2025 Edexcel IGCSE Further Pure Mathematics 模擬試題連答案詳解

To rationalize the denominator, multiply both the numerator and the denominator by the conjugate of the denominator, which is \(3\sqrt{2} + 2\sqrt{3}\). This gives: \(\frac{(\sqrt{3} + \sqrt{2})(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2} - 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3})}\). Expanding the denominator: \((3\sqrt{2})^2 - (2\sqrt{3})^2 = 18 - 12 = 6\). Expanding the numerator: \(\sqrt{3}(3\sqrt{2}) + \sqrt{3}(2\sqrt{3}) + \sqrt{2}(3\sqrt{2}) + \sqrt{2}(2\sqrt{3}) = 3\sqrt{6} + 6 + 6 + 2\sqrt{6} = 12 + 5\sqrt{6}\). Dividing the numerator by the denominator: \(\frac{12 + 5\sqrt{6}}{6} = 2 + \frac{5}{6}\sqrt{6}\). This is in the form \(a + b\sqrt{6}\) where \(a = 2\) and \(b = \frac{5}{6}\).

M1: Multiplies numerator and denominator by \(3\sqrt{2} + 2\sqrt{3}\). M1: Correctly expands numerator to \(12 + 5\sqrt{6}\) and denominator to \(6\). A1: Correct final expression \(2 + \frac{5}{6}\sqrt{6}\) or equivalent exact form.

(a)
The sum of the first \(n\) terms of an arithmetic series is given by:
\(S_n = \frac{n}{2}[2a + (n-1)d]\)

For \(S_8 = 116\):
\(\frac{8}{2}[2a + 7d] = 116\)
\(4(2a + 7d) = 116\)
\(2a + 7d = 29\) --- (Equation 1)

The \(n\)-th term of an arithmetic series is given by:
\(u_n = a + (n-1)d\)

For \(u_{15} = 46\):
\(a + 14d = 46\) --- (Equation 2)

Multiply Equation 2 by 2:
\(2a + 28d = 92\) --- (Equation 3)

Subtract Equation 1 from Equation 3:
\((2a + 28d) - (2a + 7d) = 92 - 29\)
\(21d = 63\)
\(d = 3\)

Substitute \(d = 3\) into Equation 2:
\(a + 14(3) = 46\)
\(a + 42 = 46\)
\(a = 4\)

(b)
To find the sum of the first 25 terms (\(S_{25}\)):
\(S_{25} = \frac{25}{2}[2a + 24d]\)

Substitute \(a = 4\) and \(d = 3\):
\(S_{25} = \frac{25}{2}[2(4) + 24(3)]\)
\(S_{25} = \frac{25}{2}[8 + 72]\)
\(S_{25} = \frac{25}{2}[80]\)
\(S_{25} = 25 \times 40 = 1000\)

Part (a)
M1: Attempts to write an equation for the sum of 8 terms using \(S_n\) formula or the 15th term using \(u_n\) formula (e.g., \(4(2a+7d)=116\) or \(a+14d=46\)).
M1: Forms a system of two simultaneous linear equations in \(a\) and \(d\) and attempts to solve for one variable by eliminating the other.
A1: Correct value of \(d = 3\).
A1: Correct value of \(a = 4\).

Part (b)
M1: Recalls the sum formula \(S_n = \frac{n}{2}[2a + (n-1)d]\) and attempts to use it for \(n = 25\) with their values of \(a\) and \(d\).
M1: Correct substitution of \(a = 4\) and \(d = 3\) into the sum formula.
A1: Correct final answer of \(1000\).

Part (a): The particle is instantaneously at rest when \(v = 0\). Setting \(3t^2 - 12t + 9 = 0\), we divide by 3 to get \(t^2 - 4t + 3 = 0\). Factoring gives \((t-1)(t-3) = 0\), hence \(t = 1\) or \(t = 3\). Part (b): Acceleration \(a\) is given by \(a = \frac{dv}{dt} = 6t - 12\). Substituting \(t = 4\), we get \(a = 6(4) - 12 = 12\text{ m/s}^2\). Part (c): The particle changes direction of motion at \(t = 1\) and \(t = 3\). We find the displacement function \(s(t) = \int v\,dt = t^3 - 6t^2 + 9t + C\). Taking the reference point \(s(0) = 0\), we evaluate the displacement at the key times: \(s(0) = 0\), \(s(1) = 1^3 - 6(1)^2 + 9(1) = 4\), \(s(3) = 3^3 - 6(3)^2 + 9(3) = 27 - 54 + 27 = 0\), \(s(4) = 4^3 - 6(4)^2 + 9(4) = 64 - 96 + 36 = 4\). The distance travelled in each interval is: From \(t=0\) to \(t=1\): \(|s(1) - s(0)| = |4 - 0| = 4\text{ m}\). From \(t=1\) to \(t=3\): \(|s(3) - s(1)| = |0 - 4| = 4\text{ m}\). From \(t=3\) to \(t=4\): \(|s(4) - s(3)| = |4 - 0| = 4\text{ m}\). The total distance is the sum of these distances: \(4 + 4 + 4 = 12\text{ m}\).

Part (a) [2 Marks]: M1 for setting \(v = 0\) and attempting to solve the quadratic equation. A1 for both correct values of \(t = 1\) and \(t = 3\). Part (b) [2 Marks]: M1 for differentiating \(v\) with respect to \(t\). A1 for substituting \(t=4\) to obtain \(12\). Part (c) [6 Marks]: M1 for integrating \(v\) to find an expression for displacement \(s(t)\). M1 for identifying the roots \(t=1\) and \(t=3\) as key points where direction changes. M1 for calculating values of \(s(t)\) at \(t=0, 1, 3, 4\) (at least three correct). M1 for a sum of absolute differences structure. A1 for correct distance segments of \(4\), \(4\), and \(4\). A1 for final correct total distance of \(12\text{ m}\).

(a) Using the Cosine Rule on triangle \(ABC\):
\[AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\]
Substitute the given expressions into the formula:
\[(2x + 1)^2 = (x + 1)^2 + (2x + 3)^2 - 2(x + 1)(2x + 3)\cos(60^\circ)\]
Since \(\cos(60^\circ) = \frac{1}{2}\):
\[(2x + 1)^2 = (x + 1)^2 + (2x + 3)^2 - (x + 1)(2x + 3)\]
Expand each part:
\[4x^2 + 4x + 1 = (x^2 + 2x + 1) + (4x^2 + 12x + 9) - (2x^2 + 5x + 3)\]
Simplifying both sides:
\[4x^2 + 4x + 1 = 3x^2 + 9x + 7\]
Rearranging to form a quadratic equation equal to zero:
\[x^2 - 5x - 6 = 0\quad\text{(as required)}\]

(b) Factorise the quadratic equation:
\[(x - 6)(x + 1) = 0\]
This gives \(x = 6\) or \(x = -1\).
Since \(x\) represents physical lengths, all side lengths must be positive. If \(x = -1\), then \(AB = 0\text{ cm}\) and \(AC = -1\text{ cm}\), which is impossible. Thus, we reject \(x = -1\) and conclude that \(x = 6\).

(c) When \(x = 6\):
\[AB = 6 + 1 = 7\text{ cm}\]
\[BC = 2(6) + 3 = 15\text{ cm}\]
Area of the triangle:
\[\text{Area} = \frac{1}{2}(AB)(BC)\sin(\angle ABC)\]
\[\text{Area} = \frac{1}{2}(7)(15)\sin(60^\circ) = \frac{105}{2} \times \frac{\sqrt{3}}{2} = \frac{105\sqrt{3}}{4}\text{ cm}^2\]
Thus, \(a = \frac{105}{4}\).

(a)
M1: Applies the Cosine Rule with given expressions.
M1: Substitutes \(\cos(60^\circ) = 0.5\) and attempts expansions.
M1: Correctly expands and simplifies expressions to group like terms.
A1: Fully correct algebra leading to \(x^2 - 5x - 6 = 0\).

(b)
B1: Identifies \(x = 6\) and rejects \(x = -1\) with the reason that lengths must be positive.

(c)
M1: Finds \(AB = 7\) and \(BC = 15\) and applies the area formula \(\frac{1}{2}ab\sin C\).
A1: Obtains the correct exact area \(\frac{105\sqrt{3}}{4}\text{ cm}^2\) (or equivalent).

First, solve the quadratic inequality \(2x^2 - 5x - 12 < 0\).

Factorise the quadratic expression:
\[ (2x + 3)(x - 4) < 0 \]

The critical values are \(x = -\frac{3}{2}\) and \(x = 4\).

Since we require the expression to be less than 0, the solution to this inequality is:
\[ -\frac{3}{2} < x < 4 \]

Second, solve the inequality \(x(x - 1) \ge 12 - 2x\).

Expand and rearrange the terms to form a quadratic inequality:
\[ x^2 - x \ge 12 - 2x \]
\[ x^2 + x - 12 \ge 0 \]

Factorise the quadratic expression:
\[ (x + 4)(x - 3) \ge 0 \]

The critical values are \(x = -4\) and \(x = 3\).

Since we require the expression to be greater than or equal to 0, the solution to this inequality is:
\[ x \le -4 \quad \text{or} \quad x \ge 3 \]

To find the set of values of \(x\) that satisfy both inequalities, we find the intersection of the two solution sets:
1) \(-\frac{3}{2} < x < 4\)
2) \(x \le -4\) or \(x \ge 3\)

Comparing the intervals:
- The interval \(x \le -4\) has no overlap with \(-\frac{3}{2} < x < 4\) because \(-4 < -\frac{3}{2}\).
- The interval \(x \ge 3\) overlaps with \(-\frac{3}{2} < x < 4\) for values of \(x\) from \(3\) up to (but not including) \(4\).

Thus, the set of values of \(x\) that satisfy both inequalities is:
\[ 3 \le x < 4 \]

- **M1**: For an attempt to solve the quadratic equation \(2x^2 - 5x - 12 = 0\) (by factorisation, completing the square, or using the formula).
- **A1**: For finding the correct critical values \(x = -1.5\) and \(x = 4\).
- **A1**: For the correct interval: \(-1.5 < x < 4\) (or equivalent notation).
- **M1**: For expanding and rearranging the second inequality to form a three-term quadratic \(x^2 + x - 12 \ge 0\), and attempting to solve \(x^2 + x - 12 = 0\).
- **A1**: For finding the correct critical values \(x = -4\) and \(x = 3\).
- **A1**: For the correct interval: \(x \le -4\) or \(x \ge 3\) (or equivalent notation).
- **B1**: For finding the correct intersection of both intervals: \(3 \le x < 4\) (accept equivalent notation, such as \([3, 4)\) or \(\{x : 3 \le x < 4\}\)).

