2024 Cambridge IGCSE Mathematics - Additional (0606) 模拟试题及答案详解

**Part (a)**
Expand the equation of the curve:
\( y = x^2 - 6x - 7 \)

Differentiate with respect to \( x \):
\( \frac{dy}{dx} = 2x - 6 \)

At the stationary point, \( \frac{dy}{dx} = 0 \):
\( 2x - 6 = 0 \implies x = 3 \)

Substitute \( x = 3 \) back into the original curve equation:
\( y = (3+1)(3-7) = 4 \times (-4) = -16 \)

So, the coordinates of the stationary point are \( (3, -16) \).

**Part (b)**
The \( x \)-intercepts occur where \( y = 0 \):
\( |(x+1)(x-7)| = 0 \implies x = -1 \text{ or } x = 7 \). So the intercepts are \( (-1, 0) \) and \( (7, 0) \).

The \( y \)-intercept occurs where \( x = 0 \):
\( y = |(0+1)(0-7)| = |-7| = 7 \). So the intercept is \( (0, 7) \).

The stationary point \( (3, -16) \) is reflected to \( (3, 16) \) as a local maximum.
The sketch is a reflected parabola with cusps at \( (-1, 0) \) and \( (7, 0) \), a local maximum at \( (3, 16) \), and a \( y \)-intercept at \( (0, 7) \).

**Part (c)**
The line \( y = c \) intersects the curve \( y = |(x+1)(x-7)| \) exactly twice when \( c \) is strictly greater than the maximum of the reflected part, which is \( 16 \).
Therefore, \( c > 16 \).

**Part (a)**
* **M1**: For differentiating the expanded quadratic \( y = x^2 - 6x - 7 \) (with at least one correct term) or using symmetry \( x = \frac{-1 + 7}{2} = 3 \).
* **A1**: For obtaining \( x = 3 \).
* **A1**: For obtaining \( y = -16 \).

**Part (b)**
* **B1**: For a correctly shaped modulus graph with cusps on the \( x \)-axis at \( -1 \) and \( 7 \) and a maximum in the first quadrant.
* **B1**: For correctly stating the intercepts with the axes: \((-1, 0)\), \((7, 0)\), and \((0, 7)\).

**Part (c)**
* **B1**: For \( c > 16 \) (or follow-through on their \( y \)-coordinate from part (a) as \( c > |\text{their } y| \)).

To sketch the graph of \(y = 3 + 2\cos 2\theta\):

1. **Determine the vertical range:**
The basic cosine function oscillates between \(-1\) and \(1\).
Multiplying by \(2\) changes the amplitude to \(2\) (range \([-2, 2]\)).
Adding \(3\) shifts the graph vertically upwards by \(3\) units.
Therefore, the maximum value is \(3 + 2 = 5\) and the minimum value is \(3 - 2 = 1\).

2. **Determine the period:**
The coefficient of \(\theta\) is \(2\), so the period is \(\frac{360^\circ}{2} = 180^\circ\).
In the interval \([-180^\circ, 180^\circ]\), there will be exactly two complete periods of the cosine wave.

3. **Key Points:**
- At \(\theta = 0^\circ\): \(y = 3 + 2\cos(0^\circ) = 5\) (Y-intercept is \((0, 5)\))
- At \(\theta = 90^\circ\): \(y = 3 + 2\cos(180^\circ) = 3 - 2 = 1\) (Minimum)
- At \(\theta = -90^\circ\): \(y = 3 + 2\cos(-180^\circ) = 3 - 2 = 1\) (Minimum)
- At \(\theta = 180^\circ\): \(y = 3 + 2\cos(360^\circ) = 3 + 2 = 5\) (Maximum)
- At \(\theta = -180^\circ\): \(y = 3 + 2\cos(-360^\circ) = 3 + 2 = 5\) (Maximum)
- Midline intersections \(y = 3\) occur at \(\theta = \pm 45^\circ\) and \(\pm 135^\circ\).

**B1** for correct shape: a cosine curve with two full periods within \([-180^\circ, 180^\circ]\).
**depB1** for the correct \(y\)-intercept at \((0, 5)\).
**depB1** for correct maxima at \((-180^\circ, 5)\), \((0, 5)\), and \((180^\circ, 5)\).
**depB1** for correct minima at \((-90^\circ, 1)\) and \((90^\circ, 1)\).

Rearranging the given equation into standard quadratic form:
\[x^2 + kx + (k + 3) = 0\]
Here, the coefficients are:
\[a = 1, \quad b = k, \quad c = k + 3\]
Since the equation has no real roots, the discriminant must be negative:
\[b^2 - 4ac < 0\]
Substitute the values of \(a\), \(b\), and \(c\):
\[k^2 - 4(1)(k + 3) < 0\]
\[k^2 - 4k - 12 < 0\]
Factorising the quadratic inequality:
\[(k - 6)(k + 2) < 0\]
The critical values are \(k = 6\) and \(k = -2\).
Since we require the expression to be less than zero, the solution is:
\[-2 < k < 6\]

**B1**: Rearranges the equation to \(x^2 + kx + k + 3 = 0\) (or equivalent three-term quadratic form) or identifies \(a = 1, b = k, c = k + 3\) correctly.
**M1**: Uses discriminant \(b^2 - 4ac < 0\) to obtain a quadratic inequality in terms of \(k\).
**A1**: Finds the correct critical values \(k = -2\) and \(k = 6\).
**A1**: Correct range \(-2 < k < 6\) (or equivalent notation, e.g., \(k > -2\) and \(k < 6\)). Allow strict inequalities only.

(a)
\(2 - 3\log_3 p + \log_3 q\)
\(= \log_3 (3^2) - \log_3 (p^3) + \log_3 q\)
\(= \log_3 9 - \log_3 (p^3) + \log_3 q\)
\(= \log_3 \left(\frac{9q}{p^3}\right)\)

(b)
\(\log_5 y - 6\log_y 5 = 1\)
Using the change of base rule: \(\log_y 5 = \frac{1}{\log_5 y}\)
Let \(u = \log_5 y\)
\(u - \frac{6}{u} = 1\)
\(u^2 - u - 6 = 0\)
\((u - 3)(u + 2) = 0\)
\(u = 3\) or \(u = -2\)
If \(\log_5 y = 3\), then \(y = 5^3 = 125\)
If \(\log_5 y = -2\), then \(y = 5^{-2} = \frac{1}{25} = 0.04\)

(a)
B1: For writing \(2 = \log_3 9\) or \(\log_3 3^2\) (soi)
M1: For correct use of power rule \(\log_3 (p^3)\) and at least one operation of multiplication or division for logs to base 3.
A1: For correct final answer \(\log_3 \left(\frac{9q}{p^3}\right)\)

(b)
B1: For using change of base to get \(\log_y 5 = \frac{1}{\log_5 y}\) (soi)
M1: For forming a three-term quadratic equation in \(\log_5 y\) (or a substitute variable) and solving to find two critical values.
A1: For obtaining \(y = 125\)
A1: For obtaining \(y = \frac{1}{25}\) (or \(0.04\))

(a) Since \(x-2\) is a factor of \(\mathrm{p}(x)\), we have \(\mathrm{p}(2) = 0\):
\[a(2)^3 + b(2)^2 - 12(2) + 4 = 0\]
\[8a + 4b - 20 = 0 \implies 2a + b = 5 \quad \text{--- (1)}\]

Now, differentiate \(\mathrm{p}(x)\):
\[\mathrm{p}'(x) = 3ax^2 + 2bx - 12\]
Since there is a stationary point at \(x = 1\), we have \(\mathrm{p}'(1) = 0\):
\[3a(1)^2 + 2b(1) - 12 = 0 \implies 3a + 2b = 12 \quad \text{--- (2)}\]

Solving equations (1) and (2) simultaneously:
From (1), \(b = 5 - 2a\).
Substitute into (2):
\[3a + 2(5 - 2a) = 12\]
\[3a + 10 - 4a = 12\]
\[-a = 2 \implies a = -2\]
Then, \(b = 5 - 2(-2) = 9\).

