2025 Edexcel IAL Mathematics (YMA01) Practice Paper with Answers

(a) From first principles, \(\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\).

Here \(f(x) = 3x^2 - 5x\).

\(f(x+h) = 3(x+h)^2 - 5(x+h) = 3(x^2 + 2xh + h^2) - 5x - 5h = 3x^2 + 6xh + 3h^2 - 5x - 5h\).

\(f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 5x - 5h) - (3x^2 - 5x) = 6xh + 3h^2 - 5h\).

Dividing by \(h\):

\(\frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2 - 5h}{h} = 6x + 3h - 5\).

Taking the limit as \(h \to 0\):

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{h \to 0} (6x + 3h - 5) = 6x - 5\). (as required)

(b) When \(x = 2\), \(y = 3(2)^2 - 5(2) = 12 - 10 = 2\).

The gradient of the tangent at \(P\) is \(\frac{\mathrm{d}y}{\mathrm{d}x}\Big|_{x=2} = 6(2) - 5 = 7\).

The equation of the tangent is:

\(y - 2 = 7(x - 2)\)

\(y - 2 = 7x - 14\)

\(7x - y - 12 = 0\)

(a)
M1: Writes down the definition of the derivative from first principles, or clearly sets up \(\frac{f(x+h)-f(x)}{h}\) with \(f(x) = 3x^2 - 5x\).
A1: Correctly expands \(f(x+h)\) to obtain \(3x^2 + 6xh + 3h^2 - 5x - 5h\) (or simplified equivalent).
M1: Divides the numerator expression by \(h\) to get \(6x + 3h - 5\).
A1: Shows the limit as \(h \to 0\) correctly to arrive at \(6x - 5\) (with no steps missing and correct notation throughout).

(b)
M1: Substitutes \(x = 2\) into the equation of the curve to find the y-coordinate \(y = 2\).
M1: Substitutes \(x = 2\) into the derivative to find the gradient \(m = 7\).
A1.5: Correctly writes the equation of the line as \(7x - y - 12 = 0\) or any integer multiple, e.g., \(14x - 2y - 24 = 0\) (all terms on one side, equal to 0).

(a) Express the derivative as individual terms:

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4x^3}{x^{1/2}} - \frac{6}{x^{1/2}} = 4x^{5/2} - 6x^{-1/2}\)

Integrate with respect to \(x\):

\(y = \int (4x^{5/2} - 6x^{-1/2}) \, \mathrm{d}x = \frac{4}{\frac{7}{2}}x^{7/2} - \frac{6}{\frac{1}{2}}x^{1/2} + c = \frac{8}{7}x^{7/2} - 12x^{1/2} + c\)

Using the point \((4, 20)\), substitute \(x = 4\) and \(y = 20\):

\(20 = \frac{8}{7}(4)^{7/2} - 12(4)^{1/2} + c\)

Since \(4^{7/2} = (\sqrt{4})^7 = 2^7 = 128\):

\(20 = \frac{8}{7}(128) - 12(2) + c\)

\(20 = \frac{1024}{7} - 24 + c\)

\(44 = \frac{1024}{7} + c \implies c = 44 - \frac{1024}{7} = \frac{308 - 1024}{7} = -\frac{716}{7}\)

Thus, \(f(x) = \frac{8}{7}x^{7/2} - 12x^{1/2} - \frac{716}{7}\).

(b) Since \(\frac{\mathrm{d}y}{\mathrm{d}x} = 4x^{5/2} - 6x^{-1/2}\):

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4\left(\frac{5}{2}\right)x^{3/2} - 6\left(-\frac{1}{2}\right)x^{-3/2} = 10x^{3/2} + 3x^{-3/2}\)

At \(x = 4\):

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 10(4)^{3/2} + 3(4)^{-3/2} = 10(8) + \frac{3}{8} = 80 + \frac{3}{8} = \frac{643}{8} = 80.375\).

(a)
M1: Attempts to split the fraction and write at least one term in the form \(Ax^n\), e.g., \(4x^{5/2}\) or \(-6x^{-1/2}\).
A1: Correct simplified terms for the derivative: \(4x^{5/2} - 6x^{-1/2}\).
M1: Integrates with at least one term increased in power by 1.
A1: Correct integration: \(\frac{8}{7}x^{7/2} - 12x^{1/2}\), with or without \(+ c\).
M1: Substitutes \(x = 4\) and \(y = 20\) to find the constant of integration \(c\).
A0.5: Correct value of \(c = -\frac{716}{7}\) and final equation \(f(x) = \frac{8}{7}x^{7/2} - 12x^{1/2} - \frac{716}{7}\).

(b)
M1: Differentiates \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 10x^{3/2} + 3x^{-3/2}\).
A1: Substitutes \(x=4\) to get \(80.375\) (or \(\frac{643}{8}\) or equivalent fraction).

(a) The equation of \(l_1\) is \(3x - 2y + 8 = 0\), which can be written as \(y = \frac{3}{2}x + 4\).

The gradient of \(l_1\) is \(m_1 = \frac{3}{2}\).

Since \(l_2\) is perpendicular to \(l_1\), its gradient \(m_2\) is given by \(m_2 = -\frac{1}{m_1} = -\frac{2}{3}\).

The equation of \(l_2\) passing through \(A(4, -3)\) is:

\(y - (-3) = -\frac{2}{3}(x - 4)\)

\(y + 3 = -\frac{2}{3}x + \frac{8}{3}\)

Multiplying by 3:

\(3y + 9 = -2x + 8 \implies 3y + 2x + 1 = 0\).

(b) To find the intersection of \(l_1\) and \(l_2\), solve the simultaneous equations:

1) \(3x - 2y + 8 = 0 \implies 9x - 6y + 24 = 0\)

2) \(2x + 3y + 1 = 0 \implies 4x + 6y + 2 = 0\)

Adding these two equations gives:

\(13x + 26 = 0 \implies x = -2\).

Substitute \(x = -2\) into \(3x - 2y + 8 = 0\):

\(3(-2) - 2y + 8 = 0 \implies 2 - 2y = 0 \implies y = 1\).

So the coordinates of \(P\) are \((-2, 1)\).

(c) The line \(l_1\) intersects the y-axis when \(x = 0\):

\(3(0) - 2y + 8 = 0 \implies y = 4\).

Thus, \(B\) is \((0, 4)\).

The base of triangle \(OPB\) along the y-axis is the distance from \(O(0, 0)\) to \(B(0, 4)\), which is \(4\) units.

The height of triangle \(OPB\) is the perpendicular distance from \(P(-2, 1)\) to the y-axis, which is the absolute value of the x-coordinate of \(P\), i.e., \(|-2| = 2\) units.

\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 2 = 4\).

(a)
M1: Identifies the gradient of \(l_1\) as \(\frac{3}{2}\) or uses the perpendicular condition to state the gradient of \(l_2\) is \(-\frac{2}{3}\).
M1: Uses \(y - y_1 = m(x - x_1)\) with their gradient of \(l_2\) and the coordinates of \(A(4, -3)\).
A1: Correct equation in the required form: \(3y + 2x + 1 = 0\) (or any non-zero integer multiple of this equation).

(b)
M1: Attempts to solve the simultaneous equations for \(l_1\) and \(l_2\), e.g., by multiplying to eliminate one variable or substituting.
A1: Correct \(x\)-coordinate, \(x = -2\).
A0.5: Correct \(y\)-coordinate, \(y = 1\) (Accept written as the coordinate pair \((-2, 1)\)).

(c)
M1: Finds the y-intercept of \(l_1\) to be \(B(0, 4)\) (or states that \(OB = 4\)) and uses \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\) with base \(4\) and height equal to the magnitude of the \(x\)-coordinate of \(P\).
A1: Correct area of \(4\).

(a) From the first equation, express \(y\) in terms of \(x\):

\(y = 5 - 2x\)

Substitute this into the second equation:

\(x^2 - x(5 - 2x) - (5 - 2x)^2 = -11\)

Expand the brackets and simplify:

\(x^2 - 5x + 2x^2 - (25 - 20x + 4x^2) = -11\)

\(3x^2 - 5x - 25 + 20x - 4x^2 = -11\)

\(-x^2 + 15x - 25 = -11\)

\(x^2 - 15x + 14 = 0\)

Factor the quadratic equation:

\((x - 1)(x - 14) = 0\)

This gives \(x = 1\) or \(x = 14\).

Find the corresponding values of \(y\):

- If \(x = 1 \implies y = 5 - 2(1) = 3\).
- If \(x = 14 \implies y = 5 - 2(14) = -23\).

So the solutions are \(x = 1, y = 3\) and \(x = 14, y = -23\).

(b) For the quadratic equation \(x^2 + (k+1)x + 2k = 0\) to have no real roots, its discriminant must be negative (\(b^2 - 4ac < 0\)).

Here, \(a = 1\), \(b = k+1\), and \(c = 2k\).

\(b^2 - 4ac = (k+1)^2 - 4(1)(2k) = k^2 + 2k + 1 - 8k = k^2 - 6k + 1\).

We require \(k^2 - 6k + 1 < 0\).

Find the critical values by solving \(k^2 - 6k + 1 = 0\):

\(k = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(1)}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}\).

Since the coefficient of \(k^2\) is positive, \(k^2 - 6k + 1 < 0\) is satisfied between the critical values:

\(3 - 2\sqrt{2} < k < 3 + 2\sqrt{2}\).

(a)
M1: Attempts to make \(y\) (or \(x\)) the subject of the linear equation and substitutes into the quadratic equation.
A1: Correctly expands and simplifies to obtain a 3-term quadratic in one variable, e.g., \(x^2 - 15x + 14 = 0\).
M1: Solves their quadratic equation by factoring, completing the square, or using the formula.
A1: Correct values of \(x\) (\(x = 1, 14\)) or correct values of \(y\) (\(y = 3, -23\)).
A1: Correct paired solutions: \(x = 1, y = 3\) and \(x = 14, y = -23\) (Must be correctly paired).

(b)
M1: Recognises that no real roots implies \(b^2 - 4ac < 0\) and attempts to find \(b^2 - 4ac\) in terms of \(k\).
A0.5: Finds the correct quadratic expression \(k^2 - 6k + 1\).
M1: Finds the critical values \(3 \pm 2\sqrt{2}\) (or equivalent decimals \(0.17\) and \(5.83\)).
A1: Correct inequality of \(3 - 2\sqrt{2} < k < 3 + 2\sqrt{2}\) (accept equivalent notation or rounded decimal values \(0.172 < k < 5.83\) with at least 3 significant figures).

(a) Using the trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\), substitute \(\sin^2 \theta = 1 - \cos^2 \theta\) into the given equation:

\(2(1 - \cos^2 \theta) + \cos \theta = 1\)

\(2 - 2\cos^2 \theta + \cos \theta = 1\)

Rearranging all terms to one side:

\(2\cos^2 \theta - \cos \theta - 1 = 0\) (as required)

(b) Let \(\theta = 2x - 30^\circ\).

Since \(0 \le x < 360^\circ\), the range for \(\theta\) is:

\(0 \le 2x < 720^\circ \implies -30^\circ \le 2x - 30^\circ < 690^\circ\).

We solve the quadratic equation \(2\cos^2 \theta - \cos \theta - 1 = 0\):

\((2\cos \theta + 1)(\cos \theta - 1) = 0\)

This gives \(\cos \theta = -\frac{1}{2}\) or \(\cos \theta = 1\).

Case 1: \(\cos \theta = 1\) in the interval \(-30^\circ \le \theta < 690^\circ\):

\(\theta = 0^\circ, 360^\circ\)

Case 2: \(\cos \theta = -\frac{1}{2}\) in the interval \(-30^\circ \le \theta < 690^\circ\):

\(\theta = 120^\circ, 240^\circ, 480^\circ, 600^\circ\)

Now, solve for \(x\) using \(x = \frac{\theta + 30^\circ}{2}\):

- \(\theta = 0^\circ \implies x = \frac{0^\circ + 30^\circ}{2} = 15^\circ\)
- \(\theta = 120^\circ \implies x = \frac{120^\circ + 30^\circ}{2} = 75^\circ\)
- \(\theta = 240^\circ \implies x = \frac{240^\circ + 30^\circ}{2} = 135^\circ\)
- \(\theta = 360^\circ \implies x = \frac{360^\circ + 30^\circ}{2} = 195^\circ\)
- \(\theta = 480^\circ \implies x = \frac{480^\circ + 30^\circ}{2} = 255^\circ\)
- \(\theta = 600^\circ \implies x = \frac{600^\circ + 30^\circ}{2} = 315^\circ\)

Thus, the solutions are \(x = 15^\circ, 75^\circ, 135^\circ, 195^\circ, 255^\circ, 315^\circ\).

(a)
M1: Substitutes \(\sin^2 \theta = 1 - \cos^2 \theta\) into the equation.
A1: Correctly expands and rearranges to the form \(2\cos^2 \theta - \cos \theta - 1 = 0\) with no errors.

(b)
M1: Solves the quadratic equation to find \(\cos(2x - 30^\circ) = -\frac{1}{2}\) or \(\cos(2x - 30^\circ) = 1\).
M1: Identifies the correct interval for the angle \(2x - 30^\circ\) as \([-30^\circ, 690^\circ)\).
A1: Finds at least three correct values of \(2x - 30^\circ\) (e.g., \(0^\circ, 120^\circ, 240^\circ\)).
M1: Applies the correct sequence of operations to solve for \(x\), i.e., \(x = \frac{\theta + 30^\circ}{2}\), for at least three values of \(\theta\).
A1: Gives any four of the correct solutions: \(15^\circ, 75^\circ, 135^\circ, 195^\circ, 255^\circ, 315^\circ\).
A0.5: Gives all six correct solutions, and no extras inside the interval.

(a) We identify the features of \(y = \frac{4}{x-2} + 3\):

- Vertical asymptote: as \(x \to 2\), \(y \to \pm\infty\), so the vertical asymptote is \(x = 2\).
- Horizontal asymptote: as \(x \to \pm\infty\), \(y \to 3\), so the horizontal asymptote is \(y = 3\).
- y-intercept (setting \(x = 0\)): \(y = \frac{4}{0-2} + 3 = -2 + 3 = 1\). So the curve crosses the y-axis at \((0, 1)\).
- x-intercept (setting \(y = 0\)):
\[0 = \frac{4}{x-2} + 3 \implies \frac{4}{x-2} = -3 \implies 4 = -3(x-2) \implies 4 = -3x + 6 \implies 3x = 2 \implies x = \frac{2}{3}\]
So the curve crosses the x-axis at \((\frac{2}{3}, 0)\).

(b) Translating by vector \(\begin{pmatrix} -1 \\ 2 \end{pmatrix}\) corresponds to replacing \(x\) with \(x + 1\) and \(y\) with \(y - 2\):

\(y - 2 = \frac{4}{(x+1)-2} + 3\)

\(y = \frac{4}{x-1} + 5\)

Combine into a single fraction over the denominator \(x-1\):

\(y = \frac{4 + 5(x-1)}{x-1} = \frac{4 + 5x - 5}{x-1} = \frac{5x-1}{x-1}\).

This is in the required form with \(a = 5\), \(b = -1\), \(c = 1\), and \(d = -1\).

(a)
B1: Sketch of a rectangular hyperbola in two parts in the correct quadrants (top-right and bottom-left relative to their asymptotes).
B1: Correct vertical asymptote \(x = 2\) and horizontal asymptote \(y = 3\) labeled clearly.
B1: Correct y-intercept at \((0, 1)\) or marked \(1\) on the y-axis.
B1.5: Correct x-intercept at \((\frac{2}{3}, 0)\) or marked \(\frac{2}{3}\) on the x-axis.

(b)
M1: Applies the translation to the equation: replaces \(x\) with \(x+1\) and adds \(2\) to the function (or replaces \(y\) with \(y-2\)).
M1: Attempts to write as a single fraction by finding a common denominator.
A1: Correct final equation \(y = \frac{5x-1}{x-1}\) (accept \(a=5, b=-1, c=1, d=-1\) or any non-zero integer multiple).

(a) Using the distance formula:

\(AB = \sqrt{(4 - (-2))^2 + (13 - 5)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\).

(b) The midpoint \(M\) of the line segment \(AB\) is:

\(M = \left(\frac{-2 + 4}{2}, \frac{5 + 13}{2}\right) = (1, 9)\).

The gradient of the line segment \(AB\) is:

\(m_{AB} = \frac{13 - 5}{4 - (-2)} = \frac{8}{6} = \frac{4}{3}\).

The gradient of the perpendicular bisector, \(m\), is:

\(m = -\frac{1}{m_{AB}} = -\frac{3}{4}\).

The equation of the perpendicular bisector passing through \(M(1, 9)\) is:

\(y - 9 = -\frac{3}{4}(x - 1)\)

\(y - 9 = -\frac{3}{4}x + \frac{3}{4}\)

\(y = -\frac{3}{4}x + \frac{39}{4}\).

(c) The perpendicular bisector meets the x-axis when \(y = 0\):

\(0 = -\frac{3}{4}x + \frac{39}{4}\)

\(\frac{3}{4}x = \frac{39}{4} \implies 3x = 39 \implies x = 13\).

So the coordinates of \(C\) are \((13, 0)\).

(a)
M1: Attempts to use the distance formula \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
A1: Correct simplified answer \(10\) (does not have to be written as a surd since it simplifies to an integer).

(b)
M1: Finds the midpoint of \(AB\), showing working: \((1, 9)\).
M1: Finds the gradient of \(AB\) (\(\frac{4}{3}\)) and states/uses the perpendicular gradient rule \(m_1 m_2 = -1\) to get \(-\frac{3}{4}\).
M1: Applies \(y - y_M = m(x - x_M)\) with their midpoint and perpendicular gradient.
A0.5: Correct equation \(y = -\frac{3}{4}x + \frac{39}{4}\) (or equivalent \(y = -0.75x + 9.75\)).

(c)
M1: Sets \(y = 0\) in their equation from part (b) and attempts to solve for \(x\).
A1: Correct coordinates of \(C(13, 0)\) (accept \(x=13\)).