**Part (a)**

Since \(D\) is the midpoint of \(AB\), we have:
\(\vec{OD} = \vec{OA} + \frac{1}{2}\vec{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}(\mathbf{a} + \mathbf{b})\)

Since \(C\) lies on \(OB\) such that \(OC : CB = 2 : 3\), we have:
\(\vec{OC} = \frac{2}{5}\vec{OB} = \frac{2}{5}\mathbf{b}\)

The point \(X\) lies on the line \(OD\), so there exists a scalar \(\mu\) such that:
\(\vec{OX} = \mu \vec{OD} = \frac{1}{2}\mu(\mathbf{a} + \mathbf{b})\)

The point \(X\) also lies on the line \(AC\), so there exists a scalar \(\lambda\) such that:
\(\vec{OX} = (1 - \lambda)\vec{OA} + \lambda\vec{OC} = (1 - \lambda)\mathbf{a} + \frac{2}{5}\lambda\mathbf{b}\)

Since \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel vectors, we can equate their coefficients:
1) \(\frac{1}{2}\mu = 1 - \lambda\)
2) \(\frac{1}{2}\mu = \frac{2}{5}\lambda\)

From these two equations, we get:
\(1 - \lambda = \frac{2}{5}\lambda \implies 1 = \frac{7}{5}\lambda \implies \lambda = \frac{5}{7}\)

Substituting \(\lambda = \frac{5}{7}\) into the coefficient of \(\mathbf{b}\):
\(\frac{1}{2}\mu = \frac{2}{5}\left(\frac{5}{7}\right) = \frac{2}{7} \implies \mu = \frac{4}{7}\)

Thus, substituting \(\mu = \frac{4}{7}\) into our equation for \(\vec{OX}\):
\(\vec{OX} = \frac{1}{2}\left(\frac{4}{7}\right)(\mathbf{a} + \mathbf{b}) = \frac{2}{7}(\mathbf{a} + \mathbf{b})\) (as required)

**Part (b)**

To find \(|\vec{OX}|\), we use the scalar product:
\(|\vec{OX}|^2 =
\vec{OX} \cdot \vec{OX} = \frac{2}{7}(\mathbf{a} + \mathbf{b}) \cdot \frac{2}{7}(\mathbf{a} + \mathbf{b}) = \frac{4}{49}(\mathbf{a} \cdot \mathbf{a} + 2\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b})\)

We are given:
- \(\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 = 3^2 = 9\)
- \(\mathbf{b} \cdot \mathbf{b} = |\mathbf{b}|^2 = 5^2 = 25\)
- \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(60^\circ) = 3 \times 5 \times 0.5 = 7.5\)

Substituting these values into the expression:
\(|\vec{OX}|^2 = \frac{4}{49}(9 + 2(7.5) + 25) = \frac{4}{49}(9 + 15 + 25) = \frac{4}{49}(49) = 4\)

Taking the positive square root, we get:
\(|\vec{OX}| = 2\)

**Part (a)**
- **M1**: For a correct vector expression for \(\vec{OD}\) or \(\vec{OC}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\), e.g., \(\vec{OD} =
\frac{1}{2}(\mathbf{a} + \mathbf{b})\) or \(\vec{OC} = \frac{2}{5}\mathbf{b}\).
- **M1**: For expressing \(\vec{OX}\) in two different ways using parameters (e.g., \(\mu\) and \(\lambda\)), such as \(\vec{OX} = \frac{1}{2}
\mu(\mathbf{a} + \mathbf{b})\) and \(\vec{OX} = (1-\lambda)\mathbf{a} + \frac{2}{5}\lambda\mathbf{b}\).
- **M1**: For equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to obtain two simultaneous equations: \(\frac{1}{2}\mu = 1 -
\lambda\) and \(\frac{1}{2}\mu = \frac{2}{5}\lambda\).
- **A1**: For solving the simultaneous equations to find the correct value of either parameter, e.g., \(\lambda = \frac{5}{7}\) or \(\mu =
\frac{4}{7}\).
- **A1**: For showing the complete, correct working leading to the given answer \(\vec{OX} = \frac{2}{7}(\mathbf{a} + \mathbf{b})\) with no steps omitted.

**Part (b)**
- **M1**: For using \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(60^\circ)\) to find the scalar product of \(\mathbf{a}\) and
\(\mathbf{b}\) (Accept \(7.5\) or \(\frac{15}{2}\)).
- **M1**: For expressing \(|\vec{OX}|^2\) in terms of \(\mathbf{a} \cdot \mathbf{a}\), \(\mathbf{a} \cdot \mathbf{b}\), and \(\mathbf{b}
\cdot \mathbf{b}\).
- **A1**: For substituting the values correctly and obtaining \(|\vec{OX}|^2 = 4\).
- **A1**: For obtaining the final answer \(|\vec{OX}| = 2\).

(a) Taking base 10 logarithms of both sides of \(y = A B^x\):
\(\log_{10} y = \log_{10}(A B^x)\)

Using the product law of logarithms:
\(\log_{10} y = \log_{10} A + \log_{10}(B^x)\)

Using the power law of logarithms:
\(\log_{10} y = x \log_{10} B + \log_{10} A\) (as required).

(b) Calculating \(\log_{10} y\) for each value of \(y\):
- For \(y = 15\): \(\log_{10} 15 \approx 1.18\)
- For \(y = 45\): \(\log_{10} 45 \approx 1.65\)
- For \(y = 135\): \(\log_{10} 135 \approx 2.13\)
- For \(y = 405\): \(\log_{10} 405 \approx 2.61\)
- For \(y = 1215\): \(\log_{10} 1215 \approx 3.08\)

The completed table is:

| \(x\) | 1 | 2 | 3 | 4 | 5 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| \(y\) | 15 | 45 | 135 | 405 | 1215 |
| \(\log_{10} y\) | 1.18 | 1.65 | 2.13 | 2.61 | 3.08 |

(c) The linear model is of the form \(Y = m X + c\), where \(Y = \log_{10} y\), \(X = x\), \(m = \log_{10} B\), and \(c = \log_{10} A\).

(i) Finding the gradient \(m\):
\(m = \frac{2.85 - 1.41}{4.5 - 1.5} = \frac{1.44}{3} = 0.48\)
Since \(\log_{10} B = 0.48\):
\(B = 10^{0.48} \approx 3.0199... \approx 3.0\) (to 2 s.f.)

(ii) Finding the intercept \(c\):
Using the point \((1.5, 1.41)\):
\(1.41 = 0.48(1.5) + c\)
\(1.41 = 0.72 + c\)
\(c = 0.69\)
Since \(\log_{10} A = 0.69\):
\(A = 10^{0.69} \approx 4.8977... \approx 4.9\) (to 2 s.f.)

(d) Estimating \(y\) when \(x = 3.5\):

Method 1: Using \(y = A B^x\) with the rounded values \(A = 4.9\) and \(B = 3.0\):
\(y = 4.9 \times 3.0^{3.5} \approx 4.9 \times 46.765 = 229.15\)
Rounding to the nearest integer gives \(y \approx 229\).

Method 2: Using the straight-line equation directly:
\(\log_{10} y = 0.48(3.5) + 0.69 = 1.68 + 0.69 = 2.37\)
\(y = 10^{2.37} \approx 234.42\)
Rounding to the nearest integer gives \(y \approx 234\).

Part (a):
- M1: For taking logarithms of both sides of the equation and applying the product rule of logs correctly.
- A1: Fully correct derivation showing all steps to reach the given expression.

Part (b):
- B1: At least 3 values correct to 2 decimal places.
- B1: All 5 values correct (1.18, 1.65, 2.13, 2.61, 3.08).

Part (c):
- M1: A correct method to calculate the gradient \(m = 0.48\) from the given points.
- A1: For finding \(B \approx 3.0\) (accept 3 or 3.0).
- M1: A correct method to find the intercept \(c\) (e.g., \(1.41 - 0.48 \times 1.5 = 0.69\)) and setting it equal to \(\log_{10} A\).
- A1: For finding \(A \approx 4.9\) (accept 4.9).

Part (d):
- M1: Substitute \(x = 3.5\) into either \(y = A B^{3.5}\) or the linear equation \(\log_{10} y = 0.48 x + 0.69\).
- A1: For finding either \(229\) or \(234\) (accept any integer in the range 229 to 235 inclusive, based on rounding).

(a) From the quadratic equation \(3x^2 - 4x + 6 = 0\), we have \(a = 3\), \(b = -4\), and \(c = 6\).

(i) \(\alpha + \beta = -\frac{b}{a} = \frac{4}{3}\)

(ii) \(\alpha\beta = \frac{c}{a} = \frac{6}{3} = 2\)

(b) (i) Using the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\):

\(\alpha^2 + \beta^2 = \left(\frac{4}{3}\right)^2 - 2(2) = \frac{16}{9} - 4 = -\frac{20}{9}\)

(ii) Using the identity \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\):

\(\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)^3 - 3(2)\left(\frac{4}{3}\right) = \frac{64}{27} - 8 = -\frac{152}{27}\)

(c) Let the new roots be \(\gamma = \frac{\alpha}{\beta^2}\) and \(\delta = \frac{\beta}{\alpha^2}\).

Sum of new roots, \(S\):

\(S = \gamma + \delta = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2} = \frac{\alpha^3 + \beta^3}{\alpha^2\beta^2} = \frac{\alpha^3 + \beta^3}{(\alpha\beta)^2}\)

\(S = \frac{-\frac{152}{27}}{2^2} = \frac{-\frac{152}{27}}{4} = -\frac{38}{27}\)

Product of new roots, \(P\):

\(P = \gamma\delta = \left(\frac{\alpha}{\beta^2}\right)\left(\frac{\beta}{\alpha^2}\right) = \frac{\alpha\beta}{\alpha^2\beta^2} = \frac{1}{\alpha\beta}\)

\(P = \frac{1}{2}\)

Using the formula for a quadratic equation \(x^2 - Sx + P = 0\):

\(x^2 - \left(-\frac{38}{27}\right)x + \frac{1}{2} = 0\)

\(x^2 + \frac{38}{27}x + \frac{1}{2} = 0\)

Multiplying through by 54 to get integer coefficients:

\(54x^2 + 76x + 27 = 0\)

Part (a):
- B1: Correct value of \(\alpha + \beta = \frac{4}{3}\).
- B1: Correct value of \(\alpha\beta = 2\).