(b) To find the other stationary point, set \(\mathrm{p}'(x) = 0\):
\[-6x^2 + 18x - 12 = 0\]
Dividing by \(-6\):
\[x^2 - 3x + 2 = 0\]
\[(x - 1)(x - 2) = 0\]
Since \(x = 1\) is the given stationary point, the other stationary point is at \(x = 2\).

(a)
**M1**: For substituting \(x = 2\) into \(\mathrm{p}(x)\) and setting equal to 0, leading to a linear relation between \(a\) and \(b\) (e.g. \(8a + 4b = 20\)).
**M1**: For differentiating \(\mathrm{p}(x)\) with at least two terms correct and setting \(\mathrm{p}'(1) = 0\) to get \(3a + 2b = 12\).
**M1**: For an attempt to solve their two simultaneous equations.
**A1**: For both \(a = -2\) and \(b = 9\).

(b)
**M1**: For setting their \(\mathrm{p}'(x) = 0\) and attempting to solve the quadratic equation.
**A1**: For \(x = 2\) (must reject \(x = 1\) as the final answer for the *other* point).

(a) Taking natural logarithms of both sides of \(y = a b^{\sqrt{x}}\):
\(\ln y = \ln a + \sqrt{x} \ln b\)
This matches the linear form \(Y = mX + c\) where \(Y = \ln y\) is plotted on the vertical axis and \(X = \sqrt{x}\) is plotted on the horizontal axis.
Calculating the values:
\(\sqrt{x}\): 1, 2, 3, 4, 5
\(\ln y\): \(\ln(10) \approx 2.30\), \(\ln(25) \approx 3.22\), \(\ln(62.5) \approx 4.14\), \(\ln(156) \approx 5.05\), \(\ln(391) \approx 5.97\).

(b) Using the points \((1, 2.30)\) and \((5, 5.97)\):
\(\text{Gradient } m = \frac{5.97 - 2.30}{5 - 1} = 0.9175\)
Since \(m = \ln b\):
\(\ln b = 0.9175 \implies b = \text{e}^{0.9175} \approx 2.5\).

The vertical intercept \(c\):
\(c = 2.30 - 0.9175(1) = 1.3825\)
Since \(c = \ln a\):
\(\ln a = 1.3825 \implies a = \text{e}^{1.3825} \approx 4.0\).

(c) When \(y = 100\), \(\ln y = \ln 100 \approx 4.61\).
Using the linear equation:
\(4.61 = 1.38 + 0.92 \sqrt{x} \implies 0.92 \sqrt{x} = 3.23 \implies \sqrt{x} \approx 3.51 \implies x \approx 12.3\).

(a)
- **B1**: For identifying that \(\ln y\) (or \(\log_{10} y\)) is plotted against \(\sqrt{x}\).
- **B1**: For correct calculated table of values of \(\sqrt{x}\) and \(\ln y\) (allow one calculation or rounding error).

(b)
- **M1**: For an attempt to find the gradient of the line of best fit using two points.
- **A1**: For finding gradient \(\approx 0.92\) and equating to \(\ln b\).
- **A1**: For \(b \approx 2.5\).
- **M1**: For finding the vertical intercept and equating to \(\ln a\).
- **A1**: For \(a \approx 4.0\).

(c)
- **M1**: For substituting \(\ln 100 \approx 4.61\) into their linear equation to solve for \(\sqrt{x}\).
- **A1**: For \(x \approx 12.3\) (accept answers in range \(12.1\) to \(12.5\)).

(a) Using the product rule to differentiate \(y = 2x e^{2x}\):
Let \(u = 2x\) and \(v = e^{2x}\).
Then \(\frac{du}{dx} = 2\) and \(\frac{dv}{dx} = 2e^{2x}\).
\(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\)
\(\frac{dy}{dx} = (2x)(2e^{2x}) + (e^{2x})(2) = 4xe^{2x} + 2e^{2x}\).

(b) From part (a), we have:
\(\int (4xe^{2x} + 2e^{2x}) \, dx = 2xe^{2x}\)
\(\int 4xe^{2x} \, dx + \int 2e^{2x} \, dx = 2xe^{2x}\)
\(\int 4xe^{2x} \, dx = 2xe^{2x} - \int 2e^{2x} \, dx\)
\(\int 4xe^{2x} \, dx = 2xe^{2x} - e^{2x}\)
Now we evaluate this expression between the limits of 0 and 2:
\(\int_0^2 4xe^{2x} \, dx = \left[ 2xe^{2x} - e^{2x} \right]_0^2\)
\(= \left(2(2)e^{2(2)} - e^{2(2)}\right) - \left(2(0)e^{2(0)} - e^{2(0)}\right)\)
\(= (4e^4 - e^4) - (0 - e^0)\)
\(= 3e^4 - (-1)\)
\(= 3e^4 + 1\).
Therefore, \(a = 3\) and \(b = 1\).

(a)
M1: For an attempt to differentiate using the product rule on \(2x e^{2x}\), showing at least one correct term.
A1: For \(4xe^{2x} + 2e^{2x}\) (or any equivalent form).

(b)
B1: For writing \(\int 4xe^{2x} \, dx = 2xe^{2x} - \int 2e^{2x} \, dx\) or equivalent.
M1: For integrating \(2e^{2x}\) to obtain \(e^{2x}\).
M1: For substituting the limits 2 and 0 correctly into their integrated expression \(2xe^{2x} - e^{2x}\).
A1: For \(3e^4 + 1\) (or stating \(a = 3, b = 1\)).

**(a)**

Substitute the linear equation \(y = 3x - 2\) into the equation of the curve:
\((3x - 2)^2 - 2x(3x - 2) - x^2 = 2\)

Expand the terms:
\((9x^2 - 12x + 4) - (6x^2 - 4x) - x^2 = 2\)
\(9x^2 - 12x + 4 - 6x^2 + 4x - x^2 - 2 = 0\)
\(2x^2 - 8x + 2 = 0\)
\(x^2 - 4x + 1 = 0\)

Solve the quadratic equation using the quadratic formula:
\(x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(1)}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}\)

Find the corresponding \(y\)-coordinates of \(A\) and \(B\):
\(y = 3(2 \pm \sqrt{3}) - 2 = 4 \pm 3\sqrt{3}\)

So the points of intersection are \(A(2 - \sqrt{3}, 4 - 3\sqrt{3})\) and \(B(2 + \sqrt{3}, 4 + 3\sqrt{3})\).

Find the midpoint \(M\) of \(AB\):
\(M = \left(\frac{2 - \sqrt{3} + 2 + \sqrt{3}}{2}, \frac{4 - 3\sqrt{3} + 4 + 3\sqrt{3}}{2}\right) = (2, 4)\)

Find the gradient of \(AB\):
\(m_{AB} = 3\)

Thus, the gradient of the perpendicular bisector is:
\(m_{\perp} = -\frac{1}{3}\)

Find the equation of the perpendicular bisector:
\(y - 4 = -\frac{1}{3}(x - 2)\)
\(3y - 12 = -x + 2\)
\(x + 3y = 14\)

Substitute the coordinates of \(C\left(\frac{1}{2}, k\right)\) into the equation of the perpendicular bisector:
\(\frac{1}{2} + 3k = 14\)
\(3k = 13.5\)
\(k = 4.5\) (or \(\frac{9}{2}\))


**(b)**

Since the perpendicular bisector is perpendicular to \(AB\) at its midpoint \(M(2, 4)\), the perpendicular distance of any point on the perpendicular bisector from the line \(AB\) is simply its distance to the midpoint \(M\).

