MetadataPastPaper.sampleTitle

To find the stationary points, we first differentiate \( y \) with respect to \( x \):

\( \frac{dy}{dx} = x - 3 + \frac{2}{x} \)

Setting \( \frac{dy}{dx} = 0 \):

\( x - 3 + \frac{2}{x} = 0 \)

Multiply the entire equation by \( x \) (since \( x > 0 \)):

\( x^2 - 3x + 2 = 0 \)

Factorise the quadratic equation:

\( (x - 1)(x - 2) = 0 \)

This gives \( x = 1 \) and \( x = 2 \).

Now find the corresponding exact \( y \)-coordinates:

For \( x = 1 \):

\( y = \frac{1}{2}(1)^2 - 3(1) + 2\ln(1) = 0.5 - 3 + 0 = -2.5 \)

So the first stationary point is \( (1, -2.5) \).

For \( x = 2 \):

\( y = \frac{1}{2}(2)^2 - 3(2) + 2\ln(2) = 2 - 6 + 2\ln 2 = 2\ln 2 - 4 \)

So the second stationary point is \( (2, 2\ln 2 - 4) \).

To determine the nature of these points, find the second derivative:

\( \frac{d^2y}{dx^2} = 1 - \frac{2}{x^2} \)

Evaluate the second derivative at each stationary point:

At \( x = 1 \):

\( \frac{d^2y}{dx^2} = 1 - \frac{2}{1^2} = -1 \)

Since \( \frac{d^2y}{dx^2} < 0 \), the point \( (1, -2.5) \) is a local maximum.

At \( x = 2 \):

\( \frac{d^2y}{dx^2} = 1 - \frac{2}{2^2} = 1 - 0.5 = 0.5 \)

Since \( \frac{d^2y}{dx^2} > 0 \), the point \( (2, 2\ln 2 - 4) \) is a local minimum.

M1: For attempting to differentiate the function, with at least two terms differentiated correctly.
A1: For obtaining the correct first derivative: \( \frac{dy}{dx} = x - 3 + \frac{2}{x} \).
M1: For setting their \( \frac{dy}{dx} = 0 \), forming a quadratic equation, and solving for \( x \) to get \( x = 1 \) and \( x = 2 \).
A1: For finding the correct exact coordinates of both stationary points: \( (1, -2.5) \) (or \( (1, -\frac{5}{2}) \)) and \( (2, 2\ln 2 - 4) \) (or \( (2, \ln 4 - 4) \)).
M1: For finding \( \frac{d^2y}{dx^2} = 1 - \frac{2}{x^2} \) and substituting at least one of their \( x \)-values (or using an alternative valid method, e.g., first derivative test).
A1: For correctly identifying both stationary points with their nature based on valid reasoning: \( (1, -2.5) \) is a local maximum and \( (2, 2\ln 2 - 4) \) is a local minimum.

To sketch the graph of \(y = 3\sin(2x) + 1\) for the domain \(0^\circ \le x \le 180^\circ\):

1. **Determine the period:**
The coefficient of \(x\) is 2, so the period is \(\frac{360^\circ}{2} = 180^\circ\). This means exactly one full cycle of the sine wave is completed within the given domain.

2. **Determine the amplitude and midline:**
- The midline (vertical shift) is \(y = 1\).
- The amplitude is 3, so the curve oscillates 3 units above and below the midline.
- Maximum value: \(1 + 3 = 4\).
- Minimum value: \(1 - 3 = -2\).

3. **Identify key points:**
- **\(y\)-intercept:** When \(x = 0^\circ\), \(y = 3\sin(0) + 1 = 1\). Key point is \((0, 1)\).
- **Maximum point:** The first peak of a standard sine curve occurs a quarter of the way through the cycle. \(x = \frac{180^\circ}{4} = 45^\circ\). At \(x = 45^\circ\), \(y = 3\sin(90^\circ) + 1 = 4\). Key point is \((45^\circ, 4)\).
- **Midline intersection:** The curve returns to the midline halfway through the cycle. \(x = 90^\circ\), \(y = 1\). Key point is \((90^\circ, 1)\).
- **Minimum point:** The trough occurs three-quarters of the way through the cycle. \(x = 135^\circ\). At \(x = 135^\circ\), \(y = 3\sin(270^\circ) + 1 = -2\). Key point is \((135^\circ, -2)\).
- **Endpoint:** At \(x = 180^\circ\), \(y = 3\sin(360^\circ) + 1 = 1\). Key point is \((180^\circ, 1)\).

4. **Sketching the graph:**
Plot these five key points on your axes and connect them with a smooth, continuous wave shape.

- **B1**: For a correctly shaped sine curve showing exactly one complete period from \(0^\circ\) to \(180^\circ\).
- **B1**: For indicating the correct \(y\)-intercept at \((0, 1)\) and endpoint at \((180^\circ, 1)\).
- **B1**: For indicating the correct maximum point at \((45^\circ, 4)\).
- **B1**: For indicating the correct minimum point at \((135^\circ, -2)\).

To find the set of values of \(k\) for which the line and curve do not intersect, we set their equations equal: \(kx - 3 = x^2 + 5x + k\). Rearranging into the standard quadratic form gives \(x^2 + (5 - k)x + (k + 3) = 0\). Since the line does not intersect the curve, the quadratic equation has no real roots, meaning the discriminant must be negative: \(b^2 - 4ac < 0\). Substituting the coefficients gives \((5 - k)^2 - 4(1)(k + 3) < 0\). Expanding and simplifying yields \(25 - 10k + k^2 - 4k - 12 < 0\), which simplifies to \(k^2 - 14k + 13 < 0\). Factorising this quadratic inequality gives \((k - 1)(k - 13) < 0\). Thus, the critical values are \(k = 1\) and \(k = 13\). Since the inequality is less than zero, the solution set is \(1 < k < 13\).

M1: Equating the line and curve and attempting to rearrange into a 3-term quadratic in terms of x. M1: Attempting to use the discriminant \(b^2 - 4ac < 0\) with their coefficients. A1: Obtaining the simplified quadratic inequality \(k^2 - 14k + 13 < 0\) or identifying the critical values 1 and 13. A1: Correct final range \(1 < k < 13\) (or equivalent interval notation).

From the first equation, we apply the addition rule of logarithms: \(\log_3 (xy) = 3\). Converting to exponential form gives: \(xy = 3^3 = 27\), which means \(x = \frac{27}{y}\). From the second equation, we use the change of base formula: \(\log_9 x = \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2}\). Substituting this back into the second equation gives: \(\frac{1}{2}\log_3 x - \log_3 (y - 2) = 1\). Multiplying the entire equation by 2 yields: \(\log_3 x - 2\log_3 (y - 2) = 2\). Applying the power and subtraction rules of logarithms: \(\log_3 \left(\frac{x}{(y-2)^2}\right) = 2\). Converting to exponential form gives: \(\frac{x}{(y-2)^2} = 3^2 = 9\), which means \(x = 9(y-2)^2\). Now we substitute \(x = \frac{27}{y}\) into this equation: \(\frac{27}{y} = 9(y-2)^2\). Dividing by 9: \(\frac{3}{y} = (y-2)^2\). Multiplying by \(y\) and expanding: \(3 = y(y^2 - 4y + 4)\), which simplifies to the cubic equation: \(y^3 - 4y^2 + 4y - 3 = 0\). By testing integer factors of 3, we find that \(y = 3\) is a root, because \(3^3 - 4(3)^2 + 4(3) - 3 = 27 - 36 + 12 - 3 = 0\). Factorising the cubic gives: \((y-3)(y^2 - y + 1) = 0\). The quadratic factor \(y^2 - y + 1 = 0\) has discriminant \((-1)^2 - 4(1)(1) = -3 < 0\), so there are no other real solutions. Substituting \(y = 3\) back to find \(x\): \(x = \frac{27}{3} = 9\). Checking both values in the original equations confirms they are valid since all logarithmic arguments are positive. Therefore, the solution is \(x = 9\) and \(y = 3\).

M1: For applying the addition rule to obtain \(xy = 27\) or equivalent. M1: For using the change of base formula to express \(\log_9 x\) as \(\frac{\log_3 x}{2}\). M1: For correctly removing logarithms from the second equation to obtain \(x = 9(y-2)^2\) or equivalent. M1: For substituting to form a cubic equation in terms of \(y\) (or \(x\)). A1: For obtaining the correct simplified cubic equation \(y^3 - 4y^2 + 4y - 3 = 0\) (or equivalent). M1: For solving the cubic equation to find \(y = 3\) and showing the quadratic factor has no real roots. A1: For the final correct pair \(x = 9\), \(y = 3\).