Part (b):
- M1: For attempting to use \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) with their values from (a).
- A1: Correct value of \(-\frac{20}{9}\).
- M1: For a correct identity for \(\alpha^3 + \beta^3\) in terms of sum and product, e.g., \((\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)\).
- A1ft: Correct substitution of their values into their identity.
- A1: Correct value of \(-\frac{152}{27}\).

Part (c):
- M1: For expressing the sum of new roots as \(\frac{\alpha^3 + \beta^3}{(\alpha\beta)^2}\).
- A1ft: Correct numerical sum of \(-\frac{38}{27}\) from their previous values.
- M1: For expressing the product of new roots as \(\frac{1}{\alpha\beta}\).
- A1: Correct numerical product of \(\frac{1}{2}\).
- M1: For formulating the equation \(x^2 - Sx + P = 0\) with their sum and product.
- M1: For clearing fractions by multiplying by the LCM (54).
- A1: Correct terms \(54x^2 + 76x + 27\).
- B1: Equation equated to 0, e.g., \(54x^2 + 76x + 27 = 0\) (accept any integer multiple).

(a) Using the quotient rule with \(u = 2x^2 - 2x - 7\) and \(v = e^{2x}\): \(\frac{\text{d}u}{\text{d}x} = 4x - 2\) and \(\frac{\text{d}v}{\text{d}x} = 2e^{2x}\). Then \(\frac{\text{d}y}{\text{d}x} = \frac{(4x - 2)e^{2x} - 2(2x^2 - 2x - 7)e^{2x}}{(e^{2x})^2} = \frac{e^{2x}(4x - 2 - 4x^2 + 4x + 14)}{e^{4x}} = \frac{-4x^2 + 8x + 12}{e^{2x}} = \frac{4(3 + 2x - x^2)}{e^{2x}}\). (b) For stationary points, set \(\frac{\text{d}y}{\text{d}x} = 0\), which gives \(3 + 2x - x^2 = 0\), so \((x - 3)(x + 1) = 0\). Thus, \(x = 3\) or \(x = -1\). Substituting these into the original equation: for \(x = 3\), \(y = \frac{2(9) - 2(3) - 7}{e^6} = \frac{5}{e^6} = 5e^{-6}\); for \(x = -1\), \(y = \frac{2(1) - 2(-1) - 7}{e^{-2}} = -3e^2\). The stationary points are \((3, 5e^{-6})\) and \((-1, -3e^2)\). (c) Differentiating \(\frac{\text{d}y}{\text{d}x} = (12 + 8x - 4x^2)e^{-2x}\) using the product rule: \(\frac{\text{d}^2y}{\text{d}x^2} = (8 - 8x)e^{-2x} - 2(12 + 8x - 4x^2)e^{-2x} = (8x^2 - 24x - 16)e^{-2x} = 8(x^2 - 3x - 2)e^{-2x}\). Evaluating at \(x = -1\): \(\frac{\text{d}^2y}{\text{d}x^2} = 8(1 + 3 - 2)e^2 = 16e^2 > 0\), so \((-1, -3e^2)\) is a local minimum. Evaluating at \(x = 3\): \(\frac{\text{d}^2y}{\text{d}x^2} = 8(9 - 9 - 2)e^{-6} = -16e^{-6} < 0\), so \((3, 5e^{-6})\) is a local maximum.

Part (a): M1 for applying the quotient rule or product rule to differentiate \(y\). M1 for simplifying the numerator to obtain a quadratic factor. A1 for fully correct working leading to the given expression. Part (b): M1 for setting the numerator of the derivative equal to zero and solving the quadratic. A1 for finding \(x = 3\) and \(x = -1\). M1 for substituting both values back into the original curve equation to find the exact y-values. A1 for correct coordinates \((3, 5e^{-6})\) and \((-1, -3e^2)\). Part (c): M1 for attempting to find \(\frac{\text{d}^2y}{\text{d}x^2}\) using product/quotient rule. A1 for correct second derivative values or sign evaluations leading to the correct identification of the local minimum at \((-1, -3e^2)\) and local maximum at \((3, 5e^{-6})\).

(a) Expanding \( y^2 \):
\( y^2 = (2\sin x + \cos x)^2 = 4\sin^2 x + 4\sin x\cos x + \cos^2 x \)

Using the double angle identities:
\( \sin^2 x = \frac{1 - \cos 2x}{2} \implies 4\sin^2 x = 2 - 2\cos 2x \)
\( \cos^2 x = \frac{1 + \cos 2x}{2} \)
\( 2\sin x\cos x = \sin 2x \implies 4\sin x\cos x = 2\sin 2x \)

Substituting these identities into the expansion of \( y^2 \):
\( y^2 = (2 - 2\cos 2x) + 2\sin 2x + \left(\frac{1}{2} + \frac{1}{2}\cos 2x\right) \)
\( y^2 = \frac{5}{2} - \frac{3}{2}\cos 2x + 2\sin 2x \)

Thus, \( p = \frac{5}{2} \), \( q = -\frac{3}{2} \), and \( r = 2 \).

(b) The area of region \( R \) is given by:
\( A = \int_{0}^{\frac{\pi}{4}} y \, \mathrm{d}x = \int_{0}^{\frac{\pi}{4}} (2\sin x + \cos x) \, \mathrm{d}x \)
\( A = \left[ -2\cos x + \sin x \right]_0^{\frac{\pi}{4}} \)
\( A = \left( -2\cos\frac{\pi}{4} + \sin\frac{\pi}{4} \right) - ( -2\cos 0 + \sin 0 ) \)
\( A = \left( -2\left(\frac{\sqrt{2}}{2}\right) + \frac{\sqrt{2}}{2} \right) - (-2 + 0) \)
\( A = -\sqrt{2} + \frac{\sqrt{2}}{2} + 2 = 2 - \frac{\sqrt{2}}{2} \)

(c) The volume \( V \) of the solid generated is:
\( V = \pi \int_{0}^{\frac{\pi}{4}} y^2 \, \mathrm{d}x \)
\( V = \pi \int_{0}^{\frac{\pi}{4}} \left(\frac{5}{2} - \frac{3}{2}\cos 2x + 2\sin 2x\right) \, \mathrm{d}x \)
\( V = \pi \left[ \frac{5}{2}x - \frac{3}{4}\sin 2x - \cos 2x \right]_0^{\frac{\pi}{4}} \)
\( V = \pi \left\{ \left( \frac{5}{2}\left(\frac{\pi}{4}\right) - \frac{3}{4}\sin\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi}{2}\right) \right) - \left( 0 - 0 - \cos 0 \right) \right\} \)
\( V = \pi \left\{ \left( \frac{5\pi}{8} - \frac{3}{4} - 0 \right) - ( -1 ) \right\} \)
\( V = \pi \left( \frac{5\pi}{8} - \frac{3}{4} + 1 \right) = \pi \left( \frac{5\pi}{8} + \frac{1}{4} \right) \)
\( V = \frac{\pi}{8}(5\pi + 2) \)

Thus, \( a = 5 \) and \( b = 2 \).

(a)
- M1: Attempts to expand \( (2\sin x + \cos x)^2 \) to obtain at least 3 terms, including a term in \( \sin x\cos x \).
- M1: Uses the identity \( 2\sin x\cos x = \sin 2x \) to express the cross-term as \( 2\sin 2x \).
- M1: Uses the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \) to express \( 4\sin^2 x \) in terms of \( \cos 2x \).
- M1: Uses the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \) to express \( \cos^2 x \) in terms of \( \cos 2x \).
- A1: Fully correct expression showing \( y^2 = \frac{5}{2} - \frac{3}{2}\cos 2x + 2\sin 2x \) (accept decimal equivalent coefficients, e.g., \( p=2.5 \), \( q=-1.5 \), \( r=2 \)).

(b)
- M1: Sets up the correct definite integral for area with correct limits from \( 0 \) to \( \frac{\pi}{4} \).
- M1: Integrates \( 2\sin x + \cos x \) to get \( -2\cos x + \sin x \) (allow one sign error).
- M1: Substitutes the limits \( \frac{\pi}{4} \) and \( 0 \) into their integrated expression.
- A1: Obtains the exact area \( 2 - \frac{\sqrt{2}}{2} \) or any equivalent exact simplified form (e.g., \( \frac{4 - \sqrt{2}}{2} \)).

(c)
- M1: Sets up the correct volume integral \( \pi \int y^2 \, \mathrm{d}x \) with limits from \( 0 \) to \( \frac{\pi}{4} \).
- M1: Integrates the constant term \( \frac{5}{2} \) and at least one trigonometric term of their \( y^2 \) expression correctly.
- A1: Fully correct integration to obtain \( \frac{5}{2}x - \frac{3}{4}\sin 2x - \cos 2x \) (ignore \( \pi \) for this mark).
- M1: Correct substitution of limits \( \frac{\pi}{4} \) and \( 0 \) into their integrated expression (must subtract the lower limit correctly).
- M1: Correct evaluation of all trigonometric terms at both limits (especially noting \( \cos 0 = 1 \)).
- A1: Fully correct exact volume \( \frac{\pi}{8}(5\pi + 2) \) or \( a = 5 \) and \( b = 2 \).

(a) The water inside the inverted cone forms a similar cone of height \(h\) and radius \(r\).
By similar triangles:
\(\frac{r}{h} = \frac{10}{20} = \frac{1}{2} \implies r = \frac{h}{2}\)

The volume of a cone is given by:
\(V = \frac{1}{3}\pi r^2 h\)

Substituting \(r = \frac{h}{2}\) into this formula:
\(V = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3}\pi \left(\frac{h^2}{4}\right) h = \frac{\pi h^3}{12}\) (as required)

(b) The rate of change of volume with respect to time is \(\frac{\mathrm{d}V}{\mathrm{d}t} = 18\pi\text{ cm}^3\text{ s}^{-1}\).
We want to find \(\frac{\mathrm{d}h}{\mathrm{d}t}\) when \(h = 6\text{ cm}\).

