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To find the coefficient of \(x^2\) in the expansion of \(\left(3x - \frac{2}{x}\right)^6\), we write the general term of the expansion:

\(T_{r+1} = \binom{6}{r} (3x)^{6-r} \left(-\frac{2}{x}\right)^r\)

Simplify the term to separate the constants and the variable \(x\):

\(T_{r+1} = \binom{6}{r} 3^{6-r} (-2)^r \frac{x^{6-r}}{x^r} = \binom{6}{r} 3^{6-r} (-2)^r x^{6-2r}\)

We require the term in \(x^2\), so we set the exponent of \(x\) equal to 2:

\(6 - 2r = 2\)
\(2r = 4 \implies r = 2\)

Substitute \(r = 2\) back into the term:

\(\text{Term} = \binom{6}{2} (3x)^4 \left(-\frac{2}{x}\right)^2 = 15 \times 81x^4 \times \frac{4}{x^2} = 4860x^2\)

Therefore, the coefficient of \(x^2\) is \(4860\).

**M1**: For attempting to find the general term or identifying the correct term in the expansion with a correct combination of powers.

**A1**: For obtaining \(r = 2\) or showing \(\binom{6}{2} (3)^4 (-2)^2\) (condoning sign errors in \(-2\) at this stage).

**A1**: For the correct final answer of \(4860\) (must be a single value, do not accept \(4860x^2\) as the final answer unless \(4860\) is clearly identified as the coefficient).

To find the equation of the tangent line \(L\):

First, find the center of the circle by completing the square for the circle's equation:
\( (x - 2)^2 - 4 + (y - 3)^2 - 9 + 8 = 0 \)
\( (x - 2)^2 + (y - 3)^2 = 5 \)

So, the center of the circle is \(C(2, 3)\).

Next, find the gradient of the radius \(CP\):
\( m_{\text{radius}} = \frac{2 - 3}{4 - 2} = \frac{-1}{2} = -\frac{1}{2} \)

Since the tangent line \(L\) is perpendicular to the radius at the point of contact \(P\), its gradient \(m_L\) is:
\( m_L = -\frac{1}{m_{\text{radius}}} = -\frac{1}{-1/2} = 2 \)

Now, find the equation of the tangent line \(L\) using the point-slope form with \(P(4, 2)\):
\( y - 2 = 2(x - 4) \)
\( y - 2 = 2x - 8 \)
\( y = 2x - 6 \)

Finally, reflect the line \(L\) in the line \(y = x\). This transformation interchanges \(x\) and \(y\):
\( x = 2y - 6 \)

Rearrange this equation to express it in the form \(y = mx + c\):
\( 2y = x + 6 \)
\( y = \frac{1}{2}x + 3 \)

M1: Attempt to find the center of the circle by completing the square (or other valid method), obtaining center \((2, 3)\).
M1: Attempt to find the gradient of the radius and use the perpendicular gradient rule \(m_1 m_2 = -1\) to find the gradient of the tangent, obtaining \(m = 2\).
A1: Correct equation for the tangent line \(L\), e.g., \(y = 2x - 6\).
M1: Reflect the tangent line in \(y = x\) by interchanging \(x\) and \(y\) in their equation of \(L\).
A1: Obtain the correct final equation \(y = \frac{1}{2}x + 3\) (or \(y = 0.5x + 3\)).

**(a)**

Start with the left-hand side (LHS) and express it over a common denominator:
\[\text{LHS} = \frac{1}{\cos \theta} - \frac{\cos \theta}{1 + \sin \theta} = \frac{(1 + \sin \theta) - \cos^2 \theta}{\cos \theta(1 + \sin \theta)}\]

Substitute the trigonometric identity \(\cos^2 \theta = 1 - \sin^2 \theta\) into the numerator:
\[\text{LHS} = \frac{1 + \sin \theta - (1 - \sin^2 \theta)}{\cos \theta(1 + \sin \theta)}\]
\[\text{LHS} = \frac{\sin \theta + \sin^2 \theta}{\cos \theta(1 + \sin \theta)}\]

Factorise the numerator:
\[\text{LHS} = \frac{\sin \theta(1 + \sin \theta)}{\cos \theta(1 + \sin \theta)}\]

Cancel the common factor of \((1 + \sin \theta)\), noting that \(1 + \sin \theta \neq 0\):
\[\text{LHS} = \frac{\sin \theta}{\cos \theta} = \tan \theta\]

Hence, the identity is proven.

***

**(b)**

Using the identity proven in part (a), substitute \(\theta = 2x\). The equation becomes:
\[\tan 2x = 3\tan^2 2x - 2\]

Rearrange this into a standard quadratic form in terms of \(\tan 2x\):
\[3\tan^2 2x - \tan 2x - 2 = 0\]

Factorise the quadratic equation:
\[(3\tan 2x + 2)(\tan 2x - 1) = 0\]

This gives two possible cases:
1. \(\tan 2x = 1\)
2. \(\tan 2x = -\frac{2}{3}\)

Given the domain \(0^\circ \le x \le 180^\circ\), the domain for \(2x\) is:
\[0^\circ \le 2x \le 360^\circ\]

**Case 1: \(\tan 2x = 1\)**
\[2x = 45^\circ, \quad 2x = 180^\circ + 45^\circ = 225^\circ\]
\[x = 22.5^\circ, \quad x = 112.5^\circ\]

**Case 2: \(\tan 2x = -\frac{2}{3}\)**

The basic angle is \(\tan^{-1}\left(\frac{2}{3}\right) \approx 33.69^\circ\). Since \(\tan 2x\) is negative, \(2x\) lies in the second and fourth quadrants:
\[2x = 180^\circ - 33.69^\circ = 146.31^\circ\]
\[2x = 360^\circ - 33.69^\circ = 326.31^\circ\]

Dividing by 2:
\[x \approx 73.2^\circ, \quad x \approx 163.2^\circ\quad \text{(to 1 d.p.)}\]

Thus, the solutions are:
\[x = 22.5^\circ, \; 73.2^\circ, \; 112.5^\circ, \; 163.2^\circ\]

**(a)**
* **M1**: Attempt to combine terms over a single common denominator \(\cos \theta (1 + \sin \theta)\).
* **M1**: Substitute \(\cos^2 \theta = 1 - \sin^2 \theta\) and factorise numerator to \(\sin\theta(1+\sin\theta)\).
* **A1**: Show logical steps to simplify to \(\tan \theta\) with no errors seen.

**(b)**
* **M1**: Recognise the link to part (a) to write the equation as a quadratic in \(\tan 2x\), i.e., \(3\tan^2 2x - \tan 2x - 2 = 0\), and attempt to solve for \(\tan 2x\).
* **A1**: Obtain \(x = 22.5^\circ\) and \(112.5^\circ\) (both required for this mark).
* **A1**: Obtain \(x = 73.2^\circ\) and \(163.2^\circ\) (both required for this mark; accept 1 d.p. accuracy. Deduct 1 mark overall if extra solutions within range are given).

(a) Let \(y = 2x + 5\).

Rearranging to make \(x\) the subject:
\(2x = y - 5 \implies x = \frac{y - 5}{2}\)

Thus, \(\text{f}^{-1}(x) = \frac{x - 5}{2}\).

(b) We wish to solve \(\text{f}^{-1}\text{g}(x) = 1\).

Using the result from part (a):
\(\frac{\text{g}(x) - 5}{2} = 1\)
\(\text{g}(x) - 5 = 2 \implies \text{g}(x) = 7\)

Alternatively, \(\text{f}^{-1}(\text{g}(x)) = 1 \implies \text{g}(x) = \text{f}(1) = 2(1) + 5 = 7\).

Substituting the expression for \(\text{g}(x)\):
\(\frac{8}{x - 3} = 7\)
\(8 = 7(x - 3)\)
\(8 = 7x - 21\)
\(7x = 29 \implies x = \frac{29}{7}\) (or \(4.14\) to 3 significant figures).

(c) The function is \(\text{g}(x) = \frac{8}{x-3}\) for \(x > 3\).
Since \(x - 3 > 0\) for all \(x > 3\), we must have \(\text{g}(x) > 0\).
As \(x \to 3^{+}\), \(\text{g}(x) \to \infty\), and as \(x \to \infty\), \(\text{g}(x) \to 0\).
Therefore, the range of \(\text{g}\) is \(\text{g}(x) > 0\) (or \(y > 0\)).

(a)
* **M1**: Attempt to make \(x\) the subject of \(y = 2x + 5\).
* **A1**: Correct expression, \(\text{f}^{-1}(x) = \frac{x-5}{2}\) (must be in terms of \(x\)).

(b)
* **M1**: Substitute \(\text{g}(x)\) into \(\text{f}^{-1}\) or set \(\text{g}(x) = \text{f}(1)\).
* **M1**: Set up and attempt to solve the linear equation for \(x\) (e.g. \(\frac{8}{x-3} = 7\)).
* **A1**: Correct final value of \(x = \frac{29}{7}\) (or \(4.14\) to 3 s.f.).

(c)
* **B1**: Identify that the range consists of positive values (e.g., \(> 0\)).
* **B1**: State the correct range using proper notation: \(\text{g}(x) > 0\) or \(y > 0\) (do not accept \(x > 0\)).

**(a)**
The 1st, 3rd and 11th terms of the arithmetic progression (AP) are given by:
\(T_1 = a\)
\(T_3 = a + 2d\)
\(T_{11} = a + 10d\)

The 1st, 2nd and 3rd terms of the geometric progression (GP) are given by:
\(G_1 = a\)
\(G_2 = ar\)
\(G_3 = ar^2\)

We are given that:
1) \(a + 2d = ar \implies 2d = a(r - 1)\)
2) \(a + 10d = ar^2\)

From the first equation, we can express \(d\) in terms of \(a\) and \(r\):
\(d = \frac{a(r - 1)}{2}\)

Substitute this expression for \(d\) into the second equation:
\(a + 10\left(\frac{a(r - 1)}{2}\right) = ar^2\)
\(a + 5a(r - 1) = ar^2\)

Since the sum of the AP is non-zero, \(a \neq 0\). Dividing both sides by \(a\):
\(1 + 5(r - 1) = r^2\)
\(1 + 5r - 5 = r^2\)
\(r^2 - 5r + 4 = 0\)

Factorising the quadratic equation:
\((r - 1)(r - 4) = 0\)

Since we are given that \(r \neq 1\), we have:
\(r = 4\)

**(b)**
Using \(r = 4\), we substitute this back into the relation for \(d\):
\(2d = a(4 - 1) \implies 2d = 3a \implies d = 1.5a\)

The sum of the first 20 terms of the AP is given by:
\(S_{20} = \frac{20}{2}[2a + 19d] = 1220\)
\(10(2a + 19d) = 1220\)
\(2a + 19d = 122\)

Substitute \(d = 1.5a\):
\(2a + 19(1.5a) = 122\)
\(2a + 28.5a = 122\)
\(30.5a = 122\)
\(a = 4\)

Now, find \(d\):
\(d = 1.5(4) = 6\)

**(c)**
For the GP, the first term is \(a = 4\) and the common ratio is \(r = 4\).
The sum of the first 8 terms is given by:
\(S_8 = \frac{a(r^8 - 1)}{r - 1}\)
\(S_8 = \frac{4(4^8 - 1)}{4 - 1}\)
\(S_8 = \frac{4(65536 - 1)}{3}\)
\(S_8 = \frac{4(65535)}{3}\)
\(S_8 = 4 \times 21845 = 87380\)

**(a)**
* **M1**: For writing down two correct equations linking \(a\), \(d\), and \(r\) (e.g., \(a + 2d = ar\) and \(a + 10d = ar^2\)).
* **M1**: For eliminating \(d\) to obtain a single equation in terms of \(a\) and \(r\).
* **A1**: For obtaining the correct quadratic equation in \(r\), such as \(r^2 - 5r + 4 = 0\).
* **A1**: For solving the quadratic equation and correctly showing that \(r = 4\), explaining why \(r = 1\) is rejected.

**(b)**
* **M1**: For using the AP sum formula to set up the equation \(10(2a + 19d) = 1220\) or equivalent.
* **M1**: For substituting a correct relationship between \(a\) and \(d\) (e.g., \(2d = 3a\)) into their sum equation.
* **A1**: For both correct values: \(a = 4\) and \(d = 6\).

**(c)**
* **M1**: For using the correct GP sum formula \(S_8 = \frac{a(r^8 - 1)}{r - 1}\) with their \(a\) and \(r\).
* **A1**: For obtaining the correct sum of \(87380\).

(a) First, find the \(y\)-coordinate of the point \(P\) by substituting \(x = 1\) into the equation of the curve:
\(y = 8(1)^{1/2} - 2(1) = 8 - 2 = 6\).
So the coordinates of \(P\) are \((1, 6)\).

Next, differentiate \(y = 8x^{1/2} - 2x\) to find the gradient function:
\(\frac{dy}{dx} = 8 \left(\frac{1}{2}\right) x^{-1/2} - 2 = 4x^{-1/2} - 2\).

Evaluate the gradient at \(x = 1\):
\(m = 4(1)^{-1/2} - 2 = 4 - 2 = 2\).

Using the equation of a straight line with gradient \(m = 2\) through \((1, 6)\):
\(y - 6 = 2(x - 1)\)
\(y = 2x + 4\).

(b) The shaded region is bounded above by the tangent line \(y = 2x + 4\) and below by the curve \(y = 8x^{1/2} - 2x\), from \(x = 0\) to \(x = 1\).

The area \(A\) is given by the integral:
\(A = \int_{0}^{1} \left[ (2x + 4) - (8x^{1/2} - 2x) \right] dx\)
\(A = \int_{0}^{1} (4x + 4 - 8x^{1/2}) dx\)

Integrating term by term:
\(A = \left[ \frac{4x^2}{2} + 4x - \frac{8x^{3/2}}{3/2} \right]_0^1\)
\(A = \left[ 2x^2 + 4x - \frac{16}{3}x^{3/2} \right]_0^1\)

Substitute the limits \(1\) and \(0\):
At \(x = 1\):
\(2(1)^2 + 4(1) - \frac{16}{3}(1)^{3/2} = 2 + 4 - \frac{16}{3} = 6 - \frac{16}{3} = \frac{2}{3}\).

At \(x = 0\):
\(2(0)^2 + 4(0) - \frac{16}{3}(0)^{3/2} = 0\).

Thus, the exact area of the shaded region is \(\frac{2}{3}\).

(a)
- M1: For attempting to differentiate the curve equation, resulting in a term of the form \(kx^{-1/2} - 2\).
- A1: For correct derivative \(\frac{dy}{dx} = 4x^{-1/2} - 2\).
- M1: For substituting \(x = 1\) to find the gradient and attempting to find the equation of the tangent line using \((1, 6)\).
- A1: For \(y = 2x + 4\) (or any equivalent form).

(b)
- M1: For setting up the correct difference integral \(\int_{0}^{1} (\text{tangent} - \text{curve}) dx\) with appropriate limits.
- M1: For integrating the terms correctly (increasing powers by 1), expecting to see at least two terms correct.
- A1: For correct integrated expression: \(2x^2 + 4x - \frac{16}{3}x^{3/2}\).
- M1: For substituting limits \(1\) and \(0\) into their integrated expression.
- A1: For obtaining the exact area of \(\frac{2}{3}\).

**(a)**
To find the centre and radius of \( C \), we complete the square for \( x \) and \( y \) in the equation \( x^2 + y^2 - 16x - 12y + 64 = 0 \):
\[ (x - 8)^2 - 64 + (y - 6)^2 - 36 + 64 = 0 \]
\[ (x - 8)^2 + (y - 6)^2 = 36 \]
Thus:
- The centre of the circle \( C \) is \( (8, 6) \).
- The radius of the circle \( C \) is \( \sqrt{36} = 6 \).

**(b)**
Let the equation of the tangent passing through the origin \( (0,0) \) be \( y = mx \) (where \( m \neq 0 \)). This can be rewritten as:
\[ mx - y = 0 \]
Since the line is a tangent to the circle \( C \), the perpendicular distance from the centre of the circle \( (8, 6) \) to the line must equal the radius of the circle, which is \( 6 \).
Using the perpendicular distance formula:
\[ \frac{|m(8) - 1(6)|}{\sqrt{m^2 + (-1)^2}} = 6 \]
\[ \frac{|8m - 6|}{\sqrt{m^2 + 1}} = 6 \]
Multiply both sides by \( \sqrt{m^2 + 1} \) and square both sides:
\[ (8m - 6)^2 = 36(m^2 + 1) \]
\[ 64m^2 - 96m + 36 = 36m^2 + 36 \]
\[ 28m^2 - 96m = 0 \]
\[ 4m(7m - 24) = 0 \]
Since we are given that the tangent has a non-zero gradient, we have:
\[ 7m - 24 = 0 \implies m = \frac{24}{7} \]
Thus, the equation of the tangent is:
\[ y = \frac{24}{7}x \quad \text{or} \quad 24x - 7y = 0 \]

*(Alternative Method for part b)*
Substitute \( y = mx \) into the circle equation:
\[ x^2 + (mx)^2 - 16x - 12(mx) + 64 = 0 \]
\[ (1 + m^2)x^2 - (16 + 12m)x + 64 = 0 \]
For the line to be tangent, the discriminant of this quadratic equation must be zero:
\[ b^2 - 4ac = 0 \implies (16 + 12m)^2 - 4(1+m^2)(64) = 0 \]
Dividing by 16:
\[ (4 + 3m)^2 - 16(1+m^2) = 0 \]
\[ 16 + 24m + 9m^2 - 16 - 16m^2 = 0 \]
\[ -7m^2 + 24m = 0 \implies m(24 - 7m) = 0 \]
Since \( m \neq 0 \), we obtain \( m = \frac{24}{7} \).