First, we use the factor theorem. Since \( x - 1 \) is a factor of \( p(x) \), we must have \( p(1) = 0 \): \( p(1) = 2(1)^3 + a(1)^2 + b(1) + c = 0 \) which simplifies to \( a + b + c = -2 \) (Equation 1). Next, the curve \( y = p(x) \) has a stationary point at \( (2, -3) \). This gives us two pieces of information. First, the point \( (2, -3) \) lies on the curve, so \( p(2) = -3 \): \( p(2) = 2(2)^3 + a(2)^2 + b(2) + c = -3 \) which simplifies to \( 4a + 2b + c = -19 \) (Equation 2). Second, the derivative at \( x = 2 \) is zero, so \( p'(2) = 0 \). We find the derivative: \( p'(x) = 6x^2 + 2ax + b \). Substituting \( x = 2 \) gives \( p'(2) = 6(2)^2 + 2a(2) + b = 0 \) which simplifies to \( 4a + b = -24 \) (Equation 3). Now we solve this system of equations. Subtracting Equation 1 from Equation 2 gives \( 3a + b = -17 \) (Equation 4). Subtracting Equation 4 from Equation 3 gives \( a = -7 \). Substituting \( a = -7 \) into Equation 4 gives \( 3(-7) + b = -17 \), which simplifies to \( b = 4 \). Finally, substituting \( a = -7 \) and \( b = 4 \) into Equation 1 gives \( -7 + 4 + c = -2 \), which simplifies to \( c = 1 \). Thus, the values of the constants are \( a = -7 \), \( b = 4 \), and \( c = 1 \).

M1: Attempts to use the factor theorem by setting \( p(1) = 0 \) to obtain a linear equation in \( a \), \( b \), and \( c \) (e.g., \( a + b + c = -2 \)). M1: Correctly differentiates \( p(x) \) to obtain \( p'(x) = 6x^2 + 2ax + b \). M1: Uses the stationary point condition \( p'(2) = 0 \) to obtain a second linear equation (e.g., \( 4a + b = -24 \)). M1: Uses the point coordinates \( p(2) = -3 \) to obtain a third linear equation (e.g., \( 4a + 2b + c = -19 \)). M1: Demonstrates a valid method for solving the simultaneous equations to find at least one of the constants. A1: Obtains the correct values: \( a = -7 \), \( b = 4 \), and \( c = 1 \) (all three must be correct).

Part (a): Multiply the equation \(y = ax^2 + \frac{b}{x}\) by \(x\) to obtain \(xy = ax^3 + b\). Let \(Y = xy\) and \(X = x^3\). The equation becomes the linear form \(Y = aX + b\), where the gradient is \(a\) and the vertical intercept is \(b\). Using the coordinates \((X_1, Y_1) = (2, 5)\) and \((X_2, Y_2) = (8, 23)\): \(a = \frac{23 - 5}{8 - 2} = \frac{18}{6} = 3\). Substitute \(a = 3\) and the point \((2, 5)\) into \(Y = aX + b\): \(5 = 3(2) + b \implies b = -1\). Thus, \(a = 3\) and \(b = -1\). Part (b): The relation between the variables is \(y = 3x^2 - \frac{1}{x}\). Substituting \(x = 2\) into this equation: \(y = 3(2)^2 - \frac{1}{2} = 12 - 0.5 = 11.5\). Part (c): Substitute \(y = 2x^2\) into the equation: \(2x^2 = 3x^2 - \frac{1}{x} \implies x^2 - \frac{1}{x} = 0 \implies x^2 = \frac{1}{x} \implies x^3 = 1\). Solving for the real value of \(x\) gives \(x = 1\).

Part (a): [4 marks] M1: For multiplying the given equation by \(x\) to get \(xy = ax^3 + b\). M1: For calculating the gradient, \(a = \frac{23 - 5}{8 - 2}\). A1: For obtaining \(a = 3\). A1: For obtaining \(b = -1\) by substituting a point into the linear form. Part (b): [2 marks] M1: For substituting \(x = 2\) and their values of \(a\) and \(b\) into the equation. A1: For \(y = 11.5\) (or equivalent fraction). Part (c): [3 marks] M1: For setting their \(2x^2 = 3x^2 - \frac{1}{x}\). M1: For simplifying the equation to obtain \(x^3 = 1\). A1: For \(x = 1\).

(a) To differentiate \( y = x(4x+1)^{1/2} \), we use the product rule: \( \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \) where \( u = x \) and \( v = (4x+1)^{1/2} \). This gives \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = \frac{1}{2}(4x+1)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4x+1}} \). Substituting these in, we get \( \frac{dy}{dx} = \frac{2x}{\sqrt{4x+1}} + \sqrt{4x+1} \). Expressing as a single fraction over the common denominator \( \sqrt{4x+1} \), we have \( \frac{dy}{dx} = \frac{2x + (4x+1)}{\sqrt{4x+1}} = \frac{6x+1}{\sqrt{4x+1}} \). (b) From part (a), we know that \( \int \frac{6x+1}{\sqrt{4x+1}} \, dx = x\sqrt{4x+1} + C_1 \). We can split the integral: \( \int \frac{6x}{\sqrt{4x+1}} \, dx + \int \frac{1}{\sqrt{4x+1}} \, dx = x\sqrt{4x+1} + C_1 \). The second integral is \( \int (4x+1)^{-1/2} \, dx = \frac{(4x+1)^{1/2}}{1/2 \cdot 4} = \frac{1}{2}\sqrt{4x+1} \). Substituting this back gives \( 6\int \frac{x}{\sqrt{4x+1}} \, dx + \frac{1}{2}\sqrt{4x+1} = x\sqrt{4x+1} + C_1 \). Rearranging for the required integral: \( 6\int \frac{x}{\sqrt{4x+1}} \, dx = x\sqrt{4x+1} - \frac{1}{2}\sqrt{4x+1} + C_1 = \left(x - \frac{1}{2}\right)\sqrt{4x+1} + C_1 = \frac{2x-1}{2}\sqrt{4x+1} + C_1 \). Dividing by 6, we get \( \int \frac{x}{\sqrt{4x+1}} \, dx = \frac{(2x-1)\sqrt{4x+1}}{12} + C \).

Part (a): M1: For attempting product rule differentiation of \( x(4x+1)^{1/2} \). A1: For obtaining \( \frac{2x}{\sqrt{4x+1}} + \sqrt{4x+1} \) or equivalent form. A1: For simplifying to the single fraction \( \frac{6x+1}{\sqrt{4x+1}} \). Part (b): M1: For relating the integral of their result in (a) back to \( x\sqrt{4x+1} \). M1: For correctly integrating \( (4x+1)^{-1/2} \) to obtain \( \frac{1}{2}\sqrt{4x+1} \). A1: For the final correct expression \( \frac{(2x-1)\sqrt{4x+1}}{12} + C \) (accept any equivalent simplified form; must include arbitrary constant of integration \( C \)).

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

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

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

Using the point-slope form with the midpoint \(M(4, 3)\), the equation of the perpendicular bisector is:
\(y - 3 = \frac{1}{2}(x - 4)\)
\(y = \frac{1}{2}x + 1\) (or \(x - 2y + 2 = 0\))

**(b)**
To find the coordinates of \(C\), solve the equation of the perpendicular bisector and the line \(L\) simultaneously:
1) \(y = \frac{1}{2}x + 1\)
2) \(2x - 3y = 5\)

Substitute equation (1) into (2):
\(2x - 3\left(\frac{1}{2}x + 1\right) = 5\)
\(2x - \frac{3}{2}x - 3 = 5\)
\(\frac{1}{2}x = 8 \implies x = 16\)

Substitute \(x = 16\) back into equation (1):
\(y = \frac{1}{2}(16) + 1 = 9\)

Thus, the coordinates of \(C\) are \((16, 9)\).

**(c)**
Since \(D(x, y)\) lies on the line \(L\):
\(2x - 3y = 5 \implies y = \frac{2x - 5}{3}\)

The distance \(CD = 5\sqrt{13}\), so \(CD^2 = 325\):
\((x - 16)^2 + (y - 9)^2 = 325\)

Substitute \(y = \frac{2x - 5}{3}\) into the distance equation:
\((x - 16)^2 + \left(\frac{2x - 5}{3} - 9\right)^2 = 325\)
\((x - 16)^2 + \left(\frac{2x - 32}{3}\right)^2 = 325\n\)(x - 16)^2 + \frac{4}{9}(x - 16)^2 = 325\)
\(\frac{13}{9}(x - 16)^2 = 325\)
\((x - 16)^2 = \frac{325 \times 9}{13} = 225\)
\(x - 16 = \pm 15\)

This gives two possible values for \(x\):
\(x = 31\) or \(x = 1\)

Since the problem states that the \(x\)-coordinate of \(D\) is less than 10, we select:
\(x = 1\)

Substitute \(x = 1\) to find the corresponding \(y\)-coordinate:
\(y = \frac{2(1) - 5}{3} = -1\)

Thus, the coordinates of \(D\) are \((1, -1)\).