Differentiating \(V\) with respect to \(h\):
\(\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{\mathrm{d}}{\mathrm{d}h}\left(\frac{\pi h^3}{12}\right) = \frac{3\pi h^2}{12} = \frac{\pi h^2}{4}\)

Using the chain rule:
\(\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}h} \times \frac{\mathrm{d}h}{\mathrm{d}t}\)

Substituting the known values at the instant when \(h = 6\):
\(18\pi = \frac{\pi (6)^2}{4} \times \frac{\mathrm{d}h}{\mathrm{d}t}\)
\(18\pi = 9\pi \times \frac{\mathrm{d}h}{\mathrm{d}t}\)
\(\frac{\mathrm{d}h}{\mathrm{d}t} = 2\text{ cm s}^{-1}\)

(c) The area of the horizontal surface of the water is \(A = \pi r^2\).
Since \(r = \frac{h}{2}\), we have:
\(A = \pi \left(\frac{h}{2}\right)^2 = \frac{\pi h^2}{4}\)

We want to find \(\frac{\mathrm{d}A}{\mathrm{d}t}\) when \(h = 6\text{ cm}\).

Differentiating \(A\) with respect to \(h\):
\(\frac{\mathrm{d}A}{\mathrm{d}h} = \frac{2\pi h}{4} = \frac{\pi h}{2}\)

At \(h = 6\):
\(\frac{\mathrm{d}A}{\mathrm{d}h} = 3\pi\)

Using the chain rule:
\(\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}h} \times \frac{\mathrm{d}h}{\mathrm{d}t}\)

Substituting \(\frac{\mathrm{d}A}{\mathrm{d}h} = 3\pi\) and \(\frac{\mathrm{d}h}{\mathrm{d}t} = 2\):
\(\frac{\mathrm{d}A}{\mathrm{d}t} = 3\pi \times 2 = 6\pi\text{ cm}^2\text{ s}^{-1}\)

Part (a)
* M1: For establishing the linear relationship between \(r\) and \(h\) using similar triangles, i.e., \(r = \frac{1}{2}h\).
* A1: For substituting \(r = \frac{1}{2}h\) into the volume formula and completing the derivation to show \(V = \frac{\pi h^3}{12}\) clearly.

Part (b)
* M1: For differentiating the volume formula with respect to \(h\) to obtain \(\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{\pi h^2}{4}\).
* M1: For applying the chain rule \(\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}h} \times \frac{\mathrm{d}h}{\mathrm{d}t}\) with \(\frac{\mathrm{d}V}{\mathrm{d}t} = 18\pi\) and \(h = 6\).
* A1: For obtaining the correct value of \(2\) (or \(2\text{ cm s}^{-1}\)).

Part (c)
* M1: For writing \(A = \frac{\pi h^2}{4}\) and finding \(\frac{\mathrm{d}A}{\mathrm{d}h} = \frac{\pi h}{2}\).
* M1: For applying the chain rule \(\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}h} \times \frac{\mathrm{d}h}{\mathrm{d}t}\) using their value of \(\frac{\mathrm{d}h}{\mathrm{d}t}\) from part (b).
* A1: For obtaining the correct value of \(6\pi\) (or \(6\pi\text{ cm}^2\text{ s}^{-1}\)).

First, express the key position vectors in terms of \(\mathbf{a}\) and \(\mathbf{b}\):

Since \(OP : PA = 2 : 1\), we have:
\(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\) and \(\overrightarrow{PA} = \frac{1}{3}\mathbf{a}\).

Since \(AQ : QB = 1 : 2\), we have:
\(\overrightarrow{AQ} = \frac{1}{3}\overrightarrow{AB} = \frac{1}{3}(\mathbf{b} - \mathbf{a})\).

Now, find \(\overrightarrow{PQ}\):
\(\overrightarrow{PQ} = \overrightarrow{PA} + \overrightarrow{AQ}\)
\(\overrightarrow{PQ} = \frac{1}{3}\mathbf{a} + \frac{1}{3}(\mathbf{b} - \mathbf{a})\)
\(\overrightarrow{PQ} = \frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{b} - \frac{1}{3}\mathbf{a} = \frac{1}{3}\mathbf{b}\).

Next, express \(\overrightarrow{OR}\) in terms of \(\mathbf{b}\):
Since \(OR : RB = 1 : 2\), we have:
\(\overrightarrow{OR} = \frac{1}{3}\mathbf{b}\).

Comparing the two vectors:
\(\overrightarrow{PQ} = \frac{1}{3}\mathbf{b}\) and \(\overrightarrow{OR} = \frac{1}{3}\mathbf{b}\).

Since \(\overrightarrow{PQ} = \overrightarrow{OR}\), the side \(PQ\) is parallel to the side \(OR\) and is equal in length.

Therefore, because a pair of opposite sides is both parallel and equal in length, the quadrilateral \(OPQR\) is a parallelogram.

M1: For writing a correct vector equation for \(\overrightarrow{PQ}\), such as \(\overrightarrow{PQ} = \overrightarrow{PA} + \overrightarrow{AQ}\) or \(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}\), and finding \(\overrightarrow{PA} = \frac{1}{3}\mathbf{a}\) and \(\overrightarrow{AQ} = \frac{1}{3}(\mathbf{b} - \mathbf{a})\) (or equivalent).
A1: For simplifying to show \(\overrightarrow{PQ} = \frac{1}{3}\mathbf{b}\).
B1: For stating \(\overrightarrow{OR} = \frac{1}{3}\mathbf{b}\).
A1: For stating that since \(\overrightarrow{PQ} = \overrightarrow{OR}\), the opposite sides are parallel and equal in length, thus proving \(OPQR\) is a parallelogram (must have a concluding statement based on vector equality).

**(a)**
First, find the midpoint \(M\) of \(AB\):
\(M = \left(\frac{2 + 6}{2}, \frac{7 + (-1)}{2}\right) = (4, 3)\)

Next, find the gradient of \(AB\):
\(m_{AB} = \frac{-1 - 7}{6 - 2} = \frac{-8}{4} = -2\)

The gradient of the perpendicular bisector, \(m\), is the negative reciprocal of \(m_{AB}\):
\(m = -\frac{1}{-2} = \frac{1}{2}\)

Now, find the equation of the perpendicular bisector using the midpoint \(M(4, 3)\) and gradient \(m = \frac{1}{2}\):
\(y - 3 = \frac{1}{2}(x - 4)\)
\(2(y - 3) = x - 4\)
\(2y - 6 = x - 4\)
\(x - 2y + 2 = 0\)

**(b)**
**Method 1: Using the area and height**
Calculate the length of the base \(AB\):
\(AB = \sqrt{(6 - 2)^2 + (-1 - 7)^2} = \sqrt{4^2 + (-8)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}\)

Since \(C\) lies on the perpendicular bisector of \(AB\), triangle \(ABC\) is an isosceles triangle with base \(AB\) and height \(h = MC\), where \(M\) is the midpoint of \(AB\).

The area of triangle \(ABC\) is given as \(30\):
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = 30\)
\(\frac{1}{2} \times 4\sqrt{5} \times h = 30\)
\(2\sqrt{5}h = 30\)
\(h = \frac{15}{\sqrt{5}} = 3\sqrt{5}\)

Since \(C(x, y)\) lies on the perpendicular bisector, its coordinates satisfy \(x - 2y + 2 = 0 \implies x - 4 = 2(y - 3)\).
The distance \(MC = h = 3\sqrt{5}\), so:
\((x - 4)^2 + (y - 3)^2 = (3\sqrt{5})^2\)
\((x - 4)^2 + (y - 3)^2 = 45\)

Substitute \(x - 4 = 2(y - 3)\) into the distance equation:
\((2(y - 3))^2 + (y - 3)^2 = 45\)
\(4(y - 3)^2 + (y - 3)^2 = 45\)
\(5(y - 3)^2 = 45\)
\((y - 3)^2 = 9\)
\(y - 3 = \pm 3\)

This gives two possible values for \(y\):
If \(y - 3 = 3 \implies y = 6\), then \(x - 4 = 2(3) \implies x = 10\). Thus, \(C = (10, 6)\).
If \(y - 3 = -3 \implies y = 0\), then \(x - 4 = 2(-3) \implies x = -2\). Thus, \(C = (-2, 0)\).

**Method 2: Using the coordinate geometry area formula**
Let \(C\) have coordinates \((x, y)\). Since \(C\) lies on the perpendicular bisector, we have \(x = 2y - 2\).
Using the formula for the area of a triangle with vertices \((2, 7)\), \((6, -1)\), and \((x, y)\):
\(\text{Area} = \frac{1}{2} |2(-1 - y) + 6(y - 7) + x(7 - (-1))| = 30\)
\(\frac{1}{2} |-2 - 2y + 6y - 42 + 8x| = 30\)
\(\frac{1}{2} |8x + 4y - 44| = 30\)
\(|4x + 2y - 22| = 30\)
\(|2x + y - 11| = 15\)

Substitute \(x = 2y - 2\) into the equation:
\(|2(2y - 2) + y - 11| = 15\)
\(|4y - 4 + y - 11| = 15\)
\(|5y - 15| = 15\)
\(5|y - 3| = 15\)
\(|y - 3| = 3\)

This leads to:
\(y - 3 = 3 \implies y = 6 \implies x = 2(6) - 2 = 10\)
\(y - 3 = -3 \implies y = 0 \implies x = 2(0) - 2 = -2\)

Therefore, the possible coordinates of \(C\) are \((10, 6)\) and \((-2, 0)\).

**(a)**
* **M1**: Attempts to find the midpoint of \(AB\). (e.g. \((4, 3)\))
* **M1**: Attempts to find the gradient of \(AB\) and finds the perpendicular gradient. (Gradient of \(AB = -2\) and perpendicular gradient \(= \frac{1}{2}\))
* **M1**: Attempts to write down an equation of the perpendicular bisector using their midpoint and perpendicular gradient.
* **A1**: Correct equation in the required form: \(x - 2y + 2 = 0\) (or any integer multiple thereof, e.g. \(2y - x - 2 = 0\)).