Let \(C\) be \(\left(\frac{1}{2}, \frac{9}{2}\right)\). The vector from \(M\) to \(C\) is:
\(\vec{MC} = C - M = \left(\frac{1}{2} - 2, \frac{9}{2} - 4\right) = \left(-\frac{3}{2}, \frac{1}{2}\right)\)

Since \(D\) lies on the perpendicular bisector and its distance from \(AB\) is twice that of \(C\), we have:
\(\vec{MD} = \pm 2\vec{MC}\)

Case 1:
\(\vec{MD} = 2\vec{MC} = 2\left(-\frac{3}{2}, \frac{1}{2}\right) = (-3, 1)\)
\(D_1 = M + (-3, 1) = (2 - 3, 4 + 1) = (-1, 5)\)

Case 2:
\(\vec{MD} = -2\vec{MC} = -2\left(-\frac{3}{2}, \frac{1}{2}\right) = (3, -1)\)
\(D_2 = M + (3, -1) = (2 + 3, 4 - 1) = (5, 3)\)

So the possible coordinates of \(D\) are \((-1, 5)\) and \((5, 3)\).

**(a)**
* **M1**: Substitute \(y = 3x - 2\) into the curve equation.
* **A1**: Obtain a correct 3-term quadratic equation, e.g., \(x^2 - 4x + 1 = 0\) or \(2x^2 - 8x + 2 = 0\).
* **M1**: Solve their quadratic equation to find the \(x\)-coordinates.
* **B1**: Find the correct midpoint \(M(2, 4)\).
* **B1**: Find the correct perpendicular gradient \(-\frac{1}{3}\).
* **M1**: Form the equation of the perpendicular bisector using their midpoint and perpendicular gradient.
* **A1**: Obtain the correct perpendicular bisector equation, e.g., \(x + 3y = 14\).
* **A1**: Find \(k = 4.5\) (or \(\frac{9}{2}\)) exact only.

**(b)**
* **B1**: Recognize that the perpendicular distance from \(AB\) is the distance to the midpoint \(M(2, 4)\).
* **M1**: Calculate the displacement vector \(\vec{MC} = \left(-\frac{3}{2}, \frac{1}{2}\right)\) (or use coordinate geometry distance equations).
* **M1**: Apply the ratio 2:1 to find the possible coordinates of \(D\), i.e., \(\vec{MD} = \pm 2\vec{MC}\).
* **A1**: Find both correct pairs of coordinates: \((-1, 5)\) and \((5, 3)\).

(a) \(\overrightarrow{PQ} = \overrightarrow{PO} + \overrightarrow{OR} + \overrightarrow{RQ} = -3\mathbf{p} + \mathbf{r} + 2\mathbf{p} = \mathbf{r} - \mathbf{p}\)

(b) Since \(PS : SQ = 1 : 2\), \(\overrightarrow{PS} = \frac{1}{3}\overrightarrow{PQ} = \frac{1}{3}(\mathbf{r} - \mathbf{p})\).
\(\overrightarrow{OS} = \overrightarrow{OP} + \dots = 3\mathbf{p} + \frac{1}{3}(\mathbf{r} - \mathbf{p}) = \frac{8}{3}\mathbf{p} + \frac{1}{3}\mathbf{r}\).

(c) \(\overrightarrow{OY} = \beta\overrightarrow{OS} = \beta\left(\frac{8}{3}\mathbf{p} + \frac{1}{3}\mathbf{r}\right) = \frac{8}{3}\beta\mathbf{p} + \frac{1}{3}\beta\mathbf{r}\).

(d) \(\overrightarrow{PR} = \overrightarrow{PO} + \overrightarrow{OR} = -3\mathbf{p} + \mathbf{r} = \mathbf{r} - 3\mathbf{p}\).
\(\overrightarrow{PY} = \alpha\overrightarrow{PR} = \alpha(\mathbf{r} - 3\mathbf{p}) = \alpha\mathbf{r} - 3\alpha\mathbf{p}\).

(e) \(\overrightarrow{OY} = \overrightarrow{OP} + \dots = 3\mathbf{p} + \alpha(\mathbf{r} - 3\mathbf{p}) = (3 - 3\alpha)\mathbf{p} + \alpha\mathbf{r}\).
Equating coefficients:
For \(\mathbf{r}\): \(\frac{1}{3}\beta = \alpha \implies \beta = 3\alpha\).
For \(\mathbf{p}\): \(\frac{8}{3}\beta = 3 - 3\alpha\).
Substituting \(\beta = 3\alpha\) gives:
\(\frac{8}{3}(3\alpha) = 3 - 3\alpha \implies 8\alpha = 3 - 3\alpha \implies 11\alpha = 3 \implies \alpha = \frac{3}{11}\).
Then \(\beta = 3\left(\frac{3}{11}\right) = \frac{9}{11}\).

(a) B1 for \(\mathbf{r} - \mathbf{p}\) or equivalent

(b) M1 for \(\overrightarrow{OP} + k(\text{their } \overrightarrow{PQ})\) where \(k = \frac{1}{3}\) or equivalent
A1 for \(\frac{8}{3}\mathbf{p} + \frac{1}{3}\mathbf{r}\) or equivalent

(c) B1 for \(\beta\left(\text{their } \frac{8}{3}\mathbf{p} + \frac{1}{3}\mathbf{r}\right)\) or equivalent

(d) B1 for \(\overrightarrow{PR} = \mathbf{r} - 3\mathbf{p}\)
B1 for \(\alpha(\mathbf{r} - 3\mathbf{p})\) or equivalent

(e) M1 for writing \(\overrightarrow{OY}\) in terms of \(\alpha\), \(\mathbf{p}\), \(\mathbf{r}\)
M1 for equating coefficients of \(\mathbf{p}\) and \(\mathbf{r}\) to form two equations
M1 for attempting to solve the simultaneous equations for \(\alpha\) and \(\beta\)
A1 for \(\alpha = \frac{3}{11}\) and \(\beta = \frac{9}{11}\)

**(a)**
Using the identity \(\cos^2 x = 1 - \sin^2 x\):
\(6(1 - \sin^2 x) - 5\sin x - 5 = 0\)
\(6 - 6\sin^2 x - 5\sin x - 5 = 0\)
\(6\sin^2 x + 5\sin x - 1 = 0\)

Factorizing the quadratic equation:
\((6\sin x - 1)(\sin x + 1) = 0\)

This gives \(\sin x = \frac{1}{6}\) or \(\sin x = -1\).

For \(\sin x = \frac{1}{6}\):
Basic angle \(\alpha = \sin^{-1}\left(\frac{1}{6}\right) \approx 9.59^\circ\)
In the range \(-180^\circ \le x \le 180^\circ\):
\(x = 9.6^\circ\) or \(x = 180^\circ - 9.59^\circ = 170.4^\circ\)

For \(\sin x = -1\):
\(x = -90^\circ\)

So the solutions are \(x = -90^\circ\), \(9.6^\circ\), \(170.4^\circ\).

**(b)**
Rearranging the equation:
\(4\sin(3\theta - 0.5) - 1 = 0 \implies \sin(3\theta - 0.5) = 0.25\)

Let \(\psi = 3\theta - 0.5\).
Since \(0 < \theta < \pi\), we have \(0 < 3\theta < 3\pi\), which gives the range for \(\psi\):
\(-0.5 < \psi < 3\pi - 0.5 \approx 8.92\)

Basic angle \(\alpha = \sin^{-1}(0.25) \approx 0.2527\) rad.

Since sine is positive, \(\psi\) lies in Quadrant 1 or 2:
- \(\psi_1 = 0.2527\)
- \(\psi_2 = \pi - 0.2527 \approx 2.8889\)
- \(\psi_3 = 2\pi + 0.2527 \approx 6.5359\)
- \(\psi_4 = 3\pi - 0.2527 \approx 9.1721\) (outside the range \(\psi < 8.92\))

Now, solve for \(\theta\):
- \(3\theta_1 - 0.5 = 0.2527 \implies 3\theta_1 = 0.7527 \implies \theta_1 \approx 0.251\)
- \(3\theta_2 - 0.5 = 2.8889 \implies 3\theta_2 = 3.3889 \implies \theta_2 \approx 1.13\)
- \(3\theta_3 - 0.5 = 6.5359 \implies 3\theta_3 = 7.0359 \implies \theta_3 \approx 2.35\)

So, \(\theta = 0.251, 1.13, 2.35\) radians.