**(d)**
We can find the area of the quadrilateral \(ACBD\) by using the shoelace formula with vertices ordered counter-clockwise or clockwise: \(A(2, 7)\), \(C(16, 9)\), \(B(6, -1)\), and \(D(1, -1)\):
\(\text{Area} = \frac{1}{2} | (2 \times 9 + 16 \times (-1) + 6 \times (-1) + 1 \times 7) - (7 \times 16 + 9 \times 6 + (-1) \times 1 + (-1) \times 2) |\)
\(\text{Area} = \frac{1}{2} | (18 - 16 - 6 + 7) - (112 + 54 - 1 - 2) |\)
\(\text{Area} = \frac{1}{2} | 3 - 163 | = \frac{1}{2} | -160 | = 80\)

Alternatively, we can split the quadrilateral into two triangles, \(\triangle ACB\) and \(\triangle ADB\):
- Since \(C\) lies on the perpendicular bisector of \(AB\), the altitude of \(\triangle ACB\) from \(C\) is the distance from \(C(16, 9)\) to the midpoint \(M(4, 3)\):
\(MC = \sqrt{(16-4)^2 + (9-3)^2} = \sqrt{144 + 36} = \sqrt{180}\)
The base length \(AB = \sqrt{(6-2)^2 + (-1-7)^2} = \sqrt{16 + 64} = \sqrt{80}\)
\(\text{Area}(\triangle ACB) = \frac{1}{2} \times \sqrt{80} \times \sqrt{180} = \frac{1}{2} \sqrt{14400} = 60\)

- For \(\triangle ADB\), with vertices \(A(2,7)\), \(B(6,-1)\), and \(D(1,-1)\):
\(\text{Area}(\triangle ADB) = \frac{1}{2} | 1(7 - (-1)) + 2(-1 - (-1)) + 6(-1 - 7) | = \frac{1}{2} | 8 + 0 - 48 | = 20\)

Total Area \(= 60 + 20 = 80\).

**(a)**
- **M1**: Correct method to find the midpoint of \(AB\).
- **M1**: Correct method to find the gradient of \(AB\).
- **A1**: Correct gradient of the perpendicular bisector (\(m = \frac{1}{2}\)).
- **A1**: Correct equation of the perpendicular bisector in any simplified form (e.g., \(y = \frac{1}{2}x + 1\) or \(x - 2y + 2 = 0\)).

**(b)**
- **M1**: Sensible attempt to solve the equations simultaneously (e.g., substituting \(y = \frac{1}{2}x + 1\) into \(2x - 3y = 5\)).
- **A1**: Correct \(x\)-coordinate (\(x = 16\)).
- **A1**: Correct \(y\)-coordinate (\(y = 9\)).

**(c)**
- **M1**: Setting up the distance equation using coordinates, e.g., \((x - 16)^2 + (y - 9)^2 = 325\).
- **M1**: Correctly substituting \(y = \frac{2x - 5}{3}\) and solving the quadratic equation to obtain \(x = 1\) or \(x = 31\).
- **A1**: Selecting \(x = 1\) based on the given constraint and stating the correct coordinates \(D(1, -1)\).

**(d)**
- **M1**: Any valid method to find the area of the quadrilateral \(ACBD\) (e.g., shoelace formula, splitting into two triangles, etc.).
- **A1**: Correct area of \(80\).

(a) Since \(AB\) is parallel to \(OC\) and \(AB = 3OC\), we have \(\vec{AB} = 3\vec{OC} = 3\mathbf{c}\). Therefore, \(\vec{OB} = \vec{OA} + \vec{AB} = \mathbf{a} + 3\mathbf{c}\). For \(\vec{AC}\), we have \(\vec{AC} = \vec{AO} + \vec{OC} = -\mathbf{a} + \mathbf{c} = \mathbf{c} - \mathbf{a}\). (b) Since \(\vec{OP} = \lambda\vec{OB}\), we substitute our expression from part (a): \(\vec{OP} = \lambda(\mathbf{a} + 3\mathbf{c}) = \lambda\mathbf{a} + 3\lambda\mathbf{c}\). (c) Using the vector triangle, \(\vec{OP} = \vec{OA} + \vec{AP} = \mathbf{a} + \mu\vec{AC} = \mathbf{a} + \mu(\mathbf{c} - \mathbf{a}) = (1 - \mu)\mathbf{a} + \mu\mathbf{c}\). (d) Equating the two expressions for \(\vec{OP}\) from parts (b) and (c): \(\lambda\mathbf{a} + 3\lambda\mathbf{c} = (1 - \mu)\mathbf{a} + \mu\mathbf{c}\). Since \(\mathbf{a}\) and \(\mathbf{c}\) are non-parallel vectors, we can equate their coefficients. For \(\mathbf{a}\): \(\lambda = 1 - \mu\). For \(\mathbf{c}\): \(3\lambda = \mu\). Substituting the second equation into the first: \(\lambda = 1 - 3\lambda \implies 4\lambda = 1 \implies \lambda = \frac{1}{4}\). Substituting back to find \(\mu\): \(\mu = 3\left(\frac{1}{4}\right) = \frac{3}{4}\). (e) Since \(\vec{OP} = \frac{1}{4}\vec{OB}\), the length of \(OP\) is one-quarter of the total length of \(OB\). This leaves \(PB\) as three-quarters of the length of \(OB\). Therefore, the ratio \(OP : PB = 1 : 3\).

(a) M1: For attempting to write \(\vec{OB}\) or \(\vec{AC}\) in terms of \(\mathbf{a}\) and \(\mathbf{c}\). A1: For both \(\vec{OB} = \mathbf{a} + 3\mathbf{c}\) and \(\vec{AC} = \mathbf{c} - \mathbf{a}\) correct. (b) B1: For \(\lambda\mathbf{a} + 3\lambda\mathbf{c}\) (allow factored form). (c) M1: For writing \(\vec{OP} = \vec{OA} + \mu\vec{AC}\). A1: For simplifying to \((1 - \mu)\mathbf{a} + \mu\mathbf{c}\). (d) M1: For equating the coefficients of \(\mathbf{a}\) and \(\mathbf{c}\) from the two expressions. A1: For establishing the simultaneous equations \(\lambda = 1 - \mu\) and \(3\lambda = \mu\). M1: For a correct algebraic method to solve their simultaneous equations. A1: For finding \(\lambda = \frac{1}{4}\) and \(\mu = \frac{3}{4}\). (e) B1: For the ratio \(1 : 3\) (or equivalent numerical ratio).

**(a)**
Using the trigonometric identity \(\sec^2 x = 1 + \tan^2 x\):
\(3(1 + \tan^2 x) - 5 \tan x - 5 = 0\)
\(3 \tan^2 x - 5 \tan x - 2 = 0\)

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

This gives two cases:
1. \(\tan x = 2\)
The basic angle is \(\tan^{-1}(2) \approx 63.43^\circ\).
Since tangent is positive in the 1st and 3rd quadrants:
\(x \approx 63.4^\circ\)
\(x \approx 180^\circ + 63.43^\circ = 243.4^\circ\)

2. \(\tan x = -\frac{1}{3}\)
The basic angle is \(\tan^{-1}\left(\frac{1}{3}\right) \approx 18.43^\circ\).
Since tangent is negative in the 2nd and 4th quadrants:
\(x \approx 180^\circ - 18.43^\circ = 161.6^\circ\)
\(x \approx 360^\circ - 18.43^\circ = 341.6^\circ\)

So, the solutions for part (a) are \(x = 63.4^\circ, 161.6^\circ, 243.4^\circ, 341.6^\circ\).