**(b)**
* **M1**: Correct method to find the length of \(AB\) (\(4\sqrt{5}\)) OR sets up the coordinate area formula.
* **M1**: Correctly identifies the perpendicular height \(MC = 3\sqrt{5}\) OR simplifies the coordinate area equation to \(|2x + y - 11| = 15\) (or equivalent).
* **M1**: Formulates an equation in one variable using the perpendicular bisector relation (e.g., \(5(y-3)^2 = 45\) or \(|5y - 15| = 15\)).
* **M1**: Solves their equation to find two values for \(y\) (or \(x\)) (e.g. \(y = 6\) and \(y = 0\)).
* **A1**: One correct pair of coordinates: \((10, 6)\) or \((-2, 0)\).
* **A1**: Both correct pairs of coordinates: \((10, 6)\) and \((-2, 0)\).

**(a)** Let \(C\) be the center of the inscribed circle of radius \(x \text{ cm}\).
By symmetry, the line \(OC\) bisects the angle \(AOB = \frac{2\pi}{3}\), and when extended, it meets the arc \(AB\) at its midpoint \(M\).

Since \(OM\) is the radius of the sector, \(OM = r\).
Since the circle touches the arc at \(M\), the distance from \(C\) to \(M\) is the radius of the inscribed circle, so \(CM = x\).

This gives:
\(OC = OM - CM = r - x\)

Let the inscribed circle touch the radius \(OA\) at point \(P\).
Then the radius \(CP\) is perpendicular to \(OA\), so \(\angle OPC = 90^\circ\) and \(CP = x\).

In the right-angled triangle \(OPC\):
\(\angle POC = \frac{1}{2} \angle AOB = \frac{1}{2} \left(\frac{2\pi}{3}\right) = \frac{\pi}{3}\)

Using the sine ratio in triangle \(OPC\):
\(\sin\left(\frac{\pi}{3}\right) = \frac{CP}{OC} = \frac{x}{r - x}\)

Since \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\):
\(\frac{\sqrt{3}}{2} = \frac{x}{r - x}\)

Multiplying both sides by \(2(r - x)\):
\(\sqrt{3}(r - x) = 2x\)

\(r\sqrt{3} - x\sqrt{3} = 2x\)

Rearranging to group terms in \(x\):
\(x(2 + \sqrt{3}) = r\sqrt{3}\)

\(x = \frac{r\sqrt{3}}{2 + \sqrt{3}}\)

Rationalizing the denominator by multiplying the numerator and denominator by \(2 - \sqrt{3}\):
\(x = \frac{r\sqrt{3}(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{r(2\sqrt{3} - 3)}{4 - 3} = r(2\sqrt{3} - 3)\) *(as required)*

**(b)** Given \(r = 10\):

Area of the sector \(OAB\):
\(A_{\text{sector}} = \frac{1}{2}r^2\theta = \frac{1}{2}(10)^2\left(\frac{2\pi}{3}\right) = \frac{100\pi}{3} \approx 104.72 \text{ cm}^2\)

Radius of the inscribed circle:
\(x = 10(2\sqrt{3} - 3) \approx 4.6410 \text{ cm}\)

Area of the inscribed circle:
\(A_{\text{circle}} =

\pi x^2 = \pi (4.6410)^2 \approx 67.67 \text{ cm}^2\)

Area of the shaded region outside the circle but inside the sector:
\(A_{\text{shaded}} = A_{\text{sector}} - A_{\text{circle}} \approx 104.72 - 67.67 = 37.05 \text{ cm}^2\)

Rounding to 3 significant figures gives \(37.1 \text{ cm}^2\).

**(a)**
* **M1**: Identifies that \(OC = r - x\) and recognises the right-angled triangle \(OPC\) with angle \(\frac{\pi}{3}\).
* **M1**: Sets up the trigonometric relation \(\sin\left(\frac{\pi}{3}\right) =

\frac{x}{r - x}\).
* **A1**: Resolves to a linear equation in \(x\) and \(r\), e.g., \(r\sqrt{3} - x\sqrt{3} = 2x\).
* **A1**: Shows complete, correct rationalization to arrive at the given expression \(x = r(2\sqrt{3} - 3)\).

**(b)**
* **M1**: Calculates both the area of the sector (\(104.7\dots\)) and the area of the inscribed circle (\(67.7\dots\)).
* **A1**: Subtracts the two areas to obtain the final answer \(37.1\) (accept answers in the range \(37.0 - 37.1\)).

(a) For a geometric series to be convergent, the common ratio \(R\) must satisfy \(|R| < 1\).

Here, the common ratio is \(R = \frac{2x - 1}{3x + 2}\).

So, \(\left| \frac{2x - 1}{3x + 2} \right| < 1\).

Since both sides are non-negative, we can square both sides:
\((2x - 1)^2 < (3x + 2)^2\)

\(4x^2 - 4x + 1 < 9x^2 + 12x + 4\)

\(5x^2 + 16x + 3 > 0\)

Factorizing the quadratic expression:
\((5x + 1)(x + 3) > 0\)

Thus, the critical values are \(x = -3\) and \(x = -\frac{1}{5}\).

Since the quadratic expression must be strictly greater than 0, the set of values of \(x\) is:
\(x < -3\) or \(x > -\frac{1}{5}\).

(b) When \(x = 2\):
\(R = \frac{2(2) - 1}{3(2) + 2} = \frac{3}{8}\).

Since \(|R| = \left| \frac{3}{8} \right| < 1\), the series converges.

The first term \(a\) is found by substituting \(r = 1\):
\(a = 4 \left( \frac{3}{8} \right)^{0} = 4\).

The sum to infinity is given by:
\(S_{\infty} = \frac{a}{1 - R} = \frac{4}{1 - \frac{3}{8}} = \frac{4}{\frac{5}{8}} = \frac{32}{5}\) (or \(6.4\)).

(c) Given that \(S_{\infty} = 8\):
\(\frac{a}{1 - R} = 8\)

Substitute \(a = 4\):
\(\frac{4}{1 - R} = 8 \implies 1 - R = \frac{1}{2} \implies R = \frac{1}{2}\).

Now, substitute the expression for \(R\):
\(\frac{2x - 1}{3x + 2} = \frac{1}{2}\)

Multiply both sides by \(2(3x + 2)\):
\(2(2x - 1) = 3x + 2\)

\(4x - 2 = 3x + 2\)

\(x = 4\).

Since \(x = 4\) lies in the interval \(x > -\frac{1}{5}\), this is a valid solution.

(a)
- M1: Identifies the condition for convergence as \(|R| < 1\) where \(R = \frac{2x - 1}{3x + 2}\).
- M1: Sets up the inequality \((2x - 1)^2 < (3x + 2)^2\) and expands correctly.
- A1: Obtains the correct simplified quadratic inequality: \(5x^2 + 16x + 3 > 0\) (or equivalent).
- M1: Solves the quadratic equation \((5x + 1)(x + 3) = 0\) to find critical values \(x = -3\) and \(x = -\frac{1}{5}\).
- A1: Correct final range: \(x < -3\) or \(x > -\frac{1}{5}\) (accept equivalent interval notation).

(b)
- M1: Substitutes \(x = 2\) to find \(R = \frac{3}{8}\) and states that \(|R| < 1\) (or equivalent valid reason for convergence).
- M1: Identifies the first term \(a = 4\) and applies the sum to infinity formula \(S_{\infty} = \frac{a}{1 - R}\).
- A1: Obtains the exact sum to infinity \(\frac{32}{5}\) (or \(6.4\)).

(c)
- M1: Sets up the equation \(\frac{4}{1 - R} = 8\) and attempts to solve for \(R\).
- A1: Correctly finds \(R = \frac{1}{2}\).
- A1: Equates \(\frac{2x - 1}{3x + 2} = \frac{1}{2}\) and solves to obtain \(x = 4\).

\(\textbf{Part (a)}\)
Using the factor theorem, since \( (x - 2) \) is a factor of \( P(x) \), we have:
\( P(2) = 0 \)
\( 2(2)^3 + a(2)^2 + b(2) - 6 = 0 \)
\( 16 + 4a + 2b - 6 = 0 \)
\( 2a + b = -5 \) --- (Equation 1)

Using the remainder theorem, since the remainder when \( P(x) \) is divided by \( (x + 1) \) is \( -12 \), we have:
\( P(-1) = -12 \)
\( 2(-1)^3 + a(-1)^2 + b(-1) - 6 = -12 \)
\( -2 + a - b - 6 = -12 \)
\( a - b = -4 \) --- (Equation 2)

To solve simultaneously, add Equation 1 and Equation 2:
\( (2a + b) + (a - b) = -5 + (-4) \)
\( 3a = -9 \Rightarrow a = -3 \)

Substitute \( a = -3 \) into Equation 2:
\( -3 - b = -4 \Rightarrow b = 1 \)

Thus, \( a = -3 \) and \( b = 1 \).

\(\textbf{Part (b)}\)
Using the values of \( a \) and \( b \), the polynomial is:
\( P(x) = 2x^3 - 3x^2 + x - 6 \)

Since \( (x - 2) \) is a factor, we can express \( P(x) \) in the form \( (x - 2)(2x^2 + px + q) \).
Comparing the constant terms: \( -2q = -6 \Rightarrow q = 3 \).
Comparing the \( x^2 \) coefficients: \( p - 4 = -3 \Rightarrow p = 1 \).

Therefore,
\( P(x) = (x - 2)(2x^2 + x + 3) \).

\(\textbf{Part (c)}\)
To find the roots of the equation \( P(x) = 0 \), we solve:
\( (x - 2)(2x^2 + x + 3) = 0 \)

This gives \( x = 2 \) as a real root.

For the quadratic equation \( 2x^2 + x + 3 = 0 \), we calculate the discriminant \( \Delta = b^2 - 4ac \):
\( \Delta = 1^2 - 4(2)(3) = 1 - 24 = -23 \)

Since the discriminant is less than zero (\( \Delta < 0 \)), the quadratic equation has no real roots.

Thus, the equation \( P(x) = 0 \) has only one real root, which is \( x = 2 \).