(a) The box has a square base of side length \(x\) and height \(h\).

The total surface area \(S\) of a closed rectangular box is:

\(S = 2x^2 + 4xh\)

We are given \(S = 150\), so:

\(2x^2 + 4xh = 150\)

\(4xh = 150 - 2x^2 \implies h = \frac{150 - 2x^2}{4x} = \frac{37.5}{x} - \frac{1}{2}x\).

The volume \(V\) of the box is:

\(V = x^2 h\)

Substitute the expression for \(h\) into the volume formula:

\(V = x^2 \left(\frac{150 - 2x^2}{4x}\right) = \frac{x(150 - 2x^2)}{4} = 37.5x - \frac{1}{2}x^3\) (as required)

(b) To find the maximum volume, differentiate \(V\) with respect to \(x\):

\(\frac{\mathrm{d}V}{\mathrm{d}x} = 37.5 - \frac{3}{2}x^2\)

Set \(\frac{\mathrm{d}V}{\mathrm{d}x} = 0\) to find critical points:

\(37.5 - 1.5x^2 = 0 \implies 1.5x^2 = 37.5 \implies x^2 = 25\)

Since \(x > 0\) (as it is a physical length), \(x = 5\).

To justify that \(x = 5\) corresponds to a maximum volume, calculate the second derivative:

\(\frac{\mathrm{d}^2V}{\mathrm{d}x^2} = -3x\)

At \(x = 5\):

\(\frac{\mathrm{d}^2V}{\mathrm{d}x^2} = -3(5) = -15 < 0\).

Since the second derivative is negative at \(x = 5\), the volume is indeed maximized.

The maximum volume is:

\(V = 37.5(5) - \frac{1}{2}(5)^3 = 187.5 - 62.5 = 125\text{ cm}^3\).

(a)
M1: Writes down the expression for the total surface area: \(2x^2 + 4xh = 150\) (or equivalent).
M1: Rearranges the surface area equation to make \(h\) the subject: \(h = \frac{150-2x^2}{4x}\) or equivalent.
M1: Substitutes their expression for \(h\) into \(V = x^2 h\).
A0.5: Obtains the correct given formula \(V = 37.5x - \frac{1}{2}x^3\) with no errors in the algebraic steps.

(b)
M1: Differentiates \(V\) to find \(\frac{\mathrm{d}V}{\mathrm{d}x} = 37.5 - 1.5x^2\).
M1: Sets \(\frac{\mathrm{d}V}{\mathrm{d}x} = 0\) and solves for \(x\) to get \(x = 5\) (ignore \(x = -5\)).
M1: Finds \(\frac{\mathrm{d}^2V}{\mathrm{d}x^2} = -3x\) (or uses a valid first-derivative sign test) and evaluates at \(x = 5\) to show it is negative (hence a maximum).
A1: Calculates the correct maximum volume \(V = 125\text{ cm}^3\) (units not required).

(a) First, write the equation of the curve with fractional and negative indices:
\[y = 2x^2 - 8x^{1/2} + 9x^{-1}\]
Differentiating with respect to \(x\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 2(2x) - 8\left(\frac{1}{2}x^{-1/2}\right) + 9(-1x^{-2})\]
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 4x - 4x^{-1/2} - 9x^{-2}\]

(b) At the point \(P(1, 3)\), the value of \(x\) is \(1\). Substituting \(x = 1\) into \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find the gradient of the tangent, \(m\):
\[m = 4(1) - 4(1)^{-1/2} - 9(1)^{-2}\]
\[m = 4 - 4 - 9 = -9\]
The equation of the tangent line is:
\[y - 3 = -9(x - 1)\]
\[y - 3 = -9x + 9\]
\[9x + y - 12 = 0\]

(c) To find the second derivative, differentiate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) with respect to \(x\):
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x}\left(4x - 4x^{-1/2} - 9x^{-2}\right)\]
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4 - 4\left(-\frac{1}{2}x^{-3/2}\right) - 9(-2x^{-3})\]
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4 + 2x^{-3/2} + 18x^{-3}\]
At the point \(P\), where \(x = 1\):
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4 + 2(1)^{-3/2} + 18(1)^{-3}\]
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4 + 2 + 18 = 24\]

(a)
* **M1**: Attempts to differentiate, with at least one term of the form \(x^n \to c x^{n-1}\).
* **A1**: Any two terms correct from \(4x\), \(-4x^{-1/2}\), and \(-9x^{-2}\).
* **A1**: Fully correct simplified expression: \(4x - 4x^{-1/2} - 9x^{-2}\) (accept equivalent forms, e.g. \(4x - \frac{4}{\sqrt{x}} - \frac{9}{x^2}\)).

(b)
* **M1**: Substitutes \(x = 1\) into their derivative to find the numerical gradient of the tangent.
* **M1**: Formulates a linear equation of the form \(y - 3 = m(x - 1)\) using their calculated gradient \(m\).
* **A1**: Correct equation in the required form: \(9x + y - 12 = 0\) (accept any integer multiple, such as \(-9x - y + 12 = 0\)).

(c)
* **M1**: Attempts to differentiate their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to obtain \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\).
* **A1**: Correct value of \(24\) (c.a.o.).

(a) First, calculate the gradient of the line \(l_1\):
\[m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 5}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2}\]
Now, use the point-gradient form with point \(B(4, 2)\):
\[y - 2 = -\frac{1}{2}(x - 4)\]
Multiply the entire equation by 2:
\[2(y - 2) = -(x - 4)\]
\[2y - 4 = -x + 4\]
Rearrange into the form \(ax + by + c = 0\):
\[x + 2y - 8 = 0\]

(b) Since line \(l_2\) is perpendicular to line \(l_1\), the product of their gradients is \(-1\):
\[m_2 = -\frac{1}{m_1} = -\frac{1}{-1/2} = 2\]
Using the gradient \(2\) and the point \(C(4, 7)\), the equation of \(l_2\) is:
\[y - 7 = 2(x - 4)\]
\[y - 7 = 2x - 8\]
\[y = 2x - 1\]

(c) To find the coordinates of \(D\), solve the equations of \(l_1\) and \(l_2\) simultaneously. Substitute the equation of \(l_2\) into the equation of \(l_1\):
\[x + 2(2x - 1) - 8 = 0\]
\[x + 4x - 2 - 8 = 0\]
\[5x - 10 = 0 \implies 5x = 10 \implies x = 2\]
Substitute \(x = 2\) back into the equation of \(l_2\) to find \(y\):
\[y = 2(2) - 1 = 3\]
Thus, the coordinates of the intersection point \(D\) are \((2, 3)\).

(a)
* **M1**: Attempts to find the gradient of \(l_1\) using \(\frac{y_2 - y_1}{x_2 - x_1}\).
* **M1**: Correct method to form a linear equation with their gradient and either point \(A\) or point \(B\).
* **A1**: Correct equation in the required form: \(x + 2y - 8 = 0\) (accept equivalent integer multiples, such as \(-x - 2y + 8 = 0\)).

(b)
* **M1**: Uses the perpendicular gradient rule \(m_2 = -\frac{1}{m_1}\) with their gradient from part (a).
* **A1**: Correct equation for \(l_2\), e.g., \(y = 2x - 1\) or \(2x - y - 1 = 0\).

(c)
* **M1**: Attempts to solve the simultaneous equations of \(l_1\) and \(l_2\) by elimination or substitution to find a value for \(x\) or \(y\).
* **A1**: Correct coordinates \((2, 3)\) (both values must be correct and presented clearly as coordinates or as separate statements \(x=2, y=3\)).

(a) We test each integer in the given range: For \(n = 2\): \(2^2 + 2 = 6\), which is not divisible by 4. For \(n = 3\): \(3^2 + 2 = 11\), which is not divisible by 4. For \(n = 4\): \(4^2 + 2 = 18\), which is not divisible by 4. For \(n = 5\): \(5^2 + 2 = 27\), which is not divisible by 4. For \(n = 6\): \(6^2 + 2 = 38\), which is not divisible by 4. Since all possible cases have been tested and none of the results are divisible by 4, the statement is true by proof by exhaustion. (b) Let \(n = 2\), which is a prime number. Then \(n^2 + 1 = 2^2 + 1 = 5\). Since 5 is an odd number (not even), the statement is disproved. (c) An odd integer can be represented as \(2m + 1\), where \(m\) is an integer. Squaring this expression: \((2m + 1)^2 = 4m^2 + 4m + 1 = 4m(m + 1) + 1\). For any integer \(m\), either \(m\) or \(m + 1\) must be an even integer. Thus, their product \(m(m + 1)\) must be even, so we can write \(m(m + 1) = 2k\) for some integer \(k\). Substituting this back: \((2m + 1)^2 = 4(2k) + 1 = 8k + 1\). Hence, the square of any odd integer is of the form \(8k + 1\).

(a) M1: Attempts to substitute at least three of the values \(n = 2, 3, 4, 5, 6\) into \(n^2 + 2\). A1: Correctly evaluates all five expressions: 6, 11, 18, 27, 38. A1: States a clear conclusion that none of these values are divisible by 4. (b) M1: Identifies \(n = 2\) as the candidate counterexample and attempts to evaluate \(n^2 + 1\). A1: Shows \(2^2 + 1 = 5\) and states that since 2 is prime and 5 is odd, the statement is disproved. (c) M1: Writes a general expression for an odd integer (e.g., \(2m+1\) or \(2m-1\)) and attempts to square it. M1: Factorises the quadratic terms to obtain \(4m(m+1) + 1\) and explains that \(m(m+1)\) is even as it is the product of consecutive integers. A1: Concludes with a clear and complete proof, showing the form \(8k + 1\).

(a) Using the factor theorem, we evaluate \(f(3)\): \(f(3) = 2(3)^3 - 3(3)^2 - 11(3) + 6 = 54 - 27 - 33 + 6 = 0\). Since \(f(3) = 0\), \((x - 3)\) is a factor of \(f(x)\). (b) We can divide \(f(x)\) by \((x - 3)\) using equating coefficients or algebraic long division: \(2x^3 - 3x^2 - 11x + 6 = (x - 3)(2x^2 + 3x - 2)\). Factorising the quadratic part: \(2x^2 + 3x - 2 = (2x - 1)(x + 2)\). Thus, completely factorised: \(f(x) = (x - 3)(2x - 1)(x + 2)\). (c) According to the remainder theorem, the remainder when \(f(x)\) is divided by \((2x + 3)\) is given by \(f(-1.5)\): \(f(-1.5) = 2(-1.5)^3 - 3(-1.5)^2 - 11(-1.5) + 6 = 2(-3.375) - 3(2.25) + 16.5 + 6 = -6.75 - 6.75 + 16.5 + 6 = 9\). The remainder is 9.

(a) M1: Attempts to evaluate \(f(3)\) by substituting \(x = 3\) into \(f(x)\). A1: Correctly shows that \(f(3) = 0\) and concludes that \((x-3)\) is a factor. (b) M1: Attempts algebraic long division, synthetic division, or equating coefficients to find a quadratic factor. Must get at least two terms of \(2x^2 + 3x - 2\) correct. A1: Obtains the correct quadratic factor \(2x^2 + 3x - 2\). A1: Completely factorises to \((x - 3)(2x - 1)(x + 2)\). (c) M1: Realises that the remainder is \(f(-1.5)\) and attempts to substitute \(x = -1.5\) into \(f(x)\) or uses algebraic long division with \((2x+3)\). M1: Fully evaluates the expression (allowing for minor arithmetic slips). A1: Obtains 9.

(a) We complete the square for both \(x\) and \(y\): \(x^2 - 6x + y^2 + 8y = 11 \Rightarrow (x - 3)^2 - 9 + (y + 4)^2 - 16 = 11 \Rightarrow (x - 3)^2 + (y + 4)^2 = 36\). Center coordinates: \((3, -4)\), Radius: \(\sqrt{36} = 6\). (b) Method 1: The perpendicular distance from the center of the circle \((3, -4)\) to the tangent line \(2x - y + k = 0\) must be equal to the radius, 6. Using the distance formula: \(d = \frac{|2(3) - (-4) + k|}{\sqrt{2^2 + (-1)^2}} = 6 \Rightarrow \frac{|10 + k|}{\sqrt{5}} = 6 \Rightarrow |10 + k| = 6\sqrt{5}\). This gives \(10 + k = \pm 6\sqrt{5} \Rightarrow k = -10 \pm 6\sqrt{5}\). Method 2: Substitute \(y = 2x + k\) into the circle's equation: \(x^2 + (2x + k)^2 - 6x + 8(2x + k) - 11 = 0 \Rightarrow 5x^2 + (4k + 10)x + (k^2 + 8k - 11) = 0\). For a tangent line, the discriminant of this quadratic in \(x\) must be zero (\(B^2 - 4AC = 0\)): \((4k + 10)^2 - 4(5)(k^2 + 8k - 11) = 0 \Rightarrow 16k^2 + 80k + 100 - 20k^2 - 160k + 220 = 0 \Rightarrow -4k^2 - 80k + 320 = 0 \Rightarrow k^2 + 20k - 80 = 0\). Solving this quadratic gives \(k = \frac{-20 \pm \sqrt{400 - 4(1)(-80)}}{2} = -10 \pm 6\sqrt{5}\).

(a) M1: Attempts to complete the square for both \(x\) and \(y\). Allow one sign error. A1: Center is \((3, -4)\). A1: Radius is 6. (b) M1: Substitutes \(y = 2x + k\) into the equation of the circle to form a quadratic in \(x\). M1: Expands and collects terms to get a quadratic in the form \(Ax^2 + Bx + C = 0\) with correct \(A\) and \(B\) in terms of \(k\). M1: Sets the discriminant \(B^2 - 4AC = 0\) (or uses the perpendicular distance formula equating distance to 6). A1: Obtains the simplified quadratic equation \(k^2 + 20k - 80 = 0\) (or \(|10+k| = 6\sqrt{5}\)). A1: Correct values: \(k = -10 \pm 6\sqrt{5}\) or exact equivalent simplest surd form.

(a) Since the terms are in a geometric sequence, the common ratio \(r\) must be constant: \(r = \frac{x + 4}{3x} = \frac{x - 2}{x + 4}\). Cross-multiplying gives: \((x + 4)^2 = 3x(x - 2) \Rightarrow x^2 + 8x + 16 = 3x^2 - 6x \Rightarrow 2x^2 - 14x - 16 = 0\). Dividing the entire equation by 2: \(x^2 - 7x - 8 = 0\) (as required). (b) We factorise the quadratic equation: \((x - 8)(x + 1) = 0\). Thus, the two possible values of \(x\) are \(x = 8\) and \(x = -1\). (c) For a geometric series to have a sum to infinity, the common ratio \(r\) must satisfy \(|r| < 1\). Let's check the ratio \(r\) for both possible values of \(x\): If \(x = -1\), the terms are \(-3\), \(3\), and \(-3\). The common ratio is \(r = \frac{3}{-3} = -1\). Since \(|r| = 1\), this sequence does not have a sum to infinity. If \(x = 8\), the terms are \(24\), \(12\), and \(6\). The common ratio is \(r = \frac{12}{24} = \frac{1}{2}\). Since \(|r| = \frac{1}{2} < 1\), this sequence has a sum to infinity. Therefore, the value of the common ratio is \(r = \frac{1}{2}\).

(a) M1: Sets up an equation using the common ratio, e.g., \(\frac{x+4}{3x} = \frac{x-2}{x+4}\). M1: Cross-multiplies and expands correctly to obtain \(x^2 + 8x + 16 = 3x^2 - 6x\). A1: Fully simplifies to \(x^2 - 7x - 8 = 0\) with no errors seen. (b) M1: Attempts to factorise or use the quadratic formula on the equation from (a). A1: Correct values: \(x = 8, -1\). (c) M1: Evaluates the common ratio \(r\) for at least one of their \(x\) values. M1: States the condition for the existence of the sum to infinity, \(|r| < 1\), and rejects \(x = -1\) because \(r = -1\). A1: Concludes \(r = \frac{1}{2}\) only.

(a) We use the trigonometric identity \(\sin^2 \theta = 1 - \cos^2 \theta\). Substitute this into the equation: \(2(1 - \cos^2 \theta) + 7\cos \theta - 5 = 0 \Rightarrow 2 - 2\cos^2 \theta + 7\cos \theta - 5 = 0 \Rightarrow -2\cos^2 \theta + 7\cos \theta - 3 = 0\). Multiply the entire equation by \(-1\): \(2\cos^2 \theta - 7\cos \theta + 3 = 0\) (as required). (b) From part (a), we solve: \(2\cos^2 \theta - 7\cos \theta + 3 = 0\). Factorising the quadratic in \(\cos \theta\): \((2\cos \theta - 1)(\cos \theta - 3) = 0\). This gives \(\cos \theta = \frac{1}{2}\) or \(\cos \theta = 3\). Since \(-1 \le \cos \theta \le 1\), there are no solutions for \(\cos \theta = 3\). For \(\cos \theta = \frac{1}{2}\): The principal value is \(\theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ\). The second value in the range \(0 \le \theta < 360^\circ\) is \(\theta = 360^\circ - 60^\circ = 300^\circ\). The solutions are \(\theta = 60^\circ\) and \(\theta = 300^\circ\).

(a) M1: Substitutes \(\sin^2 \theta = 1 - \cos^2 \theta\) into the equation. M1: Expands and simplifies terms to obtain a quadratic expression in \(\cos \theta\). A1: Obtains \(2\cos^2 \theta - 7\cos \theta + 3 = 0\) with no algebraic errors. (b) M1: Attempts to factorise or solve the quadratic in \(\cos\theta\). A1: Obtains \(\cos \theta = \frac{1}{2}\). B1: Mentions that \(\cos \theta = 3\) has no solutions because \(-1 \le \cos \theta \le 1\). M1: Finds at least one correct angle for their value of \(\cos\theta\). A1: Both \(\theta = 60^\circ\) and \(\theta = 300^\circ\), and no other solutions in the range.