**(b)**
Using the trigonometric identity \(\cot^2 \theta = \operatorname{cosec}^2 \theta - 1\):
\((\operatorname{cosec}^2 \theta - 1) - 3 \operatorname{cosec} \theta - 3 = 0\)
\(\operatorname{cosec}^2 \theta - 3 \operatorname{cosec} \theta - 4 = 0\)

Factorising this quadratic equation in \(\operatorname{cosec} \theta\):
\((\operatorname{cosec} \theta - 4)(\operatorname{cosec} \theta + 1) = 0\)

This gives:
\(\operatorname{cosec} \theta = 4\) or \(\operatorname{cosec} \theta = -1\)

Taking the reciprocal to express in terms of sine:
\(\sin \theta = \frac{1}{4} = 0.25\) or \(\sin \theta = -1\)

1. For \(\sin \theta = 0.25\):
The basic angle in radians is \(\sin^{-1}(0.25) \approx 0.25268\) rad.
Since sine is positive in the 1st and 2nd quadrants:
\(\theta \approx 0.253\) rad
\(\theta \approx \pi - 0.25268 \approx 2.89\) rad

2. For \(\sin \theta = -1\):
In the interval \(0 \le \theta \le 2\pi\):
\(\theta = \frac{3\pi}{2} \approx 4.71\) rad

So, the solutions for part (b) are \(\theta = 0.253\), \(2.89\), \(\frac{3\pi}{2}\) (or \(4.71\)).

**Part (a)**
* **M1**: Uses the identity \(\sec^2 x = 1 + \tan^2 x\) to write the equation in terms of \(\tan x\) only.
* **A1**: For obtaining the correct simplified quadratic equation \(3\tan^2 x - 5\tan x - 2 = 0\) and finding the correct values \(\tan x = 2\) and \(\tan x = -\frac{1}{3}\).
* **A1**: For finding any two correct solutions (usually \(63.4^\circ\) and \(243.4^\circ\), or \(161.6^\circ\) and \(341.6^\circ\)) correct to 1 decimal place.
* **A1**: For finding the remaining two correct solutions with no extra solutions in the range.

**Part (b)**
* **M1**: Uses the identity \(\cot^2 \theta = \operatorname{cosec}^2 \theta - 1\) to write the equation in terms of \(\operatorname{cosec} \theta\) only.
* **A1**: For obtaining the correct simplified quadratic equation \(\operatorname{cosec}^2 \theta - 3 \operatorname{cosec} \theta - 4 = 0\) and converting to the sine equivalents \(\sin \theta = 0.25\) and \(\sin \theta = -1\).
* **A1**: For \(\theta = \frac{3\pi}{2}\) (or 4.71) correct to 3 significant figures.
* **A1**: For \(\theta \approx 0.253\) correct to 3 significant figures.
* **A1**: For \(\theta \approx 2.89\) correct to 3 significant figures, and no extra solutions in the range.

**(a)**
Using the trigonometric identity \(\sec^2 x = 1 + \tan^2 x\):
\(3(1 + \tan^2 x) - 5 \tan x - 5 = 0\)
\(3 \tan^2 x - 5 \tan x - 2 = 0\)

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

This gives two cases:
1. \(\tan x = 2\)
The basic angle is \(\tan^{-1}(2) \approx 63.43^\circ\).
Since tangent is positive in the 1st and 3rd quadrants:
\(x \approx 63.4^\circ\)
\(x \approx 180^\circ + 63.43^\circ = 243.4^\circ\)

2. \(\tan x = -\frac{1}{3}\)
The basic angle is \(\tan^{-1}\left(\frac{1}{3}\right) \approx 18.43^\circ\).
Since tangent is negative in the 2nd and 4th quadrants:
\(x \approx 180^\circ - 18.43^\circ = 161.6^\circ\)
\(x \approx 360^\circ - 18.43^\circ = 341.6^\circ\)

So, the solutions for part (a) are \(x = 63.4^\circ, 161.6^\circ, 243.4^\circ, 341.6^\circ\).

**(b)**
Using the trigonometric identity \(\cot^2 \theta = \operatorname{cosec}^2 \theta - 1\):
\((\operatorname{cosec}^2 \theta - 1) - 3 \operatorname{cosec} \theta - 3 = 0\)
\(\operatorname{cosec}^2 \theta - 3 \operatorname{cosec} \theta - 4 = 0\)

Factorising this quadratic equation in \(\operatorname{cosec} \theta\):
\((\operatorname{cosec} \theta - 4)(\operatorname{cosec} \theta + 1) = 0\)

This gives:
\(\operatorname{cosec} \theta = 4\) or \(\operatorname{cosec} \theta = -1\)

Taking the reciprocal to express in terms of sine:
\(\sin \theta = \frac{1}{4} = 0.25\) or \(\sin \theta = -1\)

1. For \(\sin \theta = 0.25\):
The basic angle in radians is \(\sin^{-1}(0.25) \approx 0.25268\) rad.
Since sine is positive in the 1st and 2nd quadrants:
\(\theta \approx 0.253\) rad
\(\theta \approx \pi - 0.25268 \approx 2.89\) rad

2. For \(\sin \theta = -1\):
In the interval \(0 \le \theta \le 2\pi\):
\(\theta = \frac{3\pi}{2} \approx 4.71\) rad

So, the solutions for part (b) are \(\theta = 0.253\), \(2.89\), \(\frac{3\pi}{2}\) (or \(4.71\)).

**Part (a)**
* **M1**: Uses the identity \(\sec^2 x = 1 + \tan^2 x\) to write the equation in terms of \(\tan x\) only.
* **A1**: For obtaining the correct simplified quadratic equation \(3\tan^2 x - 5\tan x - 2 = 0\) and finding the correct values \(\tan x = 2\) and \(\tan x = -\frac{1}{3}\).
* **A1**: For finding any two correct solutions (usually \(63.4^\circ\) and \(243.4^\circ\), or \(161.6^\circ\) and \(341.6^\circ\)) correct to 1 decimal place.
* **A1**: For finding the remaining two correct solutions with no extra solutions in the range.

**Part (b)**
* **M1**: Uses the identity \(\cot^2 \theta = \operatorname{cosec}^2 \theta - 1\) to write the equation in terms of \(\operatorname{cosec} \theta\) only.
* **A1**: For obtaining the correct simplified quadratic equation \(\operatorname{cosec}^2 \theta - 3 \operatorname{cosec} \theta - 4 = 0\) and converting to the sine equivalents \(\sin \theta = 0.25\) and \(\sin \theta = -1\).
* **A1**: For \(\theta = \frac{3\pi}{2}\) (or 4.71) correct to 3 significant figures.
* **A1**: For \(\theta \approx 0.253\) correct to 3 significant figures.
* **A1**: For \(\theta \approx 2.89\) correct to 3 significant figures, and no extra solutions in the range.

Let the terms of the geometric progression (GP) be \(a, ar, ar^2, \dots\) and the terms of the arithmetic progression (AP) be \(a, a+d, a+2d, \dots\).

**Step 1: Use the first condition**
"The sum of the first two terms of the GP is equal to the third term of the AP."
\[a + ar = a + 2d\]
Subtracting \(a\) from both sides gives:
\[ar = 2d \implies d = \frac{ar}{2}\]

**Step 2: Use the second condition**
"The sum to infinity of the GP is equal to \(8\) times the common difference of the AP."
\[S_{\infty} = \frac{a}{1-r} = 8d\]
Substitute \(d = \frac{ar}{2}\) into this equation:
\[\frac{a}{1-r} = 8\left(\frac{ar}{2}\right)\]
\[\frac{a}{1-r} = 4ar\]
Since \(a \neq 0\), we can divide both sides by \(a\):
\[\frac{1}{1-r} = 4r\]
\[1 = 4r(1-r)\]
\[1 = 4r - 4r^2\]
\[4r^2 - 4r + 1 = 0\]
\[(2r - 1)^2 = 0 \implies r = 0.5\]
Since \(|r| = 0.5 < 1\), the sum to infinity exists.

**Step 3: Use the third condition**
"The sum of the first three terms of the AP is \(15\)."
\[a + (a+d) + (a+2d) = 15\]
\[3a + 3d = 15 \implies a + d = 5\]
Substitute \(d = \frac{ar}{2} = \frac{a(0.5)}{2} = \frac{a}{4}\) into this equation:
\[a + \frac{a}{4} = 5\]
\[\frac{5}{4}a = 5 \implies a = 4\]

**Step 4: Find the sum to infinity**
Using \(a = 4\) and \(r = 0.5\):
\[S_{\infty} = \frac{4}{1 - 0.5} = 8\]
(Alternatively, using \(d = \frac{4 \times 0.5}{2} = 1\), we have \(S_{\infty} = 8d = 8(1) = 8\).)