\(\textbf{Part (a)}\)
- M1: Formulates a linear equation in \( a \) and \( b \) using \( P(2) = 0 \).
- M1: Formulates a linear equation in \( a \) and \( b \) using \( P(-1) = -12 \).
- A1: Obtains two correct simplified linear equations, e.g., \( 2a + b = -5 \) and \( a - b = -4 \).
- A1: Solves the equations simultaneously to find both \( a = -3 \) and \( b = 1 \).

\(\textbf{Part (b)}\)
- M1: Attempts algebraic division or equates coefficients to factorise \( P(x) \).
- A1: Correctly expresses \( P(x) = (x - 2)(2x^2 + x + 3) \).

\(\textbf{Part (c)}\)
- M1: Calculates the discriminant of \( 2x^2 + x + 3 \).
- A1: Obtains \( -23 \) and explicitly states that since \( \Delta < 0 \), there are no real roots from the quadratic expression, concluding that \( x = 2 \) is the only real root.

Part (a): The base \(ABCD\) is a square with side length \(10\text{ cm}\). The diagonal \(AC\) is given by Pythagoras' Theorem: \(AC = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}\text{ cm}\). The distance from the centre \(O\) to the vertex \(A\) is half the diagonal: \(AO = 5\sqrt{2}\text{ cm}\). The angle between the slant edge \(VA\) and the base \(ABCD\) is the angle \(V\hat{A}O\). In the right-angled triangle \(VOA\): \(\cos(V\hat{A}O) = \frac{AO}{VA} = \frac{5\sqrt{2}}{13}\). Thus, \(V\hat{A}O = \arccos\left(\frac{5\sqrt{2}}{13}\right) \approx 57.049^\circ\). To 1 decimal place, the angle is \(57.0^\circ\). Part (b): Let \(M\) be the midpoint of the edge \(BC\). Since the base is a square, \(OM = 5\text{ cm}\). The angle between the plane \(VBC\) and the base \(ABCD\) is the angle \(V\hat{M}O\). In the right-angled triangle \(VMC\), where \(MC = 5\text{ cm}\) and \(VC = 13\text{ cm}\), the slant height \(VM\) is: \(VM = \sqrt{13^2 - 5^2} = \sqrt{144} = 12\text{ cm}\). In the right-angled triangle \(VOM\): \(\cos(V\hat{M}O) = \frac{OM}{VM} = \frac{5}{12}\). Thus, \(V\hat{M}O = \arccos\left(\frac{5}{12}\right) \approx 65.376^\circ\). To 1 decimal place, the angle is \(65.4^\circ\).

Part (a): [M1] For finding the half-diagonal length \(AO = 5\sqrt{2}\) (or equivalent decimal \(7.07\)). [M1] For a correct trigonometric equation for angle \(V\hat{A}O\), e.g., \(\cos(V\hat{A}O) = \frac{5\sqrt{2}}{13}\) or finding the height \(VO = \sqrt{119}\) and using sine or tangent. [A1] For the correct angle \(57.0^\circ\) (accept \(57^\circ\)). Part (b): [M1] For finding the slant height \(VM = 12\) (or finding the height \(VO = \sqrt{119} \approx 10.9\)). [M1] For a correct trigonometric equation for angle \(V\hat{M}O\), e.g., \(\cos(V\hat{M}O) = \frac{5}{12}\) or \(\tan(V\hat{M}O) = \frac{\sqrt{119}}{5}\). [A1] For the correct angle \(65.4^\circ\).

**(a)**

The volume \(V\) of the cuboid is given by:
\[V = \text{length} \times \text{width} \times \text{height}\]
\[V = 3x \times x \times h = 3x^2h\]

Given that \(V = 288\text{ cm}^3\):
\[3x^2h = 288 \implies h = \frac{288}{3x^2} = \frac{96}{x^2}\]

The total surface area \(A\) of the closed cuboid is:
\[A = 2(3x \times x) + 2(3x \times h) + 2(x \times h)\]
\[A = 6x^2 + 6xh + 2xh = 6x^2 + 8xh\]

Substituting \(h = \frac{96}{x^2}\) into the expression for \(A\):
\[A = 6x^2 + 8x \left( \frac{96}{x^2} \right)\]
\[A = 6x^2 + \frac{768}{x}\]

**(b)**

Differentiate \(A\) with respect to \(x\):
\[A = 6x^2 + 768x^{-1}\]
\[\frac{dA}{dx} = 12x - 768x^{-2} = 12x - \frac{768}{x^2}\]

To find the stationary points, set
\[\frac{dA}{dx} = 0\]
\[12x - \frac{768}{x^2} = 0 \implies 12x = \frac{768}{x^2}\]
\[12x^3 = 768 \implies x^3 = 64 \implies x = 4\]

Since \(x > 0\) represents a physical dimension, we choose \(x = 4\).

The minimum value of \(A\) is:
\[A = 6(4)^2 + \frac{768}{4} = 96 + 192 = 288\text{ cm}^2\]

To justify that this is indeed a minimum, we find the second derivative:
\[\frac{d^2A}{dx^2} = 12 + 1536x^{-3} = 12 + \frac{1536}{x^3}\]

Substituting \(x = 4\):
\[\frac{d^2A}{dx^2} = 12 + \frac{1536}{64} = 12 + 24 = 36\]

Since \(\frac{d^2A}{dx^2} = 36 > 0\) at \(x = 4\), the surface area is a minimum.

**(c)**

Substituting \(x = 4\) into the expression for \(h\):
\[h = \frac{96}{x^2} = \frac{96}{16} = 6\text{ cm}\]

**(a)**
* **M1**: For a correct expression for the volume of the cuboid in terms of \(x\) and \(h\) set equal to \(288\), i.e., \(3x^2 h = 288\).
* **A1**: For expressing \(h\) correctly in terms of \(x\): \(h = \frac{96}{x^2}\).
* **M1**: For a correct expression for the total surface area \(A\) of the closed cuboid: \(A = 6x^2 + 8xh\).
* **A1**: For substituting \(h = \frac{96}{x^2}\) into their surface area expression and simplifying to show the given formula \(A = 6x^2 + \frac{768}{x}\) (cso).

**(b)**
* **M1**: For differentiating \(A\) with respect to \(x\) to obtain a correct form of \(\frac{dA}{dx} = 12x - 768x^{-2}\).
* **M1**: For setting \(\frac{dA}{dx} = 0\) and solving to find \(x = 4\).
* **A1**: For finding the minimum area \(A = 288\).
* **M1**: For finding the second derivative \(\frac{d^2A}{dx^2} = 12 + 1536x^{-3}\) (or alternative valid method to test the nature of the stationary point).
* **A1**: For evaluating \(\frac{d^2A}{dx^2}\) at \(x = 4\) to get \(36 > 0\) and concluding it is a minimum.

**(c)**
* **M1**: For substituting \(x = 4\) into \(h = \frac{96}{x^2}\) (or using \(3x^2 h = 288\)).
* **A1**: For finding \(h = 6\text{ cm}\).

**(a)**

The volume \(V\) of the cuboid is given by:
\[V = \text{length} \times \text{width} \times \text{height}\]
\[V = 3x \times x \times h = 3x^2h\]

Given that \(V = 288\text{ cm}^3\):
\[3x^2h = 288 \implies h = \frac{288}{3x^2} = \frac{96}{x^2}\]

The total surface area \(A\) of the closed cuboid is:
\[A = 2(3x \times x) + 2(3x \times h) + 2(x \times h)\]
\[A = 6x^2 + 6xh + 2xh = 6x^2 + 8xh\]

Substituting \(h = \frac{96}{x^2}\) into the expression for \(A\):
\[A = 6x^2 + 8x \left( \frac{96}{x^2} \right)\]
\[A = 6x^2 + \frac{768}{x}\]

**(b)**

Differentiate \(A\) with respect to \(x\):
\[A = 6x^2 + 768x^{-1}\]
\[\frac{dA}{dx} = 12x - 768x^{-2} = 12x - \frac{768}{x^2}\]

To find the stationary points, set
\[\frac{dA}{dx} = 0\]
\[12x - \frac{768}{x^2} = 0 \implies 12x = \frac{768}{x^2}\]
\[12x^3 = 768 \implies x^3 = 64 \implies x = 4\]

Since \(x > 0\) represents a physical dimension, we choose \(x = 4\).

The minimum value of \(A\) is:
\[A = 6(4)^2 + \frac{768}{4} = 96 + 192 = 288\text{ cm}^2\]

To justify that this is indeed a minimum, we find the second derivative:
\[\frac{d^2A}{dx^2} = 12 + 1536x^{-3} = 12 + \frac{1536}{x^3}\]

Substituting \(x = 4\):
\[\frac{d^2A}{dx^2} = 12 + \frac{1536}{64} = 12 + 24 = 36\]

Since \(\frac{d^2A}{dx^2} = 36 > 0\) at \(x = 4\), the surface area is a minimum.

**(c)**

Substituting \(x = 4\) into the expression for \(h\):
\[h = \frac{96}{x^2} = \frac{96}{16} = 6\text{ cm}\]

**(a)**
* **M1**: For a correct expression for the volume of the cuboid in terms of \(x\) and \(h\) set equal to \(288\), i.e., \(3x^2 h = 288\).
* **A1**: For expressing \(h\) correctly in terms of \(x\): \(h = \frac{96}{x^2}\).
* **M1**: For a correct expression for the total surface area \(A\) of the closed cuboid: \(A = 6x^2 + 8xh\).
* **A1**: For substituting \(h = \frac{96}{x^2}\) into their surface area expression and simplifying to show the given formula \(A = 6x^2 + \frac{768}{x}\) (cso).

**(b)**
* **M1**: For differentiating \(A\) with respect to \(x\) to obtain a correct form of \(\frac{dA}{dx} = 12x - 768x^{-2}\).
* **M1**: For setting \(\frac{dA}{dx} = 0\) and solving to find \(x = 4\).
* **A1**: For finding the minimum area \(A = 288\).
* **M1**: For finding the second derivative \(\frac{d^2A}{dx^2} = 12 + 1536x^{-3}\) (or alternative valid method to test the nature of the stationary point).
* **A1**: For evaluating \(\frac{d^2A}{dx^2}\) at \(x = 4\) to get \(36 > 0\) and concluding it is a minimum.

**(c)**
* **M1**: For substituting \(x = 4\) into \(h = \frac{96}{x^2}\) (or using \(3x^2 h = 288\)).
* **A1**: For finding \(h = 6\text{ cm}\).