**(c)**
To find the coordinates of the point of contact \( P \), we can find the intersection of the tangent line \( y = \frac{24}{7}x \) and the normal line passing through the centre \( (8,6) \).
The gradient of the tangent is \( \frac{24}{7} \), so the gradient of the normal is \( -\frac{7}{24} \).
The equation of the normal line is:
\[ y - 6 = -\frac{7}{24}(x - 8) \implies 7x + 24y = 200 \]
Substituting \( y = \frac{24}{7}x \) into this normal equation:
\[ 7x + 24\left(\frac{24}{7}x\right) = 200 \]
\[ 49x + 576x = 1400 \]
\[ 625x = 1400 \implies x = \frac{56}{25} = 2.24 \]
Substituting back to find \( y \):
\[ y = \frac{24}{7}\left(\frac{56}{25}\right) = \frac{192}{25} = 7.68 \]
Thus, the coordinates of \( P \) are \( \left(\frac{56}{25}, \frac{192}{25}\right) \) or \( (2.24, 7.68) \).

**(a)**
* **M1**: For completing the square for both \( x \) and \( y \) to find \( (x-a)^2 + (y-b)^2 = r^2 \).
* **A1**: For obtaining the correct centre of \( (8, 6) \) and radius of \( 6 \).

**(b)**
* **M1**: For using a valid method to find the tangent. Either by setting the perpendicular distance from \( (8,6) \) to \( y=mx \) equal to the radius \( 6 \), OR by substituting \( y=mx \) into the circle equation and setting the discriminant to zero.
* **M1**: For forming a correct quadratic equation in \( m \) (e.g. \( (8m - 6)^2 = 36(m^2 + 1) \) or \( (16 + 12m)^2 - 256(1+m^2) = 0 \)).
* **A1**: For solving the quadratic to find \( m = \frac{24}{7} \) (accept showing \( m=0 \) and \( m = \frac{24}{7} \) and selecting the non-zero value).
* **A1**: For writing down the final tangent equation as \( y = \frac{24}{7}x \) (or equivalent form, e.g. \( 24x - 7y = 0 \)).

**(c)**
* **M1**: For finding the point of contact, either by solving the intersection of the tangent with the normal line through the centre, or by solving the quadratic in \( x \) from part (b) using the value of \( m = \frac{24}{7} \).
* **A1**: For obtaining the correct coordinates of \( P \) as \( \left(\frac{56}{25}, \frac{192}{25}\right) \) or \( (2.24, 7.68) \).

**(a)**
Since \(AC\) is a tangent to the circle at \(A\), the angle \(OAC = \frac{\pi}{2}\) (a right angle).
In the right-angled triangle \(OAC\):
- \(\tan\theta = \frac{AC}{OA} = \frac{AC}{r} \implies AC = r\tan\theta\)
- \(\cos\theta = \frac{OA}{OC} = \frac{r}{OC} \implies OC = \frac{r}{\cos\theta} = r\sec\theta\)

Since \(C\) lies on the line \(OB\) extended and \(OB = r\):
\(BC = OC - OB = r\sec\theta - r\)

The length of the arc \(AB\) is given by \(s = r\theta\).

The perimeter, \(P\), of the region \(R\) is the sum of the boundary lengths:
\(P = \text{arc } AB + BC + AC\)
\(P = r\theta + (r\sec\theta - r) + r\tan\theta = r(\theta + \tan\theta + \sec\theta - 1)\)

**(b)**
Substitute \(\theta = \frac{\pi}{6}\) and \(P = 10\) into the perimeter formula:
\(10 = r\left(\frac{\pi}{6} + \tan\left(\frac{\pi}{6}\right) + \sec\left(\frac{\pi}{6}\right) - 1\right)\)

Using exact trigonometric values:
\(\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\) and \(\sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}\)

\(10 = r\left(\frac{\pi}{6} + \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} - 1\right)\)
\(10 = r\left(\frac{\pi}{6} + \sqrt{3} - 1\right)\)

Calculating the value in the bracket:
\(\frac{\pi}{6} + \sqrt{3} - 1 \approx 0.5236 + 1.7321 - 1 = 1.2556...\)

Solve for \(r\):
\(r = \frac{10}{1.2556...} \approx 7.964...\)
So, \(r = 7.96\text{ cm}\) (to 3 s.f.).

**(c)**
The area of the right-angled triangle \(OAC\) is:
\(\text{Area}_{OAC} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times r \times r\tan\theta = \frac{1}{2}r^2\tan\theta\)

The area of the sector \(OAB\) is:
\(\text{Area}_{sector} = \frac{1}{2}r^2\theta\)

The area of the shaded region \(R\) is the difference between these two areas:
\(\text{Area}_R = \text{Area}_{OAC} - \text{Area}_{sector}\)
\(\text{Area}_R = \frac{1}{2}r^2\tan\theta - \frac{1}{2}r^2\theta = \frac{1}{2}r^2(\tan\theta - \theta)\)

**(a)**
* **M1**: For identifying that triangle \(OAC\) is right-angled at \(A\) and writing an expression for \(AC\) in terms of \(r\) and \(\theta\).
* **A1**: Obtaining \(AC = r\tan\theta\) and \(OC = r\sec\theta\) (or \(OC = \frac{r}{\cos\theta}\)).
* **M1**: Finding the length of \(BC = OC - r\) and writing down the sum of the three boundary lengths: \(\text{arc } AB + BC + AC\).
* **A1**: Correctly simplifying to obtain the given expression \(r(\theta + \tan\theta + \sec\theta - 1)\) with no errors seen.

**(b)**
* **M1**: Substituting \(\theta = \frac{\pi}{6}\) and \(P = 10\) into the perimeter formula and attempting to solve for \(r\).
* **A1**: Obtaining the equation \(10 = r\left(\frac{\pi}{6} + \sqrt{3} - 1\right)\) or showing the bracket is approximately \(1.26\).
* **A1**: Finding \(r = 7.96\) (accept \(7.96\) or \(7.96\text{ cm}\); do not accept \(8\) or other incorrect rounding unless \(7.96\) is also shown).

**(c)**
* **M1**: Writing an expression for the area of the right-angled triangle \(OAC\) in terms of \(r\) and \(\theta\).
* **M1**: Subtracting the area of the sector \(OAB\), \(\frac{1}{2}r^2\theta\), from their triangle area.
* **A1**: Correctly factorizing to obtain \(\frac{1}{2}r^2(\tan\theta - \theta)\) with no errors seen.

(a) To find the equation of the curve, we integrate the gradient function with respect to \(x\): \(y = \int \left(\frac{12}{\sqrt{2x+5}} - 4\right) \text{d}x = \int \left(12(2x+5)^{-\frac{1}{2}} - 4\right) \text{d}x\). Using the reverse chain rule, we get: \(y = \frac{12(2x+5)^{\frac{1}{2}}}{\frac{1}{2} \times 2} - 4x + C = 12\sqrt{2x+5} - 4x + C\). We are given that the curve passes through the point \((2, 33)\). Substituting these coordinates into the equation: \(33 = 12\sqrt{2(2)+5} - 4(2) + C \implies 33 = 12(3) - 8 + C \implies 33 = 36 - 8 + C \implies 33 = 28 + C \implies C = 5\). Therefore, the equation of the curve is \(y = 12\sqrt{2x+5} - 4x + 5\). (b) To find the stationary point, we set \(\frac{\text{d}y}{\text{d}x} = 0\): \(\frac{12}{\sqrt{2x+5}} - 4 = 0 \implies \frac{12}{\sqrt{2x+5}} = 4 \implies \sqrt{2x+5} = 3 \implies 2x+5 = 9 \implies 2x = 4 \implies x = 2\). Since \(x = 2\), we substitute this back into the curve equation to find the \(y\)-coordinate: \(y = 12\sqrt{9} - 4(2) + 5 = 33\). Thus, the stationary point is \((2, 33)\). To determine its nature, we find the second derivative: \(\frac{\text{d}^2y}{\text{d}x^2} = \frac{\text{d}}{\text{d}x}\left(12(2x+5)^{-\frac{1}{2}} - 4\right) = 12 \times \left(-\frac{1}{2}\right) \times 2 \times (2x+5)^{-\frac{3}{2}} = -12(2x+5)^{-\frac{3}{2}}\). At \(x = 2\): \(\frac{\text{d}^2y}{\text{d}x^2} = -12(9)^{-\frac{3}{2}} = -12 \times \frac{1}{27} = -\frac{4}{9}\). Since \(\frac{\text{d}^2y}{\text{d}x^2} < 0\), the stationary point \((2, 33)\) is a maximum.

(a) M1: For integrating to get a term of the form \(k(2x+5)^{\frac{1}{2}}\). A1: For the correct integrated term \(12(2x+5)^{\frac{1}{2}}\). A1: For the term \(-4x\). M1: For substituting the coordinates \((2, 33)\) into an integrated expression containing a constant of integration \(C\). A1: For obtaining the correct equation \(y = 12\sqrt{2x+5} - 4x + 5\). (b) B1: For setting gradient to 0, solving to find \(x = 2\), and stating the stationary point is \((2, 33)\). M1: For differentiating the gradient function to obtain a correct form of \(\frac{\text{d}^2y}{\text{d}x^2}\). A1: For evaluating the second derivative at \(x = 2\) as negative, and concluding it is a maximum.

(a) Let the side length of the square surface of the water be \(s\text{ cm}\) when the depth is \(x\text{ cm}\). By similar triangles, the ratio of the side length to the depth of the water is constant: \(\frac{s}{x} = \frac{10}{20} = \frac{1}{2} \implies s = \frac{1}{2}x\). The area of the square surface of the water is \(A = s^2 = \left(\frac{1}{2}x\right)^2 = \frac{1}{4}x^2\). The volume \(V\) of a pyramid with a square base of area \(A\) and height \(x\) is given by: \(V = \frac{1}{3} A x = \frac{1}{3} \left(\frac{1}{4}x^2\right) x = \frac{1}{12}x^3\).

(b) The net rate of change of volume is given by: \(\frac{\mathrm{d}V}{\mathrm{d}t} = \text{Rate in} - \text{Rate out} = 15 - 0.2x^2\). At the instant when \(x = 5\): \(\frac{\mathrm{d}V}{\mathrm{d}t} = 15 - 0.2(5)^2 = 15 - 5 = 10\text{ cm}^3\text{ s}^{-1}\). Differentiating the volume formula \(V = \frac{1}{12}x^3\) with respect to \(x\) yields: \(\frac{\mathrm{d}V}{\mathrm{d}x} = \frac{1}{4}x^2\). At \(x = 5\): \(\frac{\mathrm{d}V}{\mathrm{d}x} = \frac{1}{4}(5)^2 = 6.25\). Using the chain rule: \(\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t} \implies 10 = 6.25 \times \frac{\mathrm{d}x}{\mathrm{d}t} \implies \frac{\mathrm{d}x}{\mathrm{d}t} = 1.6\text{ cm s}^{-1}\).

(c) The surface area of the water is \(A = s^2 = \frac{1}{4}x^2\). Differentiating with respect to \(x\) yields: \(\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{1}{2}x\). At \(x = 5\): \(\frac{\mathrm{d}A}{\mathrm{d}x} = 2.5\). Using the chain rule: \(\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t} \implies \frac{\mathrm{d}A}{\mathrm{d}t} = 2.5 \times 1.6 = 4\text{ cm}^2\text{ s}^{-1}\).

(a)
- M1: For using similar triangles or linear scaling to show \(s = 0.5x\) or equivalent.
- A1: For correctly substituting this into the volume formula \(V = \frac{1}{3}s^2x\) and obtaining the given expression.

(b)
- B1: For setting up the net rate equation \(\frac{\mathrm{d}V}{\mathrm{d}t} = 15 - 0.2x^2\) and finding the value is \(10\) when \(x = 5\).
- M1: For differentiating \(V\) with respect to \(x\) to obtain \(\frac{\mathrm{d}V}{\mathrm{d}x} = \frac{1}{4}x^2\) and evaluating it at \(x = 5\).
- M1: For applying the chain rule \(\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}\).
- A1: For obtaining \(\frac{\mathrm{d}x}{\mathrm{d}t} = 1.6\) (or equivalent fraction).

(c)
- B1: For expressing \(A = \frac{1}{4}x^2\) and differentiating to get \(\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{1}{2}x\).
- M1: For evaluating \(\frac{\mathrm{d}A}{\mathrm{d}x}\) at \(x = 5\) to get \(2.5\).
- M1: For applying the chain rule \(\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}\).
- A1: For obtaining \(\frac{\mathrm{d}A}{\mathrm{d}t} = 4\).

To solve the inequality \(|2x - 3| < 3|x + 1|\), we can square both sides since both sides are non-negative. This yields \((2x - 3)^2 < 9(x + 1)^2\). Expanding both sides gives \(4x^2 - 12x + 9 < 9(x^2 + 2x + 1)\). Expanding the right-hand side, we have \(4x^2 - 12x + 9 < 9x^2 + 18x + 9\). Rearranging all terms to one side gives \(5x^2 + 30x > 0\), which factorises to \(5x(x + 6) > 0\). The critical values of this quadratic inequality are \(x = 0\) and \(x = -6\). Since we require the quadratic expression to be strictly greater than zero, the solution set lies outside the interval between these roots. Therefore, the solution is \(x < -6\) or \(x > 0\).

M1: For an attempt to solve the inequality by squaring both sides to obtain a quadratic inequality, or by using equivalent linear equations to find critical values. A1: For obtaining the correct critical values \(x = 0\) and \(x = -6\) (or a correct simplified quadratic expression such as \(5x^2 + 30x > 0\)). M1: For identifying that the solution set consists of the outside regions defined by their critical values. A1: For the correct final answer \(x < -6\) or \(x > 0\) (or equivalent interval notation).

Let \(u = 3^x\). Since \(3^{x+1} = 3 \cdot 3^x = 3u\) and \(3^{-x} = \frac{1}{u}\), we can rewrite the equation as:

\(3u - \frac{4}{u} = 11\)

Multiply the entire equation by \(u\) to clear the fraction:

\(3u^2 - 4 = 11u\)

Rearrange into a standard quadratic form:

\(3u^2 - 11u - 4 = 0\)

Factorise the quadratic expression:

\((3u + 1)(u - 4) = 0\)

This gives two potential solutions for \(u\):

\(u = -\frac{1}{3}\) or \(u = 4\)

Since \(u = 3^x\) must be strictly positive for all real values of \(x\), the solution \(u = -\frac{1}{3}\) is rejected.

Therefore, we have:

\(3^x = 4\)

Taking natural logarithms of both sides:

\(x \ln 3 = \ln 4\)

\(x = \frac{\ln 4}{\ln 3} \approx 1.26185...\)

Giving the answer to 3 significant figures, we get:

\(x = 1.26\)

**M1**: Substitute \(u = 3^x\) (or equivalent) to form a quadratic equation of the form \(3u^2 - 11u - 4 = 0\).

**A1**: Solve the quadratic equation to obtain \(u = 4\) and reject the negative root (or state clearly that \(3^x = 4\)).

**M1**: Apply logarithms correctly to solve \(3^x = k\) (where \(k > 0\)) to find \(x = \frac{\ln k}{\ln 3}\).

**A1**: Obtain \(x = 1.26\) (or any value rounding to 1.26).

Let \(u = 3^x\). Since \(3^{x+1} = 3 \cdot 3^x = 3u\) and \(3^{-x} = \frac{1}{u}\), we can rewrite the equation as:

\(3u - \frac{4}{u} = 11\)

Multiply the entire equation by \(u\) to clear the fraction:

\(3u^2 - 4 = 11u\)

Rearrange into a standard quadratic form:

\(3u^2 - 11u - 4 = 0\)

Factorise the quadratic expression:

\((3u + 1)(u - 4) = 0\)

This gives two potential solutions for \(u\):

\(u = -\frac{1}{3}\) or \(u = 4\)

Since \(u = 3^x\) must be strictly positive for all real values of \(x\), the solution \(u = -\frac{1}{3}\) is rejected.

Therefore, we have:

\(3^x = 4\)

Taking natural logarithms of both sides:

\(x \ln 3 = \ln 4\)

\(x = \frac{\ln 4}{\ln 3} \approx 1.26185...\)

Giving the answer to 3 significant figures, we get:

\(x = 1.26\)

**M1**: Substitute \(u = 3^x\) (or equivalent) to form a quadratic equation of the form \(3u^2 - 11u - 4 = 0\).

**A1**: Solve the quadratic equation to obtain \(u = 4\) and reject the negative root (or state clearly that \(3^x = 4\)).

**M1**: Apply logarithms correctly to solve \(3^x = k\) (where \(k > 0\)) to find \(x = \frac{\ln k}{\ln 3}\).

**A1**: Obtain \(x = 1.26\) (or any value rounding to 1.26).

(i) To find the intersection points with the \(x\)-axis, we set \(y = 0\):
\(e^{2x} - 5e^x + 4 = 0\)
Letting \(u = e^x\), we get:
\(u^2 - 5u + 4 = 0\)
\((u - 1)(u - 4) = 0\)
Thus, \(u = 1\) or \(u = 4\).
Since \(u = e^x\):
For \(u = 1 \implies e^x = 1 \implies x = 0\)
For \(u = 4 \implies e^x = 4 \implies x = \ln 4\).
The exact \(x\)-coordinates are \(0\) and \(\ln 4\).