**M1**: For setting up the equation for the first condition: \(a + ar = a + 2d\) (or simplified to \(ar = 2d\)).
**M1**: For setting up the sum to infinity equation: \(\frac{a}{1-r} = 8d\).
**M1**: For substituting \(d = \frac{ar}{2}\) into their sum to infinity equation to eliminate \(d\).
**A1**: For solving the quadratic equation to obtain \(r = 0.5\) (or \(\frac{1}{2}\)).
**M1**: For setting up the equation for the third condition: \(3a + 3d = 15\) (or \(a+d=5\)) and attempting to solve for \(a\) using their relation between \(a\) and \(d\).
**A1**: For obtaining \(a = 4\).
**A1**: For finding the correct sum to infinity, \(S_{\infty} = 8\).

First, rearrange the inequality to get zero on one side: \(x^3 + x^2 - 6x > 0\). Next, factorise the cubic expression to find the critical values: \(x(x^2 + x - 6) > 0\) which gives \(x(x + 3)(x - 2) > 0\). The critical values where the curve intersects the \(x\)-axis are \(x = -3\), \(x = 0\), and \(x = 2\). Since the coefficient of \(x^3\) is positive, the cubic graph starts below the \(x\)-axis, goes up through \(x = -3\), goes down through \(x = 0\), and goes up through \(x = 2\). We want the regions where the graph is strictly above the \(x\)-axis (\(y > 0\)). From our sketch, this occurs when \(-3 < x < 0\) or \(x > 2\).

**M1**: Find the critical values by factorising the cubic expression to get \(x = -3\), \(x = 0\), and \(x = 2\) (at least two correct).
**M1**: Attempt to sketch the cubic graph or use a sign table to identify the regions where the curve is above the \(x\)-axis.
**A1**: Obtain the correct intervals: \(-3 < x < 0\) or \(x > 2\) (accept equivalent interval or set notation).

(a) Express \(f(x)\) in completed square form:
\(f(x) = 2(x^2 - 6x) + 13 = 2[(x-3)^2 - 9] + 13 = 2(x-3)^2 - 5\).
For the inverse function \(f^{-1}\) to exist, \(f\) must be a one-to-one function.
Since the vertex of the quadratic curve is at \((3, -5)\), the function is one-to-one for \(x \ge 3\).
Thus, the least value of \(k\) is \(3\).

(b) Using \(k = 3\), the minimum value of \(f(x)\) occurs at the vertex where \(f(3) = -5\).
Since the curve opens upwards, the range of \(f\) is \(f(x) \ge -5\).

(c) To find the inverse function, let \(y = 2(x-3)^2 - 5\) where \(x \ge 3\).
Rearranging for \(x\):
\(y + 5 = 2(x-3)^2\)
\(\frac{y+5}{2} = (x-3)^2\)
\(x - 3 = \sqrt{\frac{y+5}{2}}\) (taking the positive square root since \(x \ge 3\))
\(x = 3 + \sqrt{\frac{y+5}{2}}\)
Therefore, \(f^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}}\).

(d) The intersection of the curve \(y = f(x)\) and its inverse \(y = f^{-1}(x)\) lies on the line \(y = x\).
Thus, we solve \(f(x) = x\) for \(x \ge 3\):
\(2x^2 - 12x + 13 = x\)
\(2x^2 - 13x + 13 = 0\)
Using the quadratic formula:
\(x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(2)(13)}}{2(2)}
\)x = \frac{13 \pm \sqrt{169 - 104}}{4}
\(x = \frac{13 \pm \sqrt{65}}{4}\)
Since the domain of \(f\) is \(x \ge 3\), we must have \(x \ge 3\).
Evaluating the two solutions:
\(x_1 = \frac{13 + \sqrt{65}}{4} \approx 5.27\) (since \(\sqrt{65} \approx 8.06\), \(x_1 \approx \frac{21.06}{4} = 5.27 \ge 3\))
\(x_2 = \frac{13 - \sqrt{65}}{4} \approx 1.23\) (which is less than 3, so reject)
Thus, the only valid solution is \(x = \frac{13 + \sqrt{65}}{4}\) (or \(x \approx 5.27\)).

(a)
- M1: For attempting to complete the square or find the vertex of the quadratic equation (e.g., using \(x = -b/(2a) = 12/4 = 3\)).
- A1: For obtaining \(k = 3\).

(b)
- B1: For stating \(f(x) \ge -5\) (accept \(y \ge -5\) or range \([-5, \infty)\)).

(c)
- M1: For attempting to rearrange \(y = f(x)\) to make \(x\) the subject (must involve a square root).
- M1: For correctly dealing with the coefficient of 2 and isolating the squared term, choosing the positive square root.
- A1: For \(f^{-1}(x) = 3 + \sqrt{\frac{x+5}{2}}\) (must be in terms of \(x\)).

(d)
- M1: For recognizing that \(f(x) = f^{-1}(x)\) is equivalent to solving \(f(x) = x\) (or equating \(f(x) = f^{-1}(x)\) and squaring correctly).
- M1: For solving the resulting quadratic equation \(2x^2 - 13x + 13 = 0\) using the quadratic formula or completing the square.
- A1: For selecting the correct solution \(x = \frac{13 + \sqrt{65}}{4}\) (or \(5.27\) correct to 3 significant figures) and rejecting \(x \approx 1.23\).

**(i)**
Start with the Left-Hand Side (LHS) and put it over a common denominator:

\(\text{LHS} = \frac{\sin^2 \theta - (1 - \cos \theta)^2}{\sin \theta (1 - \cos \theta)}\)

Expand the binomial in the numerator:

\(\text{LHS} = \frac{\sin^2 \theta - (1 - 2\cos \theta + \cos^2 \theta)}{\sin \theta (1 - \cos \theta)}\)

\(\text{LHS} = \frac{\sin^2 \theta - 1 + 2\cos \theta - \cos^2 \theta}{\sin \theta (1 - \cos \theta)}\)

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), substitute \(\sin^2 \theta - 1 = -\cos^2 \theta\):

\(\text{LHS} = \frac{-\cos^2 \theta + 2\cos \theta - \cos^2 \theta}{\sin \theta (1 - \cos \theta)}\)

\(\text{LHS} = \frac{2\cos \theta - 2\cos^2 \theta}{\sin \theta (1 - \cos \theta)}\)

Factorise the numerator:

\(\text{LHS} = \frac{2\cos \theta(1 - \cos \theta)}{\sin \theta (1 - \cos \theta)}\)

Cancel the common factor of \((1 - \cos \theta)\):

\(\text{LHS} = \frac{2\cos \theta}{\sin \theta} = 2\cot \theta = \text{RHS}\)

**(ii)**
Using the identity proven in part (i), rewrite the given equation as:

\(2\cot \theta = 3\tan \theta\)

Since \(2\cot \theta = \frac{2}{\tan \theta}\):

\(\frac{2}{\tan \theta} = 3\tan \theta\)

\(3\tan^2 \theta = 2\)

\\tan^2 \theta = \frac{2}{3}\)

Taking the square root of both sides:

\(\tan \theta = \pm\sqrt{\frac{2}{3}} \approx \pm 0.8165\)

Find the solutions within the interval \(0^\circ < \theta < 180^\circ\):

For \(\tan \theta = \sqrt{\frac{2}{3}}\):
\(\theta = \tan^{-1}\left(\sqrt{\frac{2}{3}}\right) \approx 39.2^\circ\)

For \(\tan \theta = -\sqrt{\frac{2}{3}}\):
\(\theta = 180^\circ - 39.23^\circ \approx 140.8^\circ\)

Thus, the solutions are \(\theta = 39.2^\circ\) and \(\theta = 140.8^\circ\).

**(i)**
* **M1**: For attempting to express the LHS as a single fraction over the common denominator \(\sin\theta(1-\cos\theta)\).
* **M1**: For expanding \((1-\cos\theta)^2\) correctly and correctly applying \(\sin^2\theta + \cos^2\theta = 1\) (or an equivalent identity substitution) to simplify the numerator.
* **A1**: For clear algebraic factorisation of the numerator leading to the cancellation of \(1-\cos\theta\) and concluding the proof cleanly.

**(ii)**
* **M1**: For using the identity from part (i) to rewrite the equation in terms of \(\cot\theta\) and \(\tan\theta\).
* **M1**: For obtaining the quadratic form \(\tan^2\theta = \frac{2}{3}\) and correctly recognizing both positive and negative roots: \(\tan\theta = \pm\sqrt{\frac{2}{3}}\).
* **A1**: For both correct solutions \(\theta = 39.2^\circ\) and \(\theta = 140.8^\circ\) (rounded to 1 decimal place, as required for degrees) and no extra solutions in the range.