We use the product rule, \(\frac{\mathrm{d}y}{\mathrm{d}x} = u \frac{\mathrm{d}v}{\mathrm{d}x} + v \frac{\mathrm{d}u}{\mathrm{d}x}\), where:

\(u = (2x - 1)^3\)

\(v = (3x + 2)^4\)

First, we find the derivatives of \(u\) and \(v\) using the chain rule:

\(\frac{\mathrm{d}u}{\mathrm{d}x} = 3(2x - 1)^2 \times 2 = 6(2x - 1)^2\)

\(\frac{\mathrm{d}v}{\mathrm{d}x} = 4(3x + 2)^3 \times 3 = 12(3x + 2)^3\)

Applying the product rule:

\(\frac{\mathrm{d}y}{\mathrm{d}x} = (2x - 1)^3 \cdot 12(3x + 2)^3 + (3x + 2)^4 \cdot 6(2x - 1)^2\)

Factorise out the common factor \(6(2x - 1)^2 (3x + 2)^3\):

\(\frac{\mathrm{d}y}{\mathrm{d}x} = 6(2x - 1)^2 (3x + 2)^3 [2(2x - 1) + (3x + 2)]\)

Simplify the expression inside the square brackets:

\(2(2x - 1) + (3x + 2) = 4x - 2 + 3x + 2 = 7x\)

Multiply this back into the factored expression:

\(\frac{\mathrm{d}y}{\mathrm{d}x} = 6(2x - 1)^2 (3x + 2)^3 (7x) = 42x(2x - 1)^2 (3x + 2)^3\)

Comparing this to the required form, we find \(k = 42\).

- **M1**: Correct application of the chain rule to differentiate either \((2x - 1)^3\) to get \(a(2x - 1)^2\) where \(a \neq 1\), or \((3x + 2)^4\) to get \(b(3x + 2)^3\) where \(b \neq 1\).
- **A1**: Both \(\frac{\mathrm{d}u}{\mathrm{d}x} = 6(2x - 1)^2\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = 12(3x + 2)^3\) correct.
- **M1**: Correct application of the product rule formula \(u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x}\) with their derivatives.
- **M1**: Factorising out at least \((2x-1)^2(3x+2)^3\) from their sum of two terms.
- **A1**: Correctly obtaining \(42x(2x - 1)^2 (3x + 2)^3\) (or identifying \(k = 42\)).

We use the product rule, \(\frac{\mathrm{d}y}{\mathrm{d}x} = u \frac{\mathrm{d}v}{\mathrm{d}x} + v \frac{\mathrm{d}u}{\mathrm{d}x}\), where:

\(u = (2x - 1)^3\)

\(v = (3x + 2)^4\)

First, we find the derivatives of \(u\) and \(v\) using the chain rule:

\(\frac{\mathrm{d}u}{\mathrm{d}x} = 3(2x - 1)^2 \times 2 = 6(2x - 1)^2\)

\(\frac{\mathrm{d}v}{\mathrm{d}x} = 4(3x + 2)^3 \times 3 = 12(3x + 2)^3\)

Applying the product rule:

\(\frac{\mathrm{d}y}{\mathrm{d}x} = (2x - 1)^3 \cdot 12(3x + 2)^3 + (3x + 2)^4 \cdot 6(2x - 1)^2\)

Factorise out the common factor \(6(2x - 1)^2 (3x + 2)^3\):

\(\frac{\mathrm{d}y}{\mathrm{d}x} = 6(2x - 1)^2 (3x + 2)^3 [2(2x - 1) + (3x + 2)]\)

Simplify the expression inside the square brackets:

\(2(2x - 1) + (3x + 2) = 4x - 2 + 3x + 2 = 7x\)

Multiply this back into the factored expression:

\(\frac{\mathrm{d}y}{\mathrm{d}x} = 6(2x - 1)^2 (3x + 2)^3 (7x) = 42x(2x - 1)^2 (3x + 2)^3\)

Comparing this to the required form, we find \(k = 42\).

- **M1**: Correct application of the chain rule to differentiate either \((2x - 1)^3\) to get \(a(2x - 1)^2\) where \(a \neq 1\), or \((3x + 2)^4\) to get \(b(3x + 2)^3\) where \(b \neq 1\).
- **A1**: Both \(\frac{\mathrm{d}u}{\mathrm{d}x} = 6(2x - 1)^2\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = 12(3x + 2)^3\) correct.
- **M1**: Correct application of the product rule formula \(u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x}\) with their derivatives.
- **M1**: Factorising out at least \((2x-1)^2(3x+2)^3\) from their sum of two terms.
- **A1**: Correctly obtaining \(42x(2x - 1)^2 (3x + 2)^3\) (or identifying \(k = 42\)).

From the first equation:
\(\log_y x - 6\log_x y = 1\)

Using the change of base formula, we can write \(\log_x y = \frac{1}{\log_y x}\).
Let \(u = \log_y x\). The equation becomes:
\(u - \frac{6}{u} = 1\)

Multiplying both sides by \(u\) (since \(u \neq 0\)):
\(u^2 - u - 6 = 0\)

Factoring the quadratic equation:
\((u - 3)(u + 2) = 0\)

This gives:
\(u = 3\) or \(u = -2\)

Therefore:
\(\log_y x = 3 \implies x = y^3\)
or
\(\log_y x = -2 \implies x = y^{-2}\)

From the second equation:
\(\log_2(x - 2y) - \log_2 y = 3\)

Using the subtraction law of logarithms:
\(\log_2\left(\frac{x - 2y}{y}\right) = 3\)

Converting to exponential form:
\[\frac{x - 2y}{y} = 2^3 = 8\]

Since \(y > 0\) must hold for the logarithms to be defined, we can multiply both sides by \(y\):
\(x - 2y = 8y \implies x = 10y\)

Now we substitute each of the cases from the first equation into this linear relation:

**Case 1:** \(x = y^3\)
\(y^3 = 10y\)

Since \(y > 0\) as it is the base of a logarithm, we can divide both sides by \(y\):
\(y^2 = 10 \implies y = \sqrt{10}\) (rejecting the negative root as \(y > 0\)).
Substituting this back into the linear relation gives:
\(x = 10\sqrt{10}\)

**Case 2:** \(x = y^{-2}\)
\(y^{-2} = 10y\)

Multiplying both sides by \(y^2\):
\(1 = 10y^3 \implies y^3 = \frac{1}{10} \implies y = 10^{-1/3}\)
Substituting this back into the linear relation gives:
\(x = 10 \cdot 10^{-1/3} = 10^{2/3}\)

Both pairs of solutions satisfy the constraints of the original logarithmic expressions (\(x > 0, x \neq 1, y > 0, y \neq 1, x - 2y > 0\)).

Thus, the exact solutions are:
\(x = 10\sqrt{10}, y = \sqrt{10}\) and \(x = 10^{2/3}, y = 10^{-1/3}\)

- **M1**: Uses the change of base rule \(\log_x y = \frac{1}{\log_y x}\) to write Equation 1 in terms of a single logarithmic variable.
- **M1**: Obtains and attempts to solve a three-term quadratic equation, e.g., \(u^2 - u - 6 = 0\).
- **A1**: Correctly finds the values \(\log_y x = 3\) and \(\log_y x = -2\) (or equivalent values if using \(\log_x y\)).
- **A1**: Translates these values into the correct power relations, i.e., \(x = y^3\) and \(x = y^{-2}\) (or equivalent).
- **M1**: Applies the subtraction law of logarithms to Equation 2 to obtain \(\log_2\left(\frac{x-2y}{y}\right) = 3\).
- **A1**: Converts the logarithmic equation to linear form correctly to get \(\frac{x-2y}{y} = 8\) and simplifies to \(x = 10y\).
- **M1**: Substitutes at least one of the power relations into the linear relation to form an equation in one variable and attempts to solve.
- **A1**: Obtains one correct pair of exact solutions: \(x = 10\sqrt{10}\) and \(y = \sqrt{10}\) (or equivalent index form).
- **A1**: Obtains the second correct pair of exact solutions: \(x = 10^{2/3}\) and \(y = 10^{-1/3}\) (or equivalent index form), and rejects any invalid roots (e.g., \(y = -\sqrt{10}\)).

(a) To find the y-intercept, set \( x = 0 \):
\( y = \frac{2(0) - 8}{0 + 3} = -\frac{8}{3} \)
So the point is \( (0, -\frac{8}{3}) \).

To find the x-intercept, set \( y = 0 \):
\( 2x - 8 = 0 \implies x = 4 \)
So the point is \( (4, 0) \).

(b) The vertical asymptote occurs where the denominator is zero:
\( x + 3 = 0 \implies x = -3 \)
The horizontal asymptote occurs as \( x \to \pm\infty \):
\( y = \lim_{x \to \infty} \frac{2 - \frac{8}{x}}{1 + \frac{3}{x}} = 2 \)
So the horizontal asymptote is \( y = 2 \).

(c) Differentiating \( y = \frac{2x - 8}{x + 3} \) using the quotient rule:
Let \( u = 2x - 8 \implies u' = 2 \)
Let \( v = x + 3 \implies v' = 1 \)
\( \frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{2(x + 3) - (2x - 8)(1)}{(x + 3)^2} = \frac{2x + 6 - 2x + 8}{(x + 3)^2} = \frac{14}{(x + 3)^2} \)
At \( x = 1 \):
\( \frac{dy}{dx} = \frac{14}{(1 + 3)^2} = \frac{14}{16} = \frac{7}{8} \).

(d) The sketch should show:
1. Two branches of a rectangular hyperbola located in the second and fourth quadrants relative to the asymptotes.
2. The vertical asymptote at \( x = -3 \) and horizontal asymptote at \( y = 2 \) as dashed lines.
3. The intercepts clearly marked at \( (4, 0) \) and \( (0, -\frac{8}{3}) \).

(e) To find the points of intersection, solve:
\( \frac{2x - 8}{x + 3} = x - 4 \)
\( 2x - 8 = (x - 4)(x + 3) \)
\( 2x - 8 = x^2 - x - 12 \)
\( x^2 - 3x - 4 = 0 \)
\( (x - 4)(x + 1) = 0 \)
So, \( x = 4 \) or \( x = -1 \).
Substituting back into the line equation \( y = x - 4 \):
If \( x = 4 \), \( y = 4 - 4 = 0 \implies (4, 0) \)
If \( x = -1 \), \( y = -1 - 4 = -5 \implies (-1, -5) \).
Thus, the points of intersection are \( (4, 0) \) and \( (-1, -5) \).

