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(a) Equating the equations of the line and the curve:
\[ x^2 - 4x + 4 = kx - 5 \]
Rearranging to make the right-hand side zero:
\[ x^2 - 4x - kx + 4 + 5 = 0 \]
\[ x^2 - (k + 4)x + 9 = 0 \] (as required)

(b) Since \( l \) is a tangent to \( C \), the quadratic equation of intersection has equal roots, meaning its discriminant must be zero:
\[ b^2 - 4ac = 0 \]
With \( a = 1 \), \( b = -(k + 4) \), and \( c = 9 \):
\[ [-(k + 4)]^2 - 4(1)(9) = 0 \]
\[ (k + 4)^2 - 36 = 0 \]
\[ k + 4 = \pm 6 \]
If \( k + 4 = 6 \implies k = 2 \).
If \( k + 4 = -6 \implies k = -10 \).

(c) When \( k = 2 \):
\[ x^2 - (2 + 4)x + 9 = 0 \implies x^2 - 6x + 9 = 0 \implies (x - 3)^2 = 0 \implies x = 3 \]
Substituting \( x = 3 \) into the equation of the line \( y = 2x - 5 \):
\[ y = 2(3) - 5 = 1 \]
Point of contact is \( (3, 1) \).

When \( k = -10 \):
\[ x^2 - (-10 + 4)x + 9 = 0 \implies x^2 + 6x + 9 = 0 \implies (x + 3)^2 = 0 \implies x = -3 \]
Substituting \( x = -3 \) into the equation of the line \( y = -10x - 5 \):
\[ y = -10(-3) - 5 = 25 \]
Point of contact is \( (-3, 25) \).

(a)
B1: For correctly setting the line and curve equations equal to each other and expanding/rearranging to the given form.

(b)
M1: For attempting to use \( b^2 - 4ac = 0 \) with their coefficients.
A1: Correct quadratic equation in \( k \), e.g., \( (k+4)^2 - 36 = 0 \) or \( k^2 + 8k - 20 = 0 \).
A1: \( k = 2 \)
A1: \( k = -10 \)

(c)
M1: Substituting one of their values of \( k \) back into the quadratic equation to find the corresponding value of \( x \).
A1: Finding both correct \( x \)-coordinates, or one fully correct coordinate pair, e.g., \( (3, 1) \).
A1: Both coordinate pairs fully correct: \( (3, 1) \) and \( (-3, 25) \).

(a) First, find the gradient of the line \( l_1 \):
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 5}{6 - (-2)} = \frac{-6}{8} = -\frac{3}{4} \]
Using the point-slope formula with point \( A(-2, 5) \):
\[ y - 5 = -\frac{3}{4}(x - (-2)) \]
Multiply both sides by 4:
\[ 4(y - 5) = -3(x + 2) \]
\[ 4y - 20 = -3x - 6 \]
\[ 3x + 4y - 14 = 0 \]

(b) The midpoint of \( AB \) is:
\[ M = \left(\frac{-2 + 6}{2}, \frac{5 + (-1)}{2}\right) = (2, 2) \]
Since \( l_2 \) is perpendicular to \( l_1 \), its gradient is the negative reciprocal of the gradient of \( l_1 \):
\[ m_{\perp} = -\frac{1}{m} = -\frac{1}{-3/4} = \frac{4}{3} \]
Now, find the equation of \( l_2 \) using the midpoint \( M(2, 2) \):
\[ y - 2 = \frac{4}{3}(x - 2) \]
\[ y - 2 = \frac{4}{3}x - \frac{8}{3} \]
\[ y = \frac{4}{3}x - \frac{2}{3} \]

(a)
M1: For a correct method to find the gradient of \( AB \).
M1: For using their gradient and either point \( A \) or \( B \) to form the equation of the line.
A1: \( 3x + 4y - 14 = 0 \) (or any non-zero integer multiple of this equation).

(b)
B1: For finding the correct midpoint \( (2, 2) \).
M1: For finding the negative reciprocal of their gradient from part (a).
M1: For using their midpoint and perpendicular gradient to form an equation for \( l_2 \).
A1: \( y = \frac{4}{3}x - \frac{2}{3} \) (must be in the form \( y = mx + c \)).

(a) Using the trigonometric identity \( \sin^2 \theta = 1 - \cos^2 \theta \):
\[ 4(1 - \cos^2 \theta) + 9\cos \theta - 6 = 0 \]
\[ 4 - 4\cos^2 \theta + 9\cos \theta - 6 = 0 \]
\[ -4\cos^2 \theta + 9\cos \theta - 2 = 0 \]
Multiplying the entire equation by \(-1\):
\[ 4\cos^2 \theta - 9\cos \theta + 2 = 0 \]
(as required)

(b) Letting \( \theta = 2x \), the equation is:
\[ 4\cos^2(2x) - 9\cos(2x) + 2 = 0 \]
Factorising the quadratic in \( \cos(2x) \):
\[ (4\cos(2x) - 1)(\cos(2x) - 2) = 0 \]
This gives:
\[ \cos(2x) = \frac{1}{4} \quad \text{or} \quad \cos(2x) = 2 \]
Since \( -1 \le \cos(2x) \le 1 \), the equation \( \cos(2x) = 2 \) has no solutions.

Now, solve \( \cos(2x) = 0.25 \) in the interval \( 0 \le 2x < 720^\circ \):
Primary solution:
\[ 2x = \arccos(0.25) \approx 75.52^\circ \]
Other solutions in the interval:
\[ 2x = 360^\circ - 75.52^\circ = 284.48^\circ \]
\[ 2x = 360^\circ + 75.52^\circ = 435.52^\circ \]
\[ 2x = 720^\circ - 75.52^\circ = 644.48^\circ \]

Dividing each value by 2 to solve for \( x \):
\[ x \approx 37.8^\circ, \; 142.2^\circ, \; 217.8^\circ, \; 322.2^\circ \]
All values are given to 1 decimal place.

(a)
M1: For substituting \( \sin^2 \theta = 1 - \cos^2 \theta \) into the given equation.
A1: For correct expansion and simplification to obtain the given expression without any algebraic errors.

(b)
M1: For factorising or using the quadratic formula on the quadratic expression in \( \cos(2x) \).
A1: For obtaining the correct values \( \cos(2x) = \frac{1}{4} \) (and optionally \( \cos(2x) = 2 \)).
B1: For stating or indicating that \( \cos(2x) = 2 \) has no solutions.
M1: Finding at least one correct angle for \( 2x \) from \( \cos(2x) = 0.25 \).
M1: Finding all four angles for \( 2x \) in the interval \( [0, 720^\circ) \).
A1: All four final values of \( x \) correct to 1 decimal place: \( 37.8^\circ, \; 142.2^\circ, \; 217.8^\circ, \; 322.2^\circ \). (Lose this mark if there are extra solutions within the range).

Let \( f(x) = 4x^2 - 3x \).
By the definition of the derivative from first principles:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
First, evaluate \( f(x+h) \):
\[ f(x+h) = 4(x+h)^2 - 3(x+h) \]
\[ f(x+h) = 4(x^2 + 2xh + h^2) - 3x - 3h \]
\[ f(x+h) = 4x^2 + 8xh + 4h^2 - 3x - 3h \]

Now, find the difference \( f(x+h) - f(x) \):
\[ f(x+h) - f(x) = (4x^2 + 8xh + 4h^2 - 3x - 3h) - (4x^2 - 3x) \]
\[ f(x+h) - f(x) = 8xh + 4h^2 - 3h \]

Divide this difference by \( h \):
\[ \frac{f(x+h) - f(x)}{h} = \frac{8xh + 4h^2 - 3h}{h} \]
\[ \frac{f(x+h) - f(x)}{h} = 8x + 4h - 3 \]

Take the limit as \( h \to 0 \):
\[ f'(x) = \lim_{h \to 0} (8x + 4h - 3) = 8x - 3 \]

B1: States the correct definition of the derivative from first principles: \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \) (or equivalent with \( \delta x \)).
M1: Attempts to substitute \( x+h \) into \( 4x^2 - 3x \).
A1: Obtains a correct expanded form: \( 4x^2 + 8xh + 4h^2 - 3x - 3h \).
M1: Correct subtraction of \( f(x) \) from their expanded \( f(x+h) \).
A1: Obtains \( 8xh + 4h^2 - 3h \) following correct algebra.
M1: Divides by \( h \) to obtain \( 8x + 4h - 3 \).
A1: Applies the limit \( h \to 0 \) correctly with consistent notation to conclude \( f'(x) = 8x - 3 \).

