtain \(\int \frac{h^2}{9 - h^2} \mathrm{d}h = \int 1 \mathrm{d}t\).
- M1: Rewrites \(\frac{h^2}{9-h^2}\) as \(-1 + \frac{9}{9-h^2}\).
- M1: Expresses \(\frac{9}{9-h^2}\) in partial fractions and attempts to integrate, achieving a form containing \(A h + B \ln(3-h) + C \ln(3+h)\).
- A1: Obtains \(-h + 1.5\ln\left(\frac{3+h}{3-h}\right) = t + C\) (or equivalent).
- M1: Applies the boundary condition \(t = 0, h = 1\) to find the value of the constant \(C\).
- A1: Obtains the exact value \(t = -1 + 1.5\ln(2.5)\) (or equivalent such as \(-1 + \frac{3}{2}\ln\left(\frac{5}{2}\right)\)).

Assume that there exist integers \(a\) and \(b\) such that
\[a^2 - 4b^2 = 2\]

Rearranging the equation to make \(a^2\) the subject:
\[a^2 = 4b^2 + 2\]
\[a^2 = 2(2b^2 + 1)\]

Since \(2b^2 + 1\) is an integer, \(a^2\) is an even number.
If \(a^2\) is even, then \(a\) must also be even.

Let \(a = 2k\) for some integer \(k\).

Substitute \(a = 2k\) into the original equation:
\[(2k)^2 - 4b^2 = 2\]
\[4k^2 - 4b^2 = 2\]

Dividing both sides of the equation by \(2\) gives:
\[2(k^2 - b^2) = 1\]

Since \(k\) and \(b\) are integers, \(k^2 - b^2\) is also an integer. Therefore, the left-hand side \(2(k^2 - b^2)\) must be an even integer.

However, the right-hand side is \(1\), which is an odd integer.

An even integer cannot equal an odd integer. This is a contradiction.

Hence, the assumption is false, and there are no integers \(a\) and \(b\) such that \(a^2 - 4b^2 = 2\).

**B1**: Assumes the negation of the statement: states or implies that there exist integers \(a\) and \(b\) such that \(a^2 - 4b^2 = 2\).

**M1**: Expresses \(a^2\) as a multiple of 2 (e.g. \(a^2 = 2(2b^2 + 1)\)), deduces that \(a\) is even, and writes \(a = 2k\) for some integer \(k\).

**M1**: Substitutes \(a = 2k\) into the equation and simplifies to obtain \(2(k^2 - b^2) = 1\) or \(k^2 - b^2 = \frac{1}{2}\).

**A1**: Explains why this is a contradiction (e.g. LHS is even, RHS is odd; or \(k^2 - b^2\) must be an integer but \(\frac{1}{2}\) is not) and concludes that the original statement is true.

(a) To show that \( l_1 \) and \( l_2 \) intersect, equate their parametric components:

From the \( x \)-components: \( 3 + \lambda = 2 + 2\mu \Rightarrow \lambda - 2\mu = -1 \quad (1) \)

From the \( y \)-components: \( 3 - 2\lambda = \mu \Rightarrow 2\lambda + \mu = 3 \quad (2) \)

From (2), we get \( \mu = 3 - 2\lambda \). Substituting this into (1) gives:
\( \lambda - 2(3 - 2\lambda) = -1 \Rightarrow \lambda - 6 + 4\lambda = -1 \Rightarrow 5\lambda = 5 \Rightarrow \lambda = 1 \).

Substituting \( \lambda = 1 \) back into (2) gives:
\( \mu = 3 - 2(1) = 1 \).

Now, we must verify these parameters in the \( z \)-component:
LHS (from \( l_1 \)): \( 3 + 2\lambda = 3 + 2(1) = 5 \)
RHS (from \( l_2 \)): \( 2 + 3\mu = 2 + 3(1) = 5 \)

Since LHS = RHS = 5, the lines intersect. The position vector of \( P \) is \( \mathbf{r}_P = \begin{pmatrix} 3 + 1 \\ 3 - 2 \\ 3 + 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 5 \end{pmatrix} \), so the coordinates of \( P \) are \( (4, 1, 5) \).

(b) The direction vectors of \( l_1 \) and \( l_2 \) are \( \mathbf{d}_1 = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} \) and \( \mathbf{d}_2 = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \) respectively.