(a)
- M1: Attempting to set \( x = 0 \) or \( y = 0 \) to find an intercept.
- A1: Correct y-intercept \( (0, -\frac{8}{3}) \) or equivalent.
- A1: Correct x-intercept \( (4, 0) \).

(b)
- B1: For the vertical asymptote \( x = -3 \).
- B1: For the horizontal asymptote \( y = 2 \).

(c)
- M1: Correct application of the quotient rule (or product rule) with at least one term in the numerator correct and the denominator squared.
- A1: Correct simplified derivative \( \frac{14}{(x + 3)^2} \).
- M1: Substituting \( x = 1 \) into their derivative.
- A1: Correct gradient \( \frac{7}{8} \) or equivalent fraction/decimal.

(d)
- B1: Correct shape of two branches of the rational function.
- B1: Both asymptotes correctly drawn as dashed lines and labelled with their equations.
- B1: Correct x-intercept labelled on the sketch.
- B1: Correct y-intercept labelled on the sketch.

(e)
- M1: Setting the curve equation equal to the line equation and clearing the denominator.
- A1: Obtaining the correct quadratic equation \( x^2 - 3x - 4 = 0 \) or equivalent.
- M1: Factoring or using the quadratic formula to solve their quadratic equation.
- A1: Correct x-coordinates \( x = 4 \) and \( x = -1 \).
- A1: Correct points of intersection \( (4, 0) \) and \( (-1, -5) \).

(a) To find the y-intercept, set \( x = 0 \):
\( y = \frac{2(0) - 8}{0 + 3} = -\frac{8}{3} \)
So the point is \( (0, -\frac{8}{3}) \).

To find the x-intercept, set \( y = 0 \):
\( 2x - 8 = 0 \implies x = 4 \)
So the point is \( (4, 0) \).

(b) The vertical asymptote occurs where the denominator is zero:
\( x + 3 = 0 \implies x = -3 \)
The horizontal asymptote occurs as \( x \to \pm\infty \):
\( y = \lim_{x \to \infty} \frac{2 - \frac{8}{x}}{1 + \frac{3}{x}} = 2 \)
So the horizontal asymptote is \( y = 2 \).

(c) Differentiating \( y = \frac{2x - 8}{x + 3} \) using the quotient rule:
Let \( u = 2x - 8 \implies u' = 2 \)
Let \( v = x + 3 \implies v' = 1 \)
\( \frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{2(x + 3) - (2x - 8)(1)}{(x + 3)^2} = \frac{2x + 6 - 2x + 8}{(x + 3)^2} = \frac{14}{(x + 3)^2} \)
At \( x = 1 \):
\( \frac{dy}{dx} = \frac{14}{(1 + 3)^2} = \frac{14}{16} = \frac{7}{8} \).

(d) The sketch should show:
1. Two branches of a rectangular hyperbola located in the second and fourth quadrants relative to the asymptotes.
2. The vertical asymptote at \( x = -3 \) and horizontal asymptote at \( y = 2 \) as dashed lines.
3. The intercepts clearly marked at \( (4, 0) \) and \( (0, -\frac{8}{3}) \).

(e) To find the points of intersection, solve:
\( \frac{2x - 8}{x + 3} = x - 4 \)
\( 2x - 8 = (x - 4)(x + 3) \)
\( 2x - 8 = x^2 - x - 12 \)
\( x^2 - 3x - 4 = 0 \)
\( (x - 4)(x + 1) = 0 \)
So, \( x = 4 \) or \( x = -1 \).
Substituting back into the line equation \( y = x - 4 \):
If \( x = 4 \), \( y = 4 - 4 = 0 \implies (4, 0) \)
If \( x = -1 \), \( y = -1 - 4 = -5 \implies (-1, -5) \).
Thus, the points of intersection are \( (4, 0) \) and \( (-1, -5) \).

(a)
- M1: Attempting to set \( x = 0 \) or \( y = 0 \) to find an intercept.
- A1: Correct y-intercept \( (0, -\frac{8}{3}) \) or equivalent.
- A1: Correct x-intercept \( (4, 0) \).

(b)
- B1: For the vertical asymptote \( x = -3 \).
- B1: For the horizontal asymptote \( y = 2 \).

(c)
- M1: Correct application of the quotient rule (or product rule) with at least one term in the numerator correct and the denominator squared.
- A1: Correct simplified derivative \( \frac{14}{(x + 3)^2} \).
- M1: Substituting \( x = 1 \) into their derivative.
- A1: Correct gradient \( \frac{7}{8} \) or equivalent fraction/decimal.

(d)
- B1: Correct shape of two branches of the rational function.
- B1: Both asymptotes correctly drawn as dashed lines and labelled with their equations.
- B1: Correct x-intercept labelled on the sketch.
- B1: Correct y-intercept labelled on the sketch.

(e)
- M1: Setting the curve equation equal to the line equation and clearing the denominator.
- A1: Obtaining the correct quadratic equation \( x^2 - 3x - 4 = 0 \) or equivalent.
- M1: Factoring or using the quadratic formula to solve their quadratic equation.
- A1: Correct x-coordinates \( x = 4 \) and \( x = -1 \).
- A1: Correct points of intersection \( (4, 0) \) and \( (-1, -5) \).

(a) First, find the \(y\)-coordinate of \(P\) by substituting \(x = 0\) into the equation of the curve:
\(y = 2e^{-\frac{1}{2}(0)} + 1 = 2(1) + 1 = 3\)
So, \(P\) has coordinates \((0, 3)\).

Next, differentiate \(y = 2e^{-\frac{1}{2}x} + 1\) with respect to \(x\):
\(\frac{dy}{dx} = 2 \left(-\frac{1}{2}\right) e^{-\frac{1}{2}x} = -e^{-\frac{1}{2}x}\)

At the point \(P\) where \(x = 0\), the gradient of the tangent to the curve is:
\(m_t = -e^{0} = -1\)

The gradient of the normal to the curve at \(P\) is the negative reciprocal of the tangent's gradient:
\(m_n = -\frac{1}{-1} = 1\)

Now, find the equation of the normal line using the point \(P(0, 3)\) and gradient \(m_n = 1\):
\(y - 3 = 1(x - 0) \implies y = x + 3\)

(b) The normal line intersects the \(x\)-axis where \(y = 0\):
\(0 = x + 3 \implies x = -3\)
Thus, the coordinates of \(Q\) are \((-3, 0)\).

(c) The region \(R\) is bounded by the line \(PQ\), the curve \(C\), the \(x\)-axis, and the line \(x = \ln 4\). We can split the area of \(R\) into two distinct parts at the \(y\)-axis (where \(x = 0\)):

1. For \(-3 \le x \le 0\):
The region is bounded by the normal line \(y = x + 3\) and the \(x\)-axis. This forms a right-angled triangle with vertices \((-3, 0)\), \((0, 0)\), and \((0, 3)\).
\(\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}\)

2. For \(0 \le x \le \ln 4\):
The region is bounded by the curve \(y = 2e^{-\frac{1}{2}x} + 1\) and the \(x\)-axis.
\(\text{Area}_2 = \int_{0}^{\ln 4} \left(2e^{-\frac{1}{2}x} + 1\right) dx\)

Integrating the expression:
\(\int \left(2e^{-\frac{1}{2}x} + 1\right) dx = \left[ -4e^{-\frac{1}{2}x} + x \right]_{0}^{\ln 4}\)

Substitute the upper limit \(x = \ln 4\):
\(-4e^{-\frac{1}{2}\ln 4} + \ln 4 = -4\left(e^{\ln 4}\right)^{-\frac{1}{2}} + \ln 4 = -4(4)^{-\frac{1}{2}} + \ln 4 = -4\left(\frac{1}{2}\right) + \ln 4 = -2 + \ln 4\)

Substitute the lower limit \(x = 0\):
\(-4e^{0} + 0 = -4\)

Subtract the lower limit from the upper limit:
\(\text{Area}_2 = (-2 + \ln 4) - (-4) = 2 + \ln 4\)

To find the total area of \(R\), sum \(\text{Area}_1\) and \(\text{Area}_2\):
\(\text{Area} = \frac{9}{2} + 2 + \ln 4 = \frac{13}{2} + \ln 4\)

This is in the form \(a + \ln b\), where \(a = \frac{13}{2}\) (a rational number) and \(b = 4\) (an integer).

Part (a):
- M1: For substituting \(x = 0\) into the curve's equation to find the \(y\)-coordinate of \(P\) (i.e., \(y = 3\) or coordinates \((0,3)\)).
- M1: For differentiating the exponential function correctly to obtain \(\frac{dy}{dx} = -e^{-\frac{1}{2}x}\).
- M1: For calculating the gradient of the normal by taking the negative reciprocal of the tangent gradient at \(x = 0\) (normal gradient \(m = 1\)).
- A1: For the correct equation of the normal, e.g., \(y = x + 3\) (or equivalent form).

Part (b):
- B1: For setting \(y = 0\) in their equation of the normal and showing that \(x = -3\) to conclude \(Q(-3, 0)\).

Part (c):
- M1: For recognizing the need to split the area into two regions at \(x = 0\) (or setting up the sum of two integrals: \(\int_{-3}^{0} (x+3)dx + \int_{0}^{\ln 4} (2e^{-\frac{1}{2}x} + 1)dx\)).
- A1: For obtaining the correct area of the triangular section as \(\frac{9}{2}\) (or \(4.5\)).
- M1: For integrating \(2e^{-\frac{1}{2}x} + 1\) to get an expression of the form \(ke^{-\frac{1}{2}x} + x\) where \(k \neq 0\).
- A1: For the correct integrated expression \(-4e^{-\frac{1}{2}x} + x\).
- M1: For substituting both limits \(\ln 4\) and \(0\) into their integrated expression.
- A1: For obtaining \(2 + \ln 4\) (or \(2 + 2\ln 2\)) for the area under the curve.
- A1: For adding both areas to obtain the correct final exact value of \(\frac{13}{2} + \ln 4\) (or \(6.5 + \ln 4\)).