First, write each term of the derivative \( f'(x) \) in index form:
\[ f'(x) = \frac{3x^{1/2} - 4}{x^2} = 3x^{1/2}x^{-2} - 4x^{-2} \]
\[ f'(x) = 3x^{-3/2} - 4x^{-2} \]

Now, integrate \( f'(x) \) to find \( f(x) \):
\[ f(x) = \int (3x^{-3/2} - 4x^{-2}) \, dx \]
Using the integration rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \):
\[ f(x) = \frac{3x^{-1/2}}{-1/2} - \frac{4x^{-1}}{-1} + C \]
\[ f(x) = -6x^{-1/2} + 4x^{-1} + C \]
\[ f(x) = -\frac{6}{\sqrt{x}} + \frac{4}{x} + C \]

Since the curve passes through \( P(4, 15) \), substitute \( x = 4 \) and \( y = 15 \):
\[ 15 = -\frac{6}{\sqrt{4}} + \frac{4}{4} + C \]
\[ 15 = -\frac{6}{2} + 1 + C \]
\[ 15 = -3 + 1 + C \]
\[ 15 = -2 + C \implies C = 17 \]

Thus, the equation of \( C \) is:
\[ y = -\frac{6}{\sqrt{x}} + \frac{4}{x} + 17 \]

M1: Attempts to split the fraction and rewrite at least one term of \( f'(x) \) in index form \( Ax^n \).
A1: Correct index form: \( 3x^{-3/2} - 4x^{-2} \).
M1: Integrates at least one term by increasing the power of \( x \) by 1.
A1: Correct integration of \( 3x^{-3/2} \) giving \( \frac{3x^{-1/2}}{-1/2} \) (unsimplified).
A1: Correct integration of \( -4x^{-2} \) giving \( -\frac{4x^{-1}}{-1} \) (unsimplified).
B1: Includes the constant of integration \( + C \) (can be implied by later calculation).
M1: Substitutes \( x = 4 \) and \( y = 15 \) to solve for \( C \).
A1: Correct final simplified equation: \( y = -\frac{6}{\sqrt{x}} + \frac{4}{x} + 17 \) (or equivalent simplified form).

(a) The vertical asymptote is found where the denominator is zero:
\[ x = 0 \]
The horizontal asymptote is found as \( x \to \pm\infty \):
\[ y = 3 \]

(b)(i) \( y = f(x - 2) \):
This represents a horizontal translation of 2 units to the right.
- The vertical asymptote translates from \( x = 0 \) to \( x = 2 \).
- The horizontal asymptote remains \( y = 3 \).
- To find the \( x \)-intercept, set \( y = 0 \):
\[ \frac{2}{x-2} + 3 = 0 \implies \frac{2}{x-2} = -3 \implies x - 2 = -\frac{2}{3} \implies x = \frac{4}{3} \]
So the intercept is \( (\frac{4}{3}, 0) \).
- To find the \( y \)-intercept, set \( x = 0 \):
\[ y = f(-2) = \frac{2}{-2} + 3 = 2 \]
So the intercept is \( (0, 2) \).
- Sketch shows the curve in the two regions split by the asymptotes \( x = 2 \) and \( y = 3 \) with the labelled intercepts.

(b)(ii) \( y = -f(x) \):
This is a reflection of the curve \( y = f(x) \) in the \( x \)-axis.
- The vertical asymptote remains \( x = 0 \).
- The horizontal asymptote becomes \( y = -3 \).
- To find the \( x \)-intercept, set \( y = 0 \):
\[ -f(x) = 0 \implies \frac{2}{x} + 3 = 0 \implies x = -\frac{2}{3} \]
So the intercept is \( (-\frac{2}{3}, 0) \).
- There is no \( y \)-intercept because \( x = 0 \) is an asymptote.
- Sketch shows the reflected reciprocal curves with asymptotes \( x = 0 \) and \( y = -3 \).

(a)
B1: For stating \( x = 0 \).
B1: For stating \( y = 3 \).

(b)(i)
B1: Correct shape and placement of the reciprocal curve relative to its asymptotes, which must be clearly marked and labelled as \( x = 2 \) and \( y = 3 \).
B1: Correctly identifies and labels the axis intersections at \( (0, 2) \) and \( (\frac{4}{3}, 0) \) (or equivalent decimal values).

(b)(ii)
B1: Correctly identifies and labels the asymptotes for the reflected curve: \( x = 0 \) and \( y = -3 \).
B1: Correct shape of the reflected curve (reciprocal curves in the 2nd and 4th quadrants relative to the horizontal asymptote \( y = -3 \)).
B1: Identifies and labels the intersection with the \( x \)-axis at \( (-\frac{2}{3}, 0) \).

(a) The length of \( AB \) is:
\[ AB = \sqrt{(7 - 3)^2 + (2 - 8)^2} \]
\[ AB = \sqrt{4^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} \]
Simplifying the surd:
\[ AB = \sqrt{4 \times 13} = 2\sqrt{13} \]
Thus, \( k = 2 \).

(b) Using the midpoint formula for \( B(7, 2) \) and \( C(p, q) \):
\[ \left(\frac{7 + p}{2}, \frac{2 + q}{2}\right) = (5, -3) \]
For the \( x \)-coordinate:
\[ \frac{7 + p}{2} = 5 \implies 7 + p = 10 \implies p = 3 \]
For the \( y \)-coordinate:
\[ \frac{2 + q}{2} = -3 \implies 2 + q = -6 \implies q = -8 \]
So \( C \) has coordinates \( (3, -8) \).

(c) The gradient of line segment \( AB \) is:
\[ m_{AB} = \frac{2 - 8}{7 - 3} = \frac{-6}{4} = -\frac{3}{2} \]
Since the line \( l \) is perpendicular to \( AB \), the gradient of \( l \) is:
\[ m = -\frac{1}{m_{AB}} = \frac{2}{3} \]
Using the coordinates of \( A(3, 8) \):
\[ y - 8 = \frac{2}{3}(x - 3) \]
\[ y - 8 = \frac{2}{3}x - 2 \]
\[ y = \frac{2}{3}x + 6 \]

(a)
M1: For using a correct distance formula to find the length of \( AB \).
A1: Correct simplified form \( 2\sqrt{13} \).

(b)
M1: Setting up at least one correct equation for the coordinates of \( C \) using the midpoint formula.
A1: Correct coordinates for \( C \) which are \( (3, -8) \) (or \( p = 3, q = -8 \)).

(c)
B1: Correct gradient of \( AB \) of \( -\frac{3}{2} \) (or equivalent).
M1: Attempt to use the perpendicular gradient rule \( m_1 m_2 = -1 \) with their gradient of \( AB \).
M1: Correct method to find the equation of a straight line through \( A(3, 8) \) using their perpendicular gradient.
A1: Correct simplified equation \( y = \frac{2}{3}x + 6 \).

(a) Differentiating \( y = 2x^2 - 8x + 5 \) with respect to \( x \):
\[ \frac{dy}{dx} = 4x - 8 \]

(b) At the point \( P(3, -1) \), the gradient of the tangent is:
\[ m = 4(3) - 8 = 4 \]
Using the equation of a straight line with gradient 4 and point \( P(3, -1) \):
\[ y - (-1) = 4(x - 3) \]
\[ y + 1 = 4x - 12 \]
\[ y = 4x - 13 \]

(c) Rearranging the line \( x - 4y + 8 = 0 \) into gradient-intercept form:
\[ 4y = x + 8 \implies y = \frac{1}{4}x + 2 \]
This line has gradient \( \frac{1}{4} \).
Since the normal is parallel to this line, the gradient of the normal is also \( \frac{1}{4} \).
The gradient of the tangent at this point is therefore:
\[ m_{tangent} = -\frac{1}{1/4} = -4 \]
We set our derivative equal to this gradient:
\[ 4x - 8 = -4 \]
\[ 4x = 4 \implies x = 1 \]
Substitute \( x = 1 \) back into the equation of curve \( C \) to find the \( y \)-coordinate:
\[ y = 2(1)^2 - 8(1) + 5 = 2 - 8 + 5 = -1 \]
So the coordinates of the point are \( (1, -1) \).

(a)
B1: Correct derivative \( 4x - 8 \).

(b)
M1: Substituting \( x = 3 \) into their derivative to find the gradient of the tangent.
M1: For using their gradient and point \( (3, -1) \) to find the equation of the tangent.
A1: \( y = 4x - 13 \) (must be in the form \( y = mx + c \)).

(c)
B1: Correctly identifies that the gradient of the given line (and therefore of the normal) is \( \frac{1}{4} \).
M1: Equates their derivative \( \frac{dy}{dx} \) to \( -4 \) (the perpendicular gradient) and solves for \( x \).
A1: Correct coordinates \( (1, -1) \) (or \( x = 1, y = -1 \)).

(a) We first express the terms of \(y\) in index form:

\[y = kx^{-1} + 2x^{\frac{3}{2}} - 4\]

Differentiating with respect to \(x\):

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

Since the tangent at \(x = 4\) is parallel to the line \(y = 5x - 3\), the gradient of the tangent is \(5\).