(a) Using the laws of logarithms: First, apply the power law \(k\log_a b = \log_a (b^k)\): \(\log_3(x - 2)^2 - \log_3(x + 4) = 1\). Next, apply the division law: \(\log_3\left( \frac{(x - 2)^2}{x + 4} \right) = 1\). Convert from logarithmic to exponential form: \(\frac{(x - 2)^2}{x + 4} = 3^1 \Rightarrow (x - 2)^2 = 3(x + 4)\). Expand and rearrange: \(x^2 - 4x + 4 = 3x + 12 \Rightarrow x^2 - 7x - 8 = 0\). Factorise the quadratic: \((x - 8)(x + 1) = 0\). This gives \(x = 8\) or \(x = -1\). However, the original equation requires \(x > 2\) (since the term \(\log_3(x-2)\) is only defined when \(x - 2 > 0\)). Therefore, we reject \(x = -1\). The only solution is \(x = 8\). (b) Rewrite the equation using rules of indices: \(3 \cdot (3^y)^2 - 10(3^y) + 3 = 0\). Let \(u = 3^y\). The equation becomes: \(3u^2 - 10u + 3 = 0 \Rightarrow (3u - 1)(u - 3) = 0\). This yields \(u = \frac{1}{3}\) or \(u = 3\). Substituting back: For \(3^y = \frac{1}{3} = 3^{-1} \Rightarrow y = -1\). For \(3^y = 3^1 \Rightarrow y = 1\). The solutions are \(y = -1\) and \(y = 1\).

(a) M1: Applies the power law of logarithms to write \(2\log_3(x-2)\) as \(\log_3(x-2)^2\). M1: Applies the subtraction law of logarithms to obtain \(\log_3\left(\frac{(x-2)^2}{x+4}\right)\). M1: Correctly removes the logarithm by writing \(\frac{(x-2)^2}{x+4} = 3\). A1: Obtains the correct quadratic equation \(x^2 - 7x - 8 = 0\). A1: Solves the quadratic to find \(x = 8\) and explicitly states that \(x = -1\) is rejected as a solution. (b) M1: Uses index laws to write \(3^{2y+1}\) as \(3 \cdot (3^y)^2\) and defines a substitution (e.g., \(u = 3^y\)) to form a quadratic in \(u\). A1: Correctly solves the quadratic to find \(3^y = 1/3\) and \(3^y = 3\). A1: Obtains both \(y = -1\) and \(y = 1\).

(a) The total surface area \(A\) of a solid cylinder is given by: \(A = 2\pi r^2 + 2\pi r h\). We are given that \(A = 120\pi\), so: \(2\pi r^2 + 2\pi r h = 120\pi \Rightarrow r^2 + r h = 60 \Rightarrow h = \frac{60 - r^2}{r}\). The volume \(V\) of a cylinder is: \(V = \pi r^2 h\). Substitute the expression for \(h\) into the volume formula: \(V = \pi r^2 \left( \frac{60 - r^2}{r} \right) \Rightarrow V = \pi r(60 - r^2) = 60\pi r - \pi r^3\) (as required). (b) To find the stationary points, we differentiate \(V\) with respect to \(r\): \(\frac{dV}{dr} = 60\pi - 3\pi r^2\). Set \(\frac{dV}{dr} = 0 \Rightarrow 60\pi - 3\pi r^2 = 0 \Rightarrow 3\pi r^2 = 60\pi \Rightarrow r^2 = 20 \Rightarrow r = \sqrt{20} = 2\sqrt{5}\) (since \(r > 0\)). We can check that this gives a maximum using the second derivative: \(\frac{d^2V}{dr^2} = -6\pi r\). Since \(r = 2\sqrt{5} > 0\), \(\frac{d^2V}{dr^2} = -12\pi\sqrt{5} < 0\), confirming a maximum. Substitute \(r = 2\sqrt{5}\) back into the volume formula: \(V = 60\pi(2\sqrt{5}) - \pi(2\sqrt{5})^3 \Rightarrow V = 120\pi\sqrt{5} - \pi(8 \cdot 5\sqrt{5}) \Rightarrow V = 120\pi\sqrt{5} - 40\pi\sqrt{5} = 80\pi\sqrt{5}\). Thus, the maximum volume is \(80\pi\sqrt{5}\), where \(k = 80\) and \(a = 5\).

(a) M1: Writes down the correct formula for the total surface area of a solid cylinder, \(2\pi r^2 + 2\pi rh = 120\pi\). M1: Rearranges the equation to express \(h\) in terms of \(r\). A1: Substitutes \(h\) into the volume formula \(V = \pi r^2 h\) and correctly derives \(V = 60\pi r - \pi r^3\). (b) M1: Differentiates the volume formula to get \(\frac{dV}{dr} = 60\pi - 3\pi r^2\). A1: Sets \(\frac{dV}{dr} = 0\) and solves to find \(r^2 = 20\) or \(r = 2\sqrt{5}\). M1: Substitutes their positive value of \(r\) into the expression for \(V\). M1: Shows that it is a maximum, either by evaluating \(\frac{d^2V}{dr^2}\) (must be negative) or comparing values. A1: Obtains the correct maximum volume \(80\pi\sqrt{5}\).

(a) To find the points of intersection, we set the equations of \(C\) and \(L\) equal to each other: \(6x - x^2 = 2x \Rightarrow x^2 - 4x = 0 \Rightarrow x(x - 4) = 0\). This gives \(x = 0\) (with \(y = 2(0) = 0\)) and \(x = 4\) (with \(y = 2(4) = 8\)). The points of intersection are \((0, 0)\) and \((4, 8)\). (b) The area \(A\) of the region bounded by the curve \(C\) and the line \(L\) is given by: \(A = \int_{0}^{4} (y_C - y_L) \, dx = \int_{0}^{4} ((6x - x^2) - 2x) \, dx = \int_{0}^{4} (4x - x^2) \, dx\). Now perform the integration: \(A = \left[ 2x^2 - \frac{1}{3}x^3 \right]_{0}^{4}\). Evaluate at the upper and lower limits: At \(x = 4\): \(2(4)^2 - \frac{1}{3}(4)^3 = 32 - \frac{64}{3} = \frac{32}{3}\). At \(x = 0\): \(2(0)^2 - \frac{1}{3}(0)^3 = 0\). Subtracting the limits: \(A = \frac{32}{3} - 0 = \frac{32}{3}\). The exact area of the region is \(\frac{32}{3}\).

(a) M1: Equates \(6x - x^2 = 2x\) and attempts to solve the resulting quadratic equation. A1: Correctly identifies both points of intersection: \((0,0)\) and \((4,8)\). (b) M1: Sets up a definite integral for the area, using their \(x\)-coordinates of intersection as limits, with integrand \((6x - x^2) - 2x\) or equivalent. M1: Simplifies the integrand to \(4x - x^2\). M1: Integrates the expression. Must see at least one term increased in power by 1. A1: Correct integrated expression: \(2x^2 - \frac{1}{3}x^3\). M1: Substitutes their limits \(4\) and \(0\) into their integrated expression. A1: Obtains the exact area of \(\frac{32}{3}\) (or equivalent exact fraction).

(a) The sum of the first two terms is given by:
\(S_2 = a + ar = a(1+r) = 48\) [Equation 1]

The sum to infinity is given by:
\(S_{\infty} = \frac{a}{1-r} = 64 \implies a = 64(1-r)\) [Equation 2]

Substituting Equation 2 into Equation 1 gives:
\(64(1-r)(1+r) = 48\)

Dividing both sides by 64:
\((1-r)(1+r) = \frac{48}{64}
\implies 1 - r^2 = \frac{3}{4}
\implies r^2 = 1 - \frac{3}{4} = \frac{1}{4}\) (as required)

(b) Since all terms of the series are positive, we must have \(r > 0\).

Given \(r^2 = \frac{1}{4}\), this means \(r = \frac{1}{2}\).

Using Equation 2:
\(a = 64\left(1 - \frac{1}{2}\right) = 32\)

(c) The sum of the first 8 terms, \(S_8\), is given by:
\(S_8 = \frac{a(1-r^8)}{1-r}\)

Substituting \(a = 32\) and \(r = \frac{1}{2}\):
\(S_8 = \frac{32\left(1 - \left(\frac{1}{2}\right)^8\right)}{1 - \frac{1}{2}}
= \frac{32\left(1 - \frac{1}{256}\right)}{\frac{1}{2}}
= 64\left(\frac{255}{256}\right)
= \frac{255}{4} = 63.75\)

(a)
M1: Writes down correct expressions for \(S_2\) and \(S_{\infty}\) in terms of \(a\) and \(r\). Accept \(a + ar = 48\) and \(\frac{a}{1-r} = 64\).
M1: Attempts to eliminate \(a\) to form an equation in \(r\) only, e.g., substituting \(a = 64(1-r)\) into \(a(1+r) = 48\).
A1*: Fully correct algebra leading to \(r^2 = \frac{1}{4}\) with no errors or omissions in working.

(b)
M1: Selects \(r = \frac{1}{2}\) (may be implicit) and attempts to find \(a\) using a correct equation.
A1: \(a = 32\).

(c)
M1: Attempts to use the sum formula \(S_n = \frac{a(1-r^n)}{1-r}\) with their \(a\), \(n = 8\) and \(r = 0.5\).
A1: Correct substitution of values into the formula.
A1: \(63.75\) or \(\frac{255}{4}\) (or equivalent exact fraction/decimal).

(a) Let \( y = \frac{3x - 1}{x - 2} \). Then \( y(x - 2) = 3x - 1 \implies xy - 2y = 3x - 1 \implies xy - 3x = 2y - 1 \implies x(y - 3) = 2y - 1 \implies x = \frac{2y - 1}{y - 3} \). Thus, \( \mathrm{f}^{-1}(x) = \frac{2x - 1}{x - 3} \). (b) The domain of \( \mathrm{f}^{-1} \) is the range of \( \mathrm{f} \). Since the denominator of \( \mathrm{f}^{-1}(x) \) cannot be zero, we must have \( x \neq 3 \). Hence, the domain is \( x \in \mathbb{R}, x \neq 3 \). (c) Setting \( \mathrm{f}(x) = \mathrm{f}^{-1}(x) \) gives: \( \frac{3x - 1}{x - 2} = \frac{2x - 1}{x - 3} \implies (3x - 1)(x - 3) = (2x - 1)(x - 2) \implies 3x^2 - 10x + 3 = 2x^2 - 5x + 2 \implies x^2 - 5x + 1 = 0 \). Using the quadratic formula, we find: \( x = \frac{5 \pm \sqrt{21}}{2} \).

(a) M1: Set \( y = \mathrm{f}(x) \), multiply through by \( x-2 \) and attempt to group \( x \) terms. M1: Factorise \( x \) and rearrange to make \( x \) the subject. A1: Correct expression for \( \mathrm{f}^{-1}(x) \). (b) B1: State the correct domain: \( x \in \mathbb{R}, x \neq 3 \) (or equivalent). (c) M1: Equate functions and cross-multiply. M1: Rearrange to form a 3-term quadratic equation. A1: Correct quadratic equation \( x^2 - 5x + 1 = 0 \) (or equivalent). A1: Correct exact values \( x = \frac{5 \pm \sqrt{21}}{2} \).

(a) Using the product rule with \( u = \mathrm{e}^{2x} \) and \( v = \cos(3x) \): \( \frac{\mathrm{d}u}{\mathrm{d}x} = 2\mathrm{e}^{2x} \) and \( v' = -3\sin(3x) \). Thus, \( \frac{\mathrm{d}y}{\mathrm{d}x} = -3\mathrm{e}^{2x}\sin(3x) + 2\mathrm{e}^{2x}\cos(3x) = \mathrm{e}^{2x}(2\cos(3x) - 3\sin(3x)) \). (b) At the stationary point, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \). Since \( \mathrm{e}^{2x} \neq 0 \), we have \( 2\cos(3x) - 3\sin(3x) = 0 \implies \tan(3x) = \frac{2}{3} \). Inside the given interval, this yields \( 3x = \arctan\left(\frac{2}{3}\right) \implies x = \frac{1}{3}\arctan\left(\frac{2}{3}\right) \). To find \( y \), we use the trigonometric identity \( \cos(3x) = \frac{3}{\sqrt{13}} \) since \( 3x \) is in the first quadrant. Substituting into \( y \) gives: \( y = \mathrm{e}^{2\left(\frac{1}{3}\arctan(2/3)\right)} \cdot \frac{3}{\sqrt{13}} = \frac{3}{\sqrt{13}}\mathrm{e}^{\frac{2}{3}\arctan(2/3)} \).

(a) M1: Attempt to differentiate \( \mathrm{e}^{2x} \) and \( \cos(3x) \). M1: Apply the product rule. A1: Correct derivative \( \mathrm{e}^{2x}(2\cos(3x) - 3\sin(3x)) \). (b) M1: Set their \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) and obtain \( \tan(3x) = k \). A1: Correctly find \( \tan(3x) = \frac{2}{3} \). A1: Find the exact x-coordinate \( x = \frac{1}{3}\arctan\left(\frac{2}{3}\right) \). M1: Use trigonometry to find the exact value of \( \cos(3x) \). A1: Correct exact y-coordinate \( \frac{3}{\sqrt{13}}\mathrm{e}^{\frac{2}{3}\arctan\left(\frac{2}{3}\right)} \).

(a) Using the R-formula: \( R = \sqrt{5^2 + 12^2} = 13 \). Also, \( R\cos(\alpha) = 5 \) and \( R\sin(\alpha) = 12 \), which gives \( \tan(\alpha) = \frac{12}{5} \implies \alpha = \arctan(2.4) \approx 1.176005... \approx 1.1760 \) radians. So \( 5\cos(2\theta) - 12\sin(2\theta) = 13\cos(2\theta + 1.1760) \). (b) Setting \( 13\cos(2\theta + 1.1760) = 6.5 \implies \cos(2\theta + 1.1760) = 0.5 \). Since \( 0 \le \theta < \pi \), we have \( 1.1760 \le 2\theta + 1.1760 < 7.4592 \). The solutions in this range are: \( 2\theta + 1.1760 = \frac{5\pi}{3} \approx 5.2360 \implies 2\theta = 4.0600 \implies \theta \approx 2.030 \), and \( 2\theta + 1.1760 = \frac{7\pi}{3} \approx 7.3304 \implies 2\theta = 6.1544 \implies \theta \approx 3.077 \).

(a) M1: Calculate \( R = \sqrt{5^2 + 12^2} \). A1: \( R = 13 \). A1: \( \alpha \approx 1.1760 \). (b) M1: Set \( \cos(2\theta + 1.1760) = 0.5 \). M1: Solve for one correct value of \( 2\theta + \alpha \) in the interval. A1: Obtain \( \theta \approx 2.030 \). M1: Solve for the second value of \( 2\theta + \alpha \). A1: Obtain \( \theta \approx 3.077 \) and no other solutions.

(a) \( \mathrm{f}(1.5) = 3(1.5)^3 - 5(1.5) - 4 = -1.375 \). \( \mathrm{f}(1.6) = 3(1.6)^3 - 5(1.6) - 4 = 0.288 \). Since there is a change of sign and \( \mathrm{f} \) is continuous, there is a root in the interval \( [1.5, 1.6] \). (b) \( 3x^3 - 5x - 4 = 0 \implies 3x^3 = 5x + 4 \implies x^2 = \frac{5x+4}{3x} \implies x = \sqrt{\frac{5x+4}{3x}} \). (c) Using the formula: \( x_1 = \sqrt{\frac{5(1.5)+4}{3(1.5)}} \approx 1.5986 \). \( x_2 = \sqrt{\frac{5(1.59861)+4}{3(1.59861)}} \approx 1.5814 \). \( x_3 = \sqrt{\frac{5(1.58137)+4}{3(1.58137)}} \approx 1.5855 \). (d) Choosing the interval \( [1.5835, 1.5845] \): \( \mathrm{f}(1.5835) \approx -0.0055 \) and \( \mathrm{f}(1.5845) \approx 0.0121 \). Since there is a sign change, the root \( \alpha \) must lie in \( (1.5835, 1.5845) \), meaning \( \alpha = 1.584 \) to 3 d.p.

(a) M1: Evaluate \( \mathrm{f}(1.5) \) and \( \mathrm{f}(1.6) \). A1: Correct values and conclusion. (b) M1: Rearrange the equation to isolate \( x^2 \). A1: Show correct step to obtain the root expression. (c) M1: Correct process for substituting \( x_0 = 1.5 \). A1: \( x_1 \approx 1.5986 \) and \( x_2 \approx 1.5814 \). A1: \( x_3 \approx 1.5855 \). (d) M1: Evaluate \( \mathrm{f}(x) \) at \( 1.5835 \) and \( 1.5845 \). A1: Show change of sign and write a valid concluding statement.

(a) Initial number is when \( t = 0 \): \( N = \frac{1500}{2 + 3\mathrm{e}^0} = \frac{1500}{5} = 300 \). (b) When \( N = 500 \): \( 500 = \frac{1500}{2 + 3\mathrm{e}^{-0.1t}} \implies 2 + 3\mathrm{e}^{-0.1t} = 3 \implies 3\mathrm{e}^{-0.1t} = 1 \implies \mathrm{e}^{-0.1t} = \frac{1}{3} \implies -0.1t = -\ln(3) \implies t = 10\ln(3) \approx 10.99 \) hours. (c) As \( t \to \infty \), \( \mathrm{e}^{-0.1t} \to 0 \). Therefore, \( N \to \frac{1500}{2} = 750 \).

(a) M1: Substitute \( t = 0 \). A1: 300. (b) M1: Set \( N = 500 \) and rearrange to isolate \( \mathrm{e}^{-0.1t} \). A1: \( \mathrm{e}^{-0.1t} = \frac{1}{3} \). M1: Solve using logarithms. A1: \( t \approx 10.99 \). (c) B1: 750.