(ii) To find the stationary point, we differentiate \(y\) with respect to \(x\):
\(\frac{dy}{dx} = 2e^{2x} - 5e^x\)
Set \(\frac{dy}{dx} = 0\):
\(2e^{2x} - 5e^x = 0 \implies e^x(2e^x - 5) = 0\)
Since \(e^x > 0\) for all real \(x\), we solve:
\(2e^x = 5 \implies e^x = \frac{5}{2}\)
This gives the exact \(x\)-coordinate:
\(x = \ln\left(\frac{5}{2}\right)\).
Substitute \(e^x = \frac{5}{2}\) back into the equation of the curve to find the \(y\)-coordinate:
\(y = \left(\frac{5}{2}\right)^2 - 5\left(\frac{5}{2}\right) + 4 = \frac{25}{4} - \frac{25}{2} + 4 = -\frac{9}{4}\).
Therefore, the exact coordinates of the stationary point are \(\left(\ln\left(\frac{5}{2}\right), -\frac{9}{4}\right)\).

(iii) The curve lies below the \(x\)-axis between the boundary points \(x = 0\) and \(x = \ln 4\). The exact area \(A\) is calculated by:
\(A = \int_{0}^{\ln 4} -y \, dx = \int_{0}^{\ln 4} (5e^x - e^{2x} - 4) \, dx\)
Integrating term by term:
\(A = \left[ 5e^x - \frac{1}{2}e^{2x} - 4x \right]_{0}^{\ln 4}\)
Substituting the upper limit \(x = \ln 4\):
\(5e^{\ln 4} - \frac{1}{2}e^{2\ln 4} - 4\ln 4 = 5(4) - \frac{1}{2}(16) - 4\ln 4 = 20 - 8 - 4\ln 4 = 12 - 4\ln 4 = 12 - 8\ln 2\).
Substituting the lower limit \(x = 0\):
\(5e^0 - \frac{1}{2}e^0 - 4(0) = 5 - \frac{1}{2} = \frac{9}{2}\).
Subtracting the lower limit value from the upper limit value:
\(A = (12 - 8\ln 2) - \frac{9}{2} = \frac{15}{2} - 8\ln 2\).
Here, \(a = \frac{15}{2}\) and \(b = 8\), which are rational numbers.

(i)
M1: For attempting to solve the quadratic equation in \(e^x\).
A1: For obtaining both exact values \(x = 0\) and \(x = \ln 4\) (or \(2\ln 2\)).

(ii)
M1: For differentiating to get \(\frac{dy}{dx} = 2e^{2x} - 5e^x\) and setting it to zero.
A1: For finding the correct \(x\)-coordinate \(x = \ln\left(\frac{5}{2}\right)\) (or equivalent).
A1: For finding the correct \(y\)-coordinate \(y = -\frac{9}{4}\) (or \(-2.25\)).

(iii)
M1: For integrating \(y\) (or \(-y\)) to obtain an expression of the form \(k_1 e^{2x} + k_2 e^x + k_3 x\) (where \(k_1, k_2, k_3 \neq 0\)).
A1: For obtaining the correct integrated expression: \(5e^x - \frac{1}{2}e^{2x} - 4x\) (or the negative thereof).
A1: For substituting limits correctly and obtaining the final answer \(\frac{15}{2} - 8\ln 2\) (or \(7.5 - 8\ln 2\)) from correct working.

For (i):
First, differentiate \( x \) with respect to \( t \):
\( \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{2}{t+1} \).

Next, differentiate \( y \) with respect to \( t \) using the quotient rule:
\( u = t^2 \implies \frac{\mathrm{d}u}{\mathrm{d}t} = 2t \)
\( v = t + 2 \implies \frac{\mathrm{d}v}{\mathrm{d}t} = 1 \)
\( \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{2t(t+2) - t^2(1)}{(t+2)^2} = \frac{2t^2 + 4t - t^2}{(t+2)^2} = \frac{t^2 + 4t}{(t+2)^2} = \frac{t(t+4)}{(t+2)^2} \).

Using the chain rule for parametric differentiation:
\( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} = \frac{t(t+4)}{(t+2)^2} \div \frac{2}{t+1} = \frac{t(t+4)}{(t+2)^2} \times \frac{t+1}{2} = \frac{t(t+1)(t+4)}{2(t+2)^2} \).

For (ii):
To find the value of \( t \) when \( x = \ln 9 \):
\( 2\ln(t+1) = \ln 9 \)
\( \ln(t+1)^2 = \ln 9 \)
\( (t+1)^2 = 9 \)
Since \( t > -1 \), we have:
\( t + 1 = 3 \implies t = 2 \).

Substitute \( t = 2 \) into the expression for \( \frac{\mathrm{d}y}{\mathrm{d}x} \):
\( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(2+1)(2+4)}{2(2+2)^2} = \frac{2(3)(6)}{2(16)} = \frac{36}{32} = \frac{9}{8} \).

(i)
M1: Differentiate \( x \) to obtain \( \frac{2}{t+1} \).
M1: Apply quotient rule to find \( \frac{\mathrm{d}y}{\mathrm{d}t} \).
A1: Obtain correct simplified \( \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{t^2+4t}{(t+2)^2} \) or equivalent.
A1: Correctly combine using \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t} \) to show the given result.

(ii)
M1: Set \( 2\ln(t+1) = \ln 9 \) and attempt to solve for \( t \).
A1: Obtain \( t = 2 \) (and state or imply rejection of \( t = -4 \)).
A1: Substitute \( t = 2 \) into \( \frac{\mathrm{d}y}{\mathrm{d}x} \) to obtain \( \frac{9}{8} \) (or 1.125).

(i) Using polynomial division or matching coefficients:

Dividing \(2x^3 + x^2 - 4x + 5\) by \(2x + 1\):

\(2x^3 + x^2 = x^2(2x + 1)\)

This leaves \(-4x + 5\). Dividing this by \(2x + 1\):

\(-4x + 5 = -2(2x + 1) + 7\)

Combining these results, we get:

\(\frac{2x^3 + x^2 - 4x + 5}{2x + 1} = x^2 - 2 + \frac{7}{2x + 1}\)

Thus, \(a = 1\), \(b = 0\), \(c = -2\), and \(d = 7\).

(ii) We use the expression found in part (i) to evaluate the integral:

\(\int_{0}^{2} \frac{\mathrm{p}(x)}{2x + 1} \, \mathrm{d}x = \int_{0}^{2} \left( x^2 - 2 + \frac{7}{2x + 1} \right) \mathrm{d}x\)

Integrating each term with respect to \(x\):

\(= \left[ \frac{1}{3}x^3 - 2x + \frac{7}{2} \ln|2x + 1| \right]_{0}^{2}\)

Substituting the upper limit \(x = 2\):

\(\left( \frac{1}{3}(2)^3 - 2(2) + \frac{7}{2} \ln|2(2) + 1| \right) = \frac{8}{3} - 4 + \frac{7}{2} \ln 5 = -\frac{4}{3} + \frac{7}{2} \ln 5\)

Substituting the lower limit \(x = 0\):

\(\left( \frac{1}{3}(0)^3 - 2(0) + \frac{7}{2} \ln|2(0) + 1| \right) = 0\)

Subtracting the value at the lower limit from the value at the upper limit:

\(\left(-\frac{4}{3} + \frac{7}{2} \ln 5\right) - 0 = -\frac{4}{3} + \frac{7}{2} \ln 5\)

So the exact value is \(-\frac{4}{3} + \frac{7}{2} \ln 5\).

**Part (i)**
* **M1**: For attempting algebraic division of \(2x^3 + x^2 - 4x + 5\) by \(2x + 1\), or equivalent coefficient-matching method, yielding a quotient of the form \(ax^2 + bx + c\) (with at least two of \(a, b, c\) non-zero) and a constant remainder \(d\).
* **A1**: For obtaining quotient \(x^2 - 2\) (or identifying \(a = 1\), \(b = 0\), \(c = -2\)).
* **A1**: For obtaining remainder \(7\) (or identifying \(d = 7\)).

**Part (ii)**
* **M1**: For attempting integration of their three-term expression from part (i). Must obtain terms of the form \(k x^3\), \(m x\), and \(n \ln(2x+1)\) (where \(k, m, n\) are non-zero constants).
* **A1** (FT): For correct integration of their polynomial terms to obtain \(\frac{1}{3}x^3 - 2x\) (or follow through their values of \(a, b, c\)).
* **A1** (FT): For correct integration of their remainder term to get \(\frac{7}{2} \ln|2x + 1|\) (or follow through their value of \(d\)).
* **M1**: For substituting limits \(2\) and \(0\) into an integrated expression of the form \(F(x) = kx^3 + mx + n\ln(2x+1)\) and attempting subtraction \(F(2) - F(0)\).
* **A1**: For obtaining the correct exact simplified value: \(-\frac{4}{3} + \frac{7}{2} \ln 5\) (or equivalent exact form).

(i) Using the quotient rule with \(u = \ln x\) and \(v = x^2 + 2\):
\[\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x} \quad \text{and} \quad \frac{\mathrm{d}v}{\mathrm{d}x} = 2x\]
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{1}{x}(x^2 + 2) - 2x\ln x}{(x^2 + 2)^2}\]
Set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[\frac{1}{x}(x^2 + 2) - 2x\ln x = 0\]
\[x + \frac{2}{x} - 2x\ln x = 0\]
Multiply through by \(x\) (since \(x > 0\)):
\[x^2 + 2 - 2x^2\ln x = 0\]
Rearrange to make \(\ln x\) the subject:
\[2x^2\ln x = x^2 + 2\]
\[\ln x = \frac{x^2 + 2}{2x^2} = 0.5 + \frac{1}{x^2}\]
Take exponentials of both sides:
\[x = \mathrm{e}^{0.5 + \frac{1}{x^2}}\]

(ii) Define \(f(x) = \ln x - 0.5 - \frac{1}{x^2}\).
Evaluate \(f(x)\) at the boundaries:
\[f(2.0) = \ln 2.0 - 0.5 - \frac{1}{4} = 0.6931 - 0.75 = -0.0569\]
\[f(2.3) = \ln 2.3 - 0.5 - \frac{1}{2.3^2} = 0.8329 - 0.5 - 0.1890 = +0.1439\]
Since there is a change of sign and \(f\) is continuous, there is a root between \(2.0\) and \(2.3\).

(iii) Using \(x_{n+1} = \mathrm{e}^{0.5 + \frac{1}{x_n^2}}\) with \(x_1 = 2.1\):
\[x_2 = \mathrm{e}^{0.5 + \frac{1}{2.1^2}} = 2.0684\]
\[x_3 = \mathrm{e}^{0.5 + \frac{1}{2.0684^2}} = 2.0829\]
\[x_4 = \mathrm{e}^{0.5 + \frac{1}{2.0829^2}} = 2.0761\]
\[x_5 = \mathrm{e}^{0.5 + \frac{1}{2.0761^2}} = 2.0793\]
\[x_6 = \mathrm{e}^{0.5 + \frac{1}{2.0793^2}} = 2.0778\]
\[x_7 = \mathrm{e}^{0.5 + \frac{1}{2.0778^2}} = 2.0785\]
Since successive iterations converge to \(2.08\) when rounded to 2 decimal places, the root is \(2.08\).

(i)
M1: Apply the quotient rule (or product rule) to differentiate \(y\).
A1: Obtain correct derivative \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{1}{x}(x^2 + 2) - 2x\ln x}{(x^2 + 2)^2}\) or equivalent.
M1: Set derivative to 0, clear fractions, and attempt to isolate \(\ln x\).
A1: Complete the proof convincingly to reach the given equation.

(ii)
M1: Evaluate a suitable function, e.g., \(f(x) = \ln x - 0.5 - \frac{1}{x^2}\), at \(2.0\) and \(2.3\).
A1: Obtain correct values showing a sign change (e.g., \(-0.057\) and \(+0.144\)) and write a concluding statement.

(iii)
M1: Calculate at least one iteration value correctly to 4 d.p.
A1: Obtain \(x_2 = 2.0684\) and \(x_3 = 2.0829\).
A1: Calculate sufficient iterations to justify the convergence to \(2.08\) and state the final root as \(2.08\).

**(i)**
We use the compound angle identity:
\(R \cos(2x - \alpha) = R \cos 2x \cos \alpha + R \sin 2x \sin \alpha\)

Comparing this with \(\cos 2x + \sqrt{3} \sin 2x\), we get:
\(R \cos \alpha = 1\) and \(R \sin \alpha = \sqrt{3}\)

To find \(R\):
\(R = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2\)

To find \(\alpha\):
\(\tan \alpha = \frac{\sqrt{3}}{1} = \sqrt{3} \implies \alpha = \frac{\pi}{3}\) (since \(0 < \alpha < \frac{\pi}{2}\))

Thus, \(\cos 2x + \sqrt{3} \sin 2x = 2 \cos\left(2x - \frac{\pi}{3}\right)\).

**(ii)**
Using the result from part (i), the integral becomes:
\(\int_{0}^{\frac{\pi}{6}} \frac{1}{\left[2 \cos\left(2x - \frac{\pi}{3}\right)\right]^2} \, \mathrm{d}x = \int_{0}^{\frac{\pi}{6}} \frac{1}{4 \cos^2\left(2x - \frac{\pi}{3}\right)} \, \mathrm{d}x\)

Since \(\frac{1}{\cos^2 \theta} = \sec^2 \theta\), this simplifies to:
\(\int_{0}^{\frac{\pi}{6}} \frac{1}{4} \sec^2\left(2x - \frac{\pi}{3}\right) \, \mathrm{d}x\)

Integrating with respect to \(x\):
\(\left[ \frac{1}{4} \cdot \frac{1}{2} \tan\left(2x - \frac{\pi}{3}\right) \right]_{0}^{\frac{\pi}{6}} = \left[ \frac{1}{8} \tan\left(2x - \frac{\pi}{3}\right) \right]_{0}^{\frac{\pi}{6}}\)

Now we evaluate this definite integral by substituting the upper and lower limits:

For the upper limit \(x = \frac{\pi}{6}\):
\(\frac{1}{8} \tan\left(2\left(\frac{\pi}{6}\right) - \frac{\pi}{3}\right) = \frac{1}{8} \tan(0) = 0\)

For the lower limit \(x = 0\):
\(\frac{1}{8} \tan\left(2(0) - \frac{\pi}{3}\right) = \frac{1}{8} \tan\left(-\frac{\pi}{3}\right) = \frac{1}{8}(-\sqrt{3}) = -\frac{\sqrt{3}}{8}\)

Subtracting the lower limit value from the upper limit value:
\(0 - \left(-\frac{\sqrt{3}}{8}\right) = \frac{\sqrt{3}}{8}\)

**Part (i)**
* **M1**: For using correct trigonometric method to find \(R\) or \(\alpha\).
* **A1**: For obtaining \(R = 2\) and \(\alpha = \frac{\pi}{3}\).
* **A1**: For writing the final expression as \(2 \cos\left(2x - \frac{\pi}{3}\right)\).

**Part (ii)**
* **M1**: For substituting their part (i) expression into the denominator of the integral.
* **A1**: For obtaining the simplified integrand \(\frac{1}{4}\sec^2\left(2x - \frac{\pi}{3}\right)\) (FT on their \(R\) and \(\alpha\)).
* **M1**: For integrating to obtain an expression of the form \(k \tan\left(2x - \frac{\pi}{3}\right)\).
* **A1**: For obtaining the correct integral \(\frac{1}{8} \tan\left(2x - \frac{\pi}{3}\right)\) (FT on their \(R\)).
* **M1**: For substituting both limits \(0\) and \rac{\pi}{6}\) into their integrated expression.
* **M1**: For evaluating \(\tan 0\) and \(\tan\left(-\frac{\pi}{3}\right)\) using correct exact values.
* **A1**: For obtaining the final exact answer \(\frac{\sqrt{3}}{8}\) (or equivalent single term).

To solve \(|2x + 1| < x + 5\), we can use one of two standard methods:

**Method 1: Squaring both sides**
Since both sides must be positive for solutions to exist, we can square both sides of the inequality:
\((2x + 1)^2 < (x + 5)^2\)
\(4x^2 + 4x + 1 < x^2 + 10x + 25\)

Rearranging terms to form a quadratic inequality:
\(3x^2 - 6x - 24 < 0\)

Dividing the entire inequality by 3:
\(x^2 - 2x - 8 < 0\)

Factorising the quadratic expression:
\((x - 4)(x + 2) < 0\)

This gives the critical values \(x = 4\) and \(x = -2\). Since the quadratic is less than zero, the solution lies between these critical values:
\(-2 < x < 4\)

**Method 2: Finding critical points using linear equations**
We find the boundary points by solving:
1) \(2x + 1 = x + 5 \implies x = 4\)
2) \(-(2x + 1) = x + 5 \implies -2x - 1 = x + 5 \implies 3x = -6 \implies x = -2\)

Testing values in the three regions:
- For \(x < -2\) (e.g., \(x = -3\)): \(|-5| < 2\) which is False.
- For \(-2 < x < 4\) (e.g., \(x = 0\)): \(|1| < 5\) which is True.
- For \(x > 4\) (e.g., \(x = 5\)): \(|11| < 10\) which is False.

Thus, the solution is \(-2 < x < 4\).

**M1**: For an attempt to solve by squaring both sides to obtain a three-term quadratic, or for solving the two linear equations \(2x + 1 = \pm(x + 5)\) to find two boundary values.
**A1**: For obtaining both correct critical values, \(x = 4\) and \(x = -2\).
**A1**: For the correct final inequality \(-2 < x < 4\) (or equivalent interval notation).