Substituting \(x = 4\) into \(\frac{dy}{dx} = 5\):

\[-\frac{k}{4^2} + 3\sqrt{4} = 5\]

\[-\frac{k}{16} + 6 = 5\]

\[-\frac{k}{16} = -1\]

\[k = 16\]

(b) To find the equation of the tangent, we need the \(y\)-coordinate of the point on the curve where \(x = 4\). Using \(k = 16\):

\[y = \frac{16}{4} + 2(4)\sqrt{4} - 4 = 4 + 16 - 4 = 16\]

So the point of contact is \((4, 16)\).

Using the line equation formula with gradient \(m = 5\) and point \((4, 16)\):

\[y - 16 = 5(x - 4)\]

\[y - 16 = 5x - 20\]

\[5x - y - 4 = 0\]

This is in the required form \(ax + by + c = 0\) with integers \(a = 5\), \(b = -1\), and \(c = -4\).

(a)
- M1: Attempts to write \(2x\sqrt{x}\) as \(2x^{3/2}\) or \(\frac{k}{x}\) as \(kx^{-1}\).
- M1: Differentiates to find \(\frac{dy}{dx}\), achieving at least one correct term from their indices: \(-kx^{-2}\) or \(3x^{1/2}\).
- A1: Fully correct derivative: \(\frac{dy}{dx} = -kx^{-2} + 3x^{1/2}\).
- M1: Sets their derivative at \(x = 4\) equal to \(5\) and attempts to solve for \(k\).
- A1: \(k = 16\) (cso).

(b)
- M1: Substitutes \(x = 4\) and their \(k\) into the equation of \(C\) to find the \(y\)-coordinate.
- M1: Uses their point and gradient \(m = 5\) to form the equation of the line.
- A1: \(5x - y - 4 = 0\) or equivalent integer form (e.g., \(y - 5x + 4 = 0\)).

(a) Find the gradient of \(l_1\):

\[m_1 = \frac{6 - 3}{4 - (-2)} = \frac{3}{6} = \frac{1}{2}\]

Using the point-gradient formula with point \(A(-2, 3)\):

\[y - 3 = \frac{1}{2}(x - (-2))\]

\[2(y - 3) = x + 2\]

\[2y - 6 = x + 2\]

\[x - 2y + 8 = 0\]

(b) The coordinates of the midpoint \(M\) are:

\[M = \left(\frac{-2 + 4}{2}, \frac{3 + 6}{2}\right) = \left(1, \frac{9}{2}\right) = (1, 4.5)\]

(c) Since \(l_2\) is perpendicular to \(l_1\), its gradient \(m_2\) is:

\[m_2 = -\frac{1}{m_1} = -2\]

Using the gradient \(m_2 = -2\) and the midpoint \(M\left(1, \frac{9}{2}\right)\):

\[y - \frac{9}{2} = -2(x - 1)\]

\[y - 4.5 = -2x + 2\]

\[y = -2x + 6.5\]

So, \(m = -2\) and \(c = 6.5\) (or \(c = \frac{13}{2}\)).

(a)
- M1: Attempts to find the gradient of \(l_1\) using \(\frac{y_2 - y_1}{x_2 - x_1}\).
- M1: Applies the straight line formula with their gradient and either point \(A\) or \(B\).
- A1: \(x - 2y + 8 = 0\) (or any non-zero integer multiple, e.g. \(2y - x - 8 = 0\)).

(b)
- B1: \((1, 4.5)\) or \(\left(1, \frac{9}{2}\right)\).

(c)
- M1: Identifies the perpendicular gradient relation \(m_2 = -\frac{1}{m_1}\) and finds \(m_2 = -2\) (or follow through their gradient from part a).
- M1: Uses their perpendicular gradient and their midpoint \(M\) to form an equation of the line.
- A1: \(y = -2x + 6.5\) or \(y = -2x + \frac{13}{2}\).

(a) Since the terms form a geometric series, the common ratio \(r\) is constant:
\[r = \frac{2k+2}{4k-2} = \frac{k+3}{2k+2}\]
Cross-multiplying gives:
\[(2k+2)^2 = (4k-2)(k+3)\]
\[4k^2 + 8k + 4 = 4k^2 + 10k - 6\]
Subtracting \(4k^2\) from both sides:
\[8k + 4 = 10k - 6\]
\[2k = 10 \implies k = 5\]

(b) Substituting \(k = 5\) into the expressions for the terms:
First term \(a = 4(5) - 2 = 18\)
Second term \(ar = 2(5) + 2 = 12\)
Third term \(ar^2 = 5 + 3 = 8\)

The common ratio is:
\[r = \frac{12}{18} = \frac{2}{3}\]

(c) Since \(|r| < 1\), the sum to infinity exists:
\[S_{\infty} = \frac{a}{1-r} = \frac{18}{1 - \frac{2}{3}} = \frac{18}{\frac{1}{3}} = 54\]

(a)
M1: Sets up a correct ratio equation using the terms: \(\frac{2k+2}{4k-2} = \frac{k+3}{2k+2}\) (or equivalent).
M1: Expands both sides correctly to get a linear equation in \(k\) (the \(k^2\) terms cancel).
A1*: Completes the algebra to show \(k = 5\) with no errors.

(b)
B1: State \(r = \frac{2}{3}\) (or equivalent fraction).

(c)
M1: Identifies the first term \(a = 18\) and applies the sum to infinity formula \(S_{\infty} = \frac{a}{1-r}\) with their \(a\) and \(r\).
A1: Correct substitution of \(a = 18\) and \(r = \frac{2}{3}\).
A1: Correct final value of \(54\).

(a) The curve intersects the x-axis when \(y = 0\):
\[9x - 3x^{\frac{3}{2}} = 0\]
Factoring out \(3x\):
\[3x(3 - x^{\frac{1}{2}}) = 0\]
This gives:
\[3x = 0 \implies x = 0\]
or
\[3 - x^{\frac{1}{2}} = 0 \implies x^{\frac{1}{2}} = 3 \implies x = 9\]
Thus, the curve intersects the x-axis at the origin \((0, 0)\) and at the point \((9, 0)\).

(b) The area of the region \(R\) is given by the definite integral:
\[R = \int_{0}^{9} (9x - 3x^{\frac{3}{2}}) \, dx\]
Integrating term by term:
\[\int (9x - 3x^{\frac{3}{2}}) \, dx = \left[ \frac{9}{2}x^2 - \frac{3}{\frac{5}{2}}x^{\frac{5}{2}} \right]_{0}^{9} = \left[ \frac{9}{2}x^2 - \frac{6}{5}x^{\frac{5}{2}} \right]_{0}^{9}\]
Now substitute the limits:
\[R = \left( \frac{9}{2}(9)^2 - \frac{6}{5}(9)^{\frac{5}{2}} \right) - (0)\]
\[9^{\frac{5}{2}} = (\sqrt{9})^5 = 3^5 = 243\]
\[R = \frac{729}{2} - \frac{6}{5}(243) = \frac{729}{2} - \frac{1458}{5}\]
Finding a common denominator:
\[R = \frac{5(729) - 2(1458)}{10} = \frac{3645 - 2916}{10} = \frac{729}{10} = 72.9\]

(a)
M1: Sets \(y = 0\) and attempts to factorise or solve the equation.
A1: Correctly shows that \(x = 0\) and \(x = 9\).

(b)
M1: Attempts to integrate \(9x - 3x^{\frac{3}{2}}\) (at least one term has power increased by 1).
A1: Correct integration of \(9x \to \frac{9}{2}x^2\).
A1: Correct integration of \(-3x^{\frac{3}{2}} \to -\frac{6}{5}x^{\frac{5}{2}}\).
M1: Substitutes the limit \(x = 9\) into their integrated expression.
A1: Correctly evaluates \(9^{\frac{5}{2}} = 243\).
A1: Correct exact area of \(\frac{729}{10}\) or \(72.9\).

(a) The volume of the box is given by:
\[V = x^2 h = 108 \implies h = \frac{108}{x^2}\]
The surface area of an open-topped box consists of a square base and four rectangular sides:
\[S = x^2 + 4xh\]
Substitute the expression for \(h\):
\[S = x^2 + 4x\left(\frac{108}{x^2}\right) = x^2 + \frac{432}{x}\]

(b) To find the minimum surface area, differentiate \(S\) with respect to \(x\):
\[\frac{dS}{dx} = 2x - 432x^{-2} = 2x - \frac{432}{x^2}\]
Set \(\frac{dS}{dx} = 0\) for a stationary point:
\[2x - \frac{432}{x^2} = 0 \implies 2x^3 = 432 \implies x^3 = 216 \implies x = 6\]
The minimum value of the surface area is:
\[S = (6)^2 + \frac{432}{6} = 36 + 72 = 108\text{ cm}^2\]

To justify that this is a minimum, find the second derivative:
\[\frac{d^2S}{dx^2} = 2 + 864x^{-3} = 2 + \frac{864}{x^3}\]
At \(x = 6\):
\[\frac{d^2S}{dx^2} = 2 + \frac{864}{216} = 2 + 4 = 6\]
Since \(\frac{d^2S}{dx^2} > 0\) at \(x = 6\), the value is indeed a minimum.