(a) Let \( u = 2x - 1 \implies \mathrm{d}u = 2\mathrm{d}x \implies \mathrm{d}x = \frac{1}{2}\mathrm{d}u \). Also, \( x = \frac{u+1}{2} \). Substitute into the integral: \( \int \frac{x}{\sqrt{2x-1}} \, \mathrm{d}x = \int \frac{\frac{u+1}{2}}{\sqrt{u}} \cdot \frac{1}{2} \, \mathrm{d}u = \frac{1}{4} \int (u^{1/2} + u^{-1/2}) \, \mathrm{d}u = \frac{1}{4} \left( \frac{2}{3}u^{3/2} + 2u^{1/2} \right) + C = \frac{1}{6}u^{3/2} + \frac{1}{2}u^{1/2} + C = \frac{1}{6}\sqrt{u}(u + 3) + C \). Substituting back \( u = 2x - 1 \): \( \frac{1}{6}\sqrt{2x-1}(2x + 2) + C = \frac{1}{3}(x+1)\sqrt{2x-1} + C \). (b) Evaluated from 1 to 5: \( \left[ \frac{1}{3}(x+1)\sqrt{2x-1} \right]_{1}^{5} = \left( \frac{1}{3}(6)\sqrt{9} \right) - \left( \frac{1}{3}(2)\sqrt{1} \right) = 6 - \frac{2}{3} = \frac{16}{3} \).

(a) M1: Differentiate substitution and find expression for \( x \). M1: Substitute to get integral entirely in terms of \( u \). A1: Correct simplified integrand: \( \frac{1}{4} (u^{1/2} + u^{-1/2}) \). M1: Integrate both terms correctly. A1: Fully factorise and substitute back to obtain the given expression. (b) M1: Apply limits of 5 and 1 to the given expression. A1: Correct values of \( 6 \) and \( \frac{2}{3} \). A1: \( \frac{16}{3} \) (or equivalent).

(a) We are given \( \frac{\mathrm{d}V}{\mathrm{d}t} = 15 \). Differentiating \( V \) and \( A \) with respect to \( r \): \( \frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2 \) and \( \frac{\mathrm{d}A}{\mathrm{d}r} = 8\pi r \). By the chain rule, \( \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}r} \cdot \frac{\mathrm{d}r}{\mathrm{d}V} \cdot \frac{\mathrm{d}V}{\mathrm{d}t} = 8\pi r \cdot \frac{1}{4\pi r^2} \cdot 15 = \frac{120\pi r}{4\pi r^2} = \frac{30}{r} \). Thus, \( k = 30 \). (b) When \( r = 6 \), \( \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{30}{6} = 5\,\mathrm{cm}^2\,\mathrm{s}^{-1} \).

(a) B1: Correct derivative \( \frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2 \). B1: Correct derivative \( \frac{\mathrm{d}A}{\mathrm{d}r} = 8\pi r \). M1: State or apply correct chain rule. M1: Substitute derivatives and simplify. A1: Conclude with \( \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{30}{r} \) and \( k = 30 \). (b) M1: Substitute \( r = 6 \) into their derivative. A1: Correct numerical answer 5. B1: Correct units \( \mathrm{cm}^2\,\mathrm{s}^{-1} \) (independent).

(a) The graph of \( y = |2x - 5| - 3 \) is a V-shape. The minimum (vertex) occurs when \( 2x - 5 = 0 \implies x = 2.5 \), where \( y = -3 \). So vertex is at \( (2.5, -3) \). The y-intercept is at \( x=0 \implies y = |-5|-3 = 2 \), so \( (0, 2) \). The x-intercepts occur when \( |2x - 5| = 3 \implies 2x - 5 = 3 \implies x=4 \), or \( 2x - 5 = -3 \implies x=1 \), so \( (1, 0) \) and \( (4, 0) \). (b) Solve \( |2x - 5| < x + 2 \). This can be split into two cases: \( -(x+2) < 2x - 5 < x + 2 \). 1) \( -x - 2 < 2x - 5 \implies 3x > 3 \implies x > 1 \). 2) \( 2x - 5 < x + 2 \implies x < 7 \). Combining these gives: \( 1 < x < 7 \).

(a) B1: Correctly shaped V-graph. B1: Vertex correct at \( (2.5, -3) \). B1: Y-intercept at \( (0, 2) \). B1: X-intercepts at \( (1, 0) \) and \( (4, 0) \). (b) M1: Set up and solve \( 2x - 5 = x + 2 \). A1: Find critical value \( x = 7 \). M1: Set up and solve \( -(2x - 5) = x + 2 \). A1: Correct final inequality \( 1 < x < 7 \).

(a) Starting with the Left Hand Side (LHS): \(\frac{2\sin(2\theta) - \sin(4\theta)}{2\sin(2\theta) + \sin(4\theta)}\). Using the double-angle identity \(\sin(4\theta) = 2\sin(2\theta)\cos(2\theta)\), we get: \(\frac{2\sin(2\theta) - 2\sin(2\theta)\cos(2\theta)}{2\sin(2\theta) + 2\sin(2\theta)\cos(2\theta)}\). Factoring out \(2\sin(2\theta)\) from the numerator and denominator gives: \(\frac{2\sin(2\theta)(1 - \cos(2\theta))}{2\sin(2\theta)(1 + \cos(2\theta))} = \frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}\). Using the double-angle identities \(\cos(2\theta) = 1 - 2\sin^2\theta\) (so \(1 - \cos(2\theta) = 2\sin^2\theta\)) and \(\cos(2\theta) = 2\cos^2\theta - 1\) (so \(1 + \cos(2\theta) = 2\cos^2\theta\)), we substitute these in to get: \(\frac{2\sin^2\theta}{2\cos^2\theta} = \tan^2\theta\) which is the Right Hand Side (RHS).

(b) Using the identity from part (a), the equation becomes \(\tan^2 x = 3 - \sec x\). Since \(\tan^2 x = \sec^2 x - 1\), we have \(\sec^2 x - 1 = 3 - \sec x \implies \sec^2 x + \sec x - 4 = 0\). Substituting \(\sec x = \frac{1}{\cos x}\) and multiplying by \(\cos^2 x\) gives: \(1 + \cos x - 4\cos^2 x = 0 \implies 4\cos^2 x - \cos x - 1 = 0\). Solving this quadratic using the quadratic formula: \(\cos x = \frac{1 \pm \sqrt{1 - 4(4)(-1)}}{8} = \frac{1 \pm \sqrt{17}}{8}\). This gives two cases:
Case 1: \(\cos x = \frac{1 + \sqrt{17}}{8} \approx 0.6404 \implies x = \arccos(0.6404) \approx 0.876\) radians.
Case 2: \(\cos x = \frac{1 - \sqrt{17}}{8} \approx -0.3904 \implies x = \arccos(-0.3904) \approx 1.97\) radians. Both solutions lie within the range \(0 \le x \le \pi\).

(a)
- M1: Uses the double-angle formula \(\sin(4\theta) = 2\sin(2\theta)\cos(2\theta)\) to rewrite the terms in the fraction.
- M1: Factorises and cancels \(2\sin(2\theta)\) to obtain \(\frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}\).
- M1: Uses double-angle identities for \(\cos(2\theta)\) to write the numerator as \(2\sin^2\theta\) and denominator as \(2\cos^2\theta\).
- A1*: Fully correct proof with all steps shown clearly, leading to \(\tan^2\theta\) with no errors.

(b)
- M1: Substitutes \(\tan^2 x\) for the LHS and uses \(\tan^2 x = \sec^2 x - 1\) to form a quadratic in \(\sec x\) or \(\cos x\).
- A1: Obtains a correct 3-term quadratic, e.g., \(\sec^2 x + \sec x - 4 = 0\) or \(4\cos^2 x - \cos x - 1 = 0\).
- M1: Solves their quadratic to find at least one valid value of \(\cos x\) and attempts to find at least one value of \(x\) in radians within the range.
- A1: Both answers \(x \approx 0.876\) and \(x \approx 1.97\) (accept answers rounding to 0.88 and 1.97 if working is correct).

Assume for contradiction that \(\log_2 5\) is a rational number. This means we can write it in the form \(\log_2 5 = \frac{p}{q}\), where \(p\) and \(q\) are positive integers with no common factors (or simply positive integers, since \(\log_2 5 > 0\)). From the definition of a logarithm, this equivalent to \(2^{\frac{p}{q}} = 5\). Raising both sides to the power of \(q\) gives: \(2^p = 5^q\). Since \(p \ge 1\), the left-hand side \(2^p\) is a power of 2, which is an even integer. Since \(q \ge 1\), the right-hand side \(5^q\) is a power of 5, which is an odd integer. This is a contradiction, as an even integer cannot equal an odd integer. Therefore, the original assumption must be false, and \(\log_2 5\) is an irrational number.

- **M1**: Sets up a proof by contradiction by assuming \(\log_2 5\) is rational and writing \(\log_2 5 = \frac{p}{q}\) where \(p, q \in \mathbb{Z}^+\).
- **A1**: Correctly converts to exponential form \(2^{p/q} = 5\).
- **M1**: Raises both sides to the power \(q\) to get \(2^p = 5^q\).
- **A1**: Explains clearly that the left-hand side \(2^p\) is even for positive integer \(p\).
- **M1**: Explains clearly that the right-hand side \(5^q\) is odd for positive integer \(q\).
- **A1**: Expresses that an even number cannot equal an odd number, establishing the contradiction.
- **B1**: Concludes that \(\log_2 5\) must therefore be irrational.

(a) Using the binomial expansion formula: \((1 + y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots\) with \(n = -\frac{1}{2}\) and \(y = -2x\):
\((1 - 2x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-2x) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}(-2x)^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}(-2x)^3 + \dots\)
\(= 1 + x + \frac{3}{8}(4x^2) + \frac{-15}{48}(-8x^3) = 1 + x + \frac{3}{2}x^2 + \frac{5}{2}x^3\).

(b) Substitute \(x = 0.01\) into both sides:
\((1 - 2(0.01))^{-\frac{1}{2}} \approx 1 + 0.01 + \frac{3}{2}(0.01)^2 + \frac{5}{2}(0.01)^3\)
\((0.98)^{-\frac{1}{2}} \approx 1 + 0.01 + 0.00015 + 0.0000025 = 1.0101525\).
Since \((0.98)^{-\frac{1}{2}} = \left(\frac{98}{100}\right)^{-\frac{1}{2}} = \left(\frac{49}{50}\right)^{-\frac{1}{2}} = \frac{\sqrt{50}}{7} = \frac{5\sqrt{2}}{7}\), we have:
\(\frac{5\sqrt{2}}{7} \approx 1.0101525 \implies \sqrt{2} \approx \frac{7}{5}(1.0101525) = 1.4142135\).
To 5 decimal places, \(\sqrt{2} \approx 1.41421\).

- **M1**: Attempts to use the binomial expansion with \(n = -\frac{1}{2}\) and term \((-2x)\).
- **A1**: Correctly expands the first two terms to get \(1 + x\).
- **M1**: Correct structures for the third and fourth terms with factorials.
- **A1**: Correct simplified third term \(\frac{3}{2}x^2\).
- **A1**: Correct simplified fourth term \(\frac{5}{2}x^3\).
- **M1**: Substitutes \(x = 0.01\) into their expansion to get a numerical approximation (e.g., \(1.0101525\)).
- **M1**: Recognises that \((0.98)^{-1/2} = \frac{5\sqrt{2}}{7}\) and rearranges to find \(\sqrt{2} \approx \frac{7}{5} \times 1.0101525\).
- **A1**: Correctly evaluates to \(1.41421\) (must be to 5 decimal places).

First, perform algebraic long division since the degree of the numerator (3) is greater than or equal to the degree of the denominator (2):
Dividing \(2x^3 + x^2 - 13x + 1\) by \(x^2 - x - 6\):
1. \(2x^3 \div x^2 = 2x\). Multiply: \(2x(x^2 - x - 6) = 2x^3 - 2x^2 - 12x\).
Subtracting this gives: \((x^2 - 13x + 1) - (-2x^2 - 12x) = 3x^2 - x + 1\).
2. \(3x^2 \div x^2 = 3\). Multiply: \(3(x^2 - x - 6) = 3x^2 - 3x - 18\).
Subtracting this gives the remainder: \((-x + 1) - (-3x - 18) = 2x + 19\).

Thus, we can write:
\(f(x) = 2x + 3 + \frac{2x + 19}{x^2 - x - 6}\).
So \(A = 2\) and \(B = 3\).

Next, express the remainder fraction in partial fractions:
\(\frac{2x + 19}{(x - 3)(x + 2)} = \frac{C}{x - 3} + \frac{D}{x + 2}\)
\(2x + 19 = C(x + 2) + D(x - 3)\).
- Set \(x = 3\): \(2(3) + 19 = C(5) \implies 25 = 5C \implies C = 5\).
- Set \(x = -2\): \(2(-2) + 19 = D(-5) \implies 15 = -5D \implies D = -3\).

Therefore, \(f(x) = 2x + 3 + \frac{5}{x - 3} - \frac{3}{x + 2}\).

- **M1**: Attempts algebraic division to obtain a linear quotient \(Ax + B\) and a remainder.
- **A1**: Finds \(A = 2\) correctly.
- **A1**: Finds \(B = 3\) correctly.
- **A1**: Obtains the correct remainder of \(2x + 19\).
- **M1**: Sets up a correct identity for partial fractions of their remainder: \(2x + 19 = C(x + 2) + D(x - 3)\).
- **M1**: Uses a valid method (such as substitution or equating coefficients) to solve for \(C\) and \(D\).
- **A1**: Correct value of \(C = 5\).
- **A1**: Correct value of \(D = -3\).

(a) Differentiate both parametric equations with respect to \(t\):
\(\frac{\mathrm{d}x}{\mathrm{d}t} = 2 - 2\cos(2t)\)
\(\frac{\mathrm{d}y}{\mathrm{d}t} = -4\sin(t)\)
Using the chain rule:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} = \frac{-4\sin(t)}{2 - 2\cos(2t)}\).
Using the identity \(1 - \cos(2t) = 2\sin^2(t)\), we have:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-4\sin(t)}{4\sin^2(t)} = -\frac{1}{\sin(t)} = -\csc(t)\).

(b) At the point where \(t = \frac{\pi}{3}\):
\(x = 2\left(\frac{\pi}{3}\right) - \sin\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}\)
\(y = 4\cos\left(\frac{\pi}{3}\right) = 4\left(\frac{1}{2}\right) = 2\).

The gradient of the tangent, \(m\), at this point is:
\(m = -\frac{1}{\sin(\pi/3)} = -\frac{1}{\sqrt{3}/2} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}\).

The equation of the tangent line is:
\(y - y_1 = m(x - x_1)\)
\(y - 2 = -\frac{2\sqrt{3}}{3} \left[ x - \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) \right]\)
\(y - 2 = -\frac{2\sqrt{3}}{3}x + \frac{4\sqrt{3}\pi}{9} - 1\)
\(y = -\frac{2\sqrt{3}}{3}x + 1 + \frac{4\sqrt{3}\pi}{9}\).

- **M1**: Differentiates \(x\) and \(y\) with respect to \(t\) to find \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(\frac{\mathrm{d}y}{\mathrm{d}t}\) (at least one correct).
- **A1**: Correct expressions: \(\frac{\mathrm{d}x}{\mathrm{d}t} = 2 - 2\cos(2t)\) and \(\frac{\mathrm{d}y}{\mathrm{d}t} = -4\sin(t)\).
- **A1**: Correct simplified expression for \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{\sin(t)}\) or equivalent.
- **M1**: Substitutes \(t = \frac{\pi}{3}\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to obtain the gradient \(m\).
- **A1**: Correct exact gradient \(m = -\frac{2\sqrt{3}}{3}\) or equivalent.
- **M1**: Evaluates both \(x\) and \(y\) at \(t = \frac{\pi}{3}\) to find exact coordinates.
- **M1**: Uses the straight-line equation \(y - y_1 = m(x - x_1)\) with their coordinates and gradient.
- **A1**: Simplifies to the exact form \(y = -\frac{2\sqrt{3}}{3}x + 1 + \frac{4\sqrt{3}\pi}{9}\) or equivalent.

(a) Differentiating both sides of \(e^{2x} + y^2 - 3xy = 7\) implicitly with respect to \(x\):
\(2e^{2x} + 2y\frac{\mathrm{d}y}{\mathrm{d}x} - 3\left(y + x\frac{\mathrm{d}y}{\mathrm{d}x}\right) = 0\)
\(2e^{2x} + 2y\frac{\mathrm{d}y}{\mathrm{d}x} - 3y - 3x\frac{\mathrm{d}y}{\mathrm{d}x} = 0\)
Factor out \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x}(2y - 3x) = 3y - 2e^{2x}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3y - 2e^{2x}}{2y - 3x}\).

(b) Substituting \(P(0, \sqrt{6})\) into the derivative formula:
\(m = \frac{\mathrm{d}y}{\mathrm{d}x}\Big|_{(0, \sqrt{6})} = \frac{3\sqrt{6} - 2e^{0}}{2\sqrt{6} - 3(0)} = \frac{3\sqrt{6} - 2}{2\sqrt{6}}\).
Using the equation of a straight line:
\(y - \sqrt{6} = \frac{3\sqrt{6} - 2}{2\sqrt{6}}(x - 0)\)
Multiply through by \(2\sqrt{6}\):
\(2\sqrt{6}y - 12 = (3\sqrt{6} - 2)x\)
Rearranging to the requested form:
\((3\sqrt{6} - 2)x - 2\sqrt{6}y + 12 = 0\).
Thus, \(a = 3\sqrt{6}-2\), \(b = -2\sqrt{6}\), \(c = 12\).

- **M1**: Differentiates \(e^{2x}\) to \(2e^{2x}\) and \(y^2\) to \(2y\frac{\mathrm{d}y}{\mathrm{d}x}\).
- **M1**: Uses the product rule to differentiate \(-3xy\) to obtain \(-3y - 3x\frac{\mathrm{d}y}{\mathrm{d}x}\).
- **A1**: Correct fully differentiated equation: \(2e^{2x} + 2y\frac{\mathrm{d}y}{\mathrm{d}x} - 3y - 3x\frac{\mathrm{d}y}{\mathrm{d}x} = 0\).
- **A1**: Correctly rearranges to find \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3y - 2e^{2x}}{2y - 3x}\).
- **M1**: Substitutes \(x = 0\) and \(y = \sqrt{6}\) into their expression to find the gradient.
- **M1**: Applies the line equation formula \(y - y_1 = m(x - x_1)\) with their coordinates and gradient.
- **A1**: Correct tangent equation written as \((3\sqrt{6}-2)x - 2\sqrt{6}y + 12 = 0\) or integer multiple.