Let \( \theta \) be the angle between the two lines:
\( \cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{ |\mathbf{d}_1| |\mathbf{d}_2| } \)

Calculate the dot product:
\( \mathbf{d}_1 \cdot \mathbf{d}_2 = 1(2) + (-2)(1) + 2(3) = 2 - 2 + 6 = 6 \)

Calculate the magnitudes:
\( |\mathbf{d}_1| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3 \)
\( |\mathbf{d}_2| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{14} \)

Thus, \( \cos \theta = \frac{6}{3\sqrt{14}} = \frac{2}{\sqrt{14}} \)
\( \theta = \arccos\left(\frac{2}{\sqrt{14}}\right) \approx 57.688...^\circ \approx 57.7^\circ \).

(c) For point \( A \) on \( l_1 \) where \( \lambda = 8 \):
\( \mathbf{a} = \begin{pmatrix} 3 + 8 \\ 3 - 16 \\ 3 + 16 \end{pmatrix} = \begin{pmatrix} 11 \\ -13 \\ 19 \end{pmatrix} \)

Point \( B \) lies on \( l_2 \), so its position vector for some parameter \( \mu \) is:
\( \mathbf{b} = \begin{pmatrix} 2 + 2\mu \\ \mu \\ 2 + 3\mu \end{pmatrix} \)

The vector \( \vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 2 + 2\mu - 11 \\ \mu - (-13) \\ 2 + 3\mu - 19 \end{pmatrix} = \begin{pmatrix} 2\mu - 9 \\ \mu + 13 \\ 3\mu - 17 \end{pmatrix} \)

Since \( AB \) is perpendicular to \( l_2 \), the dot product of \( \vec{AB} \) with the direction vector of \( l_2 \) must be 0:
\( \vec{AB} \cdot \mathbf{d}_2 = 0 \)
\( \begin{pmatrix} 2\mu - 9 \\ \mu + 13 \\ 3\mu - 17 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} = 0 \)
\( 2(2\mu - 9) + 1(\mu + 13) + 3(3\mu - 17) = 0 \)
\( 4\mu - 18 + \mu + 13 + 9\mu - 51 = 0 \)
\( 14\mu - 56 = 0 \Rightarrow \mu = 4 \)

Substituting \( \mu = 4 \) into the position vector of \( B \):
\( \mathbf{b} = \begin{pmatrix} 2 + 2(4) \\ 4 \\ 2 + 3(4) \end{pmatrix} = \begin{pmatrix} 10 \\ 4 \\ 14 \end{pmatrix} \)

So the coordinates of \( B \) are \( (10, 4, 14) \).

Part (a):
- M1: Sets up simultaneous equations in \( \lambda \) and \( \mu \) using any two components.
- M1: Solves the equations to find values for \( \lambda \) and \( \mu \).
- A1: Obtains \( \lambda = 1 \) and \( \mu = 1 \).
- M1: Substitutes their values of \( \lambda \) and \( \mu \) into the remaining component to check for consistency.
- A1: Correctly identifies the intersection point \( P(4, 1, 5) \).

Part (b):
- M1: Attempts to use the scalar product formula \( \cos \theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|} \) with the correct direction vectors.
- A1: Obtains \( \cos \theta = \frac{6}{3\sqrt{14}} \) or any equivalent exact form.
- A1: Obtains \( 57.7^\circ \) (accept \( 57.7 \)).

Part (c):
- B1: Correctly finds the coordinates of \( A \) as \( (11, -13, 19) \).
- M1: Expresses \( \vec{AB} \) in terms of \( \mu \).
- M1: Sets up the equation \( \vec{AB} \cdot \mathbf{d}_2 = 0 \) and attempts to solve for \( \mu \).
- A1: Correctly finds \( \mu = 4 \) and states the coordinates of \( B \) as \( (10, 4, 14) \).