(a)
B1: Writes down \(108 = x^2 h\) (or equivalent).
M1: Writes down the surface area formula \(S = x^2 + 4xh\).
A1*: Sustitutes \(h = \frac{108}{x^2}\) into the surface area formula to obtain the printed result.

(b)
M1: Differentiates \(S\) to obtain at least one term correct.
A1: Correct derivative \(\frac{dS}{dx} = 2x - \frac{432}{x^2}\).
M1: Sets their \(\frac{dS}{dx} = 0\) and solves for \(x\) to obtain \(x = 6\).
A1: Calculates the correct minimum value \(S = 108\).
M1: Differentiates again to obtain \(\frac{d^2S}{dx^2} = 2 + \frac{864}{x^3}\) and evaluates it at \(x=6\) to show it is greater than zero.

(a) Using the Factor Theorem, since \((x - 2)\) is a factor of \(f(x)\):
\[f(2) = 0 \implies 2(2)^3 + a(2)^2 + b(2) - 10 = 0\]
\[16 + 4a + 2b - 10 = 0 \implies 4a + 2b = -6 \implies 2a + b = -3 \quad \text{(Equation 1)}\]
Using the Remainder Theorem, since the remainder when divided by \((x + 1)\) is \(-12\):
\[f(-1) = -12 \implies 2(-1)^3 + a(-1)^2 + b(-1) - 10 = -12\]
\[-2 + a - b - 10 = -12 \implies a - b = 0 \implies a = b \quad \text{(Equation 2)}\]
Substitute Equation 2 into Equation 1:
\[2a + a = -3 \implies 3a = -3 \implies a = -1\]
Since \(a = b\), we also have \(b = -1\).

(b) Now we have \(f(x) = 2x^3 - x^2 - x - 10\).
Since \((x - 2)\) is a factor, we can express \(f(x)\) as:
\[f(x) = (x - 2)(2x^2 + cx + 5)\]
Expanding and comparing coefficients of \(x^2\):
\[c - 4 = -1 \implies c = 3\]
So, \(f(x) = (x - 2)(2x^2 + 3x + 5)\).
For \(f(x) = 0\), we either have \(x = 2\) or the quadratic part \(2x^2 + 3x + 5 = 0\).
We calculate the discriminant \(\Delta\) of the quadratic factor:
\[\Delta = 3^2 - 4(2)(5) = 9 - 40 = -31\]
Since \(\Delta < 0\), the quadratic equation has no real roots.
Therefore, \(f(x) = 0\) has only one real root, which is \(x = 2\).

(a)
M1: Applies the Factor Theorem correctly to set up \(f(2) = 0\).
M1: Applies the Remainder Theorem correctly to set up \(f(-1) = -12\).
M1: Attempts to solve the simultaneous equations for \(a\) and \(b\).
A1: Correct values \(a = -1\) and \(b = -1\).

(b)
M1: Attempts algebraic division or comparing coefficients to find the quadratic factor.
A1: Obtains the correct quadratic factor \(2x^2 + 3x + 5\).
M1: Calculates the discriminant of the quadratic factor and concludes that there are no real roots since \(\Delta < 0\).

(a) Using the identity \(\cos^2 x = 1 - \sin^2 x\):
\[4(1 - \sin^2 x) + 5\sin x = 5\]
\[4 - 4\sin^2 x + 5\sin x = 5\]
Rearranging to make the leading coefficient positive:
\[4\sin^2 x - 5\sin x + 1 = 0 \quad \text{(as required)}\]

(b) Factorising the quadratic equation in terms of \(\sin x\):
\[(4\sin x - 1)(\sin x - 1) = 0\]
This gives:
\[\sin x = \frac{1}{4} \quad \text{or} \quad \sin x = 1\]

Case 1: \(\sin x = 1\)
\[x = 90^\circ\]

Case 2: \(\sin x = 0.25\)
\[x = \arcsin(0.25) \approx 14.4775^\circ \approx 14.5^\circ\]
The other solution in the interval \(0 \le x < 360^\circ\) is:
\[x = 180^\circ - 14.4775^\circ = 165.5225^\circ \approx 165.5^\circ\]

Thus, the solutions are:
\[x = 14.5^\circ, \, 90^\circ, \, 165.5^\circ\]

(a)
M1: Substitute the identity \(\cos^2 x = 1 - \sin^2 x\) into the given equation.
A1*: Rearranges the terms correctly to show the printed quadratic equation.

(b)
M1: Attempts to factorise or solve the quadratic equation to find values for \(\sin x\).
A1: Correct values \(\sin x = 1\) and \(\sin x = \frac{1}{4}\).
B1: Identifies \(x = 90^\circ\) as a solution.
M1: Obtains \(x \approx 14.5^\circ\) from \(\sin x = 0.25\).
A1: Identifies \(x \approx 165.5^\circ\) as the second solution and has no extra incorrect solutions in the range.

(a) Take the natural logarithm of both sides:
\[\ln(5^{2x-1}) = \ln(3^x)\]
Using the power law of logarithms:
\[(2x - 1)\ln 5 = x\ln 3\]
\[2x\ln 5 - \ln 5 = x\ln 3\]
Rearrange to collect terms in \(x\):
\[2x\ln 5 - x\ln 3 = \ln 5\]
\[x(2\ln 5 - \ln 3) = \ln 5\]
\[x = \frac{\ln 5}{2\ln 5 - \ln 3}\]
Evaluating the values:
\[x \approx \frac{1.6094}{2(1.6094) - 1.0986} \approx \frac{1.6094}{2.1203} \approx 0.759\]

(b) Using the power law of logarithms on the second term:
\[2\log_2 w = \log_2 (w^2)\]
Substitute this back into the equation:
\[\log_2 (3w + 5) - \log_2 (w^2) = 1\]
Using the subtraction law of logarithms:
\[\log_2 \left( \frac{3w + 5}{w^2} \right) = 1\]
Convert to exponential form:
\[\frac{3w + 5}{w^2} = 2^1 = 2\]
\[3w + 5 = 2w^2\]
\[2w^2 - 3w - 5 = 0\]
Factorising the quadratic:
\[(2w - 5)(w + 1) = 0\]
So \(w = 2.5\) or \(w = -1\).
Since the term \(\log_2 w\) is only defined for \(w > 0\), we reject \(w = -1\).
Thus, the only valid solution is \(w = 2.5\).

(a)
M1: Takes logarithms of both sides and applies the power law correctly.
M1: Expands and groups the terms to solve for \(x\).
A1: Shows a correct exact expression for \(x\), e.g., \(x = \frac{\ln 5}{\ln 25 - \ln 3}\).
A1: Correctly evaluates to \(0.759\) (to 3 d.p.).

(b)
M1: Uses the power law to rewrite \(2\log_2 w\) as \(\log_2 (w^2)\).
M1: Applies the division law of logarithms to obtain \(\log_2 \left( \frac{3w + 5}{w^2} \right) = 1\).
A1: Converts to a quadratic equation: \(2w^2 - 3w - 5 = 0\).
A1: Solves the quadratic to find \(w = 2.5\) and explicitly states that \(w = -1\) is rejected.

(a) Rearranging and completing the square for the equation \(x^2 + y^2 - 6x + 8y = 0\):
\[(x - 3)^2 - 9 + (y + 4)^2 - 16 = 0\]
\[(x - 3)^2 + (y + 4)^2 = 25\]
Thus, the center of \(C\) is \((3, -4)\) and the radius of \(C\) is \(\sqrt{25} = 5\).

(b) Substitute \(x = 6\) and \(y = 0\) into the equation of \(C\):
\[\text{LHS} = 6^2 + 0^2 - 6(6) + 8(0) = 36 - 36 = 0\]
Since LHS = RHS, the point \(P(6, 0)\) lies on \(C\).

(c) Let \(M(3, -4)\) be the center of \(C\).
The gradient of the radius \(MP\) is:
\[m_1 = \frac{0 - (-4)}{6 - 3} = \frac{4}{3}\]
The tangent to the circle at \(P\) is perpendicular to the radius \(MP\), so its gradient is:
\[m_2 = -\frac{1}{m_1} = -\frac{3}{4}\]
The equation of the tangent passing through \(P(6, 0)\) is:
\[y - 0 = -\frac{3}{4}(x - 6)\]
\[4y = -3(x - 6)\]
\[4y = -3x + 18\]
\[3x + 4y - 18 = 0\]

(a)
M1: Attempts to complete the square for both \(x\) and \(y\).
A1: Correct coordinates of the center: \((3, -4)\).
A1: Correct radius: \(5\).

(b)
B1: Substitute \(x = 6, y = 0\) into the circle equation and correctly show LHS = 0.