**(a)**
**M1** for attempting to use \(\cos^2 x = 1 - \sin^2 x\) to form a three-term quadratic equation in \(\sin x\).
**M1** for solving their quadratic equation to obtain two values for \(\sin x\) (\(\sin x = \frac{1}{6}\) and \(\sin x = -1\)).
**A1** for any two correct solutions (e.g. \(-90^\circ\) and \(9.6^\circ\)).
**A1** for the third correct solution (\(170.4^\circ\)) and no extras in the range.

**(b)**
**M1** for the correct order of operations to find a basic angle \(\alpha \approx 0.253\) (or better).
**A1** for obtaining any two of the correct values of \(3\theta - 0.5\) (\(0.253\), \(2.889\), or \(6.536\)).
**M1** for solving for \(\theta\) from their values of \(3\theta - 0.5\).
**A1** for any two correct values of \(\theta\) (\(0.251\), \(1.13\), or \(2.35\)).
**A1** for the third correct value and no extras in the range.

(a) For a geometric progression with first term \(a = 1\) and common ratio \(r = 2\cos^2\theta\), a sum to infinity exists if and only if \(|r| < 1\).
Thus, we require:
\(|2\cos^2\theta| < 1 \implies 2\cos^2\theta < 1 \implies \cos^2\theta < \frac{1}{2}\).
Taking the square root gives:
\(-\frac{1}{\sqrt{2}} < \cos\theta < \frac{1}{\sqrt{2}}\).
Since \(0 < \theta < \pi\), we find that this inequality is satisfied when:
\(\frac{\pi}{4} < \theta < \frac{3\pi}{4}\).

(b) Using the formula for the sum to infinity of a geometric progression, \(S_{\infty} = \frac{a}{1-r}\):
\(\frac{1}{1 - 2\cos^2\theta} = 2\).
Rearranging this equation:
\(1 = 2(1 - 2\cos^2\theta) \implies 1 = 2 - 4\cos^2\theta \implies 4\cos^2\theta = 1 \implies \cos^2\theta = \frac{1}{4}\).
Taking the square root gives:
\(\cos\theta = \pm\frac{1}{2}\).
Since the valid range from part (a) is \(\frac{\pi}{4} < \theta < \frac{3\pi}{4}\):
For \(\cos\theta = \frac{1}{2}\), we have \(\theta = \frac{\pi}{3}\).
For \(\cos\theta = -\frac{1}{2}\), we have \(\theta = \frac{2\pi}{3}\).
Both values lie within the interval \(\frac{\pi}{4} < \theta < \frac{3\pi}{4}\).
Therefore, the exact values of \(\theta\) are \(\theta = \frac{\pi}{3}\) and \(\theta = \frac{2\pi}{3}\).

Part (a):
M1: For stating that a sum to infinity exists when \(|r| < 1\) with \(r = 2\cos^2\theta\).
A1: For obtaining \(\cos^2\theta < \frac{1}{2}\) or \(-\frac{1}{\sqrt{2}} < \cos\theta < \frac{1}{\sqrt{2}}\).
A1: For the correct set of values \(\frac{\pi}{4} < \theta < \frac{3\pi}{4}\).

Part (b):
M1: For setting up the sum to infinity equation \(\frac{1}{1 - 2\cos^2\theta} = 2\).
A1: For obtaining \(\cos^2\theta = \frac{1}{4}\) or \(\cos\theta = \pm\frac{1}{2}\).
M1: For solving to find \(\theta\) within their interval from part (a).
A1: For both exact values \(\theta = \frac{\pi}{3}\) and \(\theta = \frac{2\pi}{3}\) with no extra values in the range.

To solve the inequality \((x+3)(x+1)(x-1) > -3\), we need to find the range of values of \(x\) for which the curve \(y = (x+3)(x+1)(x-1)\) lies above the horizontal line \(y = -3\).

From the given intersection points:
- For \(x < -3.3\), the curve lies below \(y = -3\).
- For \(-3.3 < x < 0\), the curve lies above \(y = -3\).
- For \(0 < x < 0.3\), the curve lies below \(y = -3\).
- For \(x > 0.3\), the curve lies above \(y = -3\).

Therefore, the solution to the inequality is \(-3.3 < x < 0\) and \(x > 0.3\).

B1 for identifying the correct critical values of -3.3, 0, and 0.3.
B2 for both correct inequalities: \(-3.3 < x < 0\) and \(x > 0.3\) (B1 for either inequality correct).

To solve the inequality \((x+3)(x+1)(x-1) > -3\), we need to find the range of values of \(x\) for which the curve \(y = (x+3)(x+1)(x-1)\) lies above the horizontal line \(y = -3\).

From the given intersection points:
- For \(x < -3.3\), the curve lies below \(y = -3\).
- For \(-3.3 < x < 0\), the curve lies above \(y = -3\).
- For \(0 < x < 0.3\), the curve lies below \(y = -3\).
- For \(x > 0.3\), the curve lies above \(y = -3\).

Therefore, the solution to the inequality is \(-3.3 < x < 0\) and \(x > 0.3\).

B1 for identifying the correct critical values of -3.3, 0, and 0.3.
B2 for both correct inequalities: \(-3.3 < x < 0\) and \(x > 0.3\) (B1 for either inequality correct).

(a) \(\mathrm{f}(x) = 2 - 8x - x^2 = -(x^2 + 8x) + 2 = -((x+4)^2 - 16) + 2 = 18 - (x+4)^2\).
So \(a = 18\) and \(b = 4\).

(b) Since \(\mathrm{f}(x) = 18 - (x+4)^2\), the maximum value of the function is 18. Thus, the range is \(\mathrm{f}(x) \le 18\).

(c) For \(\mathrm{g}\) to have an inverse, it must be a one-to-one function. The vertex of the quadratic curve is at \(x = -4\). Since the domain is \(x \ge k\), the least possible value of \(k\) is \(-4\).

(d) For \(x \ge -4\), let \(y = 18 - (x+4)^2\).
Rearranging to express \(x\) in terms of \(y\):
\((x+4)^2 = 18 - y\)
\(x+4 = \sqrt{18-y}\) (taking the positive square root because \(x \ge -4\))
\(x = -4 + \sqrt{18-y}\)

Replacing \(y\) with \(x\) gives:
\(\mathrm{g}^{-1}(x) = -4 + \sqrt{18-x}\)

Domain of \(\mathrm{g}^{-1}\) is the range of \(\mathrm{g}\), which is \(x \le 18\).
Range of \(\mathrm{g}^{-1}\) is the domain of \(\mathrm{g}\), which is \(\mathrm{g}^{-1}(x) \ge -4\).

(a)
B2: For \(18 - (x+4)^2\)
B1: For \(-(x+4)^2\) or \((x+4)^2\) or \(a = 18\) and \(b = 4\)

(b)
B1: \(\mathrm{f}(x) \le 18\) or \(\mathrm{f} \le 18\) (FT their 18)

(c)
B1: \(-4\) (FT \(- \text{their } b\))

(d)
M1: Complete method to find inverse function: Swaps the variables and rearranges, or rearranges and swaps the variables at some point in their solution, using their part (a) provided it is in the form \(a - (x+b)^2\).
A2: \([\mathrm{g}^{-1}(x) =] -4 + \sqrt{18-x}\)
(A1 for \(-4 \pm \sqrt{18-x}\))
B1: [Domain:] \(x \le 18\) (FT their part (b) provided it is in the form \(\mathrm{f}(x) \le a\) where \(a\) is a constant)
B1: [Range:] \(\mathrm{g}^{-1}(x) \ge -4\) (FT their value of \(k\) in part (c))

(a) \(\mathrm{f}(x) = 2 - 8x - x^2 = -(x^2 + 8x) + 2 = -((x+4)^2 - 16) + 2 = 18 - (x+4)^2\).
So \(a = 18\) and \(b = 4\).