Separate the variables:
\(\int \frac{y + 1}{y} \mathrm{d}y = \int 3x^2 \ln(x) \mathrm{d}x\)
\(\int \left(1 + \frac{1}{y}\right) \mathrm{d}y = \int 3x^2 \ln(x) \mathrm{d}x\)

The left-hand side integrates to:
\(y + \ln(y)\) (since \(y > 0\)).

For the right-hand side, 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 \(u = \ln(x) \implies \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}\)
Let \(\frac{\mathrm{d}v}{\mathrm{d}x} = 3x^2 \implies v = x^3\)
Then:
\(\int 3x^2 \ln(x) \mathrm{d}x = x^3 \ln(x) - \int x^3 \left(\frac{1}{x}\right) \mathrm{d}x = x^3 \ln(x) - \int x^2 \mathrm{d}x\)
\(= x^3 \ln(x) - \frac{1}{3}x^3 + C\).

Thus, the general solution is:
\(y + \ln(y) = x^3 \ln(x) - \frac{1}{3}x^3 + C\).

Using the boundary condition \(y = 1\) when \(x = 1\):
\(1 + \ln(1) = (1)^3 \ln(1) - \frac{1}{3}(1)^3 + C\)
\(1 = -\frac{1}{3} + C \implies C = \frac{4}{3}\).

Therefore, the solution is:
\(y + \ln(y) = x^3 \ln(x) - \frac{1}{3}x^3 + \frac{4}{3}\).

- **M1**: Separates variables to obtain \(\int \frac{y+1}{y} \mathrm{d}y = \int 3x^2 \ln(x) \mathrm{d}x\).
- **A1**: Correctly integrates LHS to get \(y + \ln(y)\).
- **M1**: Applies integration by parts on RHS with correct assignment of \(u = \ln(x)\) and \(v' = 3x^2\).
- **A1**: Correct derivative of \(u\) (which is \(1/x\)) and integral of \(v'\) (which is \(x^3\)).
- **A1**: Correctly integrates RHS to obtain \(x^3 \ln(x) - \frac{1}{3}x^3\) (constant of integration not needed here).
- **M1**: Substitutes \(x = 1\) and \(y = 1\) into an equation including their constant of integration \(C\).
- **A1**: Correct value of \(C = \frac{4}{3}\).
- **A1**: Correct final equation in the required form: \(y + \ln(y) = x^3 \ln(x) - \frac{1}{3}x^3 + \frac{4}{3}\).

The volume of a solid of revolution is given by:
\(V = \pi \int y^2 \mathrm{d}x\)
Given \(x = t^2 \implies \frac{\mathrm{d}x}{\mathrm{d}t} = 2t \implies \mathrm{d}x = 2t \mathrm{d}t\).

When \(x = 0 \implies t = 0\), and when \(x = 9 \implies t = 3\).
So we can change the variable to \(t\):
\(V = \pi \int_{0}^{3} [2t(3 - t)]^2 (2t) \mathrm{d}t\)
\(V = \pi \int_{0}^{3} [4t^2(9 - 6t + t^2)] (2t) \mathrm{d}t\)
\(V = \pi \int_{0}^{3} (36t^2 - 24t^3 + 4t^4)(2t) \mathrm{d}t\)
\(V = \pi \int_{0}^{3} (72t^3 - 48t^4 + 8t^5) \mathrm{d}t\).

Now, integrate each term with respect to \(t\):
\(V = \pi \left[ \frac{72}{4}t^4 - \frac{48}{5}t^5 + \frac{8}{6}t^6 \right]_{0}^{3}\)
\(V = \pi \left[ 18t^4 - \frac{48}{5}t^5 + \frac{4}{3}t^6 \right]_{0}^{3}\).

Substitute the upper limit \(t = 3\) and lower limit \(t = 0\):
\(V = \pi \left[ 18(3)^4 - \frac{48}{5}(3)^5 + \frac{4}{3}(3)^6 - 0 \right]\)
\(V = \pi \left[ 18(81) - \frac{48(243)}{5} + \frac{4(729)}{3} \right]\)
\(V = \pi \left[ 1458 - \frac{11664}{5} + 972 \right]\)
\(V = \pi \left[ 2430 - 2332.8 \right] = 97.2\pi = \frac{486\pi}{5}\).

- **M1**: Recalls and applies the volume of revolution formula \(V = \pi \int y^2 \mathrm{d}x\).
- **M1**: Differentiates \(x = t^2\) to find \(\mathrm{d}x = 2t \mathrm{d}t\).
- **A1**: Writes the correct integrand in terms of \(t\): \(\pi \int (72t^3 - 48t^4 + 8t^5) \mathrm{d}t\) (limits of integration can be omitted here).
- **M1**: Integrates the polynomial expression in \(t\) correctly.
- **A1**: Obtains the correct integrated expression \(18t^4 - \frac{48}{5}t^5 + \frac{4}{3}t^6\).
- **M1**: Substitutes the limits \(t = 0\) and \(t = 3\) into their integrated function.
- **A1**: Simplifies to show the volume is indeed \(\frac{486\pi}{5}\).

(a) Equating the components of the two lines:
1) \(2 + 3\lambda = 2 + \mu \implies \mu = 3\lambda\)
2) \(-1 + \lambda = 6 - 2\mu\)
3) \(4 - 2\lambda = -1 + \mu\)

Substitute (1) into (2):
\(-1 + \lambda = 6 - 2(3\lambda) \implies -1 + \lambda = 6 - 6\lambda\)
\(7\lambda = 7 \implies \lambda = 1\).
Using (1), \(̖ = 3(1) = 3\).

Check if these parameters are consistent in (3):
LHS of (3): \(4 - 2(1) = 2\).
RHS of (3): \(-1 + (3) = 2\).
Since LHS = RHS, the lines intersect.

Substitute \(\lambda = 1\) into the equation of \(l_1\) to find coordinates of \(P\):
\(\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + 1 \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \\ 2 \end{pmatrix}\).
So \(P\) is \((5, 0, 2)\).

(b) The direction vectors of the lines are \(\mathbf{d}_1 = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}\) and \(\mathbf{d}_2 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\).
Compute their dot product:
\(\mathbf{d}_1 \cdot \mathbf{d}_2 = 3(1) + 1(-2) + (-2)(1) = 3 - 2 - 2 = -1\).

Compute the magnitudes of the direction vectors:
\(|\mathbf{d}_1| = \sqrt{3^2 + 1^2 + (-2)^2} = \sqrt{14}\).
\(|\mathbf{d}_2| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}\).

Therefore, the cosine of the acute angle \(\theta\) is:
\(\cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|} = \frac{|-1|}{\sqrt{14}\sqrt{6}} = \frac{1}{\sqrt{84}} = \frac{1}{2\sqrt{21}} = \frac{\sqrt{21}}{42}\).

- **M1**: Equates components to establish at least two simultaneous equations in \(\lambda\) and \(\mu\).
- **M1**: Solves the equations to find the values of both parameters (\(\lambda = 1\), \(\mu = 3\)).
- **A1**: Checks the parameters in the third component equation to prove intersection.
- **M1**: Substitutes one parameter back into the line equation to find the coordinates of \(P\).
- **A1**: Correct coordinates \(P(5, 0, 2)\) or position vector.
- **M1**: Uses \(\cos\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}\) with the correct direction vectors.
- **A1**: Correct exact value of \(\cos \theta = \frac{\sqrt{21}}{42}\) (or equivalent simplified form).

First, separate the variables to obtain:
\[\int \frac{1}{y^2} \mathrm{d}y = \int \frac{\ln x}{x} \mathrm{d}x\]
Integrate the left-hand side:
\[\int y^{-2} \mathrm{d}y = -\frac{1}{y}\]
Integrate the right-hand side using the substitution \(u = \ln x\), which gives \(\mathrm{d}u = \frac{1}{x} \mathrm{d}x\):
\[\int \frac{\ln x}{x} \mathrm{d}x = \int u \mathrm{d}u = \frac{1}{2}u^2 + C = \frac{1}{2}(\ln x)^2 + C\]
Combine these results:
\[-\frac{1}{y} = \frac{1}{2}(\ln x)^2 + C\]
Apply the boundary condition \(y = 1\) when \(x = \mathrm{e}\):
\[-\frac{1}{1} = \frac{1}{2}(\ln \mathrm{e})^2 + C \implies -1 = \frac{1}{2}(1) + C \implies C = -\frac{3}{2}\]
Substitute \(C = -\frac{3}{2}\) back into the equation:
\[-\frac{1}{y} = \frac{1}{2}(\ln x)^2 - \frac{3}{2} = \frac{(\ln x)^2 - 3}{2}\]
Rearrange to express \(y\) in terms of \(x\):
\[\frac{1}{y} = \frac{3 - (\ln x)^2}{2} \implies y = \frac{2}{3 - (\ln x)^2}\]

- **M1**: Attempts to separate variables to obtain an equation of the form \(\int \frac{1}{y^2} \mathrm{d}y = \int \frac{\ln x}{x} \mathrm{d}x\). (Condone omission of integral signs).
- **B1**: Correctly integrates the LHS to obtain \(-\frac{1}{y}\) or \(-y^{-1}\).
- **M1**: A valid attempt to integrate the RHS, such as using substitution \(u = \ln x\) or by recognition, to obtain \(k(\ln x)^2\).
- **A1**: Correct integration of the RHS: \(\frac{1}{2}(\ln x)^2\).
- **M1**: Includes a constant of integration (on either side) and substitutes \(x = \mathrm{e}\) and \(y = 1\) to find its value.
- **A1**: Correct constant of integration, e.g., \(C = -\frac{3}{2}\) (if on the RHS with \(\frac{1}{2}(\ln x)^2\)).
- **A1**: Correct particular solution in the form \(y = \frac{2}{3 - (\ln x)^2}\) or any exact equivalent.

(a) Differentiate both parametric equations with respect to \(t\):
\[\frac{\mathrm{d}x}{\mathrm{d}t} = -3 \sin t\]
\[\frac{\mathrm{d}y}{\mathrm{d}t} = 8 \cos 2t\]
Use the chain rule to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{8 \cos 2t}{-3 \sin t}\]
Substitute \(t = \frac{\pi}{6}\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{8 \cos(\frac{\pi}{3})}{-3 \sin(\frac{\pi}{6})} = \frac{8(\frac{1}{2})}{-3(\frac{1}{2})} = -\frac{8}{3}\]

(b) Use the double angle identity \(\sin 2t = 2 \sin t \cos t\):
\[y = 8 \sin t \cos t\]
Square both sides of the equation:
\[y^2 = 64 \sin^2 t \cos^2 t\]
Since \(x = 3 \cos t\), we have \(\cos t = \frac{x}{3}\), which gives \(\cos^2 t = \frac{x^2}{9}\).
Using the identity \(\sin^2 t = 1 - \cos^2 t\), we get:
\[\sin^2 t = 1 - \frac{x^2}{9}\]
Substitute these into the expression for \(y^2\):
\[y^2 = 64 \left(1 - \frac{x^2}{9}\right) \left(\frac{x^2}{9}\right)\]
Simplifying this yields:
\[y^2 = \frac{64}{81} x^2 (9 - x^2)\]

Part (a):
- **B1**: For correctly finding \(\frac{\mathrm{d}x}{\mathrm{d}t} = -3 \sin t\).
- **B1**: For correctly finding \(\frac{\mathrm{d}y}{\mathrm{d}t} = 8 \cos 2t\).
- **M1**: For using \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}\) and substituting \(t = \frac{\pi}{6}\).
- **A1**: For the correct gradient of \(-\frac{8}{3}\) or exact equivalent.

Part (b):
- **M1**: Uses the identity \(\sin 2t = 2 \sin t \cos t\) to express \(y\) in terms of \(\sin t\) and \(\cos t\).
- **M1**: Squares the expression to write \(y^2\) in terms of \(\sin^2 t\) and \(\cos^2 t\).
- **M1**: Substitutes \(\cos^2 t = \frac{x^2}{9}\) and \(\sin^2 t = 1 - \frac{x^2}{9}\) into their expression for \(y^2\).
- **A1**: For the correct cartesian equation in the required form: \(y^2 = \frac{64}{81} x^2 (9 - x^2)\) or any equivalent simplified form such as \(y^2 = \frac{64}{9}x^2 - \frac{64}{81}x^4\).

For particle \(A\), the displacement from \(O\) is given by \(s_A(t) = u_A t + \frac{1}{2} a_A t^2\).
Given \(u_A = 5\) and \(a_A = 2\):
\(s_A(t) = 5t + t^2\).

(a) When \(t = 5\):
\(s_A(5) = 5(5) + 5^2 = 25 + 25 = 50\text{ m}\).

(b) For particle \(B\), which passes \(O\) at \(t = 3\) with \(u_B = 6\) and \(a_B = 4\):
\(s_B(t) = 6(t - 3) + \frac{1}{2}(4)(t - 3)^2 = 6(t - 3) + 2(t - 3)^2\) for \(t \ge 3\).
Expanding this gives:
\(s_B(t) = 6t - 18 + 2(t^2 - 6t + 9) = 6t - 18 + 2t^2 - 12t + 18 = 2t^2 - 6t\).

When \(B\) overtakes \(A\), their displacements from \(O\) are equal:
\(s_A(t) = s_B(t)\)
\(5t + t^2 = 2t^2 - 6t\)
\(t^2 - 11t = 0\)
\(t(t - 11) = 0\).

Since \(t \ge 3\) for the overtaking to happen after \(B\) has started moving, we choose \(t = 11\).
Therefore, \(B\) overtakes \(A\) at \(t = 11\text{ seconds}\).

(a)
M1: Attempting to use \(s = ut + \frac{1}{2}at^2\) with \(u = 5\), \(a = 2\), and \(t = 5\).
A1: Correct substitution.
A1: \(50\text{ m}\) (or equivalent).

(b)
M1: Attempting to write an expression for \(s_B(t)\) using \((t - 3)\).
A1: Correct unsimplified expression for \(s_B(t)\).
M1: Equating \(s_A(t)\) to \(s_B(t)\) to form an equation in \(t\).
A1: Simplifying to a quadratic equation, e.g., \(t^2 - 11t = 0\).
M1: Solving the quadratic equation for \(t\).
A1: Selecting \(t = 11\) and rejecting \(t = 0\) with clear reasoning.

Let the acceleration of the system be \(a\). Since \(\sin \alpha = 0.6\), we have \(\cos \alpha = 0.8\).

For particle \(P\) on the inclined plane:
Normal reaction \(R = 5mg \cos \alpha = 5mg(0.8) = 4mg\).
Maximum frictional force \(F = \mu R = 0.25 \times 4mg = mg\).

Since \(Q\) has mass \(5m\), it accelerates downwards, pulling \(P\) up the incline.
Equation of motion for \(Q\):
\(5mg - T = 5ma\) (Equation 1)

Equation of motion for \(P\) moving up the incline:
\(T - 5mg \sin \alpha - F = 5ma\)
\(T - 5mg(0.6) - mg = 5ma\)
\(T - 4mg = 5ma\) (Equation 2)

(a) Adding Equation 1 and Equation 2:
\((5mg - T) + (T - 4mg) = 10ma\)
\(mg = 10ma \implies a = 0.1g\).

(b) Substituting \(a = 0.1g\) into Equation 1:
\(5mg - T = 5m(0.1g)\)
\(5mg - T = 0.5mg \implies T = 4.5mg\).

(a)
M1: Resolving forces perpendicular to the plane to find \(R = 5mg \cos \alpha\).
A1: Correct normal reaction \(R = 4mg\).
M1: Using \(F = \mu R\) to find \(F = mg\).
M1: Writing equation of motion for \(Q\).
A1: Correct equation \(5mg - T = 5ma\).
M1: Writing equation of motion for \(P\) up the incline.
A1: Correct equation \(T - 4mg = 5ma\).
A1: Solving for \(a\) to obtain \(a = 0.1g\).

(b)
M1: Substituting the value of \(a\) back into one of the motion equations.
A1: Correct calculation leading to \(T = 4.5mg\).

(a) When the rod is on the point of tilting about \(C\), the reaction at \(D\) is zero (\(R_D = 0\)).
Taking moments about \(C\) for the rod and the particle at \(A\):
\(4g \times AC = 12g \times CG\)
Since \(AC = 1.5\text{ m}\) and \(CG = x - 1.5\):
\(4(1.5) = 12(x - 1.5)\)
\(6 = 12(x - 1.5)\)
\(x - 1.5 = 0.5 \implies x = 2.0\).

(b) Let the equal reactions at \(C\) and \(D\) be \(R_C = R_D = R\).
Resolving vertically for the system (rod + particle of mass \(M\) at \(B\)):
\(2R = 12g + Mg \implies R = (6 + 0.5M)g\) (Equation 1)

Taking moments about \(C\) (with \(C\) as pivot):
- Weight of rod acting at \(G\) (distance \(0.5\text{ m}\) from \(C\)): \(12g \times 0.5\) clockwise.
- Particle of mass \(M\) at \(B\) (distance \(4.5\text{ m}\) from \(C\)): \(Mg \times 4.5\) clockwise.
- Reaction at \(D\) (distance \(3.5\text{ m}\) from \(C\)): \(R \times 3.5\) anticlockwise.

Equating moments:
\(12g(0.5) + Mg(4.5) = R(3.5)g\)
\(6 + 4.5M = 3.5R\) (Equation 2)

Substitute Equation 1 into Equation 2:
\(6 + 4.5M = 3.5(6 + 0.5M)\)
\(6 + 4.5M = 21 + 1.75M\)
\(2.75M = 15 \implies M = \frac{60}{11}\).

Now substitute \(M\) back into Equation 1:
\(R = (6 + 0.5 \times \frac{60}{11})g = (6 + \frac{30}{11})g = \frac{96}{11}g\).