We express the algebraic fraction in the form: \(\frac{x^2 + 8x - 7}{(x - 3)(x^2 + 4)} = \frac{A}{x - 3} + \frac{Bx + C}{x^2 + 4}\). Multiplying both sides by the denominator gives: \(x^2 + 8x - 7 = A(x^2 + 4) + (Bx + C)(x - 3)\). Substituting \(x = 3\) into this equation: \(3^2 + 8(3) - 7 = A(3^2 + 4)\) which simplifies to \(26 = 13A\), hence \(A = 2\). Equating the coefficients of \(x^2\): \(1 = A + B\). Since \(A = 2\), we find \(B = -1\). Equating the constant terms: \(-7 = 4A - 3C\). Since \(A = 2\), we find \(-7 = 8 - 3C\), which gives \(3C = 15\), hence \(C = 5\). Therefore, the partial fractions are \(\frac{2}{x - 3} + \frac{5 - x}{x^2 + 4}\).

M1: State or imply the form \(\frac{A}{x - 3} + \frac{Bx + C}{x^2 + 4}\). M1: Use a correct method to find a constant (e.g. substituting \(x = 3\)). A1: Obtain \(A = 2\). M1: Use a correct method to solve for the remaining constants. A1: Obtain the final answer \(\frac{2}{x - 3} + \frac{5 - x}{x^2 + 4}\) or equivalent.

We start with the given equation: \(y^2 = A e^{-kx}\). Taking the natural logarithm of both sides, we get: \(\ln(y^2) = \ln(A e^{-kx})\). Using the laws of logarithms: \(2 \ln y = \ln A - kx\). Rearranging this into the linear form \(Y = mX + c\), we obtain: \(\ln y = -\frac{k}{2}x + \frac{1}{2} \ln A\), where \(Y = \ln y\) and \(X = x\). The gradient \(m\) of the line passing through the points \((2, 4.5)\) and \((6, 1.5)\) is: \(m = \frac{1.5 - 4.5}{6 - 2} = -0.75\). Since \(m = -\frac{k}{2}\), we have: \(-\frac{k}{2} = -0.75 \implies k = 1.5\). To find the vertical intercept \(c\), we substitute the coordinates of one point, say \((2, 4.5)\), into the equation of the line: \(4.5 = -0.75(2) + c \implies 4.5 = -1.5 + c \implies c = 6\). Since \(c = \frac{1}{2}\ln A\), we have: \(\frac{1}{2}\ln A = 6 \implies \ln A = 12 \implies A = e^{12}\).

M1: Apply natural logarithms to both sides of the equation and use logarithmic laws to express \(\ln y\) in terms of \(x\).
M1: Calculate the gradient of the straight line using the given coordinates.
A1: Correctly obtain \(k = 1.5\) (or equivalent fraction).
M1: Use the gradient and a point to find the vertical intercept, and set it equal to \(\frac{1}{2}\ln A\).
A1: Correctly obtain the exact value \(A = e^{12}\).

To find the gradient of the curve at the point \((1, 1)\), we differentiate the equation implicitly with respect to \(x\). The equation is: \(2x^2 + 3xy + y^2 + \sin(\pi x) = 6\). Differentiating each term with respect to \(x\): 1. The derivative of \(2x^2\) is \(4x\). 2. Using the product rule on \(3xy\), we get \(3y + 3x\frac{dy}{dx}\). 3. The derivative of \(y^2\) is \(2y\frac{dy}{dx}\). 4. Using the chain rule on \(\sin(\pi x)\), we get \(\pi\cos(\pi x)\). 5. The derivative of the constant \(6\) on the right-hand side is \(0\). Combining these, we obtain: \(4x + 3y + 3x\frac{dy}{dx} + 2y\frac{dy}{dx} + \pi\cos(\pi x) = 0\). Now, substitute the coordinates of the point \((1, 1)\), i.e., \(x = 1\) and \(y = 1\), into this equation: \(4(1) + 3(1) + 3(1)\frac{dy}{dx} + 2(1)\frac{dy}{dx} + \pi\cos(\pi) = 0\). Since \(\cos(\pi) = -1\), this simplifies to: \(7 + 5\frac{dy}{dx} - \pi = 0\). Solving for \(\frac{dy}{dx}\): \(5\frac{dy}{dx} = \pi - 7\), which gives \(\frac{dy}{dx} = \frac{\pi - 7}{5}\). Thus, the exact gradient of the curve at the point \((1, 1)\) is \(\frac{\pi - 7}{5}\).

M1: For differentiating \(2x^2\) to obtain \(4x\) and \(y^2\) to obtain \(2y\frac{dy}{dx}\). M1: For applying the product rule correctly to differentiate \(3xy\) to obtain \(3y + 3x\frac{dy}{dx}\). B1: For differentiating \(\sin(\pi x)\) correctly to obtain \(\pi\cos(\pi x)\). M1: For substituting \(x = 1\) and \(y = 1\) into their differentiated implicit equation. A1: For obtaining a correct linear equation in \(\frac{dy}{dx}\), e.g., \(7 + 5\frac{dy}{dx} - \pi = 0\). A1: For obtaining the correct exact value \(\frac{\pi - 7}{5}\) or equivalent.

To find the gradient of the curve at the point \((1, 1)\), we differentiate the equation implicitly with respect to \(x\). The equation is: \(2x^2 + 3xy + y^2 + \sin(\pi x) = 6\). Differentiating each term with respect to \(x\): 1. The derivative of \(2x^2\) is \(4x\). 2. Using the product rule on \(3xy\), we get \(3y + 3x\frac{dy}{dx}\). 3. The derivative of \(y^2\) is \(2y\frac{dy}{dx}\). 4. Using the chain rule on \(\sin(\pi x)\), we get \(\pi\cos(\pi x)\). 5. The derivative of the constant \(6\) on the right-hand side is \(0\). Combining these, we obtain: \(4x + 3y + 3x\frac{dy}{dx} + 2y\frac{dy}{dx} + \pi\cos(\pi x) = 0\). Now, substitute the coordinates of the point \((1, 1)\), i.e., \(x = 1\) and \(y = 1\), into this equation: \(4(1) + 3(1) + 3(1)\frac{dy}{dx} + 2(1)\frac{dy}{dx} + \pi\cos(\pi) = 0\). Since \(\cos(\pi) = -1\), this simplifies to: \(7 + 5\frac{dy}{dx} - \pi = 0\). Solving for \(\frac{dy}{dx}\): \(5\frac{dy}{dx} = \pi - 7\), which gives \(\frac{dy}{dx} = \frac{\pi - 7}{5}\). Thus, the exact gradient of the curve at the point \((1, 1)\) is \(\frac{\pi - 7}{5}\).

M1: For differentiating \(2x^2\) to obtain \(4x\) and \(y^2\) to obtain \(2y\frac{dy}{dx}\). M1: For applying the product rule correctly to differentiate \(3xy\) to obtain \(3y + 3x\frac{dy}{dx}\). B1: For differentiating \(\sin(\pi x)\) correctly to obtain \(\pi\cos(\pi x)\). M1: For substituting \(x = 1\) and \(y = 1\) into their differentiated implicit equation. A1: For obtaining a correct linear equation in \(\frac{dy}{dx}\), e.g., \(7 + 5\frac{dy}{dx} - \pi = 0\). A1: For obtaining the correct exact value \(\frac{\pi - 7}{5}\) or equivalent.

(i) Let \(f(x) = x^2 + \ln x - 3\).
Evaluating at the boundaries:
\(f(1.5) = 1.5^2 + \ln(1.5) - 3 = 2.25 + 0.405465... - 3 = -0.34454... < 0\)
\(f(1.6) = 1.6^2 + \ln(1.6) - 3 = 2.56 + 0.470003... - 3 = 0.03000... > 0\)
Since \(f(x)\) is continuous on the interval \([1.5, 1.6]\) and there is a change of sign between \(f(1.5)\) and \(f(1.6)\), the equation has a root \(\alpha\) in the interval \(1.5 < \alpha < 1.6\).

(ii) Rearranging the equation:
\(x^2 + \ln x = 3\)
\(\Rightarrow x^2 = 3 - \ln x\)
Since \(1.5 < x < 1.6\), \(x > 0\). Taking the positive square root of both sides gives:
\(x = \sqrt{3 - \ln x}\).

(iii) Using the iterative formula \(x_{n+1} = \sqrt{3 - \ln x_n}\) with \(x_1 = 1.5\):
\(x_1 = 1.5\)
\(x_2 = \sqrt{3 - \ln(1.5)} = 1.61076\)
\(x_3 = \sqrt{3 - \ln(1.610756...)} = 1.58849\)
\(x_4 = \sqrt{3 - \ln(1.588489...)} = 1.59286\)
\(x_5 = \sqrt{3 - \ln(1.592861...)} = 1.59200\)
\(x_6 = \sqrt{3 - \ln(1.591998...)} = 1.59217\)
\(x_7 = \sqrt{3 - \ln(1.592168...)} = 1.59213\)
\(x_8 = \sqrt{3 - \ln(1.592134...)} = 1.59214\)

Since successive approximations round to \(1.592\) correct to 3 decimal places, we conclude that \(\alpha \approx 1.592\).

(i)
* **M1**: Evaluate \(f(x) = x^2 + \ln x - 3\) (or equivalent) at both \(1.5\) and \(1.6\).
* **A1**: Obtain correct values with correct signs (e.g., \(-0.345\) and \(0.030\)) and draw a valid conclusion mentioning the change of sign.

(ii)
* **B1**: Show the correct algebraic steps to obtain \(x = \sqrt{3 - \ln x}\) from \(x^2 + \ln x = 3\).

(iii)
* **M1**: Use the iterative formula correctly to find at least two iterations.
* **A1**: Obtain \(x_2 = 1.61076\) and \(x_3 = 1.58849\).
* **A1**: Obtain the final value \(\alpha = 1.592\) and show sufficient iterations to 5 decimal places to justify this accuracy (at least up to \(x_5\) or \(x_6\)).

(i) To find the stationary point, we differentiate the equation of the curve using the product rule:
\(y = 3x^2 \ln(2x)\)

Let \(u = 3x^2\) and \(v = \ln(2x)\).
Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = 6x\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{2x} \cdot 2 = \frac{1}{x}\).

Applying the product rule:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = u \frac{\mathrm{d}v}{\mathrm{d}x} + v \frac{\mathrm{d}u}{\mathrm{d}x}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 \left(\frac{1}{x}\right) + 6x \ln(2x) = 3x + 6x \ln(2x)\)

At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(3x(1 + 2\ln(2x)) = 0\)

Since \(x > 0\), we must have:
\(1 + 2\ln(2x) = 0\)
\(\ln(2x) = -\frac{1}{2}\)
\(2x = e^{-1/2} = \frac{1}{\sqrt{e}}\)
\(x = \frac{1}{2\sqrt{e}}\)

Now, substitute \(x = \frac{1}{2\sqrt{e}}\) back into the equation of the curve to find the \(y\)-coordinate:
\(y = 3\left(\frac{1}{2\sqrt{e}}\right)^2 \ln\left(2 \cdot \frac{1}{2\sqrt{e}}\right)\)
\(y = 3\left(\frac{1}{4e}\right) \ln\left(e^{-1/2}\right) = \frac{3}{4e} \left(-\frac{1}{2}\right) = -\frac{3}{8e}\)

So the exact coordinates of the stationary point are \(\left( \frac{1}{2\sqrt{e}}, -\frac{3}{8e} \right)\).

(ii) To find the exact value of the integral \(\int_{\frac{1}{2}}^{1} 3x^2 \ln(2x) \, \mathrm{d}x\), we use integration by parts:
\(\int u \frac{\mathrm{d}v}{\mathrm{d}x} \, \mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \, \mathrm{d}x\)

Let \(u = \ln(2x)\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = 3x^2\).
Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}\) and \(v = x^3\).

Applying the formula:
\(\int 3x^2 \ln(2x) \, \mathrm{d}x = x^3 \ln(2x) - \int x^3 \left(\frac{1}{x}\right) \, \mathrm{d}x\)
\(= x^3 \ln(2x) - \int x^2 \, \mathrm{d}x\)
\(= x^3 \ln(2x) - \frac{1}{3}x^3\)

Now apply the limits of integration from \(\frac{1}{2}\) to \(1\):
\(\left[ x^3 \ln(2x) - \frac{1}{3}x^3 \right]_{\frac{1}{2}}^{1}\)
\(= \left( 1^3 \ln(2) - \frac{1}{3}(1)^3 \right) - \left( \left(\frac{1}{2}\right)^3 \ln\left(2 \cdot \frac{1}{2}\right) - \frac{1}{3}\left(\frac{1}{2}\right)^3 \right)\)
\(= \left( \ln(2) - \frac{1}{3} \right) - \left( \frac{1}{8} \ln(1) - \frac{1}{24} \right)\)
\(= \left( \ln(2) - \frac{1}{3} \right) - \left( 0 - \frac{1}{24} \right)\)
\(= \ln(2) - \frac{8}{24} + \frac{1}{24} = \ln(2) - \frac{7}{24}\)

(i)
M1: Apply the product rule correctly to differentiate \(3x^2 \ln(2x)\).
A1: Obtain the correct derivative in any simplified form, e.g., \(3x + 6x \ln(2x)\).
M1: Equate their derivative to zero, factor out or divide by \(x\), and solve for \(x\) in terms of \(e\).
A1: Obtain the exact coordinates \(\left( \frac{1}{2\sqrt{e}}, -\frac{3}{8e} \right)\) or equivalent exact forms.

(ii)
M1: Apply integration by parts with choice of \(u = \ln(2x)\) and \(v' = 3x^2\).
A1: Obtain the correct first stage of integration, e.g., \(x^3 \ln(2x) - \int x^2 \, \mathrm{d}x\).
A1: Obtain the correct complete integral \(x^3 \ln(2x) - \frac{1}{3}x^3\).
M1: Substitute the limits \(1\) and \(\frac{1}{2}\) into their integrated expression (using \(\ln(1) = 0\) correctly).
A1: Obtain the exact final value \(\ln(2) - \frac{7}{24}\) or equivalent exact form.

Part (a): We use integration by parts, \(\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx\). Let \(u = x\) and \(\frac{dv}{dx} = \cos 2x\). This gives \(\frac{du}{dx} = 1\) and \(v = \frac{1}{2} \sin 2x\). Therefore, \(\int x \cos 2x \, dx = \frac{1}{2} x \sin 2x - \int \frac{1}{2} \sin 2x \, dx = \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + c\). Part (b): We use the double-angle identity \(\cos^2 x = \frac{1}{2}(1 + \cos 2x)\). The integral becomes \(\int_0^{\frac{1}{4}\pi} x \cos^2 x \, dx = \int_0^{\frac{1}{4}\pi} x \left( \frac{1}{2}(1 + \cos 2x) \right) \, dx = \int_0^{\frac{1}{4}\pi} \left( \frac{1}{2}x + \frac{1}{2}x \cos 2x \right) \, dx\). Integrating the terms, we get \(\left[ \frac{1}{4}x^2 \right]_0^{\frac{1}{4}\pi} + \frac{1}{2} \int_0^{\frac{1}{4}\pi} x \cos 2x \, dx\). Using the result from part (a), this is equal to \(\left[ \frac{1}{4}x^2 + \frac{1}{4}x \sin 2x + \frac{1}{8} \cos 2x \right]_0^{\frac{1}{4}\pi}\). Substituting the upper limit \(x = \frac{1}{4}\pi\): \(\frac{1}{4}\left(\frac{\pi}{4}\right)^2 + \frac{1}{4}\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{2}\right) + \frac{1}{8}\cos\left(\frac{\pi}{2}\right) = \frac{\pi^2}{64} + \frac{\pi}{16} + 0\). Substituting the lower limit \(x = 0\): \(0 + 0 + \frac{1}{8}\cos(0) = \frac{1}{8}\). Thus, the exact value is \(\left(\frac{\pi^2}{64} + \frac{\pi}{16}\right) - \frac{1}{8} = \frac{\pi^2 + 4\pi - 8}{64}\).

Part (a): M1: Apply integration by parts to \(\int x \cos 2x \, dx\) to obtain \(ax \sin 2x - b \int \sin 2x \, dx\). A1: Obtain \(\frac{1}{2}x \sin 2x - \frac{1}{2} \int \sin 2x \, dx\) or equivalent. A1: Obtain correct final expression \(\frac{1}{2}x \sin 2x + \frac{1}{4} \cos 2x\) (condone omission of \(+c\)). Part (b): M1: Use the double-angle identity \(\cos^2 x = \frac{1}{2}(1 + \cos 2x)\) to rewrite the integrand. A1: Express the integral as \(\int \left( \frac{1}{2}x + \frac{1}{2}x \cos 2x \right) \, dx\) or equivalent. B1: Correctly integrate \(\frac{1}{2}x\) to obtain \(\frac{1}{4}x^2\). M1: Obtain the complete integrated expression \(\frac{1}{4}x^2 + \frac{1}{2}(\text{their result from part a})\). M1: Substitute limits \(\frac{1}{4}\pi\) and \(0\) correctly into their integrated expression (must show substitution of both limits, particularly the non-zero contribution from \(\cos 0\)). A1: Obtain the correct final exact answer \(\frac{\pi^2 + 4\pi - 8}{64}\) or any equivalent exact form.

**(i)**
First, we examine the direction vectors of the two lines:
\( \mathbf{d}_1 = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k} \)
\( \mathbf{d}_2 = \mathbf{i} + 2\mathbf{j} - \mathbf{k} \)

Since there is no scalar \( c \) such that \( \mathbf{d}_1 = c\mathbf{d}_2 \), the direction vectors are not parallel. Thus, the lines are not parallel.