(b) Since \(\mathrm{f}(x) = 18 - (x+4)^2\), the maximum value of the function is 18. Thus, the range is \(\mathrm{f}(x) \le 18\).

(c) For \(\mathrm{g}\) to have an inverse, it must be a one-to-one function. The vertex of the quadratic curve is at \(x = -4\). Since the domain is \(x \ge k\), the least possible value of \(k\) is \(-4\).

(d) For \(x \ge -4\), let \(y = 18 - (x+4)^2\).
Rearranging to express \(x\) in terms of \(y\):
\((x+4)^2 = 18 - y\)
\(x+4 = \sqrt{18-y}\) (taking the positive square root because \(x \ge -4\))
\(x = -4 + \sqrt{18-y}\)

Replacing \(y\) with \(x\) gives:
\(\mathrm{g}^{-1}(x) = -4 + \sqrt{18-x}\)

Domain of \(\mathrm{g}^{-1}\) is the range of \(\mathrm{g}\), which is \(x \le 18\).
Range of \(\mathrm{g}^{-1}\) is the domain of \(\mathrm{g}\), which is \(\mathrm{g}^{-1}(x) \ge -4\).

(a)
B2: For \(18 - (x+4)^2\)
B1: For \(-(x+4)^2\) or \((x+4)^2\) or \(a = 18\) and \(b = 4\)

(b)
B1: \(\mathrm{f}(x) \le 18\) or \(\mathrm{f} \le 18\) (FT their 18)

(c)
B1: \(-4\) (FT \(- \text{their } b\))

(d)
M1: Complete method to find inverse function: Swaps the variables and rearranges, or rearranges and swaps the variables at some point in their solution, using their part (a) provided it is in the form \(a - (x+b)^2\).
A2: \([\mathrm{g}^{-1}(x) =] -4 + \sqrt{18-x}\)
(A1 for \(-4 \pm \sqrt{18-x}\))
B1: [Domain:] \(x \le 18\) (FT their part (b) provided it is in the form \(\mathrm{f}(x) \le a\) where \(a\) is a constant)
B1: [Range:] \(\mathrm{g}^{-1}(x) \ge -4\) (FT their value of \(k\) in part (c))

(a) \(\mathrm{f}(x) = 2 - 8x - x^2 = -(x^2 + 8x) + 2 = -((x+4)^2 - 16) + 2 = 18 - (x+4)^2\).
So \(a = 18\) and \(b = 4\).

(b) Since \(\mathrm{f}(x) = 18 - (x+4)^2\), the maximum value of the function is 18. Thus, the range is \(\mathrm{f}(x) \le 18\).

(c) For \(\mathrm{g}\) to have an inverse, it must be a one-to-one function. The vertex of the quadratic curve is at \(x = -4\). Since the domain is \(x \ge k\), the least possible value of \(k\) is \(-4\).

(d) For \(x \ge -4\), let \(y = 18 - (x+4)^2\).
Rearranging to express \(x\) in terms of \(y\):
\((x+4)^2 = 18 - y\)
\(x+4 = \sqrt{18-y}\) (taking the positive square root because \(x \ge -4\))
\(x = -4 + \sqrt{18-y}\)

Replacing \(y\) with \(x\) gives:
\(\mathrm{g}^{-1}(x) = -4 + \sqrt{18-x}\)

Domain of \(\mathrm{g}^{-1}\) is the range of \(\mathrm{g}\), which is \(x \le 18\).
Range of \(\mathrm{g}^{-1}\) is the domain of \(\mathrm{g}\), which is \(\mathrm{g}^{-1}(x) \ge -4\).

(a)
B2: For \(18 - (x+4)^2\)
B1: For \(-(x+4)^2\) or \((x+4)^2\) or \(a = 18\) and \(b = 4\)

(b)
B1: \(\mathrm{f}(x) \le 18\) or \(\mathrm{f} \le 18\) (FT their 18)

(c)
B1: \(-4\) (FT \(- \text{their } b\))

(d)
M1: Complete method to find inverse function: Swaps the variables and rearranges, or rearranges and swaps the variables at some point in their solution, using their part (a) provided it is in the form \(a - (x+b)^2\).
A2: \([\mathrm{g}^{-1}(x) =] -4 + \sqrt{18-x}\)
(A1 for \(-4 \pm \sqrt{18-x}\))
B1: [Domain:] \(x \le 18\) (FT their part (b) provided it is in the form \(\mathrm{f}(x) \le a\) where \(a\) is a constant)
B1: [Range:] \(\mathrm{g}^{-1}(x) \ge -4\) (FT their value of \(k\) in part (c))

(a) Expanding the left-hand side:
\((3\sec\theta - 2\tan\theta)(3\sec\theta + 2\tan\theta) = 9\sec^2\theta - 4\tan^2\theta\)
Using the identity \(\tan^2\theta = \sec^2\theta - 1\):
\(9\sec^2\theta - 4(\sec^2\theta - 1) = 9\sec^2\theta - 4\sec^2\theta + 4 = 5\sec^2\theta + 4\)

(b) Using the result from part (a):
\(5\sec^2\theta + 4 = 14\)
\(5\sec^2\theta = 10\)
\(\sec^2\theta = 2\)
This gives:
\(\cos^2\theta = 0.5\)
\(\cos\theta = \pm\frac{1}{\sqrt{2}}\)
Solving for \(0^\circ \le \theta \le 360^\circ\):
\(\theta = 45^\circ, 135^\circ, 225^\circ, 315^\circ\)

(a)
**M1**: For expanding the LHS to obtain \(9\sec^2\theta - 4\tan^2\theta\) or equivalent.
**A1**: For substituting \(\tan^2\theta = \sec^2\theta - 1\) and correctly completing the proof to reach the given answer \(5\sec^2\theta + 4\).

(b)
**M2**: For obtaining \(\cos^2\theta = 0.5\) or \(\cos\theta = \pm\frac{1}{\sqrt{2}}\)
(or **M1** for obtaining \(\sec^2\theta = 2\) or \(\tan^2\theta = 1\)).
**A2**: For all four correct angles: \(45^\circ, 135^\circ, 225^\circ, 315^\circ\) and no extras in the range.
(Award **A1** if at least two correct angles are found, ignoring extras).

(a) The total area of the badge consists of the area of the sector and the area of the square:
\[\text{Total Area} = \frac{1}{2}r^2\theta + r^2\]
Since the total area is \(24\text{ cm}^2\):
\[\frac{1}{2}r^2\theta + r^2 = 24\]
Multiplying by \(\frac{2}{r}\):
\[r\theta + 2r = \frac{48}{r} \implies r\theta = \frac{48}{r} - 2r\]

The perimeter, \(P\), of the badge is made up of the arc of the sector, one straight side of the sector, and three outer sides of the square:
\[P = r\theta + r + 3r = r\theta + 4r\]

Substituting \(r\theta = \frac{48}{r} - 2r\) into the perimeter equation gives:
\[P = \left(\frac{48}{r} - 2r\right) + 4r = 2r + \frac{48}{r}\]

(b) To find the stationary value of \(P\):
\[P = 2r + 48r^{-1}\]
\[\frac{\text{d}P}{\text{d}r} = 2 - 48r^{-2} = 2 - \frac{48}{r^2}\]

Setting \(\frac{\text{d}P}{\text{d}r} = 0\):
\[2 - \frac{48}{r^2} = 0 \implies r^2 = 24 \implies r = \sqrt{24} = 2\sqrt{6}\text{ (since } r > 0\text{)}\]

Substituting \(r = 2\sqrt{6}\) back into \(P\):
\[P = 2(2\sqrt{6}) + \frac{48}{2\sqrt{6}} = 4\sqrt{6} + 4\sqrt{6} = 8\sqrt{6} \approx 19.6\text{ cm}\]

To determine the nature of this stationary point:
\[\frac{\text{d}^2P}{\text{d}r^2} = \frac{96}{r^3}\]
Since \(r = 2\sqrt{6} > 0\), the second derivative is positive:
\[\frac{\text{d}^2P}{\text{d}r^2} > 0\]
Therefore, the stationary value is a minimum.