(a)
M1: Recognizing that \(R_D = 0\) when the rod is on the point of tilting about \(C\).
M1: Attempting to take moments about \(C\).
A1: Correct moments equation: \(4g(1.5) = 12g(x - 1.5)\).
A1: Simplifying to show \(x = 2.0\).

(b)
M1: Resolving vertically to find an expression for \(R\) in terms of \(M\) and \(g\).
A1: Correct relation \(2R = (12 + M)g\).
M1: Taking moments about \(C\) or \(D\) or \(A\).
A1: Correct moments equation, e.g., \(6 + 4.5M = 3.5R\).
M1: Solving the simultaneous equations to find \(M\).
A1: Correct reaction \(R = \frac{96}{11}g\) (or equivalent fraction, or \(8.73g\)).

(a) The position vector of \(P\) at time \(t\) is:
\(\mathbf{r}_P = (-2 + 3t)\mathbf{i} + (3 - t)\mathbf{j}\).
The position vector of \(Q\) at time \(t\) is:
\(\mathbf{r}_Q = (5 - t)\mathbf{i} + (-4 + 2t)\mathbf{j}\).

The position vector of \(P\) relative to \(Q\) is given by:
\(\mathbf{r}_{P/Q} = \mathbf{r}_P - \mathbf{r}_Q\)
\(\mathbf{r}_{P/Q} = [(-2 + 3t) - (5 - t)]\mathbf{i} + [(3 - t) - (-4 + 2t)]\mathbf{j}\)
\(\mathbf{r}_{P/Q} = (4t - 7)\mathbf{i} + (7 - 3t)\mathbf{j}\).

(b) Let \(d\) be the distance between \(P\) and \(Q\) at time \(t\). Then:
\(d^2 = (4t - 7)^2 + (7 - 3t)^2\).
To find the minimum distance, we differentiate \(d^2\) with respect to \(t\):
\(\frac{\text{d}}{\text{d}t}(d^2) = 2(4)(4t - 7) + 2(-3)(7 - 3t)\)
\(= 8(4t - 7) - 6(7 - 3t)\)
\(= 32t - 56 - 42 + 18t\)
\(= 50t - 98\).

Setting \(\frac{\text{d}}{\text{d}t}(d^2) = 0\) for the minimum distance:
\(50t - 98 = 0 \implies t = 1.96\text{ seconds}\).

(a)
M1: Attempting to write position vectors for \(P\) and \(Q\) at time \(t\).
M1: Finding \(\mathbf{r}_P - \mathbf{r}_Q\).
A1: Correct \(\mathbf{i}\) component.
A1: Correct \(\mathbf{j}\) component.

(b)
M1: Expressing the square of the distance \(d^2\) in terms of \(t\).
A1: Correct expansion, e.g., \(d^2 = 25t^2 - 98t + 98\).
M1: Differentiating \(d^2\) with respect to \(t\) (or completing the square).
A1: Correct derivative \(50t - 98\).
M1: Setting their derivative to 0 and solving for \(t\).
A1: \(t = 1.96\text{ s}\) (or equivalent fraction \(\frac{49}{25}\)).

Let \(T_A = 30\text{ N}\) be the tension in \(AP\) and \(T_B\) be the tension in \(BP\). Let \(g = 9.8\text{ m s}^{-2}\).

Resolving horizontally for equilibrium:
\(T_A \sin 30^\circ - T_B \sin \theta + 20 = 0\)
\(30(0.5) - T_B \sin \theta + 20 = 0\)
\(15 + 20 = T_B \sin \theta \implies T_B \sin \theta = 35\) (Equation 1)

Resolving vertically for equilibrium:
\(T_A \cos 30^\circ + T_B \cos \theta - 4g = 0\)
\(30(\frac{\sqrt{3}}{2}) + T_B \cos \theta - 4(9.8) = 0\)
\(15\sqrt{3} + T_B \cos \theta - 39.2 = 0 \implies T_B \cos \theta = 39.2 - 15\sqrt{3}\) (Equation 2)

Dividing Equation 1 by Equation 2:
\(\frac{T_B \sin \theta}{T_B \cos \theta} = \frac{35}{39.2 - 15\sqrt{3}}\)
\(\tan \theta = \frac{35}{39.2 - 15\sqrt{3}}\).

Using numerical values:
\(39.2 - 15\sqrt{3} \approx 39.2 - 25.9808 = 13.2192\)
\(\tan \theta \approx \frac{35}{13.2192} \approx 2.64766\)
\(\theta \approx 69.309^\circ \approx 69.3^\circ\).

(b) From Equation 1:
\(T_B = \frac{35}{\sin 69.309^\circ} \approx \frac{35}{0.93549} \approx 37.41\text{ N}\).
Therefore, the tension in string \(BP\) is \(37.4\text{ N}\).

(a)
M1: Resolving forces horizontally.
A1: Correct horizontal equation, e.g., \(15 - T_B \sin \theta + 20 = 0\).
M1: Resolving forces vertically.
A1: Correct vertical equation, e.g., \(15\sqrt{3} + T_B \cos \theta - 39.2 = 0\).
M1: Dividing the equations to eliminate \(T_B\).
A1: Showing the given expression for \(\tan \theta\).
A1: Finding \(\theta = 69.3^\circ\).

(b)
M1: Attempting to use \(\theta\) to find \(T_B\) (using sine, cosine or Pythagoras).
A1: Substituting values correctly.
A1: \(T_B = 37.4\text{ N}\) (allow \(37.3\text{ N}\) to \(37.5\text{ N}\)).

(a) Let the initial direction of motion of \(A\) be the positive direction.
Initial velocities:
\(v_A = 2u\)
\(v_B = -u\)

After collision:
\(v_A' = -0.5u\) (since the direction of \(A\)'s motion is reversed).
Let the velocity of \(B\) after collision be \(v_B'\).

By Conservation of Linear Momentum:
\(m_A v_A + m_B v_B = m_A v_A' + m_B v_B'\)
\(2m(2u) + 3m(-u) = 2m(-0.5u) + 3m(v_B')\)
\(4mu - 3mu = -mu + 3m v_B'\)
\(mu = -mu + 3m v_B'\)
\(2mu = 3m v_B'\)
\(v_B' = \frac{2}{3}u\).

Since \(v_B' > 0\), the velocity of \(B\) is positive, meaning its direction of motion was reversed (it now moves in the same direction as \(A\)'s original direction).

(b) The impulse exerted on \(B\) is equal to the change in momentum of \(B\):
\(I = m_B v_B' - m_B v_B\)
\(I = 3m(\frac{2}{3}u) - 3m(-u)\)
\(I = 2mu + 3mu = 5mu\).
Therefore, the magnitude of the impulse is \(5mu\).

(a)
M1: Setting up a Conservation of Linear Momentum equation with appropriate signs.
A1: Correct left-hand side: \(2m(2u) - 3m(u)\).
A1: Correct right-hand side: \(-2m(0.5u) + 3m v_B'\).
M1: Solving for \(v_B'\).
A1: Speed is \(\frac{2}{3}u\).
A1: Stating that the direction of motion is reversed.

(b)
M1: Attempting to calculate change in momentum for either particle.
A1: Correct expression for momentum change of \(B\) or \(A\).
A1: Final magnitude of \(5mu\).

(a) The speed-time graph is a trapezium starting at \((0,0)\), accelerating linearly up to \(V\), staying constant at \(V\) for \(T\) seconds, and then decelerating linearly back to the time-axis at \(t = 240\).

(b) Using \(v = u + at\):
During acceleration: \(V = 0 + 0.4 t_1 \implies t_1 = \frac{V}{0.4} = 2.5V\) seconds.
During deceleration: \(0 = V - 0.5 t_3 \implies t_3 = \frac{V}{0.5} = 2V\) seconds.

(c) The total time is \(240\) seconds:
\(t_1 + T + t_3 = 240\)
\(2.5V + T + 2V = 240 \implies T = 240 - 4.5V\).

The total distance is the area of the trapezium:
\(s = \frac{1}{2}(T + 240)V = 4800\)
Substitute \(T = 240 - 4.5V\):
\(\frac{1}{2}(240 - 4.5V + 240)V = 4800\)
\((480 - 4.5V)V = 9600\)
\(480V - 4.5V^2 = 9600\)
\(9V^2 - 960V + 19200 = 0\)
\(3V^2 - 320V + 6400 = 0\)

Factoring the quadratic:
\((3V - 80)(V - 80) = 0\)
This gives \(V = 80\) or \(V = \frac{80}{3}\).

If \(V = 80\), then \(T = 240 - 4.5(80) = -120 < 0\), which is impossible.
Therefore, \(V = \frac{80}{3}\text{ m s}^{-1}\) (or \(\approx 26.7\text{ m s}^{-1}\)).

(a)
M1: Drawing a graph starting at origin, accelerating, constant velocity, then decelerating to axis.
A1: Labeling \(V\) on speed axis and appropriate time markings.

(b)
M1: Using \(v = u + at\) for acceleration phase.
A1: Showing \(t_1 = 2.5V\).
A1: Showing \(t_3 = 2V\).

(c)
M1: Expressing \(T\) in terms of \(V\).
M1: Using the formula for the area of a trapezium to represent total distance.
A1: Correct equation in \(V\), e.g., \((480 - 4.5V)V = 9600\).
M1: Solving the quadratic equation.
A1: Finding the two potential roots \(80\) and \(\frac{80}{3}\).
A1: Explaining why \(V = 80\) is rejected (gives negative \(T\)) and concluding \(V = \frac{80}{3}\).

(a) Differentiating the position vector with respect to \(t\) gives velocity: \(\mathbf{v} = \frac{\text{d}\mathbf{r}}{\text{d}t} = (6t^2 - 18t)\mathbf{i} + (2t - 4)\mathbf{j}\). When \(t = 4\), \(\mathbf{v} = (6(16) - 18(4))\mathbf{i} + (2(4) - 4)\mathbf{j} = 24\mathbf{i} + 4\mathbf{j}\) m s\(^{-1}\). (b) Differentiating the velocity vector gives acceleration: \(\mathbf{a} = \frac{\text{d}\mathbf{v}}{\text{d}t} = (12t - 18)\mathbf{i} + 2\mathbf{j}\). If \(\mathbf{a}\) is parallel to \(2\mathbf{i} + \mathbf{j}\), then \((12t - 18)\mathbf{i} + 2\mathbf{j} = k(2\mathbf{i} + \mathbf{j})\) for some scalar \(k\). Comparing \(\mathbf{j}\) components: \(k = 2\). Comparing \(\mathbf{i}\) components: \(12t - 18 = 2k = 4 \implies 12t = 22 \implies t = \frac{11}{6}\) s. (c) When \(t = 2\), \(\mathbf{a} = (12(2) - 18)\mathbf{i} + 2\mathbf{j} = 6\mathbf{i} + 2\mathbf{j}\). Since \(\mathbf{F} = m\mathbf{a}\), with \(m = 0.5\) kg, we have \(\mathbf{F} = 0.5(6\mathbf{i} + 2\mathbf{j}) = 3\mathbf{i} + \mathbf{j}\). The magnitude of the force is \(|\mathbf{F}| = \sqrt{3^2 + 1^2} = \sqrt{10} \approx 3.16\) N.

(a) M1: Differentiates \(\mathbf{r}\) once to get velocity. A1: Correct velocity expression. A1: Substitutes \(t=4\) to get \(24\mathbf{i} + 4\mathbf{j}\). (b) M1: Differentiates velocity to get acceleration. M1: Sets up the parallel condition (either ratio or scalar multiplication). A1: Finds \(k = 2\) or sets up the correct equation \(\frac{12t - 18}{2} = \frac{2}{1}\). A1: Solves to find \(t = \frac{11}{6}\). (c) M1: Substitutes \(t = 2\) into acceleration vector. M1: Uses \(\mathbf{F} = m\mathbf{a}\) and applies Pythagoras to find magnitude. A1: Correct magnitude \(\sqrt{10}\) N or 3.16 N.

(a) Let \(F_D\) be the driving force. Power \(P = 24000\) W. At speed \(v = 15\) m s\(^{-1}\), \(F_D = \frac{24000}{15} = 1600\) N. The component of gravity acting down the slope is \(mg\sin\theta = 1200 \times 9.8 \times 0.05 = 588\) N. Using \(F = ma\) up the slope: \(F_D - R - mg\sin\theta = ma \implies 1600 - R - 588 = 1200(0.2) \implies 1012 - R = 240 \implies R = 772\). (b) Going down the hill at maximum speed \(v = 30\) m s\(^{-1}\), the acceleration is \(0\). The driving force \(F_D\) acts down the slope, as does gravity, while the resistance \(R\) acts up the slope. Resolving along the slope: \(F_D + mg\sin\theta = R \implies F_D + 588 = 772 \implies F_D = 184\) N. The power is \(P = F_D \times v = 184 \times 30 = 5520\) W \(= 5.52\) kW.

(a) M1: Uses \(P = Fv\) to find the driving force. A1: Correct driving force of 1600 N. M1: Calculates gravity component \(1200 \times 9.8 \times 0.05\). M1: Sets up equation of motion with three terms and mass times acceleration. A1: Correct equation. A1: Obtains \(R = 772\). (b) M1: Identifies that at maximum speed, acceleration is 0. M1: Sets up the equation of motion for downhill travel. A1: Correct equation \(F_D + 588 = 772\). M1: Uses \(P = F_D v\) with \(v = 30\). A1: Final answer \(5.52\) kW (accept 5.5 or 5520 W).

(a) Let \(A_1\) and \(A_2\) be the areas of the triangle and rectangle respectively. \(A_1 = \frac{1}{2} \times 8 \times 6 = 24\) cm\(^2\). The centre of mass of the triangle \(ABC\) is at \(G_1(\bar{x}_1, \bar{y}_1) = (\frac{0+0+8}{3}, \frac{0+6+0}{3}) = (\frac{8}{3}, 2)\). \(A_2 = 3 \times 2 = 6\) cm\(^2\). The centre of mass of the rectangle \(PQRS\) is at \(G_2(\bar{x}_2, \bar{y}_2) = (3.5, -1)\). Total Area \(A = 24 + 6 = 30\) cm\(^2\). Taking moments about the \(y\)-axis: \(30 \bar{x} = 24(\frac{8}{3}) + 6(3.5) = 64 + 21 = 85 \implies \bar{x} = \frac{85}{30} = \frac{17}{6} \approx 2.83\) cm. Taking moments about the \(x\)-axis: \(30 \bar{y} = 24(2) + 6(-1) = 48 - 6 = 42 \implies \bar{y} = \frac{42}{30} = 1.4\) cm. So the centre of mass is at \((\frac{17}{6}, 1.4)\). (b) When suspended from \(B(0, 6)\), the line \(BG\) is vertical. The vector \(\vec{BG} = G - B = (\frac{17}{6}, 1.4) - (0, 6) = (\frac{17}{6}, -4.6)\). The angle \(\theta\) that \(AB\) (which lies along the positive \(y\)-axis) makes with the vertical \(BG\) satisfies \(\tan\theta = \frac{|\Delta x|}{|\Delta y|} = \frac{17/6}{4.6} = \frac{85}{138} \approx 0.6159\). Therefore, \(\theta = \arctan(0.6159) \approx 31.63^\circ \approx 32^\circ\) to the nearest degree.

(a) B1: Correct areas 24 and 6. B1: Correct centre of mass for triangle \((\frac{8}{3}, 2)\). B1: Correct centre of mass for rectangle \((3.5, -1)\). M1: Applies the principle of moments for \(\bar{x}\). A1: Correct value \(\bar{x} = \frac{17}{6}\) or 2.83. M1: Applies the principle of moments for \(\bar{y}\). A1: Correct value \(\bar{y} = 1.4\). (b) M1: Correctly identifies that the line of suspension is from \(B\) to the centre of mass \(G\) and attempts to find the angle using trigonometry. A1: Shows correct fraction or ratio for tangent. A1: Evaluates angle to 32 degrees.

(a) Let the initial direction of motion of \(A\) be positive. Then \(u_A = u\) and \(u_B = -2u\). Let the velocities after the collision be \(v_A\) and \(v_B\). By conservation of linear momentum: \(3m(u) + 2m(-2u) = 3m v_A + 2m v_B \implies 3v_A + 2v_B = -u\) (Eq. 1). By Newton's law of restitution: \(v_B - v_A = e(u_A - u_B) = e(u - (-2u)) = 3eu \implies v_A = v_B - 3eu\). Substitute this into Eq. 1: \(3(v_B - 3eu) + 2v_B = -u \implies 5v_B - 9eu = -u \implies v_B = \frac{u(9e - 1)}{5}\). Since \(B\)'s initial velocity was negative, its direction is reversed if its final velocity \(v_B\) is positive. Therefore, \(v_B > 0 \implies 9e - 1 > 0 \implies e > \frac{1}{9}\). (b) If \(e = \frac{1}{2}\), substitute into the expression for \(v_B\): \(v_B = \frac{u(9(1/2) - 1)}{5} = \frac{u(4.5 - 1)}{5} = \frac{3.5u}{5} = 0.7u\). Since \(v_B > 0\), the speed of \(B\) after the collision is \(0.7u\).

(a) M1: Sets up a correct conservation of momentum equation. A1: Correct simplified equation \(3v_A + 2v_B = -u\). M1: Sets up a correct Newton's law of restitution equation. A1: Correct relation \(v_B - v_A = 3eu\). M1: Solves the simultaneous equations to express \(v_B\) in terms of \(u\) and \(e\). A1: Correct expression \(v_B = \frac{u(9e - 1)}{5}\). A1: Sets \(v_B > 0\) and deduces \(e > \frac{1}{9}\). (b) M1: Substitutes \(e = 0.5\) into their expression for \(v_B\). A1: Evaluates correctly to get \(0.7u\) (or \(\frac{7}{10}u\)).