Next, we test whether the lines intersect by equating the components:
1) \( 1 + 2\lambda = 4 + \mu \implies 2\lambda - \mu = 3 \)
2) \( 3 - \lambda = -1 + 2\mu \implies \lambda + 2\mu = 4 \)
3) \( -2 + 3\lambda = 1 - \mu \implies 3\lambda + \mu = 3 \)

From equation (1), we have \( \mu = 2\lambda - 3 \).
Substituting this into equation (2):
\( \lambda + 2(2\lambda - 3) = 4 \implies 5\lambda - 6 = 4 \implies 5\lambda = 10 \implies \lambda = 2 \).

Using \( \lambda = 2 \) in equation (1) gives:
\( \mu = 2(2) - 3 = 1 \).

Now we substitute \( \lambda = 2 \) and \( \mu = 1 \) into the LHS of equation (3):
\( \text{LHS} = -2 + 3(2) = 4 \).
However, the RHS is \( 1 - 1 = 0 \).

Since \( 4 \neq 0 \), the system of equations is inconsistent, meaning the lines do not intersect.
Since the lines are not parallel and do not intersect, they are skew.

**(ii)**
The acute angle \( \theta \) between the directions of the two lines is found using the scalar product of their direction vectors:
\( \cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|} \)

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

Calculate the magnitudes:
\( |\mathbf{d}_1| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14} \)
\( |\mathbf{d}_2| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6} \)

Therefore:
\( \cos \theta = \frac{|-3|}{\sqrt{14}\sqrt{6}} = \frac{3}{\sqrt{84}} \)
\( \theta = \arccos\left(\frac{3}{\sqrt{84}}\right) \approx 70.893^{\circ} \)

To 1 decimal place, the angle is \( 70.9^{\circ} \).

**(iii)**
Let \( N \) be the foot of the perpendicular from \( P(2, -3, 0) \) to \( l_1 \).
The position vector of a general point on \( l_1 \) is:
\( \mathbf{ON} = (1+2\lambda)\mathbf{i} + (3-\lambda)\mathbf{j} + (-2+3\lambda)\mathbf{k} \)

The vector \( \mathbf{PN} \) is:
\( \mathbf{PN} = \mathbf{ON} - \mathbf{OP} = (1+2\lambda - 2)\mathbf{i} + (3-\lambda - (-3))\mathbf{j} + (-2+3\lambda - 0)\mathbf{k} \)
\( \mathbf{PN} = (2\lambda - 1)\mathbf{i} + (6-\lambda)\mathbf{j} + (3\lambda-2)\mathbf{k} \)

Since \( \mathbf{PN} \) is perpendicular to the direction vector of \( l_1 \):
\( \mathbf{PN} \cdot (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) = 0 \)
\( 2(2\lambda - 1) - (6-\lambda) + 3(3\lambda-2) = 0 \)
\( 4\lambda - 2 - 6 + \lambda + 9\lambda - 6 = 0 \)
\( 14\lambda - 14 = 0 \implies \lambda = 1 \)

Substitute \( \lambda = 1 \) back into the expression for \( \mathbf{PN} \):
\( \mathbf{PN} = (2(1)-1)\mathbf{i} + (6-1)\mathbf{j} + (3(1)-2)\mathbf{k} = \mathbf{i} + 5\mathbf{j} + \mathbf{k} \)

The perpendicular distance is the magnitude of \( \mathbf{PN} \):
\( |\mathbf{PN}| = \sqrt{1^2 + 5^2 + 1^2} = \sqrt{27} = 3\sqrt{3} \)

**(ii)**
* **M1**: State that the direction vectors are not parallel because one is not a scalar multiple of the other.
* **M1**: Set up a system of three linear equations in terms of \( \lambda \) and \( \mu \) by equating the components.
* **A1**: Solve a pair of equations to obtain correct values for both parameters (e.g., \( \lambda = 2 \) and \( \mu = 1 \)).
* **M1**: Substitute these parameters into the third equation and show that it results in a contradiction (e.g., \( 4 \neq 0 \)).
* **A1**: Conclude correctly that since the lines are not parallel and do not intersect, they are skew.

**(ii)**
* **M1**: Apply the scalar product formula \( \cos \theta = \frac{|\mathbf{a} \cdot \mathbf{b}|}{|\mathbf{a}| |\mathbf{b}|} \) using correct direction vectors.
* **A1**: Obtain \( \cos \theta = \frac{3}{\sqrt{84}} \) or equivalent.
* **A1**: Obtain final angle of \( 70.9^{\circ} \) (or \( 1.24 \) radians; withhold this mark if the unit is missing or incorrect, but allow 1 d.p. accuracy in degrees).

**(iii)**
* **M1**: Set up an expression for the vector \( \mathbf{PN} \) from \( P \) to a general point on \( l_1 \) in terms of \( \lambda \).
* **M1**: Use the condition that \( \mathbf{PN} \cdot \mathbf{d}_1 = 0 \) to form a linear equation in \( \lambda \).
* **A1**: Solve to find \( \lambda = 1 \).
* **A1**: Substitute \( \lambda = 1 \) to find \( \mathbf{PN} \) and calculate its magnitude to obtain \( 3\sqrt{3} \) (accept \( a=3, b=3 \)).

**(a)**
We use the double-angle identities:
\( \cos 2\theta = 2\cos^2\theta - 1 \)
\( \sin 2\theta = 2\sin\theta\cos\theta \)

Substituting these into the expression for \( z \):
\( z = 1 + (2\cos^2\theta - 1) + \mathrm{i}(2\sin\theta\cos\theta) \)
\( z = 2\cos^2\theta + 2\mathrm{i}\sin\theta\cos\theta \)

Factorise out \( 2\cos\theta \):
\( z = 2\cos\theta (\cos\theta + \mathrm{i}\sin\theta) \)

Since \( \cos\theta + \mathrm{i}\sin\theta = \mathrm{e}^{\mathrm{i}\theta} \), we have:
\( z = 2\cos\theta \, \mathrm{e}^{\mathrm{i}\theta} \) (as required).

**(b)**
Using the result from part (a):
\( z^{-2} = (2\cos\theta \, \mathrm{e}^{\mathrm{i}\theta})^{-2} \)
\( z^{-2} = (2\cos\theta)^{-2} \, (\mathrm{e}^{\mathrm{i}\theta})^{-2} \)
\( z^{-2} =
\frac{1}{4\cos^2\theta} \, \mathrm{e}^{-2\mathrm{i}\theta} \)

Thus, \( R = \frac{1}{4\cos^2\theta} \) and \( \phi = -2\theta \).

**(c)**
First, observe that:
\( z - 1 = \cos 2\theta + \mathrm{i}\sin 2\theta = \mathrm{e}^{\mathrm{i}2\theta} \)

Now, substitute \( z - 1 = \mathrm{e}^{\mathrm{i}2\theta} \) and \( z = 2\cos\theta \, \mathrm{e}^{\mathrm{i}\theta} \) into the expression:
\( \frac{z-1}{z} = \frac{\mathrm{e}^{\mathrm{i}2\theta}}{2\cos\theta \, \mathrm{e}^{\mathrm{i}\theta}} \)

Using index laws for exponentials:
\( \frac{z-1}{z} = \frac{1}{2\cos\theta} \, \mathrm{e}^{\mathrm{i}(2\theta - \theta)} = \frac{\mathrm{e}^{\mathrm{i}\theta}}{2\cos\theta} \)

Convert back to trigonometric form:
\( \frac{z-1}{z} = \frac{\cos\theta + \mathrm{i}\sin\theta}{2\cos\theta} \)
\( \frac{z-1}{z} = \frac{\cos\theta}{2\cos\theta} + \mathrm{i}\frac{\sin\theta}{2\cos\theta} \)
\( \frac{z-1}{z} = \frac{1}{2} + \frac{1}{2}\mathrm{i}\tan\theta = \frac{1}{2}(1 + \mathrm{i}\tan\theta) \) (as required).

**(d)**
From part (c), we identified that:
\( z - 1 = \mathrm{e}^{\mathrm{i}2\theta} \)

Therefore:
\( \frac{1}{z-1} = \frac{1}{\mathrm{e}^{\mathrm{i}2\theta}} = \mathrm{e}^{-2\mathrm{i}\theta} \)

Using Euler's formula:
\( \mathrm{e}^{-2\mathrm{i}\theta} = \cos(-2\theta) + \mathrm{i}\sin(-2\theta) = \cos 2\theta - \mathrm{i}\sin 2\theta \)

So, the real part is \( \cos 2\theta \) and the imaginary part is \( -\sin 2\theta \).

**Part (a)**
* **M1**: Attempt to use double-angle identities for both \( \cos 2\theta \) and \( \sin 2\theta \).
* **A1**: Obtain \( 2\cos^2\theta + 2\mathrm{i}\sin\theta\cos\theta \) or equivalent.
* **A1**: Factorise and complete the proof to show \( z = 2\cos\theta \, \mathrm{e}^{\mathrm{i}\theta} \) clearly.

**Part (b)**
* **M1**: Apply index laws for powers to exponential/polar form of \( z \).
* **A1**: Obtain \( R = \frac{1}{4\cos^2\theta} \) and \( \phi = -2\theta \) (allow \( \frac{1}{4\cos^2\theta}\mathrm{e}^{-2\mathrm{i}\theta} \)).

**Part (c)**
* **M1**: Identify that \( z - 1 = \mathrm{e}^{\mathrm{i}2\theta} \) or substitute Cartesian form of \( z - 1 \).
* **M1**: Divide by \( 2\cos\theta\mathrm{e}^{\mathrm{i}\theta} \) and simplify powers of \( \mathrm{e} \) or divide Cartesian form by multiplying numerator and denominator by conjugate.
* **A1**: Complete the proof to get \( \frac{1}{2}(1 + \mathrm{i}\tan\theta) \) with no errors.

**Part (d)**
* **M1**: Use the definition of \( z-1 \) to find \( \frac{1}{z-1} \) (either in exponential or polar form).
* **A1**: Identify the real part as \( \cos 2\theta \) and the imaginary part as \( -\sin 2\theta \).

**(i)**

We start by separating the variables:
\[ \int \frac{1}{y(y+2)} \, \mathrm{d}y = \int \frac{1}{x(x+1)} \, \mathrm{d}x \]

Using partial fractions for the left-hand side:
\[ \frac{1}{y(y+2)} = \frac{A}{y} + \frac{B}{y+2} \implies 1 = A(y+2) + By \]
Setting \(y = 0 \implies A = \frac{1}{2}\).
Setting \(y = -2 \implies B = -\frac{1}{2}\).
So,
\[ \frac{1}{y(y+2)} = \frac{1}{2}\left(\frac{1}{y} - \frac{1}{y+2}\right) \]

Using partial fractions for the right-hand side:
\[ \frac{1}{x(x+1)} = \frac{C}{x} + \frac{D}{x+1} \implies 1 = C(x+1) + Dx \]
Setting \(x = 0 \implies C = 1\).
Setting \(x = -1 \implies D = -1\).
So,
\[ \frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1} \]

Now we integrate both sides:
\[ \int \frac{1}{2}\left(\frac{1}{y} - \frac{1}{y+2}\right) \mathrm{d}y = \int \left(\frac{1}{x} - \frac{1}{x+1}\right) \mathrm{d}x \]
\[ \frac{1}{2}(\ln y - \ln(y+2)) = \ln x - \ln(x+1) + c \]
\[ \frac{1}{2}\ln\left(\frac{y}{y+2}\right) = \ln\left(\frac{x}{x+1}\right) + c \]

Multiplying by 2:
\[ \ln\left(\frac{y}{y+2}\right) = 2\ln\left(\frac{x}{x+1}\right) + 2c = \ln\left(\frac{x^2}{(x+1)^2}\right) + C' \]
where \(C' = 2c\). Let \(C' = \ln k\), then:
\[ \frac{y}{y+2} = k\frac{x^2}{(x+1)^2} \]

Using the boundary condition \(x = 2\), \(y = \frac{1}{4}\):
\[ \frac{1/4}{1/4 + 2} = k\frac{2^2}{(2+1)^2} \]
\[ \frac{1/4}{9/4} = k\frac{4}{9} \implies \frac{1}{9} = \frac{4}{9}k \implies k = \frac{1}{4} \]

Thus, the equation becomes:
\[ \frac{y}{y+2} = \frac{x^2}{4(x+1)^2} \]

Now we express \(y\) in terms of \(x\):
\[ 4(x+1)^2 y = x^2(y+2) \]
\[ 4(x^2+2x+1)y = x^2 y + 2x^2 \]
\[ (4x^2+8x+4)y - x^2 y = 2x^2 \]
\[ (3x^2+8x+4)y = 2x^2 \]
\[ y = \frac{2x^2}{3x^2+8x+4} \]
which can also be factorised as \(y = \frac{2x^2}{(3x+2)(x+2)}\).

**(ii)**

To find what happens as \(x \to \infty\), we divide the numerator and the denominator of the expression for \(y\) by \(x^2\):
\[ y = \frac{2}{3 + \frac{8}{x} + \frac{4}{x^2}} \]
As \(x \to \infty\), \(\frac{8}{x} \to 0\) and \(\frac{4}{x^2} \to 0\).
Therefore, \(y \to \frac{2}{3}\).

**(i)**
* **M1**: Separate variables and attempt integration of both sides.
* **M1**: Attempt to express at least one of the integrands in partial fractions.
* **A1**: Obtain correct partial fractions on both sides, i.e., \(\frac{1}{2}\left(\frac{1}{y} - \frac{1}{y+2}\right)\) and \(\frac{1}{x} - \frac{1}{x+1}\).
* **A1**: Obtain correct integrated terms on both sides (constant of integration not required at this stage).
* **M1**: Substitute the boundary conditions \(x = 2, y = \frac{1}{4}\) to find the constant.
* **A1**: Obtain a correct value of the constant of integration (e.g., \(k = \frac{1}{4}\) or equivalent).
* **M1**: Attempt to make \(y\) the subject of the formula.
* **A1**: Obtain \(y = \frac{2x^2}{3x^2+8x+4}\) or equivalent.

**(ii)**
* **M1**: Attempt to find the limit of \(y\) as \(x \to \infty\) by dividing by \(x^2\) or an equivalent valid limit argument.
* **A1**: State that \(y \to \frac{2}{3}\) or equivalent.

First, we find the driving force \(F\) of the car's engine using the power formula \(P = Fv\):\
\(F = \frac{P}{v} = \frac{24000}{15} = 1600\text{ N}\).\
\
Next, we apply Newton's second law of motion along the direction of motion:\
\(F - R = ma\)\
where \(R = 600\text{ N}\) is the resistance force and \(m = 1200\text{ kg}\) is the mass of the car.\
\
\(1600 - 600 = 1200a\)\
\(1000 = 1200a\)\
\(a = \frac{1000}{1200} = \frac{5}{6} \approx 0.833\text{ m s}^{-2}\).

M1: For using \(P = Fv\) to find the driving force (allow if \(24\) is used instead of \(24000\)).\
M1: For applying Newton's second law \(F - R = ma\) to form an equation in \(a\) with their driving force.\
A1: For obtaining \(a = \frac{5}{6}\) or \(0.833\) (rounded to 3 significant figures).

To find the total distance travelled, we must first determine if the particle changes its direction of motion during the interval \(0 \le t \le 4\). This occurs when the velocity \(v = 0\).

Set \(v = 0\):
\[3t^2 - 14t + 8 = 0\]
\[(3t - 2)(t - 4) = 0\]

This gives \(t = \frac{2}{3}\) and \(t = 4\).

Since \(t = \frac{2}{3}\) lies within the interval \(0 < t < 4\), the particle changes direction at this time. We must calculate the distance travelled in two separate intervals: \(0 \le t \le \frac{2}{3}\) and \(\frac{2}{3} \le t \le 4\).

First, we find the displacement \(s(t)\) by integrating the velocity function with respect to time:
\[s(t) = \int (3t^2 - 14t + 8) \, dt = t^3 - 7t^2 + 8t + c\]

Taking the initial position at \(t = 0\) to be \(s(0) = 0\), we have \(c = 0\), so:
\[s(t) = t^3 - 7t^2 + 8t\]

Now, evaluate the displacement at the key times:
- At \(t = 0\):
\[s(0) = 0\]
- At \(t = \frac{2}{3}\):
\[s\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^3 - 7\left(\frac{2}{3}\right)^2 + 8\left(\frac{2}{3}\right) = \frac{8}{27} - \frac{28}{9} + \frac{16}{3} = \frac{8 - 84 + 144}{27} = \frac{68}{27} \approx 2.519\text{ m}\]
- At \(t = 4\):
\[s(4) = 4^3 - 7(4^2) + 8(4) = 64 - 112 + 32 = -16\text{ m}\]

Now, calculate the distances travelled in each interval:
- Distance from \(t = 0\) to \(t = \frac{2}{3}\):
\[d_1 = \left| s\left(\frac{2}{3}\right) - s(0) \right| = \left| \frac{68}{27} - 0 \right| = \frac{68}{27}\text{ m}\]
- Distance from \(t = \frac{2}{3}\) to \(t = 4\):
\[d_2 = \left| s(4) - s\left(\frac{2}{3}\right) \right| = \left| -16 - \frac{68}{27} \right| = \left| -\frac{500}{27} \right| = \frac{500}{27}\text{ m}\]

Total distance travelled:
\[d_{\text{total}} = d_1 + d_2 = \frac{68}{27} + \frac{500}{27} = \frac{568}{27} \approx 21.0\text{ m} \text{ (to 3 s.f.)}\]

M1: For setting \(v = 0\) and attempting to solve for \(t\).
A1: For identifying the critical value \(t = \frac{2}{3}\) (or \(0.667\)) inside the interval.
M1: For integrating the velocity function to obtain a displacement function of the form \(t^3 - 7t^2 + 8t\) (allow omission of constant).
M1: For a complete and correct method to find the total distance, summing the absolute values of displacement changes on the intervals \([0, \frac{2}{3}]\) and \([\frac{2}{3}, 4]\).
A1: For the correct final answer of \(\frac{568}{27}\) or \(21.0\) (accept \(21\) or \(21\frac{1}{27}\)).