(a) First, find the coordinates of the point where \( t = \frac{\pi}{6} \):
\( x = 3 \cos\left(\frac{\pi}{6}\right) = \frac{3\sqrt{3}}{2} \)
\( y = 4 \sin\left(2 \cdot \frac{\pi}{6}\right) = 4 \sin\left(\frac{\pi}{3}\right) = 4 \left(\frac{\sqrt{3}}{2}\right) = 2\sqrt{3} \)

Now, differentiate \( x \) and \( y \) with respect to \( t \):
\( \frac{dx}{dt} = -3 \sin t \)
\( \frac{dy}{dt} = 8 \cos 2t \)

At \( t = \frac{\pi}{6} \):
\( \frac{dx}{dt} = -3 \sin\left(\frac{\pi}{6}\right) = -\frac{3}{2} \)
\( \frac{dy}{dt} = 8 \cos\left(\frac{\pi}{3}\right) = 8 \left(\frac{1}{2}\right) = 4 \)

Using the chain rule:
\( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4}{-\frac{3}{2}} = -\frac{8}{3} \)

The equation of the tangent is:
\( y - 2\sqrt{3} = -\frac{8}{3}\left(x - \frac{3\sqrt{3}}{2}\right) \)

Multiply by 3:
\( 3y - 6\sqrt{3} = -8x + 12\sqrt{3} \)
\( 8x + 3y = 18\sqrt{3} \)

Thus, \( a = 8, b = 3, c = 18 \).

(b) Express \( y \) in terms of single-angle trigonometric functions:
\( y = 4 \sin 2t = 8 \sin t \cos t \)

From \( x = 3 \cos t \), we have \( \cos t = \frac{x}{3} \).

Since \( \sin^2 t + \cos^2 t = 1 \), we have \( \sin t = \sqrt{1 - \cos^2 t} = \sqrt{1 - \frac{x^2}{9}} \) (since \( 0 \le t \le \pi \), \( \sin t \ge 0 \)).

Substitute these into the expression for \( y \):
\( y = 8 \left(\sqrt{1 - \frac{x^2}{9}}\right) \left(\frac{x}{3}\right) = \frac{8x}{3}\sqrt{\frac{9 - x^2}{9}} = \frac{8x}{9}\sqrt{9 - x^2} \)

Squaring both sides:
\( y^2 = \frac{64x^2}{81}(9 - x^2) \)

(c) The tangent is parallel to the \( x \)-axis when \( \frac{dy}{dx} = 0 \).
This requires \( \frac{dy}{dt} = 0 \) and \( \frac{dx}{dt} \ne 0 \).

\( 8 \cos 2t = 0 \Rightarrow \cos 2t = 0 \)

Given \( 0 \le t \le \pi \), we have \( 0 \le 2t \le 2\pi \).
\( 2t = \frac{\pi}{2}, \frac{3\pi}{2} \Rightarrow t = \frac{\pi}{4}, \frac{3\pi}{4} \)

At these values of \( t \), \( \frac{dx}{dt} = -3 \sin t \ne 0 \), so these are valid critical points.

Thus, the values of \( \cos \theta \) are:
\( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \) and \( \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} \)

So \( \cos \theta = \pm \frac{\sqrt{2}}{2} \).

Part (a):
- M1: Attempts to differentiate both \( x \) and \( y \) with respect to \( t \).
- M1: Evaluates both derivatives and attempts to find the gradient \( \frac{dy}{dx} \) at \( t = \frac{\pi}{6} \).
- A1: Correctly finds \( \frac{dy}{dx} = -\frac{8}{3} \).
- M1: Finds coordinates of the point and applies the straight line equation formula.
- A1: Obtains the correct equation \( 8x + 3y = 18\sqrt{3} \) in the required form.

Part (b):
- M1: Applies double angle identity \( \sin 2t = 2 \sin t \cos t \).
- M1: Substitutes \( \cos t = \frac{x}{3} \) into their expression.
- M1: Uses the identity \( \sin^2 t + \cos^2 t = 1 \) to substitute for \( \sin t \).
- A1: Obtains a correct Cartesian equation in the form \( y^2 = f(x) \), such as \( y^2 = \frac{64}{81}x^2(9 - x^2) \).

Part (c):
- M1: Realises that a tangent parallel to the \( x \)-axis requires \( \frac{dy}{dt} = 0 \).
- M1: Solves \( \cos 2t = 0 \) to find \( t \) (or \( \theta \)) in the given interval.
- A1: Obtains both correct exact values \( \cos \theta = \pm \frac{\sqrt{2}}{2} \) (or equivalent form like \( \pm \frac{1}{\sqrt{2}} \)).