(a) Let \(R_A\) be the normal reaction of the ground on the rod at \(A\), and \(F_A\) be the friction force. Let \(R_B\) be the normal reaction of the wall on the rod at \(B\). Resolving vertically for the system: \(R_A = Mg + 2Mg = 3Mg\). Since the rod is on the point of slipping, the friction force is limiting: \(F_A = \mu R_A = 0.6(3Mg) = 1.8Mg\). Resolving horizontally: \(R_B = F_A = 1.8Mg\) (as required). (b) Taking moments about \(A\): The anticlockwise moment is due to the wall reaction \(R_B\): \(R_B \times 2a \sin\alpha\). The clockwise moments are due to the weights: \(Mg \times a \cos\alpha + 2Mg \times x \cos\alpha\). In equilibrium: \(Mg a \cos\alpha + 2Mg x \cos\alpha = R_B (2a \sin\alpha)\). Divide through by \(Mg \cos\alpha\): \(a + 2x = \frac{R_B}{Mg} \times 2a \tan\alpha\). Since \(R_B = 1.8Mg\), we have \(\frac{R_B}{Mg} = 1.8\). Also \(\tan\alpha = \frac{4}{3}\). So: \(a + 2x = 1.8 \times 2a \times \frac{4}{3} = 4.8a \implies 2x = 3.8a \implies x = 1.9a\).

(a) M1: Resolves vertically to find \(R_A = 3Mg\). M1: Uses limiting friction condition \(F_A = \mu R_A\). A1: Correct friction force \(1.8Mg\). M1: Resolves horizontally to state \(R_B = F_A\). A1: Concludes with \(R_B = 1.8Mg\). (b) M1: Takes moments about \(A\) (or any other valid point) with a correct number of terms. A2: (minus 1 for each incorrect term) Correct moments equation: \(Mg a \cos\alpha + 2Mg x \cos\alpha = R_B (2a \sin\alpha)\). M1: Substitutes \(R_B = 1.8Mg\) and \(\tan\alpha = 4/3\) into the equation. A1: Solves to get \(2x = 3.8a\). A1: Final answer \(x = 1.9a\).

Given \(m = 0.8\), \(\sin\beta = 0.6 \implies \cos\beta = 0.8\), \(u = 12\), \(\mu = 0.25\). (a) Normal reaction \(R = mg\cos\beta = 0.8 \times 9.8 \times 0.8 = 6.272\) N. Friction force \(F = \mu R = 0.25 \times 6.272 = 1.568\) N. Let \(d\) be the distance from \(A\) to \(B\). Initial KE at \(A = \frac{1}{2} m u^2 = 0.5 \times 0.8 \times 12^2 = 57.6\) J. At \(B\), KE is 0. Gain in PE \(= mgd\sin\beta = 0.8 \times 9.8 \times 0.6 \times d = 4.704d\). Work done against friction \(= Fd = 1.568d\). By Work-Energy Principle: \(57.6 = 4.704d + 1.568d \implies 57.6 = 6.272d \implies d = \frac{57.6}{6.272} = \frac{450}{49} \approx 9.18\) m. (b) For the return journey from \(B\) to \(A\): PE lost \(= mgd\sin\beta = 4.704 \times \frac{450}{49} = 43.2\) J. Work done against friction \(= Fd = 1.568 \times \frac{450}{49} = 14.4\) J. Let \(v\) be the speed of \(P\) at \(A\). By Work-Energy Principle: \(\frac{1}{2} m v^2 = \text{PE lost} - \text{Work done against friction} \implies 0.4 v^2 = 43.2 - 14.4 = 28.8 \implies v^2 = 72 \implies v = \sqrt{72} = 6\sqrt{2} \approx 8.49\) m s\(^{-1}\).

(a) M1: Calculates the normal reaction force. A1: Correct \(R = 6.272\) N. M1: Calculates the friction force \(F = 1.568\) N. M1: Sets up the work-energy equation: \(\text{Initial KE} = \text{Gain in PE} + \text{Work done against friction}\). A1: Correct equation in terms of \(d\). A1: Solves to get \(d = \frac{450}{49}\) or 9.18 m. (b) M1: Calculates PE lost or uses work-energy for the return path. M1: Sets up return work-energy equation: \(\frac{1}{2}mv^2 = \text{PE lost} - \text{Work against friction}\). A1: Correct value for final KE (28.8 J). A1: Correct speed \(6\sqrt{2}\) m s\(^{-1}\) or 8.49 m s\(^{-1}\).

(a) Let \(v_P\) and \(v_Q\) be the velocities of \(P\) and \(Q\) after their collision. By conservation of momentum: \(mu = m v_P + 2m v_Q \implies v_P + 2v_Q = u\) (Eq. 1). By law of restitution: \(v_Q - v_P = eu\) (Eq. 2). Adding Eq. 1 and Eq. 2 gives: \(3v_Q = u(1 + e) \implies v_Q = \frac{1}{3}u(1 + e)\). (b) Let \(w_Q\) and \(w_R\) be the velocities of \(Q\) and \(R\) after their collision. In this second collision, \(Q\) is moving with initial speed \(v_Q\) and \(R\) is initially at rest. By conservation of momentum: \(2m v_Q = 2m w_Q + 4m w_R \implies w_Q + 2w_R = v_Q\) (Eq. 3). By law of restitution: \(w_R - w_Q = e v_Q\) (Eq. 4). Adding Eq. 3 and Eq. 4 gives: \(3w_R = v_Q(1 + e) \implies w_R = \frac{1}{3}v_Q(1 + e)\). Substituting \(v_Q = \frac{1}{3}u(1+e)\): \(w_R = \frac{1}{3} \left[ \frac{1}{3}u(1+e) \right] (1+e) = \frac{1}{9}u(1+e)^2\).

(a) M1: Writes a correct momentum equation for the first collision. M1: Writes a correct restitution equation for the first collision. A1: Solves the equations simultaneously. A1: Obtains \(v_Q = \frac{1}{3}u(1+e)\). (b) M1: Writes a correct momentum equation for the second collision. A1: Correct equation: \(w_Q + 2w_R = v_Q\). M1: Writes a correct restitution equation for the second collision: \(w_R - w_Q = e v_Q\). A1: Solves to find \(w_R = \frac{1}{3}v_Q(1+e)\). M1: Substitutes their expression for \(v_Q\) from part (a). A1: Obtains the given expression \(\frac{1}{9}u(1+e)^2\) cleanly.

(a) Total frequency \(N = 80\). The median corresponds to the 40th value. The cumulative frequency of the first two classes is \(12 + 25 = 37\), so the 40th value lies in the class interval \(30 \le t < 40\).
Using linear interpolation:
\[ \text{Median} = 30 + \frac{40 - 37}{28} \times (40 - 30) = 30 + \frac{3}{28} \times 10 \approx 31.07 \text{ minutes} \]

(b) Midpoints (\(x\)) are 15, 25, 35, 50.
\[ \sum f = 80 \]
\[ \sum fx = (12 \times 15) + (25 \times 25) + (28 \times 35) + (15 \times 50) = 180 + 625 + 980 + 750 = 2535 \]
\[ \text{Mean } \bar{x} = \frac{2535}{80} = 31.6875 \approx 31.7 \text{ minutes} \]
\[ \sum fx^2 = (12 \times 15^2) + (25 \times 25^2) + (28 \times 35^2) + (15 \times 50^2) = 2700 + 15625 + 34300 + 37500 = 90125 \]
\[ \text{Variance } \sigma^2 = \frac{90125}{80} - (31.6875)^2 = 1126.5625 - 1004.0977 = 122.4648 \]
\[ \text{Standard deviation } \sigma = \sqrt{122.4648} \approx 11.1 \text{ minutes} \]

(c) Using the skewness measure:
\[ \text{Skewness} = \frac{3(\text{mean} - \text{median})}{\text{standard deviation}} = \frac{3(31.69 - 31.07)}{11.07} \approx 0.168 \]
Since \(0.168 > 0\), the distribution exhibits a weak positive skew.

(a)
- M1: For identifying the class interval \(30 \le t < 40\) and writing down a correct linear interpolation expression.
- A1: Correct substitution of values: \(30 + \frac{40 - 37}{28} \times 10\).
- A1: Correctly evaluated answer of 31.1 (awrt).

(b)
- M1: Attempting to calculate \(\sum fx\) and \(\sum fx^2\) using correct class midpoints.
- A1: Correct mean of 31.7.
- M1: Attempting to calculate standard deviation using a valid formula.
- A1: Correct standard deviation of 11.1 (awrt).

(c)
- M1: For attempting to evaluate the skewness using a valid formula (e.g., \(\text{mean} - \text{median}\) or Pearson's coefficient).
- A1: Correct calculation and concluding that the distribution has positive skewness.

(a) First calculate \(S_{xx}\), \(S_{yy}\), and \(S_{xy}\):
\[ S_{xx} = \sum x^2 - \frac{(\sum x)^2}{10} = 5110 - \frac{224^2}{10} = 92.4 \]
\[ S_{yy} = \sum y^2 - \frac{(\sum y)^2}{10} = 217800 - \frac{1450^2}{10} = 7550 \]
\[ S_{xy} = \sum xy - \frac{\sum x \sum y}{10} = 33120 - \frac{224 \times 1450}{10} = 640 \]
Now calculate \(r\):
\[ r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}} = \frac{640}{\sqrt{92.4 \times 7550}} \approx 0.7662 \approx 0.766 \]

(b) Calculate gradient \(b\):
\[ b = \frac{S_{xy}}{S_{xx}} = \frac{640}{92.4} \approx 6.9264 \approx 6.93 \]
Calculate intercept \(a\):
\[ \bar{x} = 22.4, \quad \bar{y} = 145 \]
\[ a = \bar{y} - b \bar{x} = 145 - 6.9264 \times 22.4 = -10.151 \approx -10.2 \]
Thus, the regression line equation is:
\[ y = -10.2 + 6.93x \]

(c) The gradient \(b = 6.93\) indicates that for each 1°C increase in maximum daily temperature, the estimated increase in daily ice cream sales is £6.93.

(a)
- M1: Attempting to calculate \(S_{xx}\), \(S_{yy}\), and \(S_{xy}\) with at least one correct formula.
- A1: Correctly obtaining \(S_{xx} = 92.4\), \(S_{yy} = 7550\), and \(S_{xy} = 640\).
- A1: Correctly evaluating \(r \approx 0.766\) (awrt).

(b)
- M1: For evaluating \(b = \frac{S_{xy}}{S_{xx}}\).
- A1: Correctly obtaining \(b \approx 6.93\).
- M1: For evaluating \(a = \bar{y} - b \bar{x}\).
- A1: Correct equation \(y = -10.2 + 6.93x\) (or equivalent with coefficients awrt \(-10.2\) and \(6.93\)).

(c)
- B1: Identification of the value £6.93.
- B1: Fully correct contextual explanation (stating that a 1°C rise in temperature yields a £6.93 sales increase).

(a) Using the addition rule:
\[ \mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B) \]
\[ 0.7 = 0.4 + 0.5 - \mathrm{P}(A \cap B) \implies \mathrm{P}(A \cap B) = 0.2 \]

(b) Check if \(\mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B)\):
\[ \mathrm{P}(A) \times \mathrm{P}(B) = 0.4 \times 0.5 = 0.2 \]
Since \(\mathrm{P}(A \cap B) = 0.2 = \mathrm{P}(A) \times \mathrm{P}(B)\), the events \(A\) and \(B\) are statistically independent.

(c) Since \(A\) and \(B\) are independent, \(A'\) and \(B\) are also independent:
\[ \mathrm{P}(A' | B) = \mathrm{P}(A') = 1 - \mathrm{P}(A) = 1 - 0.4 = 0.6 \]
Alternatively, using the formula:
\[ \mathrm{P}(A' | B) = \frac{\mathrm{P}(A' \cap B)}{\mathrm{P}(B)} = \frac{\mathrm{P}(B) - \mathrm{P}(A \cap B)}{\mathrm{P}(B)} = \frac{0.5 - 0.2}{0.5} = 0.6 \]

(d) Using the addition rule:
\[ \mathrm{P}(A \cup B') = \mathrm{P}(A) + \mathrm{P}(B') - \mathrm{P}(A \cap B') \]
Since \(A\) and \(B\) are independent, \(\mathrm{P}(A \cap B') = \mathrm{P}(A) \times \mathrm{P}(B') = 0.4 \times 0.5 = 0.2\).
\[ \mathrm{P}(A \cup B') = 0.4 + 0.5 - 0.2 = 0.7 \]

(a)
- M1: Correct use of the addition formula \(\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)\).
- A1: Correct calculation \(\mathrm{P}(A \cap B) = 0.2\).

(b)
- M1: For comparing the product \(\mathrm{P}(A) \times \mathrm{P}(B)\) to \(\mathrm{P}(A \cap B)\).
- A1: Fully correct justification showing both equal \(0.2\) and stating they are independent.

(c)
- M1: Using conditional probability formula or leveraging independence of \(A'\) and \(B\).
- A1: Correct answer \(0.6\).

(d)
- M1: Correctly expressing \(\mathrm{P}(A \cup B')\) using a formula or Venn diagram.
- M1: Correct determination of the intersection \(\mathrm{P}(A \cap B') = 0.2\).
- A1: Correct answer \(0.7\).

(a) Since the sum of the probabilities must equal 1:
\[ k + 2k + 0.4 + 1.5k = 1 \]
\[ 4.5k + 0.4 = 1 \implies 4.5k = 0.6 \]
\[ k = \frac{0.6}{4.5} = \frac{6}{45} = \frac{2}{15} \]

(b) The probability distribution is:
\[ \mathrm{P}(X=1) = \frac{2}{15}, \quad \mathrm{P}(X=2) = \frac{4}{15}, \quad \mathrm{P}(X=3) = 0.4 = \frac{6}{15}, \quad \mathrm{P}(X=4) = 1.5 \times \frac{2}{15} = \frac{3}{15} \]
Calculate expectation \(\mathrm{E}(X)\):
\[ \mathrm{E}(X) = \sum x \mathrm{P}(X=x) = 1\left(\frac{2}{15}\right) + 2\left(\frac{4}{15}\right) + 3\left(\frac{6}{15}\right) + 4\left(\frac{3}{15}\right) \]
\[ \mathrm{E}(X) = \frac{2 + 8 + 18 + 12}{15} = \frac{40}{15} = \frac{8}{3} \approx 2.67 \]

(c) First, calculate \(\mathrm{E}(X^2)\):
\[ \mathrm{E}(X^2) = 1^2\left(\frac{2}{15}\right) + 2^2\left(\frac{4}{15}\right) + 3^2\left(\frac{6}{15}\right) + 4^2\left(\frac{3}{15}\right) \]
\[ \mathrm{E}(X^2) = \frac{2 + 16 + 54 + 48}{15} = \frac{120}{15} = 8 \]
Now calculate variance \(\mathrm{Var}(X)\):
\[ \mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = 8 - \left(\frac{8}{3}\right)^2 = 8 - \frac{64}{9} = \frac{8}{9} \]
Using the properties of variance:
\[ \mathrm{Var}(3X - 2) = 3^2 \mathrm{Var}(X) = 9 \times \frac{8}{9} = 8 \]

(a)
- M1: Sets the sum of probabilities equal to 1, arriving at an algebraic equation in \(k\).
- A1: Correctly showing that \(k = \frac{2}{15}\) (or equivalent).

(b)
- M1: Correct formula for expectation used with their probabilities.
- A1: Substituting values correctly into the expectation expression.
- A1: Finding \(\mathrm{E}(X) = \frac{8}{3}\) (accept awrt 2.67).

(c)
- M1: Evaluating \(\mathrm{E}(X^2)\).
- A1: Obtaining \(\mathrm{E}(X^2) = 8\).
- M1: Using \(\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2\) and then applying \(\mathrm{Var}(3X-2) = 9\mathrm{Var}(X)\).
- A1: Fully correct evaluation of \(\mathrm{Var}(3X-2) = 8\).

(a) Let \(W \sim \mathrm{N}(150, 12^2)\).
Standardize to find the probability:
\[ \mathrm{P}(W > 165) = \mathrm{P}\left(Z > \frac{165 - 150}{12}\right) = \mathrm{P}(Z > 1.25) \]
\[ \mathrm{P}(Z > 1.25) = 1 - \Phi(1.25) = 1 - 0.8944 = 0.1056 \]

(b) We require \(w\) such that \(\mathrm{P}(W < w) = 0.90\).
From normal distribution percentage tables:
\[ \mathrm{P}\left(Z < \frac{w - 150}{12}\right) = 0.90 \implies \frac{w - 150}{12} = 1.2816 \]
\[ w = 150 + 12 \times 1.2816 = 165.3792 \approx 165.4 \text{ g} \]

(c) For the interquartile range, the critical z-values for quartiles are \(\pm 0.6745\):
\[ Q_3 = 150 + 12(0.6745) = 158.094 \]
\[ Q_1 = 150 - 12(0.6745) = 141.906 \]
\[ \mathrm{IQR} = Q_3 - Q_1 = 2 \times 12 \times 0.6745 = 16.188 \approx 16.2 \text{ g} \]

(a)
- M1: For standardizing with mean 150 and standard deviation 12.
- A1: For obtaining the standardized value \(1.25\).
- A1: Correct probability of 0.1056 (awrt).

(b)
- M1: Equating the standardized expression to a \(z\)-value in the interval \([1.28, 1.29]\).
- B1: Finding and using \(z = 1.2816\).
- A1: Correct value \(165.4\) (awrt).

(c)
- M1: Recognizing that the quartiles correspond to \(z \approx \pm 0.67\) (or \(0.6745\)).
- M1: Attempting to calculate \(2 \times 12 \times z\) or finding \(Q_3 - Q_1\).
- A1: Correct answer of 16.2 (awrt).