First, we find the normal reaction force \(R\) acting on the particle. Resolving perpendicular to the inclined plane: \(R = mg \cos\theta\). Given \(\sin\theta = 0.6\), we can find \(\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - 0.6^2} = 0.8\). Using \(m = 5\text{ kg}\) and \(g = 10\text{ m s}^{-2}\), we get \(R = 5 \times 10 \times 0.8 = 40\text{ N}\). The maximum possible frictional force is given by: \(F_{\text{max}} = \mu R = 0.25 \times 40 = 10\text{ N}\). The component of the gravitational force acting down the plane is: \(W_{\text{parallel}} = mg \sin\theta = 5 \times 10 \times 0.6 = 30\text{ N}\). There are two limiting cases for equilibrium: Case 1: The particle is on the point of sliding down the plane. In this case, the frictional force acts up the plane to oppose sliding. Resolving parallel to the plane: \(P_{\text{min}} + F_{\text{max}} = mg \sin\theta \implies P_{\text{min}} + 10 = 30 \implies P_{\text{min}} = 20\text{ N}\). Case 2: The particle is on the point of sliding up the plane. In this case, the frictional force acts down the plane to oppose motion. Resolving parallel to the plane: \(P_{\text{max}} = mg \sin\theta + F_{\text{max}} \implies P_{\text{max}} = 30 + 10 = 40\text{ N}\). Thus, for the particle to remain in equilibrium, we must have \(20 \le P \le 40\).

M1: Attempt to resolve forces perpendicular to the plane to find the normal reaction \(R\). A1: Obtain \(R = 40\text{ N}\) (or equivalent). M1: Correctly apply the limiting friction formula \(F_{\text{max}} = \mu R\). M1: Set up a parallel-to-plane equilibrium equation for the case of impending motion down the plane. A1: Obtain lower bound \(P = 20\text{ N}\). A1: Obtain upper bound \(P = 40\text{ N}\) and state the final range \(20 \le P \le 40\) (accept equivalent interval notation).

**(i)**

Let the driving force of the car's engine when traveling up the hill be \(F_1\).
Using the power formula \(P = F_1 v\):
\[24000 = F_1 \times 15\]
\[F_1 = \frac{24000}{15} = 1600 \text{ N}\]

Since the car travels up the hill at a constant speed, the acceleration is zero. Resolving forces parallel to the incline:
\[F_1 = R + mg \sin \alpha\]
\[1600 = R + 1200 \times 10 \times 0.05\]
\[1600 = R + 600\]
\[R = 1000\]

**(ii)**

Let the driving force of the car's engine when traveling on the horizontal road be \(F_2\).
At the instant when the speed \(v = 20 \text{ m s}^{-1}\):
\[F_2 = \frac{P}{v} = \frac{24000}{20} = 1200 \text{ N}\]

Using Newton's second law for horizontal motion:
\[F_2 - R = ma\]
\[1200 - 1000 = 1200 a\]
\[200 = 1200 a\]
\[a = \frac{200}{1200} = \frac{1}{6} \approx 0.167 \text{ m s}^{-2}\]

**(i)**
* **M1**: For using \(P = Fv\) to find the driving force \(F = 1600 \text{ N}\).
* **M1**: For setting up the force balance equation parallel to the slope: \(F = R + mg \sin \alpha\).
* **A1**: For obtaining \(R = 1000\).

**(ii)**
* **B1**: For calculating the new driving force: \(F_{new} = \frac{24000}{20} = 1200 \text{ N}\).
* **M1**: For applying Newton's second law on the horizontal road: \(F_{new} - R = ma\).
* **A1**: For the correct equation with values substituted: \(1200 - 1000 = 1200 a\).
* **A1**: For the correct acceleration \(a = \frac{1}{6} \approx 0.167 \text{ m s}^{-2}\) (accept \(0.167\) or \(0.17\)).

First, we identify the components of the forces acting on the block. The mass of the block is \(4\text{ kg}\), so its weight is \(W = mg = 40\text{ N}\) (taking \(g = 10\text{ m s}^{-2}\)).
s
Given \(\tan\alpha = \frac{3}{4}\), we can determine:
\(\sin\alpha = 0.6\)
\(\cos\alpha = 0.8\)

Let \(R\) be the normal reaction force between the block and the plane, and let \(F\) be the frictional force. Since the force \(P\) acts horizontally towards the incline, we resolve the forces perpendicular to the inclined plane:

\[R = W\cos\alpha + P\sin\alpha\]
\[R = 40(0.8) + P(0.6) = 32 + 0.6P\]

The maximum possible frictional force is given by:
\[F_{\text{max}} = \mu R = 0.5(32 + 0.6P) = 16 + 0.3P\]

There are two limiting cases for equilibrium:

**Case 1: The block is on the point of sliding up the plane (maximum value of \(P\))**
In this case, the friction force \(F\) acts down the plane. Resolving forces parallel to the plane:
\[P\cos\alpha = W\sin\alpha + F\]
\[0.8P = 40(0.6) + (16 + 0.3P)\]
\[0.8P = 24 + 16 + 0.3P\]
\[0.5P = 40 \implies P = 80\]

**Case 2: The block is on the point of sliding down the plane (minimum value of \(P\))**
In this case, the friction force \(F\) acts up the plane to oppose the downward motion. Resolving forces parallel to the plane:
\[P\cos\alpha + F = W\sin\alpha\]
\[0.8P + (16 + 0.3P) = 24\]
\[1.1P + 16 = 24\]
\[1.1P = 8 \implies P = \frac{80}{11} \approx 7.27\]

Thus, the range of values of \(P\) for which the block remains in equilibrium is:
\[\frac{80}{11} \le P \le 80 \quad \text{or} \quad 7.27 \le P \le 80\]

M1: Resolving forces perpendicular to the plane to express \(R\) in terms of \(P\).
A1: Obtaining \(R = 32 + 0.6P\) or equivalent.
M1: Setting up the equation of equilibrium parallel to the plane when the block is about to slide up (\(F\) acting down the plane).
A1: Substituting \(F = 0.5R\) and obtaining a correct equation in \(P\), e.g., \(0.8P = 24 + 0.5(32 + 0.6P)\).
A1: Finding the upper limit \(P = 80\).
M1: Setting up the equation of equilibrium parallel to the plane when the block is about to slide down (\(F\) acting up the plane).
A1: Substituting \(F = 0.5R\) and obtaining a correct equation in \(P\), e.g., \(0.8P + 0.5(32 + 0.6P) = 24\).
A1: Obtaining the lower limit \(P = \frac{80}{11}\) (or \(7.27\)) and stating the final range \(\frac{80}{11} \le P \le 80\).

(i) Let the direction of motion of \(P\) be the positive direction.
Using the principle of conservation of linear momentum:
\(m_P u_P + m_Q u_Q = (m_P + m_Q) v\)
\(0.4(6) + 0.6(-2) = (0.4 + 0.6) v\)
\(2.4 - 1.2 = 1.0 v\)
\(v = 1.2\text{ m s}^{-1}\).
Since \(v > 0\), the particle \(R\) moves in the direction of \(P\)'s initial motion.

(ii) Initial Kinetic Energy:
\(KE_i = \frac{1}{2} m_P u_P^2 + \frac{1}{2} m_Q u_Q^2 = \frac{1}{2}(0.4)(6)^2 + \frac{1}{2}(0.6)(-2)^2\)
\(KE_i = 7.2 + 1.2 = 8.4\text{ J}\).

Final Kinetic Energy:
\(KE_f = \frac{1}{2} (m_P + m_Q) v^2 = \frac{1}{2}(1.0)(1.2)^2\)
\(KE_f = 0.5(1.44) = 0.72\text{ J}\).

Loss in Kinetic Energy:
\(KE_{\text{loss}} = KE_i - KE_f = 8.4 - 0.72 = 7.68\text{ J}\).

(iii) The combined particle \(R\) has mass \(M = 1.0\text{ kg}\).
Given \̂\(\sin\alpha = 0.28\), we find \(\cos\alpha = \sqrt{1 - 0.28^2} = 0.96\).

Resolving perpendicular to the inclined plane:
\(R_N = Mg\cos\alpha = 1.0 \times 10 \times 0.96 = 9.6\text{ N}\).

The frictional force acting down the plane is:
\(F = \mu R_N = 0.125 \times 9.6 = 1.2\text{ N}\).

Using Newton's Second Law along the plane (taking upwards as positive):
\(-Mg\sin\alpha - F = Ma\)
\(-1.0(10)(0.28) - 1.2 = 1.0 a\)
\(-2.8 - 1.2 = a \implies a = -4\text{ m s}^{-2}\).

Using \(v^2 = u^2 + 2as\) with final velocity \(v = 0\), initial velocity \(u = 1.2\text{ m s}^{-1}\), and acceleration \(a = -4\text{ m s}^{-2}\):
\(0 = 1.2^2 + 2(-4)d\)
\(8d = 1.44 \implies d = 0.18\text{ m}\).

(i)
M1: Attempting to apply the conservation of momentum with appropriate signs for opposing directions.
A1: Obtaining the correct speed \(v = 1.2\text{ m s}^{-1}\).
B1: Stating the correct direction (e.g., "in the direction of \(P\)'s initial motion").

(ii)
M1: Attempting to calculate the total initial kinetic energy or total final kinetic energy.
M1: Finding the difference between the initial and final kinetic energies.
A1: Correctly showing that the loss is \(7.68\text{ J}\) with all steps clearly presented.

(iii)
B1: Finding \(\cos\alpha = 0.96\) or calculating the normal reaction force \(R_N = 9.6\text{ N}\).
M1: Applying \(F = \mu R_N\) and setting up Newton's second law parallel to the plane to find acceleration \(a\).
A1: Obtaining deceleration \(a = -4\text{ m s}^{-2}\).
M1: Attempting to use \(v^2 = u^2 + 2as\) with \(v = 0\) and \(u\) from part (i).
A1: Obtaining the correct distance \(d = 0.18\text{ m}\) (or equivalent fraction).

**Part (a)**

As \(B\) descends by \(2\text{ m}\), \(A\) moves up the inclined plane by \(2\text{ m}\).

- Loss in gravitational potential energy of \(B\):
\[ \Delta\text{PE}_B = m_2 g h = 3 \times 10 \times 2 = 60\text{ J} \]

- Gain in gravitational potential energy of \(A\):
\[ \Delta\text{PE}_A = m_1 g h \sin\alpha = 3 \times 10 \times 2 \times 0.6 = 36\text{ J} \]

- The normal contact force on \(A\) from the inclined plane is:
\[ R = m_1 g \cos\alpha = 3 \times 10 \times \sqrt{1 - 0.6^2} = 3 \times 10 \times 0.8 = 24\text{ N} \]

- The frictional force acting on \(A\) is:
\[ F = \mu R = 0.25 \times 24 = 6\text{ N} \]

- Work done against friction:
\[ W_f = F \times h = 6 \times 2 = 12\text{ J} \]

- Gain in kinetic energy of the system:
\[ \Delta\text{KE} = \frac{1}{2}(m_1 + m_2) v^2 = \frac{1}{2}(3 + 3)v^2 = 3v^2 \]

By the work-energy principle:
\[ \Delta\text{PE}_B = \Delta\text{PE}_A + W_f + \Delta\text{KE} \]
\[ 60 = 36 + 12 + 3v^2 \]
\[ 60 = 48 + 3v^2 \]
\[ 3v^2 = 12 \implies v^2 = 4 \implies v = 2\text{ m s}^{-1} \]

**Part (b)**

After \(B\) hits the floor, the string becomes slack. Let \(d\) be the additional distance \(A\) travels up the plane before coming to rest.

- Initial kinetic energy of \(A\) for this stage:
\[ \text{KE}_i = \frac{1}{2} m_1 v^2 = \frac{1}{2} \times 3 \times 2^2 = 6\text{ J} \]

- Gain in gravitational potential energy of \(A\):
\[ \Delta\text{PE}_A = m_1 g d \sin\alpha = 3 \times 10 \times 0.6 \times d = 18d \]

- Work done against friction:
\[ W_f = F \times d = 6 \times d = 6d \]

By the work-energy principle for \(A\) alone:
\[ \text{KE}_i = \Delta\text{PE}_A + W_f \]
\[ 6 = 18d + 6d \]
\[ 24d = 6 \implies d = 0.25\text{ m} \]

The total distance travelled by \(A\) from its initial position is:
\[ \text{Total distance} = 2 + d = 2 + 0.25 = 2.25\text{ m} \]

**Part (a)**
* **M1**: For calculating the loss in PE of \(B\) (\(60\text{ J}\)) or the gain in PE of \(A\) (\(36\text{ J}\)).
* **M1**: For calculating the normal reaction \(R = 24\text{ N}\) and the friction force \(F = 6\text{ N}\).
* **M1**: For calculating the work done against friction (\(12\text{ J}\)).
* **M1**: For writing the total kinetic energy of the system as \(3v^2\).
* **M1**: For setting up a complete work-energy equation of the form \(\Delta\text{PE}_B = \Delta\text{PE}_A + W_f + \Delta\text{KE}\).
* **A1**: For obtaining \(v = 2\text{ m s}^{-1}\).

**Part (b)**
* **M1**: For finding the initial kinetic energy of \(A\) at the start of this stage (\(6\text{ J}\)) or setting up the energy equation for the second stage.
* **M1**: For setting up the work-energy equation for \(A\): \(\frac{1}{2}m_1 v^2 = m_1 g d \sin\alpha + F d\).
* **A1**: For finding the additional distance \(d = 0.25\text{ m}\).
* **A1**: For finding the total distance \(2.25\text{ m}\).

For (a): The probability of rolling a 6 on a single roll is \(p = \frac{2}{6} = \frac{1}{3}\). The probability that the first 6 is on the 4th roll is given by the geometric probability formula \(P(X = 4) = (1-p)^3 p = \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right) = \frac{8}{81} \approx 0.0988\). For (b): Let \(Y\) be the number of times a player rolls a 6 in 5 rolls. Then \(Y \sim B(5, \frac{1}{3})\). The probability that a player rolls a 6 at least twice is \(P(Y \ge 2) = 1 - P(Y = 0) - P(Y = 1)\). Computing these individual probabilities: \(P(Y = 0) = \left(\frac{2}{3}\right)^5 = \frac{32}{243}\) and \(P(Y = 1) = \binom{5}{1} \left(\frac{1}{3}\right) \left(\frac{2}{3}\right)^4 = \frac{80}{243}\). Therefore, \(P(Y \ge 2) = 1 - \frac{32}{243} - \frac{80}{243} = \frac{131}{243} \approx 0.5391\). Now, let \(W\) be the number of players who roll a 6 at least twice in a group of 12 players. Then \(W \sim B(12, \frac{131}{243})\). The probability that exactly 3 players achieve this is \(P(W = 3) = \binom{12}{3} \left(\frac{131}{243}\right)^3 \left(\frac{112}{243}\right)^9 \approx 220 \times 0.15667 \times 0.0009668 \approx 0.0333\) (to 3 significant figures).

(a) M1 for attempting to find the probability of a first success on the 4th trial using \((1-p)^3 p\) with \(p = \frac{1}{3}\). A1 for \\frac{8}{81}\\ (or 0.0988). (b) M1 for calculating the probability that a player rolls a 6 at least twice, finding \(P(Y \ge 2) = \frac{131}{243}\\ (or 0.539). M1 for setting up the binomial term \binom{12}{3} p^3 (1-p)^9\\ with their probability \)p\). A1 for 0.0333 (accept 0.0333 to 0.0334).

Let \(R_1\) and \(G_1\) represent choosing a red sweet and a green sweet on the first draw, respectively.
Let \(R_2\) represent choosing a red sweet on the second draw.

Initially, there are 5 red sweets and 3 green sweets (total of 8 sweets).
\(P(R_1) = \frac{5}{8}\)
\(P(G_1) = \frac{3}{8}\)

**Case 1: First sweet chosen is red**
If \(R_1\) occurs, the sweet is not replaced, leaving 4 red and 3 green sweets.
Then, 3 red and 1 green sweets are added.
New bag composition: \(4 + 3 = 7\) red, \(3 + 1 = 4\) green (total of 11 sweets).
Thus, \(P(R_2 | R_1) = \frac{7}{11}\).

The probability of this path is:
\(P(R_1 \cap R_2) = \frac{5}{8} \times \frac{7}{11} = \frac{35}{88}\)

**Case 2: First sweet chosen is green**
If \(G_1\) occurs, the sweet is not replaced, leaving 5 red and 2 green sweets.
Then, 1 red and 3 green sweets are added.
New bag composition: \(5 + 1 = 6\) red, \(2 + 3 = 5\) green (total of 11 sweets).
Thus, \(P(R_2 | G_1) = \frac{6}{11}\).

The probability of this path is:
\(P(G_1 \cap R_2) = \frac{3}{8} \times \frac{6}{11} = \frac{18}{88}\)

**Total Probability of \(R_2\):**
\(P(R_2) = P(R_1 \cap R_2) + P(G_1 \cap R_2) = \frac{35}{88} + \frac{18}{88} =
\frac{53}{88}\)

**Conditional Probability:**
We want to find \(P(G_1 | R_2)\):
\(P(G_1 | R_2) = \frac{P(G_1 \cap R_2)}{P(R_2)} = \frac{\frac{18}{88}}{\frac{53}{88}} = \frac{18}{53} \approx 0.340\) (to 3 s.f.)