(a) The boundary of the template consists of the straight edge \(OA = r\), the circular arc \(AB = r\theta\), and the three outer edges of the rectangle \(OBCD\), which are \(BC = 0.5r\), \(CD = r\), and \(DO = 0.5r\).
The perimeter \(P\) of the template is:
\(P = r + r\theta + 0.5r + r + 0.5r = 3r + r\theta\).
Since \(P = 30\text{ cm}\):
\(3r + r\theta = 30\)
\(r\theta = 30 - 3r \implies \theta = \frac{30}{r} - 3\).
The area \(A\) of the template is the sum of the area of the sector and the area of the rectangle:
\(A = \frac{1}{2}r^2\theta + r(0.5r) = \frac{1}{2}r^2\theta + 0.5r^2\).
Substituting \(\theta = \frac{30}{r} - 3\) into the area formula:
\(A = \frac{1}{2}r^2\left(\frac{30}{r} - 3\right) + 0.5r^2\)
\(A = 15r - 1.5r^2 + 0.5r^2\)
\(A = 15r - r^2\).

(b) To find the stationary value of \(A\), differentiate \(A\) with respect to \(r\):
\(\frac{dA}{dr} = 15 - 2r\).
At a stationary point, \(\frac{dA}{dr} = 0\):
\(15 - 2r = 0 \implies r = 7.5\).
The stationary value of \(A\) is:
\(A = 15(7.5) - (7.5)^2 = 112.5 - 56.25 = 56.25\).
To find the nature of this stationary value, find the second derivative:
\(\frac{d^2A}{dr^2} = -2\).
Since \(\frac{d^2A}{dr^2} < 0\), the stationary value is a maximum.
The corresponding value of \(\theta\) is:
\(\theta = \frac{30}{7.5} - 3 = 4 - 3 = 1\).

(a)
M1: For writing a correct expression for the perimeter, \(3r + r\theta = 30\), and expressing \(\theta\) in terms of \(r\).
M1: For writing a correct expression for the total area, \(A = \frac{1}{2}r^2\theta + 0.5r^2\).
A1: For substituting \(\theta\) and correctly simplifying to show \(A = 15r - r^2\).

(b)
M1: For differentiating \(A\) with respect to \(r\) to obtain \(\frac{dA}{dr} = 15 - 2r\).
A1: For setting \(\frac{dA}{dr} = 0\) and finding \(r = 7.5\).
A1: For evaluating \(A = 56.25\).
M1: For finding \(\frac{d^2A}{dr^2} = -2\) and concluding it is a maximum (accept alternative valid methods).
A1: For finding the corresponding value of \(\theta = 1\).

(a) We use the quotient rule to find the derivative.
Let \(u = 3x^2 - 1 \implies \frac{du}{dx} = 6x\) and \(v = 2x + 1 \implies \frac{dv}{dx} = 2\).

Using the quotient rule formula:
\(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)

\(\frac{dy}{dx} = \frac{6x(2x + 1) - 2(3x^2 - 1)}{(2x + 1)^2}\)

\(\frac{dy}{dx} = \frac{12x^2 + 6x - 6x^2 + 2}{(2x + 1)^2} = \frac{6x^2 + 6x + 2}{(2x + 1)^2}\)

Substituting \(x = 1\):
\(\frac{dy}{dx} = \frac{6(1)^2 + 6(1) + 2}{(2(1) + 1)^2} = \frac{14}{9}\)

(b) First, find the \(y\)-coordinate of the curve at \(x = 1\):
\(y = \frac{3(1)^2 - 1}{2(1) + 1} = \frac{2}{3}\)

The equation of the tangent line with gradient \(m = \frac{14}{9}\) passing through \(\left(1, \frac{2}{3}\right)\) is:
\(y - \frac{2}{3} = \frac{14}{9}(x - 1)\)

Multiplying the entire equation by 9 to clear fractions:
\(9y - 6 = 14(x - 1)\)
\(9y - 6 = 14x - 14\)
\(14x - 9y - 8 = 0\)

(a)
M1: Attempt to apply the quotient rule with correct structure, containing at least 3 correct derivatives of \(u\) and \(v\).
A1: Correct unsimplified derivative expression: \(\frac{6x(2x+1) - 2(3x^2-1)}{(2x+1)^2}\).
A1: Correct simplified numerator: \(6x^2 + 6x + 2\).
M1: Substitute \(x = 1\) into their expression for \(\frac{dy}{dx}\).
A1: Obtain \(\frac{14}{9}\) (or equivalent fraction/decimal).

(b)
M1: Substitute \(x = 1\) into the original equation to find the \(y\)-coordinate: \(y = \frac{2}{3}\).
M1: Substitute their point and their gradient from (a) into a straight-line equation formula.
A1: Correct equation in the required form: \(14x - 9y - 8 = 0\) (or any integer multiple, e.g., \(9y - 14x + 8 = 0\)).

First, integrate the given function term-by-term: \(\int (4 \sin(2x) - 3 \cos(x)) \, dx = -2 \cos(2x) - 3 \sin(x) + C\). Now, substitute the upper limit \(x = \frac{\pi}{3}\) into the integrated function: \(-2 \cos\left(\frac{2\pi}{3}\right) - 3 \sin\left(\frac{\pi}{3}\right) = -2\left(-\frac{1}{2}\right) - 3\left(\frac{\sqrt{3}}{2}\right) = 1 - \frac{3\sqrt{3}}{2}\). Substitute the lower limit \(x = 0\) into the integrated function: \(-2 \cos(0) - 3 \sin(0) = -2(1) - 0 = -2\). Subtract the lower limit value from the upper limit value: \(\left(1 - \frac{3\sqrt{3}}{2}\right) - (-2) = 3 - \frac{3\sqrt{3}}{2}\). Thus, the exact value is \(3 - \frac{3\sqrt{3}}{2}\).

M1: For integrating \(4 \sin(2x)\) to obtain \(-2 \cos(2x)\). M1: For integrating \(-3 \cos(x)\) to obtain \(-3 \sin(x)\). M1: For correct substitution of the limits \(\frac{\pi}{3}\) and \(0\) into their integrated expression. M1: For evaluating trigonometric exact values: \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\) and \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\). A1: For the correct exact final answer \(3 - \frac{3\sqrt{3}}{2}\) or any equivalent simplified form such as \(\frac{6 - 3\sqrt{3}}{2}\).

**(a)**

Comparing the terms in the expansion of \((1 + ax)^n = 1 + nax + \frac{n(n-1)}{2}a^2x^2 + \dots\):

From the coefficient of \(x\):
\(na = 30 \implies a = \frac{30}{n}\)

From the coefficient of \(x^2\):
\(\frac{n(n-1)}{2}a^2 = 375\)

Substitute \(a = \frac{30}{n}\) into the second equation:
\(\frac{n(n-1)}{2} \left(\frac{30}{n}\right)^2 = 375\)

\(\frac{n(n-1)}{2} \cdot \frac{900}{n^2} = 375\)

\(\frac{450(n-1)}{n} = 375\)

\(450n - 450 = 375n\)

\(75n = 450\)

\(n = 6\)

Substitute \(n = 6\) back to find \(a\):
\(a = \frac{30}{6} = 5\)

**(b)**

We need to find the coefficient of \(x^2\) in the expansion of:
\((2 - 5x)(1 + 5x)^6\)

We know that:
\((1 + 5x)^6 = 1 + 30x + 375x^2 + \dots\)

So the product is:
\((2 - 5x)(1 + 30x + 375x^2 + \dots)\)

The terms containing \(x^2\) in the expansion of this product are:
\(2(375x^2) + (-5x)(30x)\)
\(= 750x^2 - 150x^2\)
\(= 600x^2\)

Thus, the coefficient of \(x^2\) is \(600\).

**(a)**
- **M1**: Attempts to write down two equations in terms of \(a\) and \(n\) from the coefficients, e.g., \(na = 30\) and \(\frac{n(n-1)}{2}a^2 = 375\).
- **M1**: Attempts to solve these equations by eliminating one variable (e.g., substituting \(a = \frac{30}{n}\)).
- **A1**: Obtains \(n = 6\).
- **A1**: Obtains \(a = 5\).

**(b)**
- **M1**: Identifies the relevant terms that contribute to the coefficient of \(x^2\), which are \(2 \times (\text{coefficient of } x^2 \text{ in } (1+ax)^n)\) and \(-5 \times (\text{coefficient of } x \text{ in } (1+ax)^n)\).
- **M1**: Substitutes correct values to get \(2(375) + (-5)(30)\) or equivalent.
- **A1**: Obtains \(600\).