(a)
**B1**: Writes a correct expression for the total area: \(\frac{1}{2}r^2\theta + r^2 = 24\) oe.
**M1**: Expresses \(r\theta\) in terms of \(r\) or substitutes \(\theta\) to eliminate it: \(r\theta = \frac{48}{r} - 2r\) oe.
**A1**: Substitutes into the perimeter expression \(P = r\theta + 4r\) and shows intermediate steps to obtain the given result \(P = 2r + \frac{48}{r}\) convincingly.

(b)
**B1**: Correctly differentiates \(P\) to obtain \(\frac{\text{d}P}{\text{d}r} = 2 - \frac{48}{r^2}\) oe.
**M1**: Sets \(\frac{\text{d}P}{\text{d}r} = 0\) and attempts to solve for \(r\).
**A1**: Obtains \(r = \sqrt{24}\) (or \(2\sqrt{6}\) or awrt \(4.90\)).
**A1**: Obtains stationary value \(P = 8\sqrt{6}\) (or awrt \(19.6\)).
**A1**: Correctly determines that the nature is a minimum by evaluating the second derivative \(\frac{\text{d}^2P}{\text{d}r^2} = \frac{96}{r^3} > 0\) or using a sign table.

At \(x = 4\), the \(y\)-coordinate is \(y = \frac{\sqrt{2(4)+1}}{4-2} = \frac{3}{2}\). To find the gradient of the tangent, differentiate \(y = \frac{\sqrt{2x+1}}{x-2}\) using the quotient rule: \(\frac{dy}{dx} = \frac{(x-2)\frac{d}{dx}(\sqrt{2x+1}) - \sqrt{2x+1}\frac{d}{dx}(x-2)}{(x-2)^2}\). This gives: \(\frac{dy}{dx} = \frac{(x-2)(2x+1)^{-1/2} - (2x+1)^{1/2}}{(x-2)^2}\). Substituting \(x = 4\) into \(\frac{dy}{dx}\) gives the gradient \(m = \frac{(4-2)(9)^{-1/2} - 9^{1/2}}{(4-2)^2} = \frac{2(\frac{1}{3}) - 3}{4} = -\frac{7}{12}\). The equation of the tangent line is: \(y - \frac{3}{2} = -\frac{7}{12}(x - 4)\), which simplifies to \(y = -\frac{7}{12}x + \frac{23}{6}\). The line meets \(x - 3y = 5\), which is \(y = \frac{1}{3}x - \frac{5}{3}\). Equating the two expressions for \(y\): \(\frac{1}{3}x - \frac{5}{3} = -\frac{7}{12}x + \frac{23}{6}\). Multiplying by 12 yields \(4x - 20 = -7x + 46\), so \(11x = 66\), giving \(x = 6\). Substituting \(x = 6\) back into \(y = \frac{1}{3}x - \frac{5}{3}\) gives \(y = \frac{1}{3}\). Thus, the coordinates of \(P\) are \((6, \frac{1}{3})\).

B1: For \(y = \frac{3}{2}\) at \(x = 4\) (seen or implied). M2: For attempt to differentiate using the quotient rule (M1 for any correct quotient rule application with one minor error; M2 for fully correct derivative structure). M1: For substituting \(x = 4\) into their derivative to find the gradient of the tangent. A1: For obtaining the correct gradient \(-\frac{7}{12}\) (or equivalent). A1: For the correct equation of the tangent (e.g. \(y = -\frac{7}{12}x + \frac{23}{6}\) or equivalent). M1: For equating their tangent equation with the given line to solve for \(x\) or \(y\). A1: For the correct coordinates \((6, \frac{1}{3})\) (dep on all previous M marks).

Integrate each term with respect to \(x\): \(\int 3\sin(3x) \, \text{d}x = -\cos(3x)\) and \(\int 4\cos(2x) \, \text{d}x = 2\sin(2x)\). Combine the integrated terms: \(\left[ -\cos(3x) + 2\sin(2x) \right]_{0}^{\frac{\pi}{6}}\). Substitute the upper limit \(x = \frac{\pi}{6}\): \(-\cos\left(3 \cdot \frac{\pi}{6}\right) + 2\sin\left(2 \cdot \frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{2}\right) + 2\sin\left(\frac{\pi}{3}\right) = 0 + 2\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}\). Substitute the lower limit \(x = 0\): \(-\cos(0) + 2\sin(0) = -1 + 0 = -1\). Subtract the lower limit value from the upper limit value: \(\sqrt{3} - (-1) = \sqrt{3} + 1\).

B1: For \(-\cos(3x)\)
B1: For \(2\sin(2x)\)
M1: For substituting the limits \(\frac{\pi}{6}\) and \(0\) into an expression of the form \(a\cos(3x) + b\sin(2x)\)
A1: For obtaining \(\sqrt{3}\) for the upper limit and \(-1\) for the lower limit (soi)
A1: For the exact final answer of \(\sqrt{3} + 1\)

(a) The general term in the expansion of \((2x^2 + \frac{k}{x})^9\) is given by:
\(T_{r+1} = \binom{9}{r}(2x^2)^{9-r}(\frac{k}{x})^r = \binom{9}{r} 2^{9-r} k^r x^{18-3r}\)

For the term in \(x^3\), we set:
\(18 - 3r = 3 \Rightarrow r = 5\)
Thus, the coefficient of \(x^3\) is:
\(C_1 = \binom{9}{5} 2^{9-5} k^5 = 126 \times 16 \times k^5 = 2016k^5\)

For the term in \(x^6\), we set:
\(18 - 3r = 6 \Rightarrow r = 4\)
Thus, the coefficient of \(x^6\) is:
\(C_2 = \binom{9}{4} 2^{9-4} k^4 = 126 \times 32 \times k^4 = 4032k^4\)

We are given that the coefficient of \(x^3\) is 4 times the coefficient of \(x^6\):
\(2016k^5 = 4 \times 4032k^4\)
Dividing both sides by \(2016k^4\) (since \(k \neq 0\)):
\(k = 4 \times 2 = 8\)

(b) The expansion of \((1 + ax)^n\) is:
\((1 + ax)^n = 1 + n(ax) + \frac{n(n-1)}{2}(ax)^2 + \dots = 1 + anx + \frac{a^2n(n-1)}{2}x^2 + \dots\)

Equating coefficients:
\(an = 15\) --- (Equation 1)
\(\frac{a^2n(n-1)}{2} = 90 \Rightarrow a^2n(n-1) = 180\) --- (Equation 2)

From Equation 2:
\((an) \times a(n-1) = 180\)
Substitute \(an = 15\):
\(15 a(n-1) = 180 \Rightarrow a(n-1) = 12\)
\(an - a = 12\)
Substitute \(an = 15\):
\(15 - a = 12 \Rightarrow a = 3\)

Substitute \(a = 3\) into Equation 1:
\(3n = 15 \Rightarrow n = 5\)

(a)
**M1**: For identifying the general term exponent equation \(18-3r = 3\) or \(18-3r=6\) to find the correct value of \(r\) (5 or 4).
**M1**: For calculating \(\binom{9}{5} 2^4 = 2016\) and \(\binom{9}{4} 2^5 = 4032\) (or a simplified ratio of their coefficients).
**M1**: For setting up the equation \(2016k^5 = 4 \times 4032k^4\) and attempting to solve for \(k\).
**A1**: For obtaining \(k = 8\).

(b)
**M1**: For setting up the equations \(an = 15\) and \(\frac{a^2n(n-1)}{2} = 90\).
**M1**: For solving the simultaneous equations to find either \(a\) or \(n\).
**A1**: For obtaining \(a = 3\) and \(n = 5\) (both correct).