**M1**: For identifying the two correct scenarios for the first and second draw, with correct conditional denominators of 8 and 11.
**A1**: For finding either \(P(R_1 \cap R_2) = \frac{35}{88}\) or \(P(G_1 \cap R_2) = \frac{18}{88}\) (or equivalent decimals).
**M1**: For finding the total probability \(P(R_2)\) by adding two product probabilities, i.e., \(\frac{35}{88} + \frac{18}{88} = \frac{53}{88}\) (or approximately \(0.602\)).
**M1**: For using Bayes' Theorem formula correctly, dividing their \(P(G_1 \cap R_2)\) by their total \(P(R_2)\).
**A1**: For obtaining the correct final answer of \(\frac{18}{53}\) (or \(0.340\) correct to 3 significant figures).

We are given that \(X \sim N(\mu, \sigma^2)\). Using the standard normal distribution \(Z \sim N(0, 1)\):

1) Standardise \(P(X < 24) = 0.05\):
\(P\left(Z < \frac{24 - \mu}{\sigma}\right) = 0.05\)
Since the probability is less than 0.5, the z-value must be negative. From the normal tables, the critical value for 0.95 is 1.645.
Therefore, we have:
\(24 - \mu = -1.645\sigma\) (Equation 1)

2) Standardise \(P(X < 60) = 0.90\):
\(P\left(Z < \frac{60 - \mu}{\sigma}\right) = 0.90\)
From the normal tables, the critical value for 0.90 is 1.282.
Therefore, we have:
\(60 - \mu = 1.282\sigma\) (Equation 2)

3) Solve Equations 1 and 2 simultaneously:
Subtract Equation 1 from Equation 2:
\((60 - \mu) - (24 - \mu) = 1.282\sigma - (-1.645\sigma)\)

\(36 = 2.927\sigma\)

\(\sigma = \frac{36}{2.927} \approx 12.299\)

Substitute \(\sigma = 12.299\) back into Equation 2:
\(60 - \mu = 1.282 \times 12.299\)

\(60 - \mu = 15.767\)

\(\mu = 44.233\)

To 3 significant figures, we obtain:
\(\mu = 44.2\) and \(\sigma = 12.3\).

M1: For standardising and equating to a z-value (either \(\pm 1.645\) or \(\pm 1.282\))
A1: For obtaining a correct equation in \(\mu\) and \(\sigma\), e.g. \(24 - \mu = -1.645\sigma\) (allow \(z = -1.64\))
A1: For obtaining a second correct equation in \(\mu\) and \(\sigma\), e.g. \(60 - \mu = 1.282\sigma\) (allow \(z = 1.28\))
M1: For an attempt to solve their two linear simultaneous equations to find one variable
A1: For \(\sigma = 12.3\) (or \(12.299...\))
A1: For \(\mu = 44.2\) (or \(44.23...\))

(a)
There are $n = 15$ data values in total.
- The median is the $\frac{15+1}{2} = 8\text{th}$ value in the ordered list. Counting from the smallest: $21, 24, 27, 28, 30, 32, 35, 35, \dots$, so the 8th value is $35$.
Thus, the median is $35$ (or $$35\,000$).

- The lower quartile ($Q_1$) is the $\frac{15+1}{4} = 4\text{th}$ value.
The 4th value is $28$.
Thus, $Q_1 = 28$ (or $$28\,000$).

- The upper quartile ($Q_3$) is the $\frac{3(15+1)}{4} = 12\text{th}$ value.
The 12th value is $46$ (from $21, 24, 27, 28, 30, 32, 35, 35, 39, 41, 43, 46$).
Thus, $Q_3 = 46$ (or $$46\,000$).

(b)
First, find the interquartile range (IQR):
$$\text{IQR} = Q_3 - Q_1 = 46 - 28 = 18 \text{ (or } \$18\,000\text{)}$$

Now calculate $1.5 \times \text{IQR}$:
$$1.5 \times 18 = 27 \text{ (or } \$27\,000\text{)}$$

The boundaries for outliers are:
- Lower boundary:
$$Q_1 - 1.5 \times \text{IQR} = 28 - 27 = 1 \text{ (or } \$1\,000\text{)}$$
- Upper boundary:
$$Q_3 + 1.5 \times \text{IQR} = 46 + 27 = 73 \text{ (or } \$73\,000\text{)}$$

Comparing the data set's minimum and maximum values to these boundaries:
- The minimum value in the data set is $21$ (or $$21\,000$), which is greater than the lower boundary of $1$.
- The maximum value in the data set is $65$ (or $$65\,000$), which is less than the upper boundary of $73$.

Since all values in the data set lie between the lower boundary of $1$ and the upper boundary of $73$, there are no outliers.

(a)
- B1: For Median = $35$ (or $$35\,000$)
- B1: For $Q_1 = 28$ (or $$28\,000$)
- B1: For $Q_3 = 46$ (or $$46\,000$)
[Accept values with or without the thousands/currency unit, provided they are consistent.]

(b)
- M1: For calculating $\text{IQR} = Q_3 - Q_1$ using their values from part (a)
- M1: For a method to calculate both the lower boundary ($Q_1 - 1.5 \times \text{IQR}$) and upper boundary ($Q_3 + 1.5 \times \text{IQR}$)
- A1ft: For correct boundaries of $1$ and $73$ (or $$1\,000$ and $$73\,000$) follow-through their quartiles
- A1: For comparing the data extremes ($21$ and $65$) to these boundaries and concluding that there are no outliers.

**(a)**
The sum of the probabilities in a probability distribution table must equal 1:
\[ a + b + 2b + (a + 0.1) = 1 \]
\[ 2a + 3b + 0.1 = 1 \]
\[ 2a + 3b = 0.9 \quad \text{--- (Equation 1)} \]

Using the formula for expectation \(\text{E}(X) = \sum x \cdot \text{P}(X=x)\):
\[ \text{E}(X) = 1(a) + 2(b) + 3(2b) + 4(a + 0.1) = 2.7 \]
\[ a + 2b + 6b + 4a + 0.4 = 2.7 \]
\[ 5a + 8b + 0.4 = 2.7 \]
\[ 5a + 8b = 2.3 \quad \text{--- (Equation 2)} \]

We can solve these simultaneous equations. From Equation 1:
\[ 2a = 0.9 - 3b \implies a = 0.45 - 1.5b \]

Substitute this into Equation 2:
\[ 5(0.45 - 1.5b) + 8b = 2.3 \]
\[ 2.25 - 7.5b + 8b = 2.3 \]
\[ 2.25 + 0.5b = 2.3 \]
\[ 0.5b = 0.05 \implies b = 0.1 \]

Substitute \(b = 0.1\) back to find \(a\):
\[ a = 0.45 - 1.5(0.1) = 0.3 \]

Thus, \(a = 0.3\) and \(b = 0.1\).

**(b)**
Using the values \(a = 0.3\) and \(b = 0.1\), the probability distribution is:
- \(\text{P}(X=1) = 0.3\)
- \(\text{P}(X=2) = 0.1\)
- \(\text{P}(X=3) = 0.2\)
- \(\text{P}(X=4) = 0.4\)

First, find \(\text{E}(X^2)\):
\[ \text{E}(X^2) = \sum x^2 \cdot \text{P}(X=x) \]
\[ \text{E}(X^2) = 1^2(0.3) + 2^2(0.1) + 3^2(0.2) + 4^2(0.4) \]
\[ \text{E}(X^2) = 1(0.3) + 4(0.1) + 9(0.2) + 16(0.4) \]
\[ \text{E}(X^2) = 0.3 + 0.4 + 1.8 + 6.4 = 8.9 \]

Now, find \(\text{Var}(X)\):
\[ \text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 \]
\[ \text{Var}(X) = 8.9 - 2.7^2 \]
\[ \text{Var}(X) = 8.9 - 7.29 = 1.61 \]

**(a)**
* **M1**: Write an equation for the sum of probabilities equal to 1: \(2a + 3b = 0.9\) (or equivalent).
* **M1**: Write an equation for the expectation \(\text{E}(X) = 2.7\): \(5a + 8b = 2.3\) (or equivalent).
* **M1**: Solve the simultaneous equations to find one variable.
* **A1**: Obtain \(a = 0.3\) and \(b = 0.1\) (both correct).

**(b)**
* **M1**: Attempt to find \(\text{E}(X^2)\) using \(\sum x^2 \cdot \text{P}(X=x)\) with their values of \(a\) and \(b\).
* **A1**: Obtain \(\text{E}(X^2) = 8.9\) (or correct value based on their \(a\) and \(b\)).
* **A1**: Correctly calculate \(\text{Var}(X) = 1.61\) (or correct FT variance from their \(a\) and \(b\), provided the probabilities are valid).

**(a)**
The sum of the probabilities in a probability distribution table must equal 1:
\[ a + b + 2b + (a + 0.1) = 1 \]
\[ 2a + 3b + 0.1 = 1 \]
\[ 2a + 3b = 0.9 \quad \text{--- (Equation 1)} \]

Using the formula for expectation \(\text{E}(X) = \sum x \cdot \text{P}(X=x)\):
\[ \text{E}(X) = 1(a) + 2(b) + 3(2b) + 4(a + 0.1) = 2.7 \]
\[ a + 2b + 6b + 4a + 0.4 = 2.7 \]
\[ 5a + 8b + 0.4 = 2.7 \]
\[ 5a + 8b = 2.3 \quad \text{--- (Equation 2)} \]

We can solve these simultaneous equations. From Equation 1:
\[ 2a = 0.9 - 3b \implies a = 0.45 - 1.5b \]

Substitute this into Equation 2:
\[ 5(0.45 - 1.5b) + 8b = 2.3 \]
\[ 2.25 - 7.5b + 8b = 2.3 \]
\[ 2.25 + 0.5b = 2.3 \]
\[ 0.5b = 0.05 \implies b = 0.1 \]

Substitute \(b = 0.1\) back to find \(a\):
\[ a = 0.45 - 1.5(0.1) = 0.3 \]

Thus, \(a = 0.3\) and \(b = 0.1\).

**(b)**
Using the values \(a = 0.3\) and \(b = 0.1\), the probability distribution is:
- \(\text{P}(X=1) = 0.3\)
- \(\text{P}(X=2) = 0.1\)
- \(\text{P}(X=3) = 0.2\)
- \(\text{P}(X=4) = 0.4\)

First, find \(\text{E}(X^2)\):
\[ \text{E}(X^2) = \sum x^2 \cdot \text{P}(X=x) \]
\[ \text{E}(X^2) = 1^2(0.3) + 2^2(0.1) + 3^2(0.2) + 4^2(0.4) \]
\[ \text{E}(X^2) = 1(0.3) + 4(0.1) + 9(0.2) + 16(0.4) \]
\[ \text{E}(X^2) = 0.3 + 0.4 + 1.8 + 6.4 = 8.9 \]

Now, find \(\text{Var}(X)\):
\[ \text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 \]
\[ \text{Var}(X) = 8.9 - 2.7^2 \]
\[ \text{Var}(X) = 8.9 - 7.29 = 1.61 \]

**(a)**
* **M1**: Write an equation for the sum of probabilities equal to 1: \(2a + 3b = 0.9\) (or equivalent).
* **M1**: Write an equation for the expectation \(\text{E}(X) = 2.7\): \(5a + 8b = 2.3\) (or equivalent).
* **M1**: Solve the simultaneous equations to find one variable.
* **A1**: Obtain \(a = 0.3\) and \(b = 0.1\) (both correct).

**(b)**
* **M1**: Attempt to find \(\text{E}(X^2)\) using \(\sum x^2 \cdot \text{P}(X=x)\) with their values of \(a\) and \(b\).
* **A1**: Obtain \(\text{E}(X^2) = 8.9\) (or correct value based on their \(a\) and \(b\)).
* **A1**: Correctly calculate \(\text{Var}(X) = 1.61\) (or correct FT variance from their \(a\) and \(b\), provided the probabilities are valid).

(a) Let \(X\) be the number of seeds that germinate, where \(X \sim B(120, 0.35)\). Here, \(n = 120\) and \(p = 0.35\). We calculate: \(np = 120 \times 0.35 = 42\) and \(n(1-p) = 120 \times 0.65 = 78\). Since both \(np > 5\) and \(n(1-p) > 5\), a normal distribution is a suitable approximation. (b) We approximate \(X\) using \(Y \sim N(\mu, \sigma^2)\), where: \(\mu = np = 42\) and \(\sigma^2 = np(1-p) = 120 \times 0.35 \times 0.65 = 27.3\). We want to find \(P(36 < X < 50) = P(37 \le X \le 49)\). Applying the continuity correction: \(P(36.5 \le Y \le 49.5)\). Standardizing: \(z_1 = \frac{36.5 - 42}{\sqrt{27.3}} = -1.053\) and \(z_2 = \frac{49.5 - 42}{\sqrt{27.3}} = 1.435\). The required probability is \(\Phi(1.435) - \Phi(-1.053) = \Phi(1.435) - (1 - \Phi(1.053)) = 0.9243 - (1 - 0.8538) = 0.7781 \approx 0.778\). (c) Let \(W\) be the number of germinating seeds in a sample of 8, so \(W \sim B(8, 0.35)\). We want \(P(W \ge 2) = 1 - P(W = 0) - P(W = 1)\). \(P(W = 0) = (0.65)^8 = 0.03186\). \(P(W = 1) = \binom{8}{1} (0.35)^1 (0.65)^7 = 8 \times 0.35 \times 0.04902 = 0.13726\). \(P(W \ge 2) = 1 - 0.03186 - 0.13726 = 0.83088 \approx 0.831\).

(a) M1: For calculating both \(np\) and \(nq\). A1: For comparing both to 5 and concluding normal approximation is justified. (b) B1: For stating mean \(\mu = 42\) and variance \(\sigma^2 = 27.3\). M1: For applying continuity correction correctly (using \(36.5\) and \(49.5\)). M1: For standardizing at least one boundary using their mean and standard deviation. M1: For finding correct area region (e.g., \(\Phi(z_2) - (1 - \Phi(z_1))\)). A1: For obtaining \(0.778\) (accept \(0.778\) or \(0.779\)). (c) M1: For stating or using \(1 - P(W=0) - P(W=1)\). A1: For correct calculation of \(P(W=0)\) and \(P(W=1)\). A1: For obtaining \(0.831\).

**Part (i)** We want to arrange the 11 letters of CATASTROPHE such that the two A's are together, but the two T's are not together. First, we treat the two A's as a single block: \( (AA) \). This leaves us with 10 items to arrange: \( (AA) \), T, T, C, S, R, O, P, H, E. We first arrange the 8 items other than T: \( (AA) \), C, S, R, O, P, H, E. Since these 8 items are all distinct, they can be arranged in \( 8! = 40320 \) ways. This creates 9 possible slots (including the ends) to place the two identical T's. We choose 2 of these 9 slots: \( \binom{9}{2} = 36 \) ways. Thus, the number of successful arrangements is \( 8! \times \binom{9}{2} = 40320 \times 36 = 1451520 \). *Alternative Method:* Total arrangements with \( AA \) together is \( \frac{10!}{2!} = 1814400 \). Subtract the arrangements where both \( AA \) and \( TT \) are together, which is \( 9! = 362880 \) ways. This gives \( 1814400 - 362880 = 1451520 \). **Part (ii)** The arrangement must start with T and end with A, leaving 9 positions for the remaining 9 letters. After using one T and one A, the remaining letters are A, T, C, S, R, O, P, H, E. Since these 9 letters are all distinct, the number of arrangements is \( 9! = 362880 \). **Part (iii)** The letters are two A's, two T's, and seven other distinct letters. We have 9 distinct letter types: A, T, C, S, R, O, P, H, E. We consider three mutually exclusive cases for selecting 4 letters: Case 1: All 4 letters are distinct. We choose 4 types from 9: \( \binom{9}{4} = 126 \). Case 2: One pair and two distinct letters. We choose 1 type for the pair (A or T): \( \binom{2}{1} = 2 \) ways. We choose 2 types from the remaining 8: \( \binom{8}{2} = 28 \) ways. This gives \( 2 \times 28 = 56 \) ways. Case 3: Two pairs (AA and TT). There is only \( 1 \) way. Total selections = \( 126 + 56 + 1 = 183 \).

**Part (i)** M1: For treating AA as a single unit and arranging the other 8 items, or for finding the total arrangements with AA together. M1: For a correct method to exclude arrangements with T's together (e.g. multiplying by \( \binom{9}{2} \) or subtracting \( 9! \)). A1: For identifying either \( 8! \) and \( \binom{9}{2} \) correctly, or \( 1814400 \) and \( 362880 \) correctly. A1: For the correct final answer of \( 1451520 \\. **Part (ii)** M1: For identifying the 9 remaining positions and letters. A1: For recognizing that the 9 remaining letters are all distinct. A1: For the correct final answer of \) 362880 \\. **Part (iii)** M1: For identifying the three cases (all distinct, one pair, two pairs). A1: For correctly calculating at least two of the cases (e.g. 126 and 56). A1: For the correct final answer of \( 183 \\.