To find the area of the region enclosed by the curve \(y = 9 - x^2\) and the line \(y = x + 3\), we first find their points of intersection.

Equating the two equations:
\[9 - x^2 = x + 3\]

Rearranging to form a quadratic equation:
\[x^2 + x - 6 = 0\]

Factoring the quadratic:
\[(x + 3)(x - 2) = 0\]

Thus, the points of intersection occur at \(x = -3\) and \(x = 2\).

The area \(A\) is given by the definite integral of the upper function (the curve) minus the lower function (the line) from \(x = -3\) to \(x = 2\):
\[A = \int_{-3}^{2} \left( (9 - x^2) - (x + 3) \right) \text{d}x\]
\[A = \int_{-3}^{2} (6 - x - x^2) \text{d}x\]

Integrating with respect to \(x\):
\[A = \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-3}^{2}\]

Now, substitute the upper and lower limits:

For \(x = 2\):
\[\left( 6(2) - \frac{2^2}{2} - \frac{2^3}{3} \right) = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3}\]

For \(x = -3\):
\[\left( 6(-3) - \frac{(-3)^2}{2} - \frac{(-3)^3}{3} \right) = -18 - \frac{9}{2} + 9 = -9 - 4.5 = -13.5 = -\frac{27}{2}\]

Subtract the lower limit result from the upper limit result:
\[A = \frac{22}{3} - \left( -\frac{27}{2} \right) = \frac{22}{3} + \frac{27}{2}\]

Find a common denominator of 6:
\[A = \frac{44}{6} + \frac{81}{6} = \frac{125}{6}\]

Thus, the exact area is \(\frac{125}{6}\) (or \(20\frac{5}{6}\)).

**M1**: Equating the curve and line equations to find the points of intersection: \(9 - x^2 = x + 3\).
**A1**: Solving the quadratic equation to obtain correct boundaries: \(x = -3\) and \(x = 2\).
**M1**: Setting up the correct definite integral of \(y_{\text{curve}} - y_{\text{line}}\):
\(\int_{-3}^{2} (6 - x - x^2) \text{d}x\).
**M1**: Attempting to integrate \((6 - x - x^2)\), with at least two terms integrated correctly.
**A1**: Correct integrated expression: \(\left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]\).
**M1**: Substituting limits \(-3\) and \(2\) into their integrated expression.
**A1**: Correct final exact area of \(\frac{125}{6}\) or \(20\frac{5}{6}\) (or equivalent exact fraction).

To find the area of the region enclosed by the curve \(y = 9 - x^2\) and the line \(y = x + 3\), we first find their points of intersection.

Equating the two equations:
\[9 - x^2 = x + 3\]

Rearranging to form a quadratic equation:
\[x^2 + x - 6 = 0\]

Factoring the quadratic:
\[(x + 3)(x - 2) = 0\]

Thus, the points of intersection occur at \(x = -3\) and \(x = 2\).

The area \(A\) is given by the definite integral of the upper function (the curve) minus the lower function (the line) from \(x = -3\) to \(x = 2\):
\[A = \int_{-3}^{2} \left( (9 - x^2) - (x + 3) \right) \text{d}x\]
\[A = \int_{-3}^{2} (6 - x - x^2) \text{d}x\]

Integrating with respect to \(x\):
\[A = \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-3}^{2}\]

Now, substitute the upper and lower limits:

For \(x = 2\):
\[\left( 6(2) - \frac{2^2}{2} - \frac{2^3}{3} \right) = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3}\]

For \(x = -3\):
\[\left( 6(-3) - \frac{(-3)^2}{2} - \frac{(-3)^3}{3} \right) = -18 - \frac{9}{2} + 9 = -9 - 4.5 = -13.5 = -\frac{27}{2}\]

Subtract the lower limit result from the upper limit result:
\[A = \frac{22}{3} - \left( -\frac{27}{2} \right) = \frac{22}{3} + \frac{27}{2}\]

Find a common denominator of 6:
\[A = \frac{44}{6} + \frac{81}{6} = \frac{125}{6}\]

Thus, the exact area is \(\frac{125}{6}\) (or \(20\frac{5}{6}\)).

**M1**: Equating the curve and line equations to find the points of intersection: \(9 - x^2 = x + 3\).
**A1**: Solving the quadratic equation to obtain correct boundaries: \(x = -3\) and \(x = 2\).
**M1**: Setting up the correct definite integral of \(y_{\text{curve}} - y_{\text{line}}\):
\(\int_{-3}^{2} (6 - x - x^2) \text{d}x\).
**M1**: Attempting to integrate \((6 - x - x^2)\), with at least two terms integrated correctly.
**A1**: Correct integrated expression: \(\left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]\).
**M1**: Substituting limits \(-3\) and \(2\) into their integrated expression.
**A1**: Correct final exact area of \(\frac{125}{6}\) or \(20\frac{5}{6}\) (or equivalent exact fraction).

Let \(h\) be the perpendicular height of the triangle.

Using the formula for the area of a triangle:
\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\]

Substitute the given values into the formula:
\[2 + 2\sqrt{5} = \frac{1}{2} (3 - \sqrt{5}) h\]

Multiply both sides of the equation by 2:
\[4 + 4\sqrt{5} = (3 - \sqrt{5}) h\]

Rearrange to solve for \(h\):
\[h = \frac{4 + 4\sqrt{5}}{3 - \sqrt{5}}\]

To rationalise the denominator, multiply the numerator and the denominator by the conjugate of the denominator, which is \(3 + \sqrt{5}\):
\[h = \frac{(4 + 4\sqrt{5})(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})}\]

Expand the numerator:
\[(4 + 4\sqrt{5})(3 + \sqrt{5}) = 12 + 4\sqrt{5} + 12\sqrt{5} + 4(5) = 12 + 16\sqrt{5} + 20 = 32 + 16\sqrt{5}\]

Expand the denominator:
\[(3 - \sqrt{5})(3 + \sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4\]

Simplify the fraction:
\[h = \frac{32 + 16\sqrt{5}}{4} = 8 + 4\sqrt{5}\]

Thus, the perpendicular height is \(8 + 4\sqrt{5}\).

**M1**: Sets up a correct equation for the area of the triangle and attempts to make \(h\) the subject, obtaining \(h = \frac{4 + 4\sqrt{5}}{3 - \sqrt{5}}\) or equivalent.

**M1**: Multiplies the numerator and the denominator by the conjugate \(3 + \sqrt{5}\).

**A1**: Correctly expands the numerator to \(32 + 16\sqrt{5}\) and denominator to \(4\).

**A1**: Obtains the final answer \(8 + 4\sqrt{5}\) (or identifies \(a = 8\) and \(b = 4\)).

**(a)**

To divide 10 students into Team Red (4 members), Team Blue (3 members), and Team Green (3 members):

- First, choose 4 students for Team Red from the 10 available: \(\binom{10}{4} = 210\) ways.
- Next, choose 3 students for Team Blue from the remaining 6: \(\binom{6}{3} = 20\) ways.
- Finally, the remaining 3 students must join Team Green: \(\binom{3}{3} = 1\) way.

Total number of ways is: \(210 \times 20 \times 1 = 4200\).

**(b)**

Since there are 4 girls and each team must contain at least one girl, we analyze the distribution of girls. The only possible distributions of the 4 girls among Team Red (size 4), Team Blue (size 3), and Team Green (size 3) are:

- **Case 1: Team Red has 2 girls, Team Blue has 1 girl, Team Green has 1 girl**
- Ways to choose girls: \(\binom{4}{2} \times \binom{2}{1} \times \binom{1}{1} = 6 \times 2 \times 1 = 12\)
- Ways to choose boys (Red needs 2, Blue needs 2, Green needs 2): \(\binom{6}{2} \times \binom{4}{2} \times \binom{2}{2} = 15 \times 6 \times 1 = 90\)
- Total for Case 1: \(12 \times 90 = 1080\)

- **Case 2: Team Red has 1 girl, Team Blue has 2 girls, Team Green has 1 girl**
- Ways to choose girls: \(\binom{4}{1} \times \binom{3}{2} \times \binom{1}{1} = 4 \times 3 \times 1 = 12\)
- Ways to choose boys (Red needs 3, Blue needs 1, Green needs 2): \(\binom{6}{3} \times \binom{3}{1} \times \binom{2}{2} = 20 \times 3 \times 1 = 60\)
- Total for Case 2: \(12 \times 60 = 720\)

- **Case 3: Team Red has 1 girl, Team Blue has 1 girl, Team Green has 2 girls**
- Ways to choose girls: \(\binom{4}{1} \times \binom{3}{1} \times \binom{2}{2} = 4 \times 3 \times 1 = 12\)
- Ways to choose boys (Red needs 3, Blue needs 2, Green needs 1): \(\binom{6}{3} \times \binom{3}{2} \times \binom{1}{1} = 20 \times 3 \times 1 = 60\)
- Total for Case 3: \(12 \times 60 = 720\)

Summing the ways from all three mutually exclusive cases:
\(1080 + 720 + 720 = 2520\).