(a) The general term in the expansion of \((2x^2 + \frac{k}{x})^9\) is given by:
\(T_{r+1} = \binom{9}{r}(2x^2)^{9-r}(\frac{k}{x})^r = \binom{9}{r} 2^{9-r} k^r x^{18-3r}\)

For the term in \(x^3\), we set:
\(18 - 3r = 3 \Rightarrow r = 5\)
Thus, the coefficient of \(x^3\) is:
\(C_1 = \binom{9}{5} 2^{9-5} k^5 = 126 \times 16 \times k^5 = 2016k^5\)

For the term in \(x^6\), we set:
\(18 - 3r = 6 \Rightarrow r = 4\)
Thus, the coefficient of \(x^6\) is:
\(C_2 = \binom{9}{4} 2^{9-4} k^4 = 126 \times 32 \times k^4 = 4032k^4\)

We are given that the coefficient of \(x^3\) is 4 times the coefficient of \(x^6\):
\(2016k^5 = 4 \times 4032k^4\)
Dividing both sides by \(2016k^4\) (since \(k \neq 0\)):
\(k = 4 \times 2 = 8\)

(b) The expansion of \((1 + ax)^n\) is:
\((1 + ax)^n = 1 + n(ax) + \frac{n(n-1)}{2}(ax)^2 + \dots = 1 + anx + \frac{a^2n(n-1)}{2}x^2 + \dots\)

Equating coefficients:
\(an = 15\) --- (Equation 1)
\(\frac{a^2n(n-1)}{2} = 90 \Rightarrow a^2n(n-1) = 180\) --- (Equation 2)

From Equation 2:
\((an) \times a(n-1) = 180\)
Substitute \(an = 15\):
\(15 a(n-1) = 180 \Rightarrow a(n-1) = 12\)
\(an - a = 12\)
Substitute \(an = 15\):
\(15 - a = 12 \Rightarrow a = 3\)

Substitute \(a = 3\) into Equation 1:
\(3n = 15 \Rightarrow n = 5\)

(a)
**M1**: For identifying the general term exponent equation \(18-3r = 3\) or \(18-3r=6\) to find the correct value of \(r\) (5 or 4).
**M1**: For calculating \(\binom{9}{5} 2^4 = 2016\) and \(\binom{9}{4} 2^5 = 4032\) (or a simplified ratio of their coefficients).
**M1**: For setting up the equation \(2016k^5 = 4 \times 4032k^4\) and attempting to solve for \(k\).
**A1**: For obtaining \(k = 8\).

(b)
**M1**: For setting up the equations \(an = 15\) and \(\frac{a^2n(n-1)}{2} = 90\).
**M1**: For solving the simultaneous equations to find either \(a\) or \(n\).
**A1**: For obtaining \(a = 3\) and \(n = 5\) (both correct).

(a) To find the points of intersection, equate the two equations:
\(\sqrt{x+2} = \frac{1}{2}x + 1\)

Square both sides:
\(x + 2 = \left(\frac{1}{2}x + 1\right)^2\)
\(x + 2 = \frac{1}{4}x^2 + x + 1\)

Subtract \(x + 2\) from both sides:
\(0 = \frac{1}{4}x^2 - 1\)
\(x^2 = 4\)
\(x = 2\) or \(x = -2\)

Substituting back into either equation to find the \(y\)-coordinates:
For \(x = 2\), \(y = \frac{1}{2}(2) + 1 = 2\)
For \(x = -2\), \(y = \frac{1}{2}(-2) + 1 = 0\)

So the coordinates of the points of intersection are \((-2, 0)\) and \((2, 2)\).

(b) The area \(A\) of the enclosed region is given by the integral of the upper curve minus the lower line from \(x = -2\) to \(x = 2\):
\(A = \int_{-2}^{2} \left( \sqrt{x+2} - \left(\frac{1}{2}x + 1\right) \right) dx\)

Integrate each term:
\(A = \left[ \frac{2}{3}(x+2)^{3/2} - \frac{1}{4}x^2 - x \right]_{-2}^{2}\)

Substitute the upper limit \(x = 2\):
\(\left( \frac{2}{3}(2+2)^{3/2} - \frac{1}{4}(2)^2 - 2 \right) = \frac{2}{3}(4)^{3/2} - 1 - 2 = \frac{2}{3}(8) - 3 = \frac{16}{3} - 3 = \frac{7}{3}\)

Substitute the lower limit \(x = -2\):
\(\left( \frac{2}{3}(-2+2)^{3/2} - \frac{1}{4}(-2)^2 - (-2) \right) = 0 - 1 + 2 = 1\)

Subtract the lower limit evaluation from the upper limit evaluation:
\(A = \frac{7}{3} - 1 = \frac{4}{3}\)

(a)
**M1**: Equating \(\sqrt{x+2}\) and \(\frac{1}{2}x+1\) and attempting to solve by squaring both sides.
**M1**: Solving the resulting quadratic equation to find two values of \(x\).
**A1**: Correct coordinates \((-2, 0)\) and \((2, 2)\).

(b)
**M1**: Writing down a correct integral expression for the area with limits from part (a).
**M1**: Attempting to integrate to obtain an expression of the form \(a(x+2)^{3/2} - bx^2 - cx\).
**A1**: Correct integration: \(\frac{2}{3}(x+2)^{3/2} - \frac{1}{4}x^2 - x\).
**A1**: Correct substitution of limits to obtain \(\frac{4}{3}\) or equivalent fraction.

To simplify the expression without using a calculator:

First, rationalise the fraction by multiplying the numerator and the denominator by the conjugate of the denominator, which is \(3+\sqrt{7}\):

\[\frac{11+5\sqrt{7}}{3-\sqrt{7}} = \frac{(11+5\sqrt{7})(3+\sqrt{7})}{(3-\sqrt{7})(3+\sqrt{7})}\]

Expand the numerator:
\[(11+5\sqrt{7})(3+\sqrt{7}) = 11(3) + 11\sqrt{7} + 15\sqrt{7} + 5(7)\]
\[= 33 + 26\sqrt{7} + 35\]
\[= 68 + 26\sqrt{7}\]

Expand the denominator:
\[(3-\sqrt{7})(3+\sqrt{7}) = 3^2 - (\sqrt{7})^2 = 9 - 7 = 2\]

Write the rationalised fraction:
\[\frac{68 + 26\sqrt{7}}{2} = 34 + 13\sqrt{7}\]

Subtract 2 as required by the original expression:
\[(34 + 13\sqrt{7}) - 2 = 32 + 13\sqrt{7}\]

Comparing this with the form \(p+q\sqrt{7}\), we get \(p = 32\) and \(q = 13\).

**M1** for an attempt to multiply the numerator and denominator by the conjugate \(3+\sqrt{7}\)
**A1** for obtaining the correct expanded numerator \(68 + 26\sqrt{7}\) or denominator \(2\)
**A1** for obtaining \(34 + 13\sqrt{7}\) from simplifying the fraction
**A1** for the final answer \(32 + 13\sqrt{7}\) (or stating \(p=32\) and \(q=13\))

**(a)**
The annual waste forms a geometric sequence with first term \(a = 800\) and common ratio \(r = 1 - 0.15 = 0.85\).
The sum of the first \(n\) terms is given by:
\[S_n = \frac{a(1 - r^n)}{1 - r}\]
We require:
\[\frac{800(1 - 0.85^n)}{1 - 0.85} > 4000\]
\[\frac{800(1 - 0.85^n)}{0.15} > 4000\]
\[1 - 0.85^n > \frac{4000 \times 0.15}{800}\]
\[1 - 0.85^n > 0.75\]
\[0.85^n < 0.25\]
Taking natural logarithms on both sides:
\[n \ln(0.85) < \ln(0.25)\]
Since \(\ln(0.85) < 0\), dividing by it reverses the inequality:
\[n > \frac{\ln(0.25)}{\ln(0.85)} \approx 8.53\]
Since \(n\) must be an integer, the smallest value is \(n = 9\).