Let \(X\) be the number of email queries received in a 24-hour period.
Since the number of queries in 1 hour has a Poisson distribution with mean \(2.4\), the distribution of \(X\) is:
\[X \sim \text{Po}(24 \times 2.4) \implies X \sim \text{Po}(57.6)\]

Since \(\lambda = 57.6 > 15\), we can approximate \(X\) using a normal distribution:
\[X \approx Y \sim \text{N}(57.6, 57.6)\]

We want to find the probability:
\[\text{P}(50 \le X \le 65)\]

Applying a continuity correction, this becomes:
\[\text{P}(49.5 < Y < 65.5)\]

Standardizing the values:
\[z_1 = \frac{49.5 - 57.6}{\sqrt{57.6}} = \frac{-8.1}{7.5895} \approx -1.067\]
\[z_2 = \frac{65.5 - 57.6}{\sqrt{57.6}} = \frac{7.9}{7.5895} \approx 1.041\]

Now, calculate the probability using the standard normal distribution table:
\[\text{P}(-1.067 < Z < 1.041) = \Phi(1.041) - \Phi(-1.067)\]
\[= \Phi(1.041) - (1 - \Phi(1.067))\]
\[\approx 0.8511 - (1 - 0.8570) = 0.8511 - 0.1430 = 0.7081\]

Thus, the probability is \(0.708\) (to 3 significant figures).

M1: State or imply \(\lambda = 57.6\) and identify the approximating distribution as \(\text{N}(57.6, 57.6)\).
M1: Apply continuity correction to find \(\text{P}(49.5 < Y < 65.5)\) (using both \(49.5\) and \(65.5\)).
M1: Standardize with their mean and standard deviation to find both z-values.
M1: Subtract \((1 - \Phi(z_1))\) from \Γ(\Phi(z_2)\) using their calculated z-values.
A1: Correct final answer of \(0.708\) (accept answers in the range \(0.707\) to \(0.709\)).

Let \(y = x - 12.5\). We have \(n = 50\), \(\sum y = 100\), and \(\sum y^2 = 935\).

**(a)**
First, find the mean:
\[ \bar{y} = \frac{\sum y}{n} = \frac{100}{50} = 2 \]
The unbiased estimate of the population mean, \(\bar{x}\), is:
\[ \bar{x} = \bar{y} + 12.5 = 2 + 12.5 = 14.5 \text{ cm} \]

Alternatively, we can find \(\sum x\):
\[ \sum x = \sum (y + 12.5) = \sum y + 50 \times 12.5 = 100 + 625 = 725 \]
\[ \bar{x} = \frac{725}{50} = 14.5 \text{ cm} \]

Next, find the unbiased estimate of the population variance, \(s^2\):
Since the variance is independent of a shift in origin:
\[ s^2 = \frac{1}{n-1} \left( \sum y^2 - \frac{(\sum y)^2}{n} \right) \]
\[ s^2 = \frac{1}{49} \left( 935 - \frac{100^2}{50} \right) \]
\[ s^2 = \frac{1}{49} (935 - 200) = \frac{735}{49} = 15 \]

**(b)**
A statistic is an unbiased estimator of a population parameter if the expected value (or the mean of the sampling distribution) of the statistic is equal to the true value of that population parameter, i.e., \(E(T) = \theta\), where \(T\) is the estimator and \(\theta\) is the population parameter.

**(a)**
* **M1** for a correct method to find the unbiased estimate of the population mean (e.g., finding \(\bar{y} = 2\) and adding 12.5, or finding \(\sum x = 725\) and dividing by 50).
* **A1** for the correct mean of \(14.5\).
* **M1** for a correct substitution into a valid formula for the unbiased estimate of the population variance with divisor \(n-1 = 49\).
* **A1** for the correct variance of \(15\).

**(b)**
* **B1** for stating that the expected value (or average value over many samples) of the estimator is equal to the parameter.
* **B1** for clearly identifying that this refers to the relationship between the estimator and the true population parameter (or expressing as \(E(T) = \theta\) with defined terms).

The width of the confidence interval is:
\[56.256 - 51.744 = 4.512\]

The margin of error is half the width of the confidence interval:
\[E = \frac{4.512}{2} = 2.256\]

The formula for the margin of error is:
\[E = z \times \frac{\sigma}{\sqrt{n}}\]

Substitute \(\sigma = 12.0\), \(n = 100\), and \(E = 2.256\):
\[z \times \frac{12}{\sqrt{100}} = 2.256\]
\[z \times 1.2 = 2.256\]
\[z = \frac{2.256}{1.2} = 1.88\]

Let the confidence level be \(\alpha\). Since the confidence interval is symmetric around the sample mean, we find the probability:
\[\alpha = 2\Phi(z) - 1\]

Using standard normal tables for \(z = 1.88\):
\[\Phi(1.88) = 0.9699\]

Thus:
\[\alpha = 2(0.9699) - 1 = 0.9398\]

This corresponds to a confidence level of \(93.98\%\), which to 3 significant figures is \(94.0\%\) (or \(94\%\)).

M1: For calculating the margin of error as \(2.256\) (or width as \(4.512\)) and setting up a correct equation for \(z\).
A1: For obtaining \(z = 1.88\).
M1: For calculating \(2\Phi(z) - 1\) using their \(z\) value.
A1: For obtaining \(94\%\) or \(94.0\%\) (accept answers in the range \(93.9\%\) to \(94.1\%\)).

Let \(S\) be the weight of a small bag of flour and \(L\) be the weight of a large bag of flour. We are given: \(S \sim N(250, 3^2)\) and \(L \sim N(515, 8^2)\). Define a new normal random variable \(W = L - 2S\). We want to find \(\text{P}(L > 2S) = \text{P}(L - 2S > 0) = \text{P}(W > 0)\). First, calculate the expectation of \(W\): \(\text{E}(W) = \text{E}(L) - 2\text{E}(S) = 515 - 2(250) = 15\). Next, calculate the variance of \(W\): \(\text{Var}(W) = \text{Var}(L) + 2^2 \text{Var}(S) = 8^2 + 4(3^2) = 64 + 36 = 100\). Thus, \(W \sim N(15, 100)\). Standardizing to find the probability: \(\text{P}(W > 0) = \text{P}\left(Z > \frac{0 - 15}{\sqrt{100}}\right) = \text{P}(Z > -1.5) = \text{P}(Z < 1.5) = \Phi(1.5)\). From the normal distribution table, \(\Phi(1.5) = 0.9332\) (or \(0.933\) to 3 significant figures).

M1: For defining a linear combination of the form \(L - 2S\) or \(2S - L\).
B1: For finding the correct mean of the combined distribution (\(15\) or \(-15\)).
M1: For attempting to find the variance of the combined distribution using \(\text{Var}(L) + 4\text{Var}(S)\).
A1: For obtaining the correct variance of \(100\).
M1: For standardizing with their mean and standard deviation and attempting to find the correct area.
A1: For the correct probability of \(0.9332\) (or \(0.933\)).

(a) Let \(X_1\) be the number of goals scored by Team A in the first half, so \(X_1 \sim \text{Po}(0.8)\).
Let \(X_2\) be the number of goals scored by Team A in the second half, so \(X_2 \sim \text{Po}(1.2)\).
Since the goals in the first and second halves are independent, the total goals scored by Team A in a match, \(X = X_1 + X_2\), has the distribution:
\(X \sim \text{Po}(0.8 + 1.2) \Rightarrow X \sim \text{Po}(2.0)\).

We require:
\(P(X = 2) = \frac{e^{-2} \times 2^2}{2!} = 2e^{-2} \approx 0.27067 \approx 0.271\) (to 3 s.f.).

(b) We want to find the conditional probability:
\(P(X_1 = 1 \cap X_2 = 1 \mid X = 2) = \frac{P(X_1 = 1 \cap X_2 = 1)}{P(X = 2)}\).

Since \(X_1\) and \(X_2\) are independent:
\(P(X_1 = 1 \cap X_2 = 1) = P(X_1 = 1) \times P(X_2 = 1) = \left(e^{-0.8} \times 0.8\right) \times \left(e^{-1.2} \times 1.2\right) = 0.96 e^{-2}\).

From part (a), \(P(X = 2) = 2 e^{-2}\).

Therefore, the conditional probability is:
\(\frac{0.96 e^{-2}}{2 e^{-2}} = \frac{0.96}{2} = 0.48\).

*Alternative method:*
Given \(X = 2\), the distribution of \(X_1\) is binomial \(B(2, p)\) where \(p = \frac{0.8}{0.8 + 1.2} = 0.4\).
Thus, \(P(X_1 = 1 \mid X = 2) = \binom{2}{1} (0.4)^1 (0.6)^1 = 2 \times 0.4 \times 0.6 = 0.48\).

(c) Let \(Y_1\) and \(Y_2\) be the goals scored by Team B in the first and second halves respectively.
\(Y_1 \sim \text{Po}(0.6)\) and \(Y_2 \sim \text{Po}(0.9)\).
The total goals scored by Team B in a match is \(Y = Y_1 + Y_2 \sim \text{Po}(0.6 + 0.9) = \text{Po}(1.5)\).

Let \(T = X + Y\) be the total goals scored by both teams combined.
Since \(X\) and \(Y\) are independent:
\(T \sim \text{Po}(2.0 + 1.5) \Rightarrow T \sim \text{Po}(3.5)\).

We require \(P(T > 3) = 1 - P(T \le 3)\):
\(P(T \le 3) = P(T = 0) + P(T = 1) + P(T = 2) + P(T = 3)\)
\(P(T \le 3) = e^{-3.5} \left( 1 + 3.5 + \frac{3.5^2}{2!} + \frac{3.5^3}{3!} \right)\)
\(P(T \le 3) = e^{-3.5} \left( 1 + 3.5 + 6.125 + 7.145833 \right)\)
\(P(T \le 3) = e^{-3.5} (17.770833) \approx 0.53663\).

Thus, \(P(T > 3) = 1 - 0.53663 = 0.46337 \approx 0.463\) (to 3 s.f.).

(a)
M1: For adding the means of the independent Poisson variables (mean = 2.0) and substituting into the Poisson probability formula.
A1: For obtaining 0.271 (or 2e^{-2}).

(b)
M1: For finding the probability of the intersection \(P(X_1 = 1) \times P(X_2 = 1)\) (or recognizing and defining the correct binomial model \(B(2, 0.4)\)).
M1: For dividing by \(P(X = 2)\) to find the conditional probability.
A1: For obtaining the correct exact value of 0.48.

(c)
M1: For calculating the combined mean of both teams as \(2.0 + 1.5 = 3.5\).
M1: For expressing \(P(T > 3)\) as \(1 - P(T \le 3)\) with four terms in the expansion.
A1: For correct calculation of the sum \(P(T \le 3) \approx 0.537\).
A1: For obtaining the final answer of 0.463 (accept 0.463 to 0.464).

(a) Let \(X\) be the lifetime of a lightbulb. We test the hypotheses: \(H_0: \mu = 1200\) and \(H_1: \mu < 1200\). Under \(H_0\), the sample mean \(\bar{X}\) has the distribution \(\bar{X} \sim N\left(1200, \frac{80^2}{64}\right) = N(1200, 10^2)\). For a 5% significance level one-tailed test, the critical value of \(Z\) is \(-1.645\). Thus, the critical region is given by: \(\frac{\bar{X} - 1200}{10} < -1.645\), which simplifies to \(\bar{X} - 1200 < -16.45\), yielding \(\bar{X} < 1183.55\) (or \(\bar{X} \le 1184\) to 4 s.f.). (b) A Type II error occurs when we fail to reject \(H_0\) given that \(H_0\) is false. We fail to reject \(H_0\) if \(\bar{X} \ge 1183.55\). If the true mean is \(\mu = 1170\), then \(\bar{X} \sim N(1170, 10^2)\). The probability of a Type II error is: \(P(\bar{X} \ge 1183.55 \mid \mu = 1170) = P\left(Z \ge \frac{1183.55 - 1170}{10}\right) = P(Z \ge 1.355) = 1 - \Phi(1.355)\). From the normal distribution tables, \(\Phi(1.355) \approx 0.9123\). Thus, the probability of a Type II error is \(1 - 0.9123 = 0.0877\) (to 3 s.f.). (c) The sample mean is \(\bar{x} = 1181.2\). Since \(1181.2 < 1183.55\), the sample mean lies in the critical region, so we reject \(H_0\). Since we reject the null hypothesis, we could only have made a Type I error (rejecting \(H_0\) when it is in fact true). We could not have made a Type II error.

(a) B1: State both hypotheses correctly in terms of \(\mu\). B1: Correctly state or use the standard error of the mean as 10. M1: Standardise and equate to \(-1.645\) (or \(-1.64\)). A1: Correct critical region: \(\bar{X} < 1183.55\) (or \(\bar{X} < 1184\)). (b) M1: Identify that Type II error is \(P(\bar{X} \ge \text{their critical value} \mid \mu = 1170)\). M1: Standardise with the new mean 1170 and standard error 10. A1: Correctly obtain \(Z = 1.355\) (or equivalent from their critical value). A1: Correct probability of \(0.0877\) (accept values from \(0.0805\) to \(0.0880\) depending on early rounding). (c) B1: Conclude that \(H_0\) is rejected because the sample mean of 1181.2 lies in the critical region. B1: State that a Type I error could have been made, and a Type II error could not have been made.

(a) To show that \(k = \frac{1}{3}\), we use the condition that the total area under the probability density function must equal 1:

\(\int_{0}^{2} kx \, dx + \int_{2}^{3} 2k(3-x) \, dx = 1\)

Integrating both terms:

\(\left[ \frac{1}{2}kx^2 \right]_{0}^{2} + 2k \left[ 3x - \frac{1}{2}x^2 \right]_{2}^{3} = 1\)

Evaluating the limits:

\(\left( \frac{1}{2}k(4) - 0 \right) + 2k \left( \left( 9 - 4.5 \right) - \left( 6 - 2 \right) \right) = 1\)

\(2k + 2k(4.5 - 4) = 1\)

\(2k + k = 1 \implies 3k = 1 \implies k = \frac{1}{3}\).

Alternative geometric method: The total area is the sum of the areas of two triangles:

- The first triangle from \(x = 0\) to \(x = 2\) has base 2 and height \(2k\), giving Area = \(\frac{1}{2} \times 2 \times 2k = 2k\).

- The second triangle from \(x = 2\) to \(x = 3\) has base 1 and height \(2k\), giving Area = \(\frac{1}{2} \times 1 \times 2k = k\).

Total area = \(2k + k = 3k = 1 \implies k = \frac{1}{3}\).

(b) The expectation \(\text{E}(X)\) is defined as:

\(\text{E}(X) = \int_{-\infty}^{\infty} x f(x) \, dx\)

Substituting \(k = \frac{1}{3}\):

\(\text{E}(X) = \int_{0}^{2} \frac{1}{3}x^2 \, dx + \int_{2}^{3} \frac{2}{3}x(3-x) \, dx\)

\(\text{E}(X) = \left[ \frac{1}{9}x^3 \right]_{0}^{2} + \frac{2}{3} \int_{2}^{3} (3x - x^2) \, dx\)

\(\text{E}(X) = \frac{8}{9} + \frac{2}{3} \left[ \frac{3}{2}x^2 - \frac{1}{3}x^3 \right]_{2}^{3}\)

\(\text{E}(X) = \frac{8}{9} + \frac{2}{3} \left[ \left( \frac{27}{2} - 9 \right) - \left( 6 - \frac{8}{3} \right) \right]\)

\(\text{E}(X) = \frac{8}{9} + \frac{2}{3} \left[ 4.5 - \frac{10}{3} \right]\)

\(\text{E}(X) = \frac{8}{9} + \frac{2}{3} \left[ \frac{9}{2} - \frac{10}{3} \right] = \frac{8}{9} + \frac{2}{3} \left[ \frac{7}{6} \right] = \frac{8}{9} + \frac{7}{9} = \frac{15}{9} = \frac{5}{3}\).

(c) Let \(m\) be the median of \(X\). First, check the area from 0 to 2:

\(\text{P}(X \le 2) = \int_{0}^{2} \frac{1}{3}x \, dx = \left[ \frac{1}{6}x^2 \right]_{0}^{2} = \frac{4}{6} = \frac{2}{3}\).

Since \(\frac{2}{3} \approx 0.667 > 0.5\), the median \(m\) must lie in the interval \([0, 2]\).

\(\int_{0}^{m} \frac{1}{3}x \, dx = 0.5\)

\(\left[ \frac{1}{6}x^2 \right]_{0}^{m} = 0.5 \implies \frac{1}{6}m^2 = \frac{1}{2} \implies m^2 = 3\).

Since \(m\) must be positive, \(m = \sqrt{3}\).

(a)
- **M1**: For attempting to integrate both parts of the PDF and equate the sum to 1, or for summing the areas of both triangles geometrically.
- **A1**: For correct integration of both terms (or correct geometric areas \(2k\) and \(k\)).
- **A1**: For obtaining \(3k = 1\) and showing clearly that \(k = \frac{1}{3}\) (given answer).

(b)
- **M1**: For attempting to find \(\text{E}(X)\) using \(\int x f(x) \, dx\) as the sum of two integral expressions.
- **A1**: For correct integration of both terms: \(\left[\frac{1}{9}x^3\right]\) and \(\frac{2}{3}\left[\frac{3}{2}x^2 - \frac{1}{3}x^3\right]\).
- **M1**: For substituting limits correctly in both integrated expressions.
- **A1**: For obtaining the correct exact value \(\frac{5}{3}\) (or \(1.67\) to 3 s.f.).

(c)
- **M1**: For showing \(\text{P}(X \le 2) = \frac{2}{3}\) or otherwise establishing that the median \(m\) lies in the interval \([0, 2]\).
- **M1**: For setting up and integrating \(\int_{0}^{m} \frac{1}{3}x \, dx = 0.5\).
- **A1**: For obtaining the exact value \(m = \sqrt{3}\) (must discard the negative root).