(c)
M1: Calculates the gradient of the radius \(MP\).
M1: Finds the perpendicular gradient for the tangent.
M1: Writes down the equation of the line using \(P(6, 0)\) and their perpendicular gradient.
A1: Correct equation in the specified form: \(3x + 4y - 18 = 0\) (or any integer multiple thereof).

(a) We start with the fact that the square of any real number is non-negative:
\[(a - b)^2 \ge 0\]
Expanding the left-hand side:
\[a^2 - 2ab + b^2 \ge 0\]
Adding \(2ab\) to both sides gives:
\[a^2 + b^2 \ge 2ab\quad \text{(as required)}\]

(b) We test each integer in the given interval \(1 \le n \le 4\):
- For \(n = 1\): \(1^2 + 1 + 11 = 13\) (which is prime)
- For \(n = 2\): \(2^2 + 2 + 11 = 17\) (which is prime)
- For \(n = 3\): \(3^2 + 3 + 11 = 23\) (which is prime)
- For \(n = 4\): \(4^2 + 4 + 11 = 31\) (which is prime)
Since the statement holds true for all integers in the defined domain, the statement is proven by exhaustion.

(c) To find a counterexample, we look for an integer \(n\) such that \(n^2 + n + 11\) is composite.
- Let \(n = 10\): \(10^2 + 10 + 11 = 121\). Since \(121 = 11^2\), it is not a prime number.
- Alternatively, let \(n = 11\): \(11^2 + 11 + 11 = 11(11 + 1 + 1) = 11 \times 13 = 143\). Since \(143\) is divisible by \(11\) and \(13\), it is not a prime number.
Either case serves as a valid counterexample.

(a)
M1: Realises the need to start with \((a - b)^2 \ge 0\) (or equivalent algebraic setup).
A1*: Expands and rearranges correctly with no errors to reach the required inequality.

(b)
M1: Evaluates \(n^2 + n + 11\) for all four values of \(n = 1, 2, 3, 4\).
A1: Shows correct values of \(13, 17, 23, 31\).
A1: Explicitly states that all four results are prime numbers, thereby completing the proof by exhaustion.

(c)
M1: Selects a positive integer \(n\) (e.g., \(n = 10\) or \(n = 11\)) and calculates \(n^2 + n + 11\).
A1: Correctly evaluates the composite result and explains why it is not prime (e.g., \(121 = 11 \times 11\) or \(143 = 11 \times 13\)) to disprove the statement.

(a) The curve crosses the \(x\)-axis when \(y = 0\).
Setting \(y = 0\) gives:
\[2^{2x+1} - 17(2^x) + 8 = 0\]
Using the laws of indices:
\[2^{2x+1} = 2^1 \times 2^{2x} = 2(2^x)^2\]
Substituting this back into the equation yields:
\[2(2^x)^2 - 17(2^x) + 8 = 0 \quad \text{(as required)}\]

(b) Let \(u = 2^x\).
The equation becomes:
\[2u^2 - 17u + 8 = 0\]
Factorising this quadratic equation:
\[(2u - 1)(u - 8) = 0\]
This gives:
\[u = \frac{1}{2} \quad \text{or} \quad u = 8\]
Now substitute \(u = 2^x\) back:

1) \(2^x = \frac{1}{2} \implies 2^x = 2^{-1} \implies x = -1\)

2) \(2^x = 8 \implies 2^x = 2^3 \implies x = 3\)

Thus, the solutions are \(x = -1\) and \(x = 3\).

(a)
M1: Recognises that \(2^{2x+1} = 2 \times 2^{2x}\) or \(2 \times (2^x)^2\).
A1*: Complete and convincing proof showing all steps to reach the given equation.

(b)
M1: Introduces a substitution (e.g., \(u = 2^x\)) and writes a quadratic in terms of that variable, or factorises directly in terms of \(2^x\).
M1: Solves the quadratic equation to find two values for \(u\) (or \(2^x\)) by factorisation, quadratic formula, or completing the square.
A1: Obtains \(2^x = \frac{1}{2}\) and \(2^x = 8\) (or equivalent).
A1: Correctly deduces \(x = -1\) from \(2^x = \frac{1}{2}\).
A1: Correctly deduces \(x = 3\) from \(2^x = 8\).

(a) First, find the radius squared, \(r^2\), which is the square of the distance between \(A(3, -2)\) and \(B(7, 1)\):
\[r^2 = (7 - 3)^2 + (1 - (-2))^2 = 4^2 + 3^2 = 16 + 9 = 25\]
The general equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the centre.
With centre \((3, -2)\) and \(r^2 = 25\), the equation is:
\[(x - 3)^2 + (y + 2)^2 = 25\]

(b) The gradient of the radius \(AB\) is:
\[m_{\text{radius}} = \frac{1 - (-2)}{7 - 3} = \frac{3}{4}\]
The tangent is perpendicular to the radius at the point of contact, so the gradient of the tangent, \(m_{\text{tangent}}\), is:
\[m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -\frac{4}{3}\]
The equation of the tangent line passing through \(B(7, 1)\) is:
\[y - 1 = -\frac{4}{3}(x - 7)\]
Multiplying both sides by 3 to eliminate the fraction:
\[3(y - 1) = -4(x - 7)\]
\[3y - 3 = -4x + 28\]
Rearranging into the form \(ax + by + c = 0\):
\[4x + 3y - 31 = 0\]

(a)
M1: Attempts to find the radius or \(r^2\) using the distance formula between \(3, -2\) and \(7, 1\). Accept \(r^2 = (7-3)^2 + (1+2)^2\) or \(r = 5\).
M1: Correct form for the equation of the circle using their centre \(3, -2\) and their \(r^2\), i.e., \((x - 3)^2 + (y + 2)^2 = k\) where \(k > 0\).
A1: \((x - 3)^2 + (y + 2)^2 = 25\) (or equivalent expanded form).

(b)
M1: Attempts to find the gradient of the radius \(AB\) using \(\frac{y_2 - y_1}{x_2 - x_1}\).
A1: Gradient of \(AB\) is \(\frac{3}{4}\).
M1: Uses the perpendicular gradient rule \(m_1 m_2 = -1\) to find the gradient of the tangent. Follow through on their gradient of \(AB\).
M1: Applies their perpendicular gradient and the coordinates of point \(B(7, 1)\) to form the equation of a line, e.g., \(y - 1 = m(x - 7)\).
A1: Correct equation in the required form with integer coefficients, e.g., \(4x + 3y - 31 = 0\) or any integer multiple (such as \(-4x - 3y + 31 = 0\)).

(a) Acceleration is given by the derivative of velocity: \(a = \frac{\text{d}v}{\text{d}t} = 6t - 12\).
When \(t = 3\): \(a = 6(3) - 12 = 6\text{ m s}^{-2}\).

(b) \(P\) is instantaneously at rest when \(v = 0\):
\(3t^2 - 12t + 9 = 0\)
\(3(t^2 - 4t + 3) = 0\)
\(3(t-1)(t-3) = 0\)
So \(t = 1\) and \(t = 3\).

(c) The particle changes direction of motion at \(t = 1\) and \(t = 3\).
We calculate the displacement in each interval:
From \(t = 0\) to \(t = 1\):
\(\int_0^1 (3t^2 - 12t + 9) \text{d}t = \left[ t^3 - 6t^2 + 9t \right]_0^1 = 1 - 6 + 9 = 4\text{ m}\)
From \(t = 1\) to \(t = 3\):
\(\int_1^3 (3t^2 - 12t + 9) \text{d}t = \left[ t^3 - 6t^2 + 9t \right]_1^3 = (27 - 54 + 27) - 4 = -4\text{ m}\)
From \(t = 3\) to \(t = 4\):
\(\int_3^4 (3t^2 - 12t + 9) \text{d}t = \left[ t^3 - 6t^2 + 9t \right]_3^4 = (64 - 96 + 36) - 0 = 4\text{ m}\)
Total distance travelled is: \(|4| + |-4| + |4| = 12\text{ m}\).

(a)
M1: Differentiating the velocity expression to find acceleration.
A1: Correct acceleration expression \(6t - 12\).
A1: Correct acceleration of \(6\text{ m s}^{-2}\).

(b)
M1: Setting \(v = 0\) and attempting to solve the quadratic equation.
A1: Correct values of \(t = 1\) and \(t = 3\).

(c)
M1: Realising the need to integrate \(v\) in intervals where the sign changes, or using absolute values.
M1: Integrating the expression to get \(t^3 - 6t^2 + 9t\) (ignore limits for this mark).
A1: Correct evaluation of at least two of the interval integrals (magnitudes of 4, 4, or 4).
M1: Adding the absolute values of the three displacements.
A1: Correct final answer of \(12\text{ m}\).

(a) For particle \(A\):
The normal reaction \(R\) is: \(R = 3mg\).
The maximum frictional force \(F\) is: \(F = \mu R = \frac{1}{3}(3mg) = mg\).
The equation of motion for \(A\) is: \(T - mg = 3ma\).
For particle \(B\):
The equation of motion is: \(2mg - T = 2ma\).

(b) Adding the two equations of motion:
\((T - mg) + (2mg - T) = 3ma + 2ma\)
\(mg = 5ma\)
\(a = \frac{1}{5}g = 0.2g\).

(c) Substituting \(a\) back into the equation for \(B\):
\(2mg - T = 2m(0.2g)\)
\(T = 2mg - 0.4mg = 1.6mg\).

(d) The tension \(T\) has the same magnitude throughout the length of the string.

(a)
M1: Resolving forces vertically for \(A\) to find \(R = 3mg\) and finding \(F = mg\).
A1: Correct equation for \(A\): \(T - mg = 3ma\).
A1: Correct equation for \(B\): \(2mg - T = 2ma\).

(b)
M1: Attempting to solve the simultaneous equations to eliminate \(T\).
A1: Showing \(mg = 5ma\) or equivalent.
A1: Correct acceleration \(a = 0.2g\) (or \(\frac{1}{5}g\)).

(c)
M1: Substituting \(a\) into either equation of motion to solve for \(T\).
A1: Correct tension \(T = 1.6mg\) (or \(\frac{8}{5}mg\)).

(d)
B2: Stating that the tension is constant throughout the string (accept 'same tension at both ends of the string' or 'tension does not vary').

Resolving horizontally:
\(T_1 \sin(30^\circ) = 25 \sin(\theta)\)
\(0.5 T_1 = 25 \sin(\theta) \implies T_1 = 50 \sin(\theta)\)

Resolving vertically:
\(T_1 \cos(30^\circ) + 25 \cos(\theta) = 4g\)
Since \(g = 9.8\text{ m s}^{-2}\), \(4g = 39.2\text{ N}\).
\(T_1 \cos(30^\circ) + 25 \cos(\theta) = 39.2\)

Substitute \(T_1 = 50 \sin(\theta)\):
\(50 \sin(\theta) \cos(30^\circ) + 25 \cos(\theta) = 39.2\)
Since \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\):
\(25\sqrt{3} \sin(\theta) + 25 \cos(\theta) = 39.2\)
Using the identity \(R \sin(\theta + \alpha) = R \sin\theta \cos\alpha + R \cos\theta \sin\alpha\), we identify:
\(25\sqrt{3} \sin(\theta) + 25 \cos(\theta) = 50 \sin(\theta + 30^\circ)\)
Therefore:
\(50 \sin(\theta + 30^\circ) = 39.2\)
\(\sin(\theta + 30^\circ) = 0.784\)
Since \(\theta\) is acute:
\(\theta + 30^\circ = \arcsin(0.784) \approx 51.63^\circ\)
\(\theta \approx 21.63^\circ\)
So, to the nearest tenth of a degree, \(\theta = 21.6^\circ\).

(b) Substitute \(\theta = 21.63^\circ\) into the expression for \(T_1\):
\(T_1 = 50 \sin(21.63^\circ) \approx 18.43\text{ N}\).
To 3 significant figures, \(T_1 = 18.4\text{ N}\).

(a)
M1: Resolving forces horizontally to get an equation in \(T_1\) and \(\theta\).
A1: Correct horizontal equation: \(T_1 \sin(30^\circ) = 25 \sin(\theta)\).
M1: Resolving forces vertically to get an equation in \(T_1\), \(\theta\), and \(4g\).
A1: Correct vertical equation: \(T_1 \cos(30^\circ) + 25 \cos(\theta) = 39.2\).
M1: Eliminating \(T_1\) and using trigonometric identities to solve for \(\theta\).
A1: Correct acute angle \(\theta = 21.6^\circ\) (to 1 d.p.).

(b)
M1: Substituting the value of \(\theta\) back to find \(T_1\).
A1: Correct calculation.
A1: Correct final answer \(T_1 = 18.4\text{ N}\) (to 3 s.f.).

Let \(G\) be the centre of mass of the rod at distance \(x\) from \(A\).
The support at \(C\) is at \(1\text{ m}\) from \(A\).
The support at \(D\) is at \(1.5\text{ m}\) from \(B\), which is \(6 - 1.5 = 4.5\text{ m}\) from \(A\).

(a) When the particle of mass \(M\) is at \(A\) and the rod is on the point of tilting about \(C\), the reaction force at \(D\) is zero (\(R_D = 0\)).
Taking moments about \(C\):
\(Mg \times AC = 15g \times CG\)
\(M(1) = 15(x - 1) \implies M = 15(x - 1)\) --- (Equation 1)

When the particle of mass \(M\) is at \(B\) and the rod is on the point of tilting about \(D\), the reaction force at \(C\) is zero (\(R_C = 0\)).
Taking moments about \(D\):
\(15g \times GD = Mg \times DB\)
\(15(4.5 - x) = M(1.5) \implies 1.5M = 15(4.5 - x) \implies M = 10(4.5 - x)\) --- (Equation 2)

Equating the two expressions for \(M\):
\(15(x - 1) = 10(4.5 - x)\)
\(1.5(x - 1) = 4.5 - x\)
\(1.5x - 1.5 = 4.5 - x\)
\(2.5x = 6 \implies x = 2.4\)

(b) Substitute \(x = 2.4\) into Equation 1:
\(M = 15(2.4 - 1) = 15(1.4) = 21\).

(a)
M1: Identifies that tilting about \(C\) means \(R_D = 0\) and attempts to take moments about \(C\).
A1: Correct equation: \(M = 15(x - 1)\).
M1: Identifies that tilting about \(D\) means \(R_C = 0\) and attempts to take moments about \(D\).
A1: Correct equation: \(1.5M = 15(4.5 - x)\) or equivalent.
M1: Equates the two expressions for \(M\) and attempts to solve for \(x\).
A1: Correct value of \(x = 2.4\).

(b)
M1: Substituting their \(x\) value back into either equation for \(M\).
A1: Correct value of \(M = 21\).

(a) The position vector \(\mathbf{r}_P\) of \(P\) at time \(t\) is:
\(\mathbf{r}_P = (2\mathbf{i} + 5\mathbf{j}) + t(4\mathbf{i} - \mathbf{j}) = (2 + 4t)\mathbf{i} + (5 - t)\mathbf{j}\)

(b) The position vector \(\mathbf{r}_Q\) of \(Q\) at time \(t\) is:
\(\mathbf{r}_Q = (10\mathbf{i} + \mathbf{j}) + t(\mathbf{i} + 3\mathbf{j}) = (10 + t)\mathbf{i} + (1 + 3t)\mathbf{j}\)
The position vector of \(Q\) relative to \(P\), \(\mathbf{r}_{Q/P}\), is:
\(\mathbf{r}_{Q/P} = \mathbf{r}_Q - \mathbf{r}_P = (10 + t - (2 + 4t))\mathbf{i} + (1 + 3t - (5 - t))\mathbf{j}\)
\(\mathbf{r}_{Q/P} = (8 - 3t)\mathbf{i} + (4t - 4)\mathbf{j}\)

(c) The distance squared \(d^2\) between the ships is given by:
\(d^2 = (8 - 3t)^2 + (4t - 4)^2\)
\(d^2 = (64 - 48t + 9t^2) + (16t^2 - 32t + 16)\)
\(d^2 = 25t^2 - 80t + 80\)
To find the minimum, we differentiate with respect to \(t\):
\(\frac{\text{d}(d^2)}{\text{d}t} = 50t - 80 = 0\)
\(t = 1.6\text{ hours}\)
Since \(1.6\text{ hours} = 1\text{ hour } 36\text{ minutes}\), the ships are closest at 13:36.

(a)
M1: Attempt to write down a general position vector equation for \(P\).
A1: Correct expression \((2+4t)\mathbf{i} + (5-t)\mathbf{j}\).

(b)
M1: Attempt to find position vector of \(Q\) at time \(t\).
M1: Subtracting \(\mathbf{r}_P\) from \(\mathbf{r}_Q\).
A1: Correctly showing the given expression \((8 - 3t)\mathbf{i} + (4t - 4)\mathbf{j}\).

(c)
M1: Expressing the distance squared \(d^2\) or distance \(d\) in terms of \(t\).
M1: Differentiating with respect to \(t\) and setting to zero (or completing the square).
A1: Solving to get \(t = 1.6\).
M1: Converting \(1.6\) hours into hours and minutes.
A1: Correct time of 13:36.

(a) The graph should show:
- A straight line starting from \((0, 12)\) rising to \((6, 24)\).
- A horizontal line at speed \(24\) from \(t = 6\) to \(t = 6 + T\).
- A straight line falling from \((6 + T, 24)\) to \((22 + T, 0)\).

(b) Using \(v = u + at\) for the deceleration stage:
\(0 = 24 - 1.5t_3\)
\(1.5t_3 = 24 \implies t_3 = 16\text{ seconds}\)

(c) The total distance is the area under the speed-time graph:
Distance in first stage (acceleration):
\(s_1 = \frac{12 + 24}{2} \times 6 = 18 \times 6 = 108\text{ m}\)
Distance in second stage (constant speed):
\(s_2 = 24 \times T = 24T\text{ m}\)
Distance in third stage (deceleration):
\(s_3 = \frac{24 + 0}{2} \times 16 = 12 \times 16 = 192\text{ m}\)
Total distance:
\(s_1 + s_2 + s_3 = 720\)
\(108 + 24T + 192 = 720\)
\(300 + 24T = 720\)
\(24T = 420 \implies T = 17.5\text{ seconds}\).

(a)
B1: Straight line of positive gradient starting at \((0, 12)\) and ending at speed \(24\).
B1: Horizontal line segment at speed \(24\).
B1: Straight line of negative gradient ending on the time axis.

(b)
M1: Use of \(v = u + at\) with \(v=0\), \(u=24\), and \(a=-1.5\).
A1: Correctly showing \(t = 16\).

(c)
M1: Attempting to find the area under the graph for the first stage (trapezium or triangle + rectangle).
A1: Correct area of \(108\).
M1: Expressing the area of the second stage as \(24T\).
M1: Attempting to find the area of the third stage (triangle).
A1: Correct area of \(192\).
M1: Setting up an equation for the total area equal to \(720\).
A1: Correct value of \(T = 17.5\).

Since \(\tan\alpha = \frac{3}{4}\), we have \(\sin\alpha = 0.6\) and \(\cos\alpha = 0.8\).

(a) The diagram must show:
- Weight of \(8g\) (or \(78.4\text{ N}\)) acting vertically downwards.
- Normal reaction \(R\) acting perpendicular to the plane, away from the plane.
- Pulling force \(P\) acting up parallel to the plane.
- Frictional force \(F\) acting down parallel to the plane (opposing motion).

(b) Resolving forces perpendicular to the plane:
\(R = mg \cos\alpha\)
\(R = 8 \times 9.8 \times 0.8 = 62.72\text{ N}\).
To 3 significant figures, \(R = 62.7\text{ N}\).

(c) The maximum frictional force is:
\(F = \mu R = 0.25 \times 62.72 = 15.68\text{ N}\).
Resolving forces parallel to the plane and using Newton's second law:
\(P - mg \sin\alpha - F = ma\)
\(P - (8 \times 9.8 \times 0.6) - 15.68 = 8 \times 1.5\)
\(P - 47.04 - 15.68 = 12\)
\(P - 62.72 = 12\)
\(P = 74.72\text{ N}\).
To 3 significant figures, \(P = 74.7\text{ N}\) (or \(75\text{ N}\) to 2 significant figures).

(a)
B2: Award 1 mark for showing at least three forces correctly, and 2 marks for showing all four forces correctly with correct directions (Weight, Reaction, Pull, Friction).

(b)
M1: Resolving perpendicular to the plane to get \(R = mg \cos\alpha\).
A1: Substituting \(\cos\alpha = 0.8\).
A1: Correct reaction \(R = 62.7\text{ N}\) or \(62.72\text{ N}\).

(c)
M1: Use of \(F = \mu R\) to calculate friction.
A1: Correct friction force of \(15.68\text{ N}\).
M1: Setting up equation of motion parallel to the plane: \(P - mg \sin\alpha - F = ma\).
A1: Correctly substituting the numerical values: \(P - 47.04 - 15.68 = 12\).
A1: Correct final value of \(P = 74.7\text{ N}\) or \(75\text{ N}\).

(a) Let \(Z\) be the standard normal distribution \(Z \sim \mathrm{N}(0, 1)\).
From \(P(W < 110) = 0.0918\), we standardise:
\(P\left(Z < \frac{110 - \mu}{\sigma}\right) = 0.0918\)
Using the normal table, \(\Phi(-1.33) = 1 - 0.9082 = 0.0918\), so:
\(\frac{110 - \mu}{\sigma} = -1.33 \implies 110 - \mu = -1.33\sigma\) [Equation 1]

From \(P(W > 185) = 0.1210\):
\(P\left(Z > \frac{185 - \mu}{\sigma}\right) = 0.1210 \implies P\left(Z < \frac{185 - \mu}{\sigma}\right) = 0.8790\)
Using the normal table, \(\Phi(1.17) = 0.8790\), so:
\(\frac{185 - \mu}{\sigma} = 1.17 \implies 185 - \mu = 1.17\sigma\) [Equation 2]

Subtracting Equation 1 from Equation 2:
\(75 = 2.5\sigma \implies \sigma = 30\)

Substituting \(\sigma = 30\) back into Equation 1:
\(110 - \mu = -1.33(30) \implies \mu = 110 + 39.9 = 149.9 \approx 150\) (to 3 s.f.)

(b) Using \(\mu = 150\) and \(\sigma = 30\):
\(P(W > 140) = P\left(Z > \frac{140 - 150}{30}\right) = P(Z > -0.33)\)
\(P(Z > -0.33) = P(Z < 0.33) = \Phi(0.33) = 0.6293\)

(c) Let \(X\) be the number of apples weighing more than 140 g out of 3, so \(X \sim \mathrm{B}(3, 0.6293)\).
\(P(X = 2) = \binom{3}{2} (0.6293)^2 (1 - 0.6293)\)
\(P(X = 2) = 3 \times 0.39602 \times 0.3707 = 0.4404\) (to 4 d.p.)

(a)
M1: Standardising 110 and 185, setting equal to z-values.
A1: Correct z-values of \(-1.33\) and \(1.17\).
M1: Subtracting equations to find \(\sigma\).
A1: \(\sigma = 30\) and \(\mu = 150\) (or 149.9) shown.

(b)
M1: Standardising 140 with \(\mu = 150\) and \(\sigma = 30\).
A1: 0.6293 (allow 0.629).

(c)
M1: Applying the binomial formula with \(n=3\), \(r=2\) and their probability from part (b).
A1: 0.4404 (allow 0.440).

(a) Find the midpoints \(x\) for each class:
\(12.5, 17.5, 22.5, 27.5, 35.0\)
Calculate \(\sum fx\):
\(\sum fx = (8 \times 12.5) + (17 \times 17.5) + (25 \times 22.5) + (18 \times 27.5) + (12 \times 35.0)\)
\(\sum fx = 100 + 297.5 + 562.5 + 495 + 420 = 1875\)
Mean \(\bar{x} = \frac{1875}{80} = 23.4375 \approx 23.4\) minutes (3 s.f.).

(b) The median position is \(80 / 2 = 40\).
Cumulative frequencies are: 8, 25, 50, 68, 80.
The median lies in the interval \(20 \le t < 25\).
Using linear interpolation:
\(\text{Median} = 20 + \frac{40 - 25}{25} \times 5 = 20 + 3 = 23\) minutes.

(c) Calculate \(\sum fx^2\):
\(\sum fx^2 = 8(12.5)^2 + 17(17.5)^2 + 25(22.5)^2 + 18(27.5)^2 + 12(35.0)^2\)
\(\sum fx^2 = 1250 + 5206.25 + 12656.25 + 13612.5 + 14700 = 47425\)
Variance \(\sigma^2 = \frac{47425}{80} - 23.4375^2 = 592.8125 - 549.3164 = 43.4961\)
Standard deviation \(\sigma = \sqrt{43.4961} = 6.595 \approx 6.60\) minutes (3 s.f.).

(d) Comparing the mean and the median:
\(\text{Mean} (23.44) > \text{Median} (23.00)\)
Therefore, the distribution is positively skewed.

(a)
M1: Finding midpoints and attempting \(\sum fx\).
A1: Mean = 23.4 (or 23.4375).

(b)
M1: Identifying the interval \(20 \le t < 25\) and attempting linear interpolation.
A1: Median = 23.

(c)
M1: Attempting to calculate \(\sum fx^2\).
M1: Using standard deviation formula with their \(\bar{x}\).
A1: Standard deviation = 6.60 (or 6.595).

(d)
B1: Concluding positive skewness with a correct comparison (Mean > Median).

(a)
Equation 1 (Sum of probabilities = 1):
\(a + b + 0.3 + b + c = 1 \implies a + 2b + c = 0.7\)

Equation 2 (Expectation):
\(\mathrm{E}(X) = -2a - b + 0 + b + 2c = 0.2 \implies -2a + 2c = 0.2\)

Equation 3 (Variance):
\(\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = 1.56 \implies \mathrm{E}(X^2) - 0.04 = 1.56 \implies \mathrm{E}(X^2) = 1.6\)
\(\mathrm{E}(X^2) = 4a + b + 0 + b + 4c = 1.6 \implies 4a + 2b + 4c = 1.6\)

(b) From Equation 2:
\(-2a + 2c = 0.2 \implies c - a = 0.1 \implies c = a + 0.1\)

From Equation 1:
\(2b = 0.7 - a - c = 0.7 - a - (a + 0.1) = 0.6 - 2a \implies b = 0.3 - a\)

Substitute these into Equation 3:
\(4a + 2(0.3 - a) + 4(a + 0.1) = 1.6\)
\(4a + 0.6 - 2a + 4a + 0.4 = 1.6\)
\(6a + 1.0 = 1.6 \implies 6a = 0.6 \implies a = 0.1\)

Then:
\(b = 0.3 - 0.1 = 0.2\)
\(c = 0.1 + 0.1 = 0.2\)

(c) \(\mathrm{E}(3X - 1) = 3\mathrm{E}(X) - 1 = 3(0.2) - 1 = 0.6 - 1 = -0.4\)

(d) \(\mathrm{Var}(2 - 3X) = (-3)^2 \mathrm{Var}(X) = 9 \times 1.56 = 14.04\)

(a)
B1: Sum of probabilities equation \(a+2b+c=0.7\).
B1: Expectation equation \(-2a+2c=0.2\).
M1: Variance equation using \(\mathrm{E}(X^2) - (\mathrm{E}(X))^2\).

(b)
M1: Substituting to eliminate variables.
A1: Correct values for \(a\), \(b\) and \(c\).

(c)
M1: Attempting expectation transformation.
A1: \(-0.4\).

(d)
M1: Attempting variance transformation \(9 \times \mathrm{Var}(X)\).
A1: \(14.04\).

(a) Since \(A\) and \(B\) are independent:
\(P(A \cap B) = P(A) \times P(B) = 0.4 \times 0.3 = 0.12\)

(b) Since \(B\) and \(C\) are mutually exclusive, there is no intersection between \(B\) and \(C\).
- The intersection \(P(A \cap B) = 0.12\).
- The intersection \(P(A \cap C) = 0.15\).
- \(P(B \text{ only}) = P(B) - P(A \cap B) = 0.3 - 0.12 = 0.18\).
- \(P(C \text{ only}) = P(C) - P(A \cap C) = 0.5 - 0.15 = 0.35\).
- \(P(A \text{ only}) = P(A) - P(A \cap B) - P(A \cap C) = 0.4 - 0.12 - 0.15 = 0.13\).
- Sum of all probabilities in \(A, B, C\) is \(0.13 + 0.12 + 0.18 + 0.15 + 0.35 = 0.93\).
- The probability outside all sets is \(1 - 0.93 = 0.07\).

(c) \(P(A \cup B \cup C) = 0.13 + 0.12 + 0.18 + 0.15 + 0.35 = 0.93\).

(d) \(P(A' \cap C' | B') = \frac{P(A' \cap C' \cap B')}{P(B')}\)
- \(P(B') = 1 - 0.3 = 0.7\)
- \(A' \cap C' \cap B'\) represents the region outside \(A\), \(B\), and \(C\), which has probability \(0.07\).
- \(P(A' \cap C' | B') = \frac{0.07}{0.7} = 0.1\)

(a)
B1: Correctly calculating \(P(A \cap B) = 0.12\).

(b)
M1: Correct structure of Venn diagram with \(B\) and \(C\) separate.
A1: Correct probabilities inside circles.
A1: Correct outer probability (0.07).

(c)
B1: 0.93.

(d)
M1: Correct formula for conditional probability.
A1: 0.1.

(a)
\(S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 2720 - \frac{160^2}{10} = 160\)
\(S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 1040000 - \frac{3200^2}{10} = 16000\)
\(S_{xy} = \sum xy - \frac{\sum x \sum y}{n} = 50000 - \frac{160 \times 3200}{10} = -1200\)

(b)
\(r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}} = \frac{-1200}{\sqrt{160 \times 16000}} = \frac{-1200}{1600} = -0.75\)

(c) A PMCC of \(-0.75\) indicates a strong negative correlation, meaning that as daily average temperature increases, electricity consumption decreases.

(d)
\(b = \frac{S_{xy}}{S_{xx}} = \frac{-1200}{160} = -7.5\)
\(\bar{x} = 16\), \(\bar{y} = 320\)
\(a = \bar{y} - b\bar{x} = 320 - (-7.5 \times 16) = 320 + 120 = 440\)
So, \(y = 440 - 7.5x\)

(e)
At \(x = 18\):
\(y = 440 - 7.5(18) = 305\) kWh.
Since 18 is within the range of the given data (mean temperature is 16), it is an interpolation and likely to be reliable.

(a)
M1: Attempting formula for any of \(S_{xx}\), \(S_{yy}\) or \(S_{xy\).
A1: All three calculated correctly.

(b)
M1: Applying the formula for \(r\).
A1: \(-0.75\).

(c)
B1: Contextual explanation of negative correlation.

(d)
M1: Correct method for \(b\).
M1: Correct method for \(a\).
A1: \(y = 440 - 7.5x\).

(e)
M1: Substituting 18 into their regression line.
B1: 305 kWh and comment on reliability due to interpolation.

(a)
- Median is the 8th value: \(Q_2 = 16.5\).
- Lower quartile: Since \(15 \times 0.25 = 3.75\), we round up to the 4th value: \(Q_1 = 15.2\).
- Upper quartile: Since \(15 \times 0.75 = 11.25\), we round up to the 12th value: \(Q_3 = 18.2\).

(b)
\(\mathrm{IQR} = Q_3 - Q_1 = 18.2 - 15.2 = 3.0\)
- Lower outlier boundary: \(Q_1 - 1.5 \times \mathrm{IQR} = 15.2 - 4.5 = 10.7\).
- Upper outlier boundary: \(Q_3 + 1.5 \times \mathrm{IQR} = 18.2 + 4.5 = 22.7\).

Since \(24.5 > 22.7\), 24.5 is an outlier. The next highest value is 20.1, which is not an outlier. The minimum value is 12.1, which is greater than 10.7, so there are no lower outliers. Thus, 24.5 is the only outlier.

(c) Draw a box plot showing:
- Median line at 16.5.
- Box boundaries at 15.2 and 18.2.
- Lower whisker ending at 12.1.
- Upper whisker ending at 20.1 (the highest value within the boundary).
- An outlier marked with 'x' at 24.5.

(d) Comparison:
- Averages: The second brand has a higher median lifetime (18.0 hours vs 16.5 hours), indicating it generally lasts longer.
- Spread: The second brand has a lower interquartile range (1.5 hours vs 3.0 hours), indicating its lifetimes are more consistent/less variable.

(a)
B1: Median = 16.5.
B1: Lower quartile = 15.2, Upper quartile = 18.2.

(b)
M1: Attempting to calculate \(\mathrm{IQR}\) and outlier boundaries.
A1: Correct boundaries (10.7 and 22.7) and concluding 24.5 is the only outlier.

(c)
M1: Correct box boundaries and median.
A1: Whiskers drawn to 12.1 and 20.1, with outlier at 24.5 clearly marked.

(d)
B1: Comparing medians (averages) correctly.
B1: Comparing IQRs (consistency) correctly.

(a)
Stage 1 probabilities:
- Win Stage 1 (\(S_1\)): \(P(S_1) = 2/6 = 1/3\)
- Lose Stage 1 (\(S_1'\)): \(P(S_1') = 4/6 = 2/3\)

Stage 2 probabilities:
- From \(S_1\): Win (\(W\)) = \(0.4\), Lose (\(L\)) = \(0.6\)
- From \(S_1'\): Win (\(W\)) = \(0.2\), Lose (\(L\)) = \(0.8\)

(b)
\(P(W) = P(S_1 \cap W) + P(S_1' \cap W)\)
\(P(W) = \left(\frac{1}{3} \times 0.4\right) + \left(\frac{2}{3} \times 0.2\right)\)
\(P(W) = \frac{4}{30} + \frac{4}{30} = \frac{8}{30} = \frac{4}{15} \approx 0.267\)

(c)
\(P(S_1 | W) = \frac{P(S_1 \cap W)}{P(W)} = \frac{4/30}{8/30} = 0.5\)

(d)
Probability that a player does not win Stage 2 is \(1 - \frac{4}{15} = \frac{11}{15}\).
For three players, probability that none of them wins:
\(P(\text{None}) = \left(\frac{11}{15}\right)^3 = \frac{1331}{3375} \approx 0.3944\)

\(P(\text{At least one wins}) = 1 - 0.3944 = 0.6056 \approx 0.606\) (to 3 s.f.)

(a)
M1: Correct structure of tree diagram.
A1: Correct branch probabilities.

(b)
M1: Summing products along the two winning branches.
A1: \(\frac{4}{15}\) or 0.267.

(c)
M1: Applying conditional probability formula.
A1: 0.5.

(d)
M1: Attempting \(1 - P(\text{None})\).
A1: 0.606 (accept 0.61).