**(b)**
Let the arithmetic progression have first term \(A\) and common difference \(d\).
The 1st, 3rd, and 5th terms of the geometric progression are \(u_1 = a\), \(u_3 = a r^2\), and \(u_5 = a r^4\).
These correspond to the 2nd, 5th, and 11th terms of the arithmetic progression respectively:
\[T_2 = A + d = a\]
\[T_5 = A + 4d = a r^2\]
\[T_{11} = A + 10d = a r^4\]
Subtracting the equations to eliminate \(A\):
\[T_5 - T_2 = 3d = a r^2 - a = a(r^2 - 1)\]
\[T_{11} - T_5 = 6d = a r^4 - a r^2 = a r^2(r^2 - 1)\]
Since \(6d = 2(3d)\), we can write:
\[a r^2(r^2 - 1) = 2a(r^2 - 1)\]
Given that \(a \neq 0\) and \(r > 1\) (so \(r^2 - 1 \neq 0\)), we can divide both sides by \(a(r^2 - 1)\):
\[r^2 = 2\]
Since \(r > 1\), we have:
\[r = \sqrt{2}\]

**(a)**
* **B1** for identifying \(r = 0.85\) or writing a correct equation for the sum \(S_n\).
* **M1** for setting up the inequality \(\frac{800(1 - 0.85^n)}{1 - 0.85} > 4000\).
* **M1** for correct use of logarithms to solve an inequality of the form \(b^n < c\).
* **A1** for \(n = 9\).

**(b)**
* **B1** for expressing the terms of the arithmetic progression in terms of \(a\) and \(r\) (e.g., \(A+d=a\), \(A+4d=ar^2\), \(A+10d=ar^4\)).
* **M1** for subtracting terms to obtain expressions for \(d\) (e.g., \(3d = a(r^2-1)\) and \(6d = ar^2(r^2-1)\)).
* **M1** for eliminating \(d\) to obtain a quadratic equation in terms of \(r^2\) (e.g., \(ar^2(r^2-1) = 2a(r^2-1)\)).
* **A1** for simplifying to \(r^2 = 2\) and stating \(r = \sqrt{2}\) only (rejecting \(-\sqrt{2}\) since \(r > 1\)).

(a)
**Method 1: Direct selection**
Since Alex and Ben must be in different groups, we can place Alex in Group 1 and Ben in Group 2.
We then choose 4 more players from the remaining 8 players to join Alex in Group 1.
The remaining 4 players automatically join Ben in Group 2.
Number of ways = \(\binom{8}{4} = 70\).

**Method 2: Complement method**
Total number of ways to divide 10 players into two groups of 5 is:
\(\frac{\binom{10}{5}}{2!} = \frac{252}{2} = 126\).

Number of ways where Alex and Ben are in the same group:
We choose 3 more players from the remaining 8 players to join Alex and Ben.
Number of ways = \(\binom{8}{3} = 56\).

Number of ways where they are in different groups = \(126 - 56 = 70\).

(b)
Let \(d\) be the number of doctors and \(n\) be the number of nurses chosen.
We require:
1. \(d + n = 5\)
2. \(n \ge 2\)
3. \(d \le n\)

We analyze the possible combinations of \((d, n)\):
- If \(n = 2\): then \(d = 3\). This violates \(d \le n\) (since \(3 > 2\)).
- If \(n = 3\): then \(d = 2\). This satisfies both conditions (\(3 \ge 2\) and \(2 \le 3\)).
- If \(n = 4\): then \(d = 1\). This satisfies both conditions (\(4 \ge 2\) and \(1 \le 4\)).
- If \(n = 5\): impossible as there are only 4 nurses available.

So the valid combinations are:
- **Case 1**: 2 doctors and 3 nurses
Number of ways = \(\binom{6}{2} \times \binom{4}{3} = 15 \times 4 = 60\).

- **Case 2**: 1 doctor and 4 nurses
Number of ways = \(\binom{6}{1} \times \binom{4}{4} = 6 \times 1 = 6\).

Total number of ways = \(60 + 6 = 66\).

(a)
**M1** for an attempt to find \(\binom{8}{4}\) OR for finding total number of ways and subtracting the number of ways they are in the same group.
**M1** for \(\binom{8}{4}\) or \(126 - 56\) seen.
**A1** for 70.

(b)
**M1** for identifying the two correct cases: (2 doctors, 3 nurses) and (1 doctor, 4 nurses).
**M1** for calculating the number of ways for at least one correct case: \(\binom{6}{2} \times \binom{4}{3}\) or \(\binom{6}{1} \times \binom{4}{4}\).
**A1** for 66.

(a) Acceleration \(a\) is given by the derivative of velocity \(v\) with respect to time \(t\):
\[a = \frac{\text{d}v}{\text{d}t} = -9\sin(3t) - 12\cos(3t).\]
When \(t = \frac{\pi}{6}\):
\[a = -9\sin\left(\frac{\pi}{2}\right) - 12\cos\left(\frac{\pi}{2}\right) = -9(1) - 12(0) = -9\text{ m s}^{-2}.\]

(b) The particle changes direction when \(v = 0\):
\[3\cos(3t) - 4\sin(3t) = 0 \implies \tan(3t) = \frac{3}{4}.\]
Since \(t > 0\), the first positive solution is when \(3t = \arctan(0.75)\) (in the first quadrant), which gives:
\[\sin(3t) = \frac{3}{5} \quad \text{and} \quad \cos(3t) = \frac{4}{5}.\]
We integrate the velocity to find the displacement \(s\):
\[s = \int v \text{ d}t = \int (3\cos(3t) - 4\sin(3t)) \text{ d}t = \sin(3t) + \frac{4}{3}\cos(3t) + c.\]
Given that the particle passes through \(O\) at \(t = 0\), we have \(s = 0\) when \(t = 0\):
\[0 = \sin(0) + \frac{4}{3}\cos(0) + c \implies c = -\frac{4}{3}.\]
Thus, the displacement expression is:
\[s = \sin(3t) + \frac{4}{3}\cos(3t) - \frac{4}{3}.\]
At the instant when the particle first changes direction, we substitute \(\sin(3t) = \frac{3}{5}\) and \(\cos(3t) = \frac{4}{5}\):
\[s = \frac{3}{5} + \frac{4}{3}\left(\frac{4}{5}\right) - \frac{4}{3} = \frac{9}{15} + \frac{16}{15} - \frac{20}{15} = \frac{5}{15} = \frac{1}{3}\text{ m}.\]

(c) From part (b), we have:
\[s = \sin(3t) + \frac{4}{3}\cos(3t) - \frac{4}{3} \implies \sin(3t) + \frac{4}{3}\cos(3t) = s + \frac{4}{3}.\]
Using the expression for acceleration \(a\) from part (a):
\[a = -9\sin(3t) - 12\cos(3t) = -9\left(\sin(3t) + \frac{4}{3}\cos(3t)\right).\]
Substituting the expression for \(s\):
\[a = -9\left(s + \frac{4}{3}\right) = -9s - 12.\]

(a)
- **M1**: For differentiating \(v\) with at least one correct term.
- **A1**: For obtaining the correct acceleration \(a = -9\).

(b)
- **B1**: For setting \(v = 0\) to find \(\tan(3t) = \frac{3}{4}\) (or \(t \approx 0.215\)).
- **M1**: For integrating \(v\) to obtain a form of \(A\sin(3t) + B\cos(3t)\).
- **A1**: For correct integration to get \(s = \sin(3t) + \frac{4}{3}\cos(3t) + c\).
- **M1**: For using \(t = 0, s = 0\) to find the correct value of \(c = -\frac{4}{3}\).
- **M1**: For substituting the critical values of \(\sin(3t)\) and \(\cos(3t)\) (or \(t\)) into their expression for \(s\).
- **A1**: For obtaining the exact displacement \(\frac{1}{3}\) (or awrt \(0.333\)).

(c)
- **B1**: For obtaining \(a = -9s - 12\) (or any equivalent form).