*Alternative Method (Complementary Counting):*
Total unrestricted ways = \(4200\).
Subtract cases where at least one team has 0 girls:
- Team Red has 0 girls: \(\binom{6}{4} \times \binom{6}{3} = 15 \times 20 = 300\) ways.
- Team Blue has 0 girls: \(\binom{6}{3} \times \binom{7}{4} = 20 \times 35 = 700\) ways.
- Team Green has 0 girls: \(\binom{6}{3} \times \binom{7}{4} = 20 \times 35 = 700\) ways.
- Both Team Blue and Team Green have 0 girls: \(\binom{6}{3} \times \binom{3}{3} \times \binom{4}{4} = 20\) ways.

Applying inclusion-exclusion for the invalid ways: \(300 + 700 + 700 - 20 = 1680\) ways.
Total valid ways: \(4200 - 1680 = 2520\).

**(a)**
* **M1**: For a product of correct combinations, e.g., \(\binom{10}{4} \times \binom{6}{3}\) or equivalent.
* **A1**: For \(4200\).

**(b)**
* **M1**: For identifying the three valid girl distributions: (2, 1, 1), (1, 2, 1), and (1, 1, 2).
* **M1**: For calculating the number of ways for one case correctly (e.g., \(12 \times 90 = 1080\) or \(12 \times 60 = 720\)).
* **M1**: For summing all three valid cases (or equivalent complete subtraction method using inclusion-exclusion).
* **A1**: For \(2520\).

**(a)**

To divide 10 students into Team Red (4 members), Team Blue (3 members), and Team Green (3 members):

- First, choose 4 students for Team Red from the 10 available: \(\binom{10}{4} = 210\) ways.
- Next, choose 3 students for Team Blue from the remaining 6: \(\binom{6}{3} = 20\) ways.
- Finally, the remaining 3 students must join Team Green: \(\binom{3}{3} = 1\) way.

Total number of ways is: \(210 \times 20 \times 1 = 4200\).

**(b)**

Since there are 4 girls and each team must contain at least one girl, we analyze the distribution of girls. The only possible distributions of the 4 girls among Team Red (size 4), Team Blue (size 3), and Team Green (size 3) are:

- **Case 1: Team Red has 2 girls, Team Blue has 1 girl, Team Green has 1 girl**
- Ways to choose girls: \(\binom{4}{2} \times \binom{2}{1} \times \binom{1}{1} = 6 \times 2 \times 1 = 12\)
- Ways to choose boys (Red needs 2, Blue needs 2, Green needs 2): \(\binom{6}{2} \times \binom{4}{2} \times \binom{2}{2} = 15 \times 6 \times 1 = 90\)
- Total for Case 1: \(12 \times 90 = 1080\)

- **Case 2: Team Red has 1 girl, Team Blue has 2 girls, Team Green has 1 girl**
- Ways to choose girls: \(\binom{4}{1} \times \binom{3}{2} \times \binom{1}{1} = 4 \times 3 \times 1 = 12\)
- Ways to choose boys (Red needs 3, Blue needs 1, Green needs 2): \(\binom{6}{3} \times \binom{3}{1} \times \binom{2}{2} = 20 \times 3 \times 1 = 60\)
- Total for Case 2: \(12 \times 60 = 720\)

- **Case 3: Team Red has 1 girl, Team Blue has 1 girl, Team Green has 2 girls**
- Ways to choose girls: \(\binom{4}{1} \times \binom{3}{1} \times \binom{2}{2} = 4 \times 3 \times 1 = 12\)
- Ways to choose boys (Red needs 3, Blue needs 2, Green needs 1): \(\binom{6}{3} \times \binom{3}{2} \times \binom{1}{1} = 20 \times 3 \times 1 = 60\)
- Total for Case 3: \(12 \times 60 = 720\)

Summing the ways from all three mutually exclusive cases:
\(1080 + 720 + 720 = 2520\).

*Alternative Method (Complementary Counting):*
Total unrestricted ways = \(4200\).
Subtract cases where at least one team has 0 girls:
- Team Red has 0 girls: \(\binom{6}{4} \times \binom{6}{3} = 15 \times 20 = 300\) ways.
- Team Blue has 0 girls: \(\binom{6}{3} \times \binom{7}{4} = 20 \times 35 = 700\) ways.
- Team Green has 0 girls: \(\binom{6}{3} \times \binom{7}{4} = 20 \times 35 = 700\) ways.
- Both Team Blue and Team Green have 0 girls: \(\binom{6}{3} \times \binom{3}{3} \times \binom{4}{4} = 20\) ways.

Applying inclusion-exclusion for the invalid ways: \(300 + 700 + 700 - 20 = 1680\) ways.
Total valid ways: \(4200 - 1680 = 2520\).

**(a)**
* **M1**: For a product of correct combinations, e.g., \(\binom{10}{4} \times \binom{6}{3}\) or equivalent.
* **A1**: For \(4200\).

**(b)**
* **M1**: For identifying the three valid girl distributions: (2, 1, 1), (1, 2, 1), and (1, 1, 2).
* **M1**: For calculating the number of ways for one case correctly (e.g., \(12 \times 90 = 1080\) or \(12 \times 60 = 720\)).
* **M1**: For summing all three valid cases (or equivalent complete subtraction method using inclusion-exclusion).
* **A1**: For \(2520\).

(a) Since \(a = \frac{dv}{dt}\), we differentiate \(v\) with respect to \(t\): \(a = \frac{d}{dt}(3t^2 - 14t + 8) = 6t - 14\). When \(t = 3\): \(a = 6(3) - 14 = 4\text{ m s}^{-2}\). (b) The particle is at instantaneous rest when \(v = 0\): \(3t^2 - 14t + 8 = 0\), which factors to \((3t - 2)(t - 4) = 0\). Therefore, \(t = \frac{2}{3}\text{ s}\) (or \(0.667\text{ s}\)) and \(t = 4\text{ s}\). (c) The particle changes direction at \(t = \frac{2}{3}\text{ s}\) and comes to rest again at \(t = 4\text{ s}\). To find the total distance travelled in the first 4 seconds, we find the displacement \(s(t)\) from \(O\): \(s(t) = \int v \, dt = \int (3t^2 - 14t + 8) \, dt = t^3 - 7t^2 + 8t + c\). Since the particle passes through \(O\) at \(t = 0\), we have \(s(0) = 0 \implies c = 0\), so \(s(t) = t^3 - 7t^2 + 8t\). We calculate the displacement at the critical times: At \(t = 0\), \(s(0) = 0\text{ m}\). At \(t = \frac{2}{3}\), \(s\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^3 - 7\left(\frac{2}{3}\right)^2 + 8\left(\frac{2}{3}\right) = \frac{68}{27}\text{ m}\). At \(t = 4\), \(s(4) = 4^3 - 7(4)^2 + 8(4) = -16\text{ m}\). The total distance travelled is: \(|s(\frac{2}{3}) - s(0)| + |s(4) - s(\frac{2}{3})| = \frac{68}{27} + |-16 - \frac{68}{27}| = \frac{68}{27} + \frac{500}{27} = \frac{568}{27}\text{ m}\) (or \(21.0\text{ m}\) to 3 s.f.).

(a) M1: For attempting to differentiate \(v\) to find \(a\) (at least one power decreased by 1). A1: For \(4\text{ m s}^{-2}\) or equivalent. (b) M1: For setting \(v = 0\) and attempting to solve the quadratic equation. A1: For both \(t = \frac{2}{3}\) (or \(0.667\)) and \(t = 4\). (c) M1: For integration of \(v\) to find \(s(t)\) (at least two terms integrated correctly). A1: For \(s(t) = t^3 - 7t^2 + 8t\) (constant of integration not required if using limits). M1: For substituting \(t = \frac{2}{3}\) (or their non-zero resting time < 4) and \(t = 4\) into their \(s(t)\). M1: For a correct method to find the total distance, summing the absolute changes in displacement. A1: For \(\frac{568}{27}\) or \(21.0\) (or \(21\frac{1}{27}\)).