(a) (i) Using expectation properties:
\[ \mathrm{E}(Y) = \mathrm{E}(3X - 5) = 3\mathrm{E}(X) - 5 = 3(4) - 5 = 7 \]
(ii) Using variance properties:
\[ \mathrm{Var}(Y) = \mathrm{Var}(3X - 5) = 3^2 \mathrm{Var}(X) = 9(2) = 18 \]

(b) We know:
\[ \mathrm{Var}(W) = a^2 \mathrm{Var}(X) = 2a^2 = 1 \]
Since \(a > 0\):
\[ a = \frac{1}{\sqrt{2}} \approx 0.707 \]
Next, for expectation:
\[ \mathrm{E}(W) = a\mathrm{E}(X) + b = 4a + b = 0 \]
Substituting \(a\):
\[ b = -4a = -4\left(\frac{1}{\sqrt{2}}\right) = -2\sqrt{2} \approx -2.83 \]

(c) Coding is used to simplify arithmetic or data representation, reducing the sizes of numbers without altering basic linear relationships, which makes standard deviation and mean calculations easier.

(a)
- M1: For applying \(\mathrm{E}(3X-5) = 3\mathrm{E}(X) - 5\).
- A1: Correct value \(7\).
- B1: Correct value \(18\).

(b)
- M1: Setting up \(a^2 \mathrm{Var}(X) = 1\) and solving for \(a\).
- A1: \(a = \frac{1}{\sqrt{2}}\) (or awrt 0.707).
- M1: Setting up \(a \mathrm{E}(X) + b = 0\) and solving for \(b\).
- A1: \(b = -2\sqrt{2}\) (or awrt -2.83).

(c)
- B1: Mentioning 'simplifying calculations' or 'making numbers smaller/more manageable' as the main reason.

(a) The tree diagram has two stages:
Stage 1: Machine selected.
- Branch to \(A\) with probability 0.6
- Branch to \(B\) with probability 0.4
Stage 2: Kit status.
- From \(A\), branch to Defective (\(D\)) with 0.02 and Not Defective (\(D'\)) with 0.98
- From \(B\), branch to Defective (\(D\)) with 0.05 and Not Defective (\(D'\)) with 0.95

(b) Use the law of total probability:
\[ \mathrm{P}(D) = \mathrm{P}(A \cap D) + \mathrm{P}(B \cap D) \]
\[ \mathrm{P}(A \cap D) = 0.6 \times 0.02 = 0.012 \]
\[ \mathrm{P}(B \cap D) = 0.4 \times 0.05 = 0.020 \]
\[ \mathrm{P}(D) = 0.012 + 0.020 = 0.032 \]

(c) We require the conditional probability \(\mathrm{P}(B | D)\):
\[ \mathrm{P}(B | D) = \frac{\mathrm{P}(B \cap D)}{\mathrm{P}(D)} = \frac{0.020}{0.032} = \frac{20}{32} = 0.625 \]

(a)
- M1: For drawing a tree diagram with correct branches representing machines and defects.
- A1: For labelling all branches with correct probabilities.

(b)
- M1: Attempting to calculate \(\mathrm{P}(A \cap D)\) and \(\mathrm{P}(B \cap D)\).
- A1: Correctly obtaining \(0.012\) and \(0.020\).
- A1: Summing the probabilities to obtain \(0.032\).

(c)
- M1: For writing or applying the conditional probability formula \(\mathrm{P}(B | D) = \frac{\mathrm{P}(B \cap D)}{\mathrm{P}(D)}\).
- A1: Substituting their values correctly.
- A1: Correct final answer of 0.625 (or \(\frac{5}{8}\)).

(a) The data is sorted with \(n = 15\).
Using standard position rules:
\[ \text{Median} = 8\text{th value} = 37 \]
\[ Q_1 = 4\text{th value} = 28 \]
\[ Q_3 = 12\text{th value} = 52 \]

(b) Calculate interquartile range (\(\mathrm{IQR}\)):
\[ \mathrm{IQR} = Q_3 - Q_1 = 52 - 28 = 24 \]
Compute boundaries:
\[ \text{Upper Boundary} = Q_3 + 1.5 \times \mathrm{IQR} = 52 + 1.5(24) = 52 + 36 = 88 \]
\[ \text{Lower Boundary} = Q_1 - 1.5 \times \mathrm{IQR} = 28 - 1.5(24) = -8 \]
Since \(95 > 88\), the age 95 is an outlier.
Since \(21 > -8\), there are no lower outliers. Hence, 95 is the only outlier.

(c) Compare distances between quartiles and median:
\[ Q_3 - \text{Median} = 52 - 37 = 15 \]
\[ \text{Median} - Q_1 = 37 - 28 = 9 \]
Since \(Q_3 - \text{Median} > \text{Median} - Q_1\) (\(15 > 9\)), the distribution of ages is positively skewed.

(a)
- B1: Correctly identifying the Median as 37.
- B1: Correctly identifying \(Q_1 = 28\).
- B1: Correctly identifying \(Q_3 = 52\).

(b)
- M1: For finding the \(\mathrm{IQR} = 24\).
- M1: Evaluating the upper limit \(88\) and the lower limit \(-8\).
- A1: Showing clearly that \(95 > 88\) is the only value falling outside the boundaries.

(c)
- M1: For calculating both differences: \(Q_3 - \text{Median}\) and \(\text{Median} - Q_1\).
- A1: Comparing \(15\) and \(9\) correctly.
- A1: Concluding positive skewness.

Let \(X\) be the number of emails in a 1-hour period. \(X \sim \text{Po}(4.5)\). (a) \(P(X = 3) = \frac{e^{-4.5} 4.5^3}{3!} \approx 0.1687\). (b) Let \(Y\) be the number of emails in a 2-hour period. \(Y \sim \text{Po}(9)\). \(P(Y > 10) = 1 - P(Y \leq 10) = 1 - 0.7060 = 0.2940\). (c) A busy hour occurs when \(X \geq 6\). \(p = P(X \geq 6) = 1 - P(X \leq 5) = 1 - 0.7029 = 0.2971\). Let \(W\) be the number of busy hours in 8 hours. \(W \sim \text{B}(8, 0.2971)\). \(P(W \leq 2) = P(W=0) + P(W=1) + P(W=2) = (0.7029)^8 + 8(0.2971)(0.7029)^7 + 28(0.2971)^2(0.7029)^6 \approx 0.0595 + 0.2011 + 0.2974 = 0.5580\).

(a) M1 for setting up \(\text{Po}(4.5)\) and finding \(P(X=3)\). A1 for 0.1687. (b) M1 for identifying \(Y \sim \text{Po}(9)\). M1 for writing \(1 - P(Y \leq 10)\). A1 for 0.2940. (c) M1 for finding \(p = 0.2971\). M1 for identifying Binomial distribution \(W \sim \text{B}(8, p)\). M1 for setting up the sum for \(P(W \leq 2)\). A1 for \(P(W=0) \approx 0.0595\), A1 for \(P(W=1) \approx 0.2011\), A1 for final answer 0.5580 (accept 0.557 to 0.559).

(a) \(\int_0^4 k(4x^2 - x^3) \text{d}x = 1 \implies k \left[ \frac{4x^3}{3} - \frac{x^4}{4} \right]_0^4 = 1 \implies k \left( \frac{256}{3} - 64 \right) = 1 \implies k \left( \frac{64}{3} \right) = 1 \implies k = \frac{3}{64}\). (b) For \(0 \leq x \leq 4\), \(\text{F}(x) = \int_0^x \frac{3}{64}(4t^2 - t^3) \text{d}t = \frac{3}{64} \left[ \frac{4t^3}{3} - \frac{t^4}{4} \right]_0^x = \frac{x^3}{16} - \frac{3x^4}{256}\). Thus, \(\text{F}(x) = 0\) for \(x < 0\), \(\text{F}(x) = \frac{x^3}{16} - \frac{3x^4}{256}\) for \(0 \leq x \leq 4\), and \(\text{F}(x) = 1\) for \(x > 4\). (c) Evaluate \(\text{F}(2.4) = \frac{2.4^3}{16} - \frac{3 \times 2.4^4}{256} = 0.4752\). Evaluate \(\text{F}(2.5) = \frac{2.5^3}{16} - \frac{3 \times 2.5^4}{256} = 0.5188\). Since \(\text{F}(2.4) < 0.5\) and \(\text{F}(2.5) > 0.5\), the median \(m\) must lie in the interval \([2.4, 2.5]\).

(a) M1 for setting up the integral of \(f(x)\) from 0 to 4 and equating to 1. M1 for integrating correctly. A1 for substituting limits and showing the intermediate step. A1 for concluding \(k = \frac{3}{64}\). (b) M1 for setting up the integral \(\int_0^x f(t) \text{d}t\). A1 for correct integration to find \(\frac{x^3}{16} - \frac{3x^4}{256}\). A1 for fully defining \(\text{F}(x)\) with piecewise cases (including 0 and 1). A1 for correct bounds. (c) M1 for attempting to evaluate \(\text{F}(2.4)\). A1 for obtaining \(\text{F}(2.4) = 0.4752\) and \(\text{F}(2.5) = 0.5188\). A1 for a clear concluding statement referencing that 0.5 lies between these values.

(a) Let \(p\) be the probability of germination. \(H_0: p = 0.75\), \(H_1: p < 0.75\). (b) Let \(X \sim \text{B}(30, 0.75)\). We need to find \(c\) such that \(P(X \leq c) \leq 0.05\). Let \(Y = 30 - X \sim \text{B}(30, 0.25)\). \(P(X \leq 18) = P(Y \geq 12) = 1 - P(Y \leq 11) = 1 - 0.9493 = 0.0507\). \(P(X \leq 17) = P(Y \geq 13) = 1 - P(Y \leq 12) = 1 - 0.9784 = 0.0216\). Since \(0.0216 \leq 0.05\) and \(0.0507 > 0.05\), the critical region is \(X \leq 17\). (c) The observed number of germinated seeds is 17. Since \(17 \leq 17\), this value lies in the critical region. Therefore, we reject \(H_0\). There is sufficient evidence at the 5% significance level to suggest that the germination rate is less than 75%. (d) The actual significance level is \(P(X \leq 17) = 0.0216\) or 2.16%.

(a) B1 for \(H_0: p = 0.75\). B1 for \(H_1: p < 0.75\). (b) M1 for attempting to find \(P(X \leq c)\) using Binomial tables or formulas. A1 for showing \(P(X \leq 18) = 0.0507\). A1 for showing \(P(X \leq 17) = 0.0216\). A1 for the critical region \(X \leq 17\). (c) M1 for comparing the test statistic (17) to the critical region. A1 for rejecting \(H_0\). A1 for concluding in context (germination rate is less than 75%). (d) B1 for stating 0.0216 (or 2.16%).

(a) For \(X \sim U(\alpha, \beta)\), \(\text{Var}(X) = \frac{(\beta - \alpha)^2}{12}\). Given \(\text{Var}(X) = 12\), \(\frac{(\beta - \alpha)^2}{12} = 12 \implies (\beta - \alpha)^2 = 144 \implies \beta - \alpha = 12\) (since \(\beta > \alpha\)). (b) \(\text{E}(X) = \frac{\alpha + \beta}{2} = 5 \implies \alpha + \beta = 10\). Adding \(\beta - \alpha = 12\) and \(\beta + \alpha = 10\) gives \(2\beta = 22 \implies \beta = 11\). Subtracting gives \(2\alpha = -2 \implies \alpha = -1\). (c) \(P(X > 8) = \frac{11 - 8}{11 - (-1)} = \frac{3}{12} = 0.25\). (d) \(P(X_1 > 8 \cap X_2 > 8) = (0.25)^2 = 0.0625\). (e) Let \(M = \max(X_1, X_2)\). \(P(M > 8) = 1 - P(M \leq 8) = 1 - [P(X \leq 8)]^2 = 1 - (0.75)^2 = 1 - 0.5625 = 0.4375\).

(a) M1 for using the variance formula for a uniform distribution. A1 for setting up the equation. A1 for solving and justifying \(\beta - \alpha = 12\). (b) M1 for using the expectation formula \(\frac{\alpha + \beta}{2} = 5\). M1 for solving the simultaneous equations. A1 for \(\alpha = -1\) and \(\beta = 11\). (c) M1 for setting up the probability calculation. A1 for 0.25. (d) B1 for 0.0625. (e) M1 for setting up \(1 - P(M \leq 8)\) or equivalent. A1 for 0.4375.

(a) Let \(\lambda\) be the mean number of potholes in 5 km. Under \(H_0\), \(\lambda = 1.2 \times 5 = 6\). \(H_0: \lambda = 6\), \(H_1: \lambda > 6\). (b) Under \(H_0\), \(Y \sim \text{Po}(6)\). The observed number of potholes is 11. \(p\text{-value} = P(Y \geq 11) = 1 - P(Y \leq 10)\). From Poisson tables with \(\lambda = 6\), \(P(Y \leq 10) = 0.9574\). Thus, \(p\text{-value} = 1 - 0.9574 = 0.0426\). Since \(0.0426 < 0.05\), we reject \(H_0\). There is sufficient evidence to suggest that the rate of pothole occurrence has increased. (c) For a 10-kilometre stretch, let \(W\) be the number of potholes. Under \(H_0\), \(W \sim \text{Po}(12)\). We require \(P(W \geq c) \leq 0.05 \implies P(W \leq c-1) \geq 0.95\). From tables with \(\lambda = 12\): \(P(W \leq 17) = 0.9370\), so \(P(W \geq 18) = 0.0630 > 0.05\). \(P(W \leq 18) = 0.9626\), so \(P(W \geq 19) = 0.0374 \leq 0.05\). The critical region is \(W \geq 19\).

(a) B1 for \(H_0: \lambda = 6\) (or rate = 1.2/km). B1 for \(H_1: \lambda > 6\) (or rate > 1.2/km). (b) M1 for identifying \(Y \sim \text{Po}(6)\). M1 for attempting to calculate \(P(Y \geq 11)\). A1 for \(p\text{-value} = 0.0426\). A1 for rejecting \(H_0\). A1 for concluding in context. (c) M1 for identifying \(W \sim \text{Po}(12)\). M1 for attempting to find critical values using the Poisson tables. A1 for showing cumulative probabilities for 17 and 18. A1 for the critical region \(W \geq 19\).

(a) Let \(X \sim \text{B}(200, 0.08)\). Mean \(\mu = np = 200 \times 0.08 = 16\). Variance \(\sigma^2 = np(1-p) = 16 \times 0.92 = 14.72\). (b) We approximate \(X\) using \(Y \sim \text{N}(16, 14.72)\). (i) \(P(X \leq 12) \approx P(Y \leq 12.5)\). \(z = \frac{12.5 - 16}{\sqrt{14.72}} = \frac{-3.5}{3.8367} \approx -0.912\). \(P(Z \leq -0.912) = 1 - \Phi(0.912) \approx 1 - 0.8186 = 0.1814\). (ii) \(P(10 \leq X \leq 20) \approx P(9.5 \leq Y \leq 20.5)\). \(z_1 = \frac{9.5 - 16}{\sqrt{14.72}} = \frac{-6.5}{3.8367} \approx -1.694\). \(z_2 = \frac{20.5 - 16}{\sqrt{14.72}} = \frac{4.5}{3.8367} \approx 1.173\). Using Normal tables: \(\Phi(1.173) \approx 0.8796\) and \(\Phi(-1.694) \approx 0.0451\). Thus, the probability is \(0.8796 - 0.0451 = 0.8345\) (using \(z_1 = -1.69, z_2 = 1.17\) gives \(0.8790 - 0.0455 = 0.8335\)).

(a) B1 for Mean = 16. B1 for Variance = 14.72. (b)(i) M1 for applying continuity correction to get \(P(Y \leq 12.5)\). M1 for standardising with their mean and variance. A1 for \(z = -0.912\) (accept \(-0.91\)). A1 for 0.1814 (accept 0.180 to 0.183). (ii) M1 for applying continuity corrections to get \(P(9.5 \leq Y \leq 20.5)\). M1 for standardising both boundaries. A1 for both z-values. M1 for subtracting areas. A1 for 0.8335 or 0.8345 (accept 0.833 to 0.835).

(a) We require \(\int_1^3 c(x^2 - 1) \text{d}x = 1 \implies c \left[ \frac{x^3}{3} - x \right]_1^3 = 1 \implies c \left( (9 - 3) - (\frac{1}{3} - 1) \right) = 1 \implies c \left( 6 + \frac{2}{3} \right) = 1 \implies c \left( \frac{20}{3} \right) = 1 \implies c = \frac{3}{20} = 0.15\). (b) \(\text{E}(X) = \int_1^3 x f(x) \text{d}x = \int_1^3 \frac{3}{20} (x^3 - x) \text{d}x = \frac{3}{20} \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_1^3 = \frac{3}{20} \left( (\frac{81}{4} - \frac{9}{2}) - (\frac{1}{4} - \frac{1}{2}) \right) = \frac{3}{20} \left( \frac{63}{4} + \frac{1}{4} \right) = \frac{3}{20} \times 16 = 2.4\). (c) \(\text{E}(X^2) = \int_1^3 x^2 f(x) \text{d}x = \int_1^3 \frac{3}{20} (x^4 - x^2) \text{d}x = \frac{3}{20} \left[ \frac{x^5}{5} - \frac{x^3}{3} \right]_1^3 = \frac{3}{20} \left( (\frac{243}{5} - 9) - (\frac{1}{5} - \frac{1}{3}) \right) = \frac{3}{20} \left( \frac{198}{5} + \frac{2}{15} \right) = \frac{3}{20} \left( \frac{594}{15} + \frac{2}{15} \right) = \frac{3}{20} \times \frac{596}{15} = \frac{596}{100} = 5.96\). \(\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 = 5.96 - (2.4)^2 = 5.96 - 5.76 = 0.2\).

(a) M1 for attempting integration of \(x^2 - 1\) from 1 to 3 and setting equal to 1. A1 for correct integration. A1 for finding \(c = 3/20\) or 0.15. (b) M1 for attempting integration of \(x(x^2 - 1)\). A1 for correct integration of \(x^3 - x\). M1 for substituting limits 1 and 3. A1 for finding \(\text{E}(X) = 2.4\). (c) M1 for attempting integration of \(x^2(x^2 - 1)\). A1 for correct integration of \(x^4 - x^2\). M1 for calculating \(\text{E}(X^2) - [\text{E}(X)]^2\) with their values. A1 for finding \(\text{Var}(X) = 0.2\) (